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Problem 3. For what least natural $k$ is the expression $2016 \cdot 2017 \cdot 2018 \cdot 2019 + k$ a square of a natural number?
|
We will prove that $k=1$ already works. Let $n=2017$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1\right)+1=\left(n^{2}+n-1\right)^{2}
\end{aligned}
$$
|
1
|
Number Theory
|
olympiads
| null | null |
\(\boxed{1}\)
|
Problem 3. For what least natural $k$ is the expression $2016 \cdot 2017 \cdot 2018 \cdot 2019 + k$ a square of a natural number?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
We will prove that $k=1$ already works. Let $n=2017$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)
|
Problem 2. $n$ mushroom pickers went to the forest and brought a total of 200 mushrooms (it is possible that some of them did not bring any mushrooms home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify your answer.
|
First, let's prove that when $n \leqslant 20$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $0,1, \ldots, n-2$ mushrooms, and the $n$-th collected all the rest. Since
$$
0+1+\ldots+(n-2) \leqslant 0+1+\ldots+18=171=200-29
$$
the last mushroom picker collected at least 29 mushrooms, i.e., more than each of the others. Thus, when $n \leqslant 20$, there exists an example where Petya could be wrong.
Let's show that when $n=21$, Petya will always be right. Suppose he is wrong and let the mushroom pickers have collected $a_{0}<a_{1}<\ldots<a_{20}$ mushrooms. It is easy to see that $a_{i} \geqslant i$, hence
$$
200=a_{0}+a_{1}+\ldots+a_{20} \geqslant 0+1+\ldots+20=210
$$
a contradiction.
|
21
|
Combinatorics
|
olympiads
| null | null |
\(\boxed{21}\)
|
Problem 2. $n$ mushroom pickers went to the forest and brought a total of 200 mushrooms (it is possible that some of them did not bring any mushrooms home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify your answer.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
First, let's prove that when $n \leqslant 20$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $0,1, \ldots, n-2$ mushrooms, and the $n$-th collected all the rest. Since
$$
0+1+\ldots+(n-2) \leqslant 0+1+\ldots+18=171=200-29
$$
the last mushroom picker collected at least 29 mushrooms, i.e., more than each of the others. Thus, when $n
|
Problem 2. $n$ mushroom pickers went to the forest and brought a total of 338 mushrooms (it is possible that some of them did not bring any mushrooms home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify your answer.
|
First, let's prove that when $n \leqslant 26$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $0,1, \ldots, n-2$ mushrooms, and the $n$-th collected all the rest. Since
$$
0+1+\ldots+(n-2) \leqslant 0+1+\ldots+24=300=338-38
$$
the last mushroom picker collected at least 38 mushrooms, i.e., more than each of the others. Thus, when $n \leqslant 26$, there exists an example where Petya could be wrong.
Let's show that when $n=27$, Petya will always be right. Suppose he is wrong and let the mushroom pickers have collected $a_{0}<a_{1}<\ldots<a_{26}$ mushrooms. It is easy to see that $a_{i} \geqslant i$, hence
$$
338=a_{0}+a_{1}+\ldots+a_{25} \geqslant 0+1+\ldots+25=351
$$
a contradiction
|
27
|
Combinatorics
|
olympiads
| null | null |
\(\boxed{27}\)
|
Problem 2. $n$ mushroom pickers went to the forest and brought a total of 338 mushrooms (it is possible that some of them did not bring any mushrooms home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify your answer.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
First, let's prove that when $n \leqslant 26$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $0,1, \ldots, n-2$ mushrooms, and the $n$-th collected all the rest. Since
$$
0+1+\ldots+(n-2) \leqslant 0+1+\ldots+24=300=338-38
$$
the last mushroom picker collected at least 38 mushrooms, i.e., more than each of the others. Thus, when $n
|
. $n$ mushroom pickers went to the forest and brought a total of 450 mushrooms (each brought at least one mushroom home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify your answer.
|
First, let's prove that when $n \leqslant 29$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $1, \ldots, n-1$ mushrooms, and the $n$-th collected all the rest. Since
$$
1+\ldots+(n-1) \leqslant 1+\ldots+28=406=450-44
$$
the last mushroom picker collected at least 44 mushrooms, i.e., more than each of the others. Thus, when $n \leqslant 29$, there is an example where Petya could be wrong.
Let's show that when $n=30$, Petya will always be right. Suppose he is wrong and let the mushroom pickers have collected $a_{1}<\ldots<a_{30}$ mushrooms. It is easy to see that $a_{i} \geqslant i$, hence
$$
450=a_{1}+\ldots+a_{30} \geqslant 1+\ldots+30=465
$$
a contradiction
|
30
|
Combinatorics
|
olympiads
| null | null |
\(\boxed{30}\)
|
. $n$ mushroom pickers went to the forest and brought a total of 450 mushrooms (each brought at least one mushroom home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify your answer.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
First, let's prove that when $n \leqslant 29$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $1, \ldots, n-1$ mushrooms, and the $n$-th collected all the rest. Since
$$
1+\ldots+(n-1) \leqslant 1+\ldots+28=406=450-44
$$
the last mushroom picker collected at least 44 mushrooms, i.e., more than each of the others. Thus, w
|
Problem 2. $n$ mushroom pickers went to the forest and brought a total of 162 mushrooms (each brought at least one mushroom home). Boy Petya, upon learning this, declared: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify your answer.
|
First, let's prove that when $n \leqslant 17$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $1, \ldots, n-1$ mushrooms, and the $n$-th collected all the rest. Since
$$
1+\ldots+(n-1) \leqslant 1+\ldots+16=136=162-26
$$
the last mushroom picker collected at least 26 mushrooms, i.e., more than each of the others. Thus, when $n \leqslant 17$, there is an example where Petya could be wrong.
Let's show that when $n=18$, Petya will always be right. Suppose he is wrong and let the mushroom pickers have collected $a_{1}<\ldots<a_{18}$ mushrooms. It is easy to see that $a_{i} \geqslant i$, hence
$$
162=a_{1}+\ldots+a_{18} \geqslant 1+\ldots+18=171
$$
a contradiction
|
18
|
Combinatorics
|
olympiads
| null | null |
\(\boxed{18}\)
|
Problem 2. $n$ mushroom pickers went to the forest and brought a total of 162 mushrooms (each brought at least one mushroom home). Boy Petya, upon learning this, declared: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify your answer.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
First, let's prove that when $n \leqslant 17$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $1, \ldots, n-1$ mushrooms, and the $n$-th collected all the rest. Since
$$
1+\ldots+(n-1) \leqslant 1+\ldots+16=136=162-26
$$
the last mushroom picker collected at least 26 mushrooms, i.e., more than each of the others. Thus, w
|
Problem 6. The function $f$ is such that $f(2 x-3 y)-f(x+y)=-2 x+8 y$ for all $x, y$. Find all possible values of the expression $\frac{f(5 t)-f(t)}{f(4 t)-f(3 t)}$.
|
By substituting $y=-x$, we get that $f(5 x)=-10 x+f(0)$, i.e., $f(t)=-2 t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 4.
|
4
|
Algebra
|
olympiads
| null | null |
\(\boxed{4}\)
|
Problem 6. The function $f$ is such that $f(2 x-3 y)-f(x+y)=-2 x+8 y$ for all $x, y$. Find all possible values of the expression $\frac{f(5 t)-f(t)}{f(4 t)-f(3 t)}$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
By substituting $y=-x$, we get that $f(5 x)=-10 x+f(0)$, i.e., $f(t)=-2 t+c$ (where $c$ is s
|
Problem 6. The function $f$ is such that $f(2 x-3 y)-f(x+y)=-2 x+8 y$ for all $x, y$. Find all possible values of the expression $\frac{f(4 t)-f(t)}{f(3 t)-f(2 t)}$.
|
By substituting $y=-x$, we get that $f(5 x)=-10 x+f(0)$, i.e., $f(t)=-2 t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 3.
|
3
|
Algebra
|
olympiads
| null | null |
\(\boxed{3}\)
|
Problem 6. The function $f$ is such that $f(2 x-3 y)-f(x+y)=-2 x+8 y$ for all $x, y$. Find all possible values of the expression $\frac{f(4 t)-f(t)}{f(3 t)-f(2 t)}$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
By substituting $y=-x$, we get that $f(5 x)=-10 x+f(0)$, i.e., $f(t)=-2 t+c$ (where $c$ is s
|
Problem 6. The function $f$ is such that $f(x+2 y)-f(3 x-2 y)=2 y-x$ for all $x, y$. Find all possible values of the expression $\frac{f(5 t)-f(t)}{f(4 t)-f(3 t)}$.
|
Substituting $x=-2 y$, we get that $f(0)-f(-8 y)=-4 y$, i.e., $f(t)=\frac{1}{2} t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 4.
|
4
|
Algebra
|
olympiads
| null | null |
\(\boxed{4}\)
|
Problem 6. The function $f$ is such that $f(x+2 y)-f(3 x-2 y)=2 y-x$ for all $x, y$. Find all possible values of the expression $\frac{f(5 t)-f(t)}{f(4 t)-f(3 t)}$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Substituting $x=-2 y$, we get that $f(0)-f(-8 y)=-4 y$, i.e., $f(t)=\frac{1}{2} t+c$ (where $c$
|
Problem 6. The function $f$ is such that $f(x+2 y)-f(3 x-2 y)=2 y-x$ for all $x, y$. Find all possible values of the expression $\frac{f(4 t)-f(t)}{f(3 t)-f(2 t)}$.
|
Substituting $x=-2 y$, we get that $f(0)-f(-8 y)=-4 y$, i.e., $f(t)=\frac{1}{2} t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 3.
|
3
|
Algebra
|
olympiads
| null | null |
\(\boxed{3}\)
|
Problem 6. The function $f$ is such that $f(x+2 y)-f(3 x-2 y)=2 y-x$ for all $x, y$. Find all possible values of the expression $\frac{f(4 t)-f(t)}{f(3 t)-f(2 t)}$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Substituting $x=-2 y$, we get that $f(0)-f(-8 y)=-4 y$, i.e., $f(t)=\frac{1}{2} t+c$ (where $c$
|
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 13 adventurers have rubies; exactly 9 have emeralds; exactly 15 have sapphires; exactly 6 have diamonds. In addition, it is known that
- if an adventurer has sapphires, then they have either emeralds or diamonds (but not both at the same time);
- if an adventurer has emeralds, then they have either rubies or sapphires (but not both at the same time).
What is the minimum number of adventurers that can be in such a group?
|
Note that the number of adventurers who have sapphires is equal to the total number of adventurers who have emeralds or diamonds. Then, from the first condition, it follows that 9 adventurers have sapphires and emeralds, and 6 have sapphires and diamonds. That is, every adventurer who has emeralds must also have sapphires. Then, from the second condition, there cannot be an adventurer who has both emeralds and rubies. Therefore, there must be at least $13+9=22$ adventurers.
This number of adventurers is indeed possible: let's say we have 9 adventurers who have sapphires and emeralds, 6 adventurers who have sapphires, diamonds, and rubies, and 7 adventurers who have only rubies. One can verify that this example meets all the conditions.
|
22
|
Combinatorics
|
olympiads
| null | null |
\(\boxed{22}\)
|
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 13 adventurers have rubies; exactly 9 have emeralds; exactly 15 have sapphires; exactly 6 have diamonds. In addition, it is known that
- if an adventurer has sapphires, then they have either emeralds or diamonds (but not both at the same time);
- if an adventurer has emeralds, then they have either rubies or sapphires (but not both at the same time).
What is the minimum number of adventurers that can be in such a group?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that the number of adventurers who have sapphires is equal to the total number of adventurers who have emeralds or diamonds. Then, from the first condition, it follows that 9 adventurers have sapphires and emeralds, and 6 have sapphires and diamonds. That is, every adventurer who has emeralds must also have sapphires. Then, from the second condition, there cannot be
|
Problem 9. The function $F$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $F(n a, n b, n c)=n \cdot F(a, b, c)$, $F(a+n, b+n, c+n)=F(a, b, c)+n$, $F(a, b, c)=F(c, b, a)$ hold. Find $F(58,59,60)$.
|
Note that $F(-1,0,1)=F(1,0,-1)=(-1) \cdot F(-1,0,1)$, from which $F(-1,0,1)=0$. Then $F(58,59,60)=F(-1,0,1)+59=59$.
Comment. The function $F$ cannot be uniquely determined. For example, the functions $F(a, b, c)=(a+b+c) / 3$, $F(a, b, c)=b$, and $F(a, b, c)=$ median of the numbers $\{a, b, c\}$ all fit.
|
59
|
Algebra
|
olympiads
| null | null |
\(\boxed{59}\)
|
Problem 9. The function $F$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $F(n a, n b, n c)=n \cdot F(a, b, c)$, $F(a+n, b+n, c+n)=F(a, b, c)+n$, $F(a, b, c)=F(c, b, a)$ hold. Find $F(58,59,60)$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that $F(-1,0,1)=F(1,0,-1)=(-1) \cdot F(-1,0,1)$, from which $F(-1,0,1)=0$. Then $F(58,59,60)=F(-1,0,1)+59=59$.
Comment. The function $F$ cannot be
|
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 5 adventurers have rubies; exactly 11 have emeralds; exactly 10 have sapphires; exactly 6 have diamonds. In addition, it is known that
- if an adventurer has diamonds, then they have either emeralds or sapphires (but not both at the same time);
- if an adventurer has emeralds, then they have either rubies or diamonds (but not both at the same time).
What is the minimum number of adventurers that can be in such a group?
|
Note that the number of adventurers who have emeralds is equal to the total number of adventurers who have rubies or diamonds. Then, from the second condition, it follows that 5 adventurers have rubies and emeralds, and 6 have emeralds and diamonds. That is, every adventurer who has diamonds must also have emeralds. Then, from the first condition, there cannot be an adventurer who has both sapphires and diamonds. Therefore, there are at least $10+6=16$ adventurers.
Indeed, there can be this many adventurers: let's say we have 6 adventurers who have emeralds and diamonds, 5 adventurers who have rubies, emeralds, and sapphires, and 5 adventurers who have only sapphires. One can verify that this example meets all the conditions.
|
16
|
Combinatorics
|
olympiads
| null | null |
\(\boxed{16}\)
|
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 5 adventurers have rubies; exactly 11 have emeralds; exactly 10 have sapphires; exactly 6 have diamonds. In addition, it is known that
- if an adventurer has diamonds, then they have either emeralds or sapphires (but not both at the same time);
- if an adventurer has emeralds, then they have either rubies or diamonds (but not both at the same time).
What is the minimum number of adventurers that can be in such a group?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that the number of adventurers who have emeralds is equal to the total number of adventurers who have rubies or diamonds. Then, from the second condition, it follows that 5 adventurers have rubies and emeralds, and 6 have emeralds and diamonds. That is, every adventurer who has diamonds must also have emeralds. Then, from the first condition, there cannot be an
|
Problem 9. The function $f$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $f(n a, n b, n c)=n \cdot f(a, b, c)$, $f(a+n, b+n, c+n)=f(a, b, c)+n$, $f(a, b, c)=f(c, b, a)$ hold. Find $f(24,25,26)$.
|
Note that $f(-1,0,1)=f(1,0,-1)=(-1) \cdot f(-1,0,1)$, from which $f(-1,0,1)=0$. Then $f(24,25,26)=f(-1,0,1)+25=25$.
Comment. The function $f$ cannot be uniquely determined. For example, the functions $f(a, b, c)=(a+b+c) / 3$, $f(a, b, c)=b$, and $f(a, b, c)=$ median of the numbers $\{a, b, c\}$ all fit.
|
25
|
Algebra
|
olympiads
| null | null |
\(\boxed{25}\)
|
Problem 9. The function $f$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $f(n a, n b, n c)=n \cdot f(a, b, c)$, $f(a+n, b+n, c+n)=f(a, b, c)+n$, $f(a, b, c)=f(c, b, a)$ hold. Find $f(24,25,26)$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that $f(-1,0,1)=f(1,0,-1)=(-1) \cdot f(-1,0,1)$, from which $f(-1,0,1)=0$. Then $f(24,25,26)=f(-1,0,1)+25=25$.
Comment. The function $f$ cannot be
|
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 4 adventurers have rubies; exactly 10 have emeralds; exactly 6 have sapphires; exactly 14 have diamonds. Moreover, it is known that
- if an adventurer has rubies, then they have either emeralds or diamonds (but not both at the same time)
- if an adventurer has emeralds, then they have either rubies or sapphires (but not both at the same time).
What is the minimum number of adventurers that can be in such a group?
|
Note that the number of adventurers who have emeralds is equal to the total number of adventurers who have rubies or sapphires. Then, from the second condition, it follows that 4 adventurers have rubies and emeralds, and 6 have emeralds and sapphires. That is, every adventurer who has rubies must also have emeralds. Then, from the first condition, there cannot be an adventurer who has both rubies and diamonds. Therefore, there must be at least $4+14=18$ adventurers.
Indeed, there can be this many adventurers: suppose we have 4 adventurers who have emeralds and rubies, 6 adventurers who have diamonds, emeralds, and sapphires, and 10 adventurers who have only sapphires. One can verify that this example meets all the conditions.
|
18
|
Combinatorics
|
olympiads
| null | null |
\(\boxed{18}\)
|
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 4 adventurers have rubies; exactly 10 have emeralds; exactly 6 have sapphires; exactly 14 have diamonds. Moreover, it is known that
- if an adventurer has rubies, then they have either emeralds or diamonds (but not both at the same time)
- if an adventurer has emeralds, then they have either rubies or sapphires (but not both at the same time).
What is the minimum number of adventurers that can be in such a group?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that the number of adventurers who have emeralds is equal to the total number of adventurers who have rubies or sapphires. Then, from the second condition, it follows that 4 adventurers have rubies and emeralds, and 6 have emeralds and sapphires. That is, every adventurer who has rubies must also have emeralds. Then, from the first condition, there cannot be an
|
Problem 9. The function $G$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$, the equalities $G(n a, n b, n c)=n \cdot G(a, b, c)$, $G(a+n, b+n, c+n)=G(a, b, c)+n$, $G(a, b, c)=G(c, b, a)$ hold. Find $G(89,90,91)$.
|
Note that $G(-1,0,1)=G(1,0,-1)=(-1) \cdot G(-1,0,1)$, from which $G(-1,0,1)=0$. Then $G(89,90,91)=G(-1,0,1)+90=90$.
Comment. The function $G$ cannot be uniquely determined. For example, the functions $G(a, b, c)=(a+b+c) / 3, G(a, b, c)=b$ and $G(a, b, c)=$ median of the numbers $\{a, b, c\}$ are suitable.
|
90
|
Algebra
|
olympiads
| null | null |
\(\boxed{90}\)
|
Problem 9. The function $G$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$, the equalities $G(n a, n b, n c)=n \cdot G(a, b, c)$, $G(a+n, b+n, c+n)=G(a, b, c)+n$, $G(a, b, c)=G(c, b, a)$ hold. Find $G(89,90,91)$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that $G(-1,0,1)=G(1,0,-1)=(-1) \cdot G(-1,0,1)$, from which $G(-1,0,1)=0$. Then $G(89,90,91)=G(-1,0,1)+90=90$.
Comment. The function $G$ cannot be u
|
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 9 adventurers have rubies; exactly 8 have emeralds; exactly 2 have sapphires; exactly 11 have diamonds. Moreover, it is known that
- if an adventurer has diamonds, then they have either rubies or sapphires (but not both at the same time);
- if an adventurer has rubies, then they have either emeralds or diamonds (but not both at the same time).
What is the smallest number of adventurers that can be in such a group?
|
Note that the number of adventurers who have diamonds is equal to the total number of adventurers who have rubies or sapphires. Then, from the first condition, it follows that 9 adventurers have rubies and diamonds, and 2 have sapphires and diamonds. That is, every adventurer who has rubies must also have diamonds. Then, from the second condition, there cannot be an adventurer who has both rubies and emeralds. Therefore, there must be at least $9+8=17$ adventurers.
Indeed, there can be this many adventurers: let's say we have 9 adventurers who have rubies and diamonds, 2 adventurers who have emeralds, sapphires, and diamonds, and 6 adventurers who have only emeralds. One can verify that this example meets all the conditions.
|
17
|
Combinatorics
|
olympiads
| null | null |
\(\boxed{17}\)
|
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 9 adventurers have rubies; exactly 8 have emeralds; exactly 2 have sapphires; exactly 11 have diamonds. Moreover, it is known that
- if an adventurer has diamonds, then they have either rubies or sapphires (but not both at the same time);
- if an adventurer has rubies, then they have either emeralds or diamonds (but not both at the same time).
What is the smallest number of adventurers that can be in such a group?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that the number of adventurers who have diamonds is equal to the total number of adventurers who have rubies or sapphires. Then, from the first condition, it follows that 9 adventurers have rubies and diamonds, and 2 have sapphires and diamonds. That is, every adventurer who has rubies must also have diamonds. Then, from the second condition, there cannot be an
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Problem 9. The function $g$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $g(n a, n b, n c)=$ $n \cdot g(a, b, c), g(a+n, b+n, c+n)=g(a, b, c)+n, g(a, b, c)=g(c, b, a)$ hold. Find $g(14,15,16)$.
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Note that $g(-1,0,1)=g(1,0,-1)=(-1) \cdot g(-1,0,1)$, from which $g(-1,0,1)=0$. Then $g(14,15,16)=g(-1,0,1)+15=15$.
Comment. The function $g$ cannot be uniquely determined. For example, the functions $g(a, b, c)=(a+b+c) / 3$, $g(a, b, c)=b$, and $g(a, b, c)=$ median of the numbers $\{a, b, c\}$ all fit.
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15
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Algebra
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olympiads
| null | null |
\(\boxed{15}\)
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Problem 9. The function $g$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $g(n a, n b, n c)=$ $n \cdot g(a, b, c), g(a+n, b+n, c+n)=g(a, b, c)+n, g(a, b, c)=g(c, b, a)$ hold. Find $g(14,15,16)$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that $g(-1,0,1)=g(1,0,-1)=(-1) \cdot g(-1,0,1)$, from which $g(-1,0,1)=0$. Then $g(14,15,16)=g(-1,0,1)+15=15$.
Comment. The function $g$ cannot be
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Problem 3. How many natural numbers $n>1$ exist, for which there are $n$ consecutive natural numbers, the sum of which is equal to 2016?
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Suppose that for some $n$ there exist $n$ consecutive natural numbers $a$, $a+1, \ldots, a+(n-1)$, the sum of which is 2016. Then $n a+(n-1) n / 2=2016$, or, after algebraic transformations, $n(2 a+n-1)=4032=2^{6} \cdot 3^{2} \cdot 7$.
Note that $n$ and $2 a+n-1$ have different parity. Therefore, if $n$ is even, then $n$ must be divisible by $2^{6}=64$, from which $n \geqslant 64, 2 a+n-1 \leqslant 4032 / 64=63$, which contradicts $2 a+n-1>n$.
Thus, $n$ is an odd divisor of the number 4032, i.e., a divisor of the number 63. Let's check that each of them fits. If $n=3: 2 a+3-1=1344, a=671$; if $n=7: 2 a+7-1=576, a=285$; if $n=9$ : $2 a+9-1=448, a=220$; if $n=21: 2 a+21-1=192, a=86$; if $n=63: 2 a+63-1=64, a=1$. In total, there are five suitable $n$.
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5
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Number Theory
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olympiads
| null | null |
\(\boxed{5}\)
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Problem 3. How many natural numbers $n>1$ exist, for which there are $n$ consecutive natural numbers, the sum of which is equal to 2016?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Suppose that for some $n$ there exist $n$ consecutive natural numbers $a$, $a+1, \ldots, a+(n-1)$, the sum of which is 2016. Then $n a+(n-1) n / 2=2016$, or, after algebraic transformations, $n(2 a+n-1)=4032=2^{6} \cdot 3^{2} \cdot 7$.
Note that $n$ and $2 a+n-1$ have different parity. Therefore, if $n$ is even, then $n$ must be divisible by $2^{6}=64$, from which $n \geqslant
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7.1. On a circular route, two buses operate with the same speed and a movement interval of 21 minutes. What will be the movement interval if 3 buses operate on this route at the same constant speed?
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Since the interval of movement with two buses on the route is 21 minutes, the length of the route in "minutes" is 42 minutes. Therefore, the interval of movement with three buses on the route is $42: 3=14$ minutes.
Criteria: only answer, answer with verification - 3 points.
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14
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Logic and Puzzles
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olympiads
| null | null |
\(\boxed{14}\)
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7.1. On a circular route, two buses operate with the same speed and a movement interval of 21 minutes. What will be the movement interval if 3 buses operate on this route at the same constant speed?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since the interval of movement with two buses on the route is 21 minutes, the length of the route in "minutes" is 42 minutes. Therefore, t
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7.4. Natasha and Inna bought the same box of tea bags. It is known that one tea bag is enough for two or three cups of tea. Natasha's box was enough for only 41 cups of tea, while Inna's was enough for 58 cups. How many tea bags were in the box?
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Let there be $n$ tea bags in the box. Then the number of brewings can vary from $2n$ to $3n$. Therefore, 58 is not greater than $3n$, which means $19<n$. Additionally, 41 is not less than $2n$, so $n<21$. Since the number of tea bags must be a natural number that is less than 21 but greater than 19, there are exactly 20 tea bags in the box.
Criteria: answer only - 0 points, answer with verification - 1 point. Correct system of inequalities without conclusions - 2 points, correct system of inequalities and correct answer - 4 points. Proof of one of the estimates for the number of packets in the pack (that there are not fewer than 20 or not more than 20) - 3 points regardless of whether the correct answer is given. These points DO NOT add up.
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20
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Number Theory
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olympiads
| null | null |
\(\boxed{20}\)
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7.4. Natasha and Inna bought the same box of tea bags. It is known that one tea bag is enough for two or three cups of tea. Natasha's box was enough for only 41 cups of tea, while Inna's was enough for 58 cups. How many tea bags were in the box?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let there be $n$ tea bags in the box. Then the number of brewings can vary from $2n$ to $3n$. Therefore, 58 is not greater than $3n$, which means $19<n$. Additionally, 41 is not less than $2n$, so $n<21$. Since the number of tea bags must be a natural number that is less than 21 but greater than 19, there are exactly 20 tea bags in the box.
Criteria: answer only - 0 points
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7.1. Provide an example of a natural number that is a multiple of 2020, such that the sum of its digits is also a multiple of 2020.
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For example, the number 20202020...2020, where the fragment 2020 repeats 505 times, fits. Such a number is obviously divisible by 2020, and the sum of its digits is $505 \times 4=2020$.
Criteria: Any correct answer with or without verification - 7 points.
Correctly conceived example, but with a calculation error (for example, 2020 repeats 500 times) - 3 points.
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2020
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Number Theory
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olympiads
| null | null |
\(\boxed{2020}\)
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7.1. Provide an example of a natural number that is a multiple of 2020, such that the sum of its digits is also a multiple of 2020.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
For example, the number 20202020...2020, where the fragment 2020 repeats 505 times, fits. Such a number is obviously divisible by 2020, and the sum of its digits is $505 \times 4=202
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9.1. What is the minimum sum of digits in the decimal representation of the number $f(n)=17 n^{2}-11 n+1$, where $n$ runs through all natural numbers?
# Answer. 2.
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When $n=8$, the number $f(n)$ is 1001, so the sum of its digits is 2. If $f(n)$ for some $n$ had a sum of digits equal to 1, it would have the form $100, \ldots 00$ and would either be equal to 1 or divisible by 10. The function of a real variable $f(x)$ reaches its minimum at $x=\frac{11}{34}1$, and $f(n)$ cannot take the value 1. Furthermore, it is easy to notice that $f(n)$ is always an odd number, so it cannot be divisible by 10. Therefore, the minimum sum of the digits of the number $f(n)=17 n^{2}-11 n+1$ is 2, and it is achieved when $n=8$.
Grading Criteria. Correct answer with verification for $n=8$: 2 points. Proven that $f(n)>1$ and that $f(n)$ cannot take the value 1: 2 points. Proven that $f(n)$ is always an odd number, so it cannot be divisible by 10: 3 points.
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2
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Number Theory
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olympiads
| null | null |
\(\boxed{2}\)
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9.1. What is the minimum sum of digits in the decimal representation of the number $f(n)=17 n^{2}-11 n+1$, where $n$ runs through all natural numbers?
# Answer. 2.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
When $n=8$, the number $f(n)$ is 1001, so the sum of its digits is 2. If $f(n)$ for some $n$ had a sum of digits equal to 1, it would have the form $100, \ldots 00$ and would either be equal to 1 or divisible by 10. The function of a real variable $f(x)$ reaches its minimum at $x=\frac{11}{34}1$, and $f(n)$ cannot take the value 1. Furthermore, it is easy to notice that $f(n)$ is always an
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8.2. It is known that $70 \%$ of mathematicians who have moved to IT regret their change of activity. At the same time, only $7 \%$ of all people who have moved to IT regret the change. What percentage of those who have moved to IT are mathematicians, if only they regret the change of activity?
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Let a total of $x$ people went into IT, and $y$ of them are mathematicians. According to the condition of the change in activity, on the one hand, $0.07 x$ people regret, and on the other - $0.7 y$. From this, we get that $0.07 x=0.7 y$, from which $y / x=0.1$, that is, $10 \%$.
Criteria. Only the answer - 1 point.
Solution in a particular case (let there be 100 people in total...) without reference to the general - no more than 5 points.
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10
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Other
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olympiads
| null | null |
\(\boxed{10}\)
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8.2. It is known that $70 \%$ of mathematicians who have moved to IT regret their change of activity. At the same time, only $7 \%$ of all people who have moved to IT regret the change. What percentage of those who have moved to IT are mathematicians, if only they regret the change of activity?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let a total of $x$ people went into IT, and $y$ of them are mathematicians. According to the condition of the change in activity, on the one hand, $0.07 x$ people regret, and on the other - $0.7 y$. From this, we get that
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7.2. Ellie and Toto painted daisies in the field. On the first day, Ellie painted one fourteenth of the entire field. On the second day, she painted twice as much as on the first day, and on the third day, she painted twice as much as on the second day. Toto, in total, painted 7000 daisies. How many daisies were there in total on the field, given that all of them were painted? (Find all possible answers and prove that there are no others)
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Let there be $n$ daisies in the field. Then Ellie painted $n / 14 + 2 n / 14 + 4 n / 14 = 7 n / 14 = n / 2$ daisies in total. This means Toto also painted half of the field, from which it follows that half of the field is 7000 daisies, and the entire field is 14000.
Criteria. Only the answer - 1 point.
The answer with a check that it is correct - 2 points.
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14000
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Algebra
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olympiads
| null | null |
\(\boxed{14000}\)
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7.2. Ellie and Toto painted daisies in the field. On the first day, Ellie painted one fourteenth of the entire field. On the second day, she painted twice as much as on the first day, and on the third day, she painted twice as much as on the second day. Toto, in total, painted 7000 daisies. How many daisies were there in total on the field, given that all of them were painted? (Find all possible answers and prove that there are no others)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let there be $n$ daisies in the field. Then Ellie painted $n / 14 + 2 n / 14 + 4 n / 14 = 7 n / 14 = n / 2$ daisies in total. This means Toto also painted half of the field, from w
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10.4. On the cells of an 8 by 8 board, chips are placed such that for each chip, the row or column of the board in which it lies contains only one chip. What is the maximum possible number of chips on the board?
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Let's match each chip to the row or column of the board in which it is the only one. If it is the only one in both, we match it to the row. From the condition, it follows that different chips are matched to different rows and columns. If not all rows and columns are matched, then their total number does not exceed 14, in each of them there is no more than one chip, so the total number of chips does not exceed 14. If, however, all rows, say, are matched, then there is a chip in each row - a total of 8.
Example of placing 14 chips: all cells of the left column and the bottom row are filled, except for the bottom left corner cell.
Evaluation: 5 points. Example: 2 points. Any incorrect estimate with any reasoning - 0 points.
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14
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Combinatorics
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olympiads
| null | null |
\(\boxed{14}\)
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10.4. On the cells of an 8 by 8 board, chips are placed such that for each chip, the row or column of the board in which it lies contains only one chip. What is the maximum possible number of chips on the board?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's match each chip to the row or column of the board in which it is the only one. If it is the only one in both, we match it to the row. From the condition, it follows that different chips are matched to different rows and columns. If not all rows and columns are matched, then their total number does not exceed 14, in each of them there is no more than one chip
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11.1. Find the value of the expression $\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{2}{1+x y}$, given that $x \neq y$ and the sum of the first two terms of the expression is equal to the third.
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Let's write the condition of the equality of the sum of the first two terms to the third one as: $\frac{1}{1+x^{2}}-\frac{1}{1+x y}=\frac{1}{1+x y}-\frac{1}{1+y^{2}} \quad$ and bring it to a common denominator:
$\frac{x(y-x)}{(1+x^{2})(1+x y)}=\frac{y(y-x)}{(1+y^{2})(1+x y)}$. Given $x \neq y$, we cancel $x-y$ and $1+x y$: $x(1+y^{2})=y(1+x^{2})$. The latter is equivalent to $(x-y)(x y-1)=0$, and again we cancel $x-y$, obtaining $x y=1$ and the desired expression equals $\frac{2}{1+x y} \cdot 2=2$.
Grading criteria. Any suitable $x \neq y$ guessed and the answer: 1 point.
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2
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Algebra
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olympiads
| null | null |
\(\boxed{2}\)
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11.1. Find the value of the expression $\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{2}{1+x y}$, given that $x \neq y$ and the sum of the first two terms of the expression is equal to the third.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's write the condition of the equality of the sum of the first two terms to the third one as: $\frac{1}{1+x^{2}}-\frac{1}{1+x y}=\frac{1}{1+x y}-\frac{1}{1+y^{2}} \quad$ and bring it to a common denominator:
$\frac{x(y-x)}{(1+x^{2})(1+x y)}=\frac{y(y-x)}{(1+y^{2})(1+x y)}$. Given $x \ne
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8.5. In Nastya's room, 16 people gathered, each pair of whom either are friends or enemies. Upon entering the room, each of them wrote down the number of friends who had already arrived, and upon leaving - the number of enemies remaining in the room. What can the sum of all the numbers written down be, after everyone has first arrived and then left?
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Consider any pair of friends. Their "friendship" was counted exactly once, as it is included in the sum by the person who arrived later than their friend. Therefore, after everyone has arrived, the sum of the numbers on the door will be equal to the total number of friendships between people. Similarly, each "enmity" will be counted exactly once by the person who left earlier. Therefore, after everyone has left, the sum of the numbers on the door will be increased by the sum of all "enmities." In total, the overall sum of the numbers on the door will be equal to the sum of the total number of friendships and enmities, which is precisely the number of pairs of people who arrived, i.e., $16 * 15 / 2 = 120$.
Criteria: answer only - 0 points.
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120
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Combinatorics
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olympiads
| null | null |
\(\boxed{120}\)
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8.5. In Nastya's room, 16 people gathered, each pair of whom either are friends or enemies. Upon entering the room, each of them wrote down the number of friends who had already arrived, and upon leaving - the number of enemies remaining in the room. What can the sum of all the numbers written down be, after everyone has first arrived and then left?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Consider any pair of friends. Their "friendship" was counted exactly once, as it is included in the sum by the person who arrived later than their friend. Therefore, after everyone has arrived, the sum of the numbers on the door will be equal to the total number of friendships between people. Similarly, each "enmity" will be counted exactly once by the person who left ear
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7.2. Losharik is going to visit Sovunya along the river at a speed of 4 km/h. Every half hour, he launches paper boats that float to Sovunya at a speed of 10 km/h. With what time interval do the boats arrive at Sovunya? (Provide a complete solution, not just the answer.)
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If Losyash launched the boats from one place, they would arrive every half hour. But since he is walking, the next boat has to travel a shorter distance than the previous one. In half an hour, the distance between Losyash and the last boat will be $(10-4) \cdot 0.5=3$. This means that the distance between adjacent boats is 3 km, and the next boat will have to travel exactly this distance when the previous one has already arrived at Sovunya. The speed of the boat is 10 km/h, which means 1 km in 6 minutes, so 3 km will be covered by the boat in 18 minutes, which is the answer.
Criteria: Correctly found the distance of 3 km - 3 points.
Only the answer or the answer with verification - 1 point.
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18
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Algebra
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olympiads
| null | null |
\(\boxed{18}\)
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7.2. Losharik is going to visit Sovunya along the river at a speed of 4 km/h. Every half hour, he launches paper boats that float to Sovunya at a speed of 10 km/h. With what time interval do the boats arrive at Sovunya? (Provide a complete solution, not just the answer.)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If Losyash launched the boats from one place, they would arrive every half hour. But since he is walking, the next boat has to travel a shorter distance than the previous one. In half an hour, the distance between Losyash and the last boat will be $(10-4) \cdot 0.5=3$. This means that the distance between adjacent boats is 3 km, and the next boat w
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7.5. A square box 3 by 3 is divided into 9 cells. It is allowed to place balls in some cells (possibly a different number in different cells). What is the minimum number of balls that need to be placed in the box so that each row and each column of the box contains a different number of balls?
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Add up the number of balls in all rows and columns. Since these are 6 different non-negative numbers, this sum is at least $0+1+\ldots+5=15$.
Now notice that the sum of the numbers in the rows is equal to the sum of the numbers in the columns, since these sums are equal to the total number of balls in the box. Therefore, the sum of the six numbers is twice the sum of the numbers in the rows, which means it is even. Thus, twice the number of balls in the box is greater than 15, so the number of balls is greater than 7. An example with 8 is shown alongside.
Criteria: only the answer - 0 points, only the example - 2 points, only noted that the sum of 6 numbers is not less than $15-1$ point. Only the correct estimate - 3 points.
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8
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Combinatorics
|
olympiads
| null | null |
\(\boxed{8}\)
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7.5. A square box 3 by 3 is divided into 9 cells. It is allowed to place balls in some cells (possibly a different number in different cells). What is the minimum number of balls that need to be placed in the box so that each row and each column of the box contains a different number of balls?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Add up the number of balls in all rows and columns. Since these are 6 different non-negative numbers, this sum is at least $0+1+\ldots+5=15$.
Now notice that the sum of the numbers in the rows is equal to the sum of the numbers in the columns, since these sums are equal to the total number of balls in the box. Therefore, the sum of the six numbers is twice the sum
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9.1. Petya wrote 10 integers on the board (not necessarily distinct).
Then he calculated the pairwise products (that is, he multiplied each of the written numbers by each other). Among them, there were exactly 15 negative products. How many zeros were written on the board?
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Let there be $A$ positive numbers and $B$ negative numbers on the board. Then $A+B \leq 10$ and $A \cdot B=15$. Since a negative product is obtained when we multiply a negative and a positive number. From this, it is easy to understand that the numbers $A$ and $B$ are 3 and 5 (1). Therefore, $A+B=8$ and there are exactly two zeros on the board.
Criteria. Correct answer without justification - 1 point. Correctly found intermediate relationship (1) - 4 points. Correct answer with justification - 7 points.
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2
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Combinatorics
|
olympiads
| null | null |
\(\boxed{2}\)
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9.1. Petya wrote 10 integers on the board (not necessarily distinct).
Then he calculated the pairwise products (that is, he multiplied each of the written numbers by each other). Among them, there were exactly 15 negative products. How many zeros were written on the board?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let there be $A$ positive numbers and $B$ negative numbers on the board. Then $A+B \leq 10$ and $A \cdot B=15$. Since a negative product is obtained when we multiply a negative and a positive number. From this, it is easy to understand that the numbers $
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9.2. Find the maximum odd natural number that cannot be represented as the sum of three distinct composite numbers.
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An odd number gives a remainder of 1 or 3 when divided by 4. In the first case, the desired representation has the form \( n = 4k + 1 = 4(k-4) + 8 + 9, k \geq 5, n \geq 21 \), in the second case - \( n = 4k + 3 = 4(k-3) + 6 + 9, k \geq 4, n \geq 19 \). On the other hand, the three smallest composite numbers are 4, 6, 8, the sum of which is 18, so the number 17 cannot be represented in the required form. It is the answer to the problem.
Grading: Simply providing the answer with verification: 0 points. Failure to justify the non-representability of 17 in the desired form - minus 2 points.
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17
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Number Theory
|
olympiads
| null | null |
\(\boxed{17}\)
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9.2. Find the maximum odd natural number that cannot be represented as the sum of three distinct composite numbers.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
An odd number gives a remainder of 1 or 3 when divided by 4. In the first case, the desired representation has the form \( n = 4k + 1 = 4(k-4) + 8 + 9, k \geq 5, n \geq 21 \), in the second case - \( n = 4k + 3 = 4(k-3) + 6 + 9, k \geq 4, n \geq 19 \). On the other hand, the three smallest compos
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11.1. Find all positive integer solutions of the equation $(n+2)!-(n+1)!-(n)!=n^{2}+n^{4}$. Answer. $n=3$.
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Rewrite the equation as $n!=\left(n^{*}\left(n^{2}+1\right)\right) /(n+2)$. Transforming the right side, we get $n!=n^{2}-2 n+5-10:(n+2)$. The last fraction will be an integer for $n=3$ and $n=8$, but the latter number is not a solution (substitute and check!)
Grading criteria. Acquiring extraneous solutions: minus 3 points. Guessed and verified answer: 1 point.
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3
|
Algebra
|
olympiads
| null | null |
\(\boxed{3}\)
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11.1. Find all positive integer solutions of the equation $(n+2)!-(n+1)!-(n)!=n^{2}+n^{4}$. Answer. $n=3$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Rewrite the equation as $n!=\left(n^{*}\left(n^{2}+1\right)\right) /(n+2)$. Transforming the right side, we get $n!=n^{2}-2 n+5-10:(n+2)$. The last fraction will be an integer for $n
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9.3. Find the value of the expression $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}$, given that $\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}=5$ and $x+y+z=2$.
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Transform: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{x+y+z}{y+z}+\frac{y+x+z}{x+z}+\frac{z+x+y}{x+y}-3=$ $=(x+y+z)\left(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}\right)-3=2 \cdot 5-3=7$.
Grading Criteria. Presence of arithmetic errors: minus 1-2 points. Correct answer calculated on some example: 1 point.
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7
|
Algebra
|
olympiads
| null | null |
\(\boxed{7}\)
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9.3. Find the value of the expression $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}$, given that $\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}=5$ and $x+y+z=2$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Transform: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{x+y+z}{y+z}+\frac{y+x+z}{x+z}+\frac{z+x+y}{x+y}-3=$ $=(x+y+z)\left(\frac{1}{y+z}+\frac{1}{x+z}+\
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8.3. There is a steamship route between the cities of Dzerzhinsk and Lviv. Every midnight, a steamship departs from Dzerzhinsk, arriving exactly eight days later in Lviv. How many steamships will the steamship "Raritet" meet on its way to Dzerzhinsk if it departs from Lviv exactly at midnight and spends the same eight days on the journey?
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As "Raritet" departs from Lviv, a steamer arrives there, which left Dzerzhinsk 8 days ago. By the time "Raritet" arrives at its final destination, 8 days have passed since the initial moment, and at this moment, the last steamer departs from Dzerzhinsk, which "Raritet" meets on its way. Thus, "Raritet" will meet all the steamers that departed from Dzerzhinsk, starting 8 days before the initial time and ending 8 days after the initial time - a total of 17 steamers (if it is considered that the ships meet when one departs and the other arrives at the same moment).
Criteria: only the answer, answer with verification - 0 points.
If it is considered that "Raritet" does not meet the first and/or last steamers - do not deduct points.
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17
|
Combinatorics
|
olympiads
| null | null |
\(\boxed{17}\)
|
8.3. There is a steamship route between the cities of Dzerzhinsk and Lviv. Every midnight, a steamship departs from Dzerzhinsk, arriving exactly eight days later in Lviv. How many steamships will the steamship "Raritet" meet on its way to Dzerzhinsk if it departs from Lviv exactly at midnight and spends the same eight days on the journey?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
As "Raritet" departs from Lviv, a steamer arrives there, which left Dzerzhinsk 8 days ago. By the time "Raritet" arrives at its final destination, 8 days have passed since the initial moment, and at this moment, the last steamer departs from Dzerzhinsk, which "Raritet" meets on its way. Thus, "Raritet" will meet all the steamers that departed from Dzerzhinsk, startin
|
8.5. Egor, Nikita, and Innokentiy took turns playing chess with each other (two play, one watches). After each game, the loser gave up their place at the board to the spectator (there were no draws). In the end, it turned out that Egor participated in 13 games, and Nikita in 27. How many games did Innokentiy play?
|
On the one hand, there were no fewer than 27 games. On the other hand, a player cannot skip two games in a row, meaning each player participates in at least every other game. Therefore, if there were at least 28 games, Egor would have participated in at least 14, which contradicts the condition. Thus, exactly 27 games were played, and Nikita participated in all of them. In 13 of these, his opponent was Egor, so in the remaining 14, it was Innokentiy, and this is the answer.
Criteria: Only the answer, answer with verification - 0 points.
|
14
|
Logic and Puzzles
|
olympiads
| null | null |
\(\boxed{14}\)
|
8.5. Egor, Nikita, and Innokentiy took turns playing chess with each other (two play, one watches). After each game, the loser gave up their place at the board to the spectator (there were no draws). In the end, it turned out that Egor participated in 13 games, and Nikita in 27. How many games did Innokentiy play?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
On the one hand, there were no fewer than 27 games. On the other hand, a player cannot skip two games in a row, meaning each player participates in at least every other game. Therefore, if there were at least 28 games, Egor would have participated in at least 14, which co
|
7.5. Egor, Nikita, and Innokentiy took turns playing chess with each other (two play, one watches). After each game, the loser would give up their place at the board to the spectator (there were no draws). In the end, it turned out that Egor participated in 13 games, and Nikita in 27. How many games did Innokentiy play?
|
On the one hand, there were no fewer than 27 games. On the other hand, a player cannot skip two games in a row, meaning each player participates in at least every other game. Therefore, if there were at least 28 games, Egor would have participated in at least 14, which contradicts the condition. Thus, exactly 27 games were played, and Nikita participated in all of them. In 13 of these, his opponent was Egor, so in the remaining 14, it was Innokentiy, and this is the answer.
Criteria: Only the answer, answer with verification - 0 points.
|
14
|
Logic and Puzzles
|
olympiads
| null | null |
\(\boxed{14}\)
|
7.5. Egor, Nikita, and Innokentiy took turns playing chess with each other (two play, one watches). After each game, the loser would give up their place at the board to the spectator (there were no draws). In the end, it turned out that Egor participated in 13 games, and Nikita in 27. How many games did Innokentiy play?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
On the one hand, there were no fewer than 27 games. On the other hand, a player cannot skip two games in a row, meaning each player participates in at least every other game. Therefore, if there were at least 28 games, Egor would have participated in at least 14, which co
|
8.1. Large sandglasses measure an hour, and small ones measure 11 minutes. How can you use these sandglasses to measure a minute?
|
We will run the large hourglass twice in a row and the small one eleven times in a row. A minute will be measured between the second time the large hourglass finishes (120 minutes) and the 11th time the small one finishes (121 minutes).
Criteria: Any correct example - 7 points.
|
1
|
Logic and Puzzles
|
olympiads
| null | null |
\(\boxed{1}\)
|
8.1. Large sandglasses measure an hour, and small ones measure 11 minutes. How can you use these sandglasses to measure a minute?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
We will run the large hourglass twice in a row and the small one eleven times in a row. A minute will be measured between the second time th
|
9.2. Several consecutive natural numbers are written on the board. It is known that $48 \%$ of them are even, and $36 \%$ of them are less than 30. Find the smallest of the written numbers.
|
$\frac{48}{100}=\frac{12}{25}, \frac{36}{100}=\frac{9}{25}$ - these are irreducible fractions, so the total number of numbers is divisible by 25. If there were 50 or more, then, by the condition, there would be at least 2 fewer even numbers than odd numbers, which is impossible for consecutive natural numbers. Therefore, there are 25, and exactly 9 of them are less than 30. Thus, the first one is 21.
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21
|
Number Theory
|
olympiads
| null | null |
\(\boxed{21}\)
|
9.2. Several consecutive natural numbers are written on the board. It is known that $48 \%$ of them are even, and $36 \%$ of them are less than 30. Find the smallest of the written numbers.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
$\frac{48}{100}=\frac{12}{25}, \frac{36}{100}=\frac{9}{25}$ - these are irreducible fractions, so the total number of numbers is divisible by 25. If there were 50 or more, then, by the condition, there
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8.2. Losharik is going to visit Sovunya along the river at a speed of 4 km/h. Every half hour, he launches paper boats that float to Sovunya at a speed of 10 km/h. With what time interval do the boats arrive at Sovunya?
|
If Losyash launched the boats from one place, they would arrive every half hour. But since he is walking, the next boat has to travel a shorter distance than the previous one. In half an hour, the distance between Losyash and the last boat will be $(10-4) \cdot 0.5=3$. This means that the distance between adjacent boats is 3 km, and the next boat will have to travel exactly this distance when the previous one has already reached Sovunya. The speed of the boat is 10 km/h, which means 1 km in 6 minutes, so 3 km will be covered by the boat in 18 minutes, which is the answer.
Criteria: Correctly found the distance of 3 km - 3 points.
Only the answer or the answer with verification - 1 point.
|
18
|
Other
|
olympiads
| null | null |
\(\boxed{18}\)
|
8.2. Losharik is going to visit Sovunya along the river at a speed of 4 km/h. Every half hour, he launches paper boats that float to Sovunya at a speed of 10 km/h. With what time interval do the boats arrive at Sovunya?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If Losyash launched the boats from one place, they would arrive every half hour. But since he is walking, the next boat has to travel a shorter distance than the previous one. In half an hour, the distance between Losyash and the last boat will be $(10-4) \cdot 0.5=3$. This means that the distance between adjacent boats is 3 km, and the next boat
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8.3. Find the largest four-digit number, all digits of which are different, and which is divisible by each of its digits. Of course, zero cannot be used.
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Since the number of digits is fixed, the number will be larger the larger the digits in its higher places are. We will look for the number in the form $\overline{98 a}$. It must be divisible by 9. Therefore, the sum $a+b$ must give a remainder of 1 when divided by 9. At the same time, this sum does not exceed 13, as it consists of different single-digit addends, which are less than 8. Therefore, this sum is either 1 or 10. The largest variant that fits these conditions is 9873, but it does not work since it is not divisible by 8. The next highest in seniority, 9864, clearly works.
Solution: Only proven that the number 9864 works - 2 points. The idea to look for the answer in the form $\overline{98 a b}-1$ - 1 point.
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9864
|
Number Theory
|
olympiads
| null | null |
\(\boxed{9864}\)
|
8.3. Find the largest four-digit number, all digits of which are different, and which is divisible by each of its digits. Of course, zero cannot be used.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since the number of digits is fixed, the number will be larger the larger the digits in its higher places are. We will look for the number in the form $\overline{98 a}$. It must be divisible by 9. Therefore, the sum $a+b$ must give a remainder of 1 when divided by 9. At the same time, this sum does not exceed 13, as it consists of different single-digit addends
|
7.5. In Nastya's room, 16 people gathered, each pair of whom either are friends or enemies. Upon entering the room, each of them wrote down the number of friends who had already arrived, and upon leaving - the number of enemies remaining in the room. What can the sum of all the numbers written down be, after everyone has first arrived and then left?
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Consider any pair of friends. Their "friendship" was counted exactly once, as it was included in the sum by the person who arrived later than their friend. Therefore, after everyone has arrived, the sum of the numbers on the door will be equal to the total number of friendships between people. Similarly, each "enmity" will be counted exactly once by the person who left earlier. Therefore, after everyone has left, the sum of the numbers on the door will be increased by the total number of "enmities." In total, the overall sum of the numbers on the door will be equal to the sum of the total number of friendships and enmities, which is precisely the number of pairs of people who arrived, i.e., $16 * 15 / 2 = 120$.
Criteria: only answer - 0 points.
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120
|
Combinatorics
|
olympiads
| null | null |
\(\boxed{120}\)
|
7.5. In Nastya's room, 16 people gathered, each pair of whom either are friends or enemies. Upon entering the room, each of them wrote down the number of friends who had already arrived, and upon leaving - the number of enemies remaining in the room. What can the sum of all the numbers written down be, after everyone has first arrived and then left?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Consider any pair of friends. Their "friendship" was counted exactly once, as it was included in the sum by the person who arrived later than their friend. Therefore, after everyone has arrived, the sum of the numbers on the door will be equal to the total number of friendships between people. Similarly, each "enmity" will be counted exactly once by the person who left earlie
|
7.4. Arseny sat down at the computer between 4 and 5 PM, when the hour and minute hands were pointing in opposite directions, and got up from it on the same day between 10 and 11 PM, when the hands coincided. How long did Arseny sit at the computer?
|
Let's see where the hands will be 6 hours after Arseny sat down at the computer. The minute hand will go around the clock 6 times and return to its place. The hour hand will move exactly half a circle. Therefore, the angle between the hands will change by 180 degrees, i.e., the hands will coincide. It is obvious that there is exactly one moment between 22 and 23 hours when the hands coincide, so this will be the time when Arseny gets up from the computer. Thus, Arseny sat at the computer for 6 hours.
## Criteria:
Only the answer - 1 point.
If in a similar solution it is not noted that such a moment between 22 and 23 is exactly one, deduct 1 point.
|
6
|
Logic and Puzzles
|
olympiads
| null | null |
\(\boxed{6}\)
|
7.4. Arseny sat down at the computer between 4 and 5 PM, when the hour and minute hands were pointing in opposite directions, and got up from it on the same day between 10 and 11 PM, when the hands coincided. How long did Arseny sit at the computer?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's see where the hands will be 6 hours after Arseny sat down at the computer. The minute hand will go around the clock 6 times and return to its place. The hour hand will move exactly half a circle. Therefore, the angle between the hands will change by 180 degrees, i.e., the hands will coincide. It is obvious that there is e
|
8.1. Arrange the digits from 1 to 9 in a circle such that any two adjacent digits, when read clockwise, form a composite two-digit number. One example is sufficient.
Place the digits from 1 to 9 in a circle so that any two adjacent digits, when read clockwise, form a composite two-digit number. One example is sufficient.
|
For example, the arrangement 12745639 8 is suitable. Other solutions are possible. Criteria:
Any correct example without explanation - 7 points.
|
127456398
|
Number Theory
|
olympiads
| null | null |
\(\boxed{127456398}\)
|
8.1. Arrange the digits from 1 to 9 in a circle such that any two adjacent digits, when read clockwise, form a composite two-digit number. One example is sufficient.
Place the digits from 1 to 9 in a circle so that any two adjacent digits, when read clockwise, form a composite two-digit number. One example is sufficient.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
For example, the arrangement 12745639 8 is suitable. Other solutions are
|
8.2. In a competition, there are 2018 Dota teams, all of different strengths. In a match between two teams, the stronger one always wins. All teams paired up and played one game. Then they paired up differently and played another game. It turned out that exactly one team won both games. How could this be
|
Let's number the teams in ascending order of strength from 1 to 2018. In the first round, we will have the matches 1 - 2, $3-4, \ldots, 2017$ - 2018, and in the second round - $2018-1, 2-3, 4$ - 5, ..., 2016 - 2017. It is obvious that only the team with the number 2018 will win in both rounds.
## Criteria:
Any correct example without explanation - 7 points.
|
2018
|
Logic and Puzzles
|
olympiads
| null | null |
\(\boxed{2018}\)
|
8.2. In a competition, there are 2018 Dota teams, all of different strengths. In a match between two teams, the stronger one always wins. All teams paired up and played one game. Then they paired up differently and played another game. It turned out that exactly one team won both games. How could this be
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's number the teams in ascending order of strength from 1 to 2018. In the first round, we will have the matches 1 - 2, $3-4, \ldots, 2017$ - 2018, and in the second round - $201
|
6.
Find the roots of the equation $f(x)=8$, if $4 f(3-x)-f(x)=3 x^{2}-4 x-3$ for any real value of $x$. In your answer, specify the product of the found roots. #
|
# Solution:
Notice that when $x$ is replaced by $3-x$, the expression $3-x$ changes to $x$. That is, the pair $f(x)$ and $f(3-x)$ is invariant under this substitution. Replace $x$ with $3-x$ in the equation given in the problem. We get:
$4 f(x) - f(3-x) = 3(3-x)^2 - 4(3-x) - 3 = 3x^2 - 14x + 12$. Express $f(x)$ from the system: $\left\{\begin{array}{l}4 f(3-x) - f(x) = 3x^2 - 4x - 3 \\ 4 f(x) - f(3-x) = 3x^2 - 14x + 12\end{array}\right.$. Multiply the second equation of the system by 4 and add it to the first, we get: $15 f(x) = 15x^2 - 60x + 45$; $f(x) = x^2 - 4x + 3$. Form the equation: $x^2 - 4x + 3 = 8 ; x^2 - 4x - 5 = 0$; its roots are $x_1 = -1 ; x_2 = 5$. The product of the roots is ( -5 ).
|
-5
|
Algebra
|
olympiads
| null | null |
\(\boxed{-5}\)
|
6.
Find the roots of the equation $f(x)=8$, if $4 f(3-x)-f(x)=3 x^{2}-4 x-3$ for any real value of $x$. In your answer, specify the product of the found roots. #
The following text is the beginning part of the answer, which you can refer to for solving the problem:
# Solution:
Notice that when $x$ is replaced by $3-x$, the expression $3-x$ changes to $x$. That is, the pair $f(x)$ and $f(3-x)$ is invariant under this substitution. Replace $x$ with $3-x$ in the equation given in the problem. We get:
$4 f(x) - f(3-x) = 3(3-x)^2 - 4(3-x) - 3 = 3x^2 - 14x + 12$. Express $f(x)$ from the system: $\left\{\begin{array}{
|
1. According to the inverse theorem of Vieta's theorem, we form a quadratic equation. We get $x^{2}-\sqrt{2019} x+248.75=0$.
Next, solving it, we find the roots $a$ and $b$: $a=\frac{\sqrt{2019}}{2}+\frac{32}{2}$ and $b=\frac{\sqrt{2019}}{2}-\frac{32}{2}$, and consequently, the distance between the points $a$ and $b$: $a-b=32$.
|
| 15 points | The correct answer is obtained justifiably |
| :---: | :---: |
| 10 points | The quadratic equation is solved, but an arithmetic error is made or the distance between the points is not found |
| 5 points | The quadratic equation is correctly formulated according to the problem statement. |
| 0 points | The solution does not meet any of the criteria listed above |
|
32
|
Algebra
|
olympiads
| null | null |
\(\boxed{32}\)
|
1. According to the inverse theorem of Vieta's theorem, we form a quadratic equation. We get $x^{2}-\sqrt{2019} x+248.75=0$.
Next, solving it, we find the roots $a$ and $b$: $a=\frac{\sqrt{2019}}{2}+\frac{32}{2}$ and $b=\frac{\sqrt{2019}}{2}-\frac{32}{2}$, and consequently, the distance between the points $a$ and $b$: $a-b=32$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
| 15 points | The correct answer is obtained justifiably |
| :---: | :---: |
| 10 points | The quadratic equation is solved, but an arithmetic error is made or the distance between the point
|
3. What is the smallest value that the sum $S_{n}$ of the first $n$ terms of an arithmetic progression can take, given that the sum $S_{3}=-141$ and the sum $S_{35}=35$?
|
# Solution:
If $a-$ is the first term and $d-$ is the common difference of the arithmetic progression,
$$
\left\{\begin{array} { l }
{ \frac { a + a + 2 d } { 2 } \cdot 3 = - 141 , } \\
{ \frac { a + a + 34 d } { 2 } \cdot 35 = 35 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a+d=-47, \\
a+17 d=1
\end{array} \Rightarrow 16 d=48 ; d=3, a=-50\right.\right.
$$
The sum $S_{n}$ of the first $n$ terms of the arithmetic progression takes the minimum value if $a_{n}<0$, and $a_{n+1} \geq 0$. Since $a_{n}=a+d(n-1)$, from the inequality $-50+3(n-1)<0$ we find $n=[53 / 3]=[172 / 3]=17$. Then $\min S_{n}=S_{17}=0.5 \cdot(-50-50+3 \cdot 16) \cdot 17=-442$. -442.
|
-442
|
Algebra
|
olympiads
| null | null |
\(\boxed{-442}\)
|
3. What is the smallest value that the sum $S_{n}$ of the first $n$ terms of an arithmetic progression can take, given that the sum $S_{3}=-141$ and the sum $S_{35}=35$?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
# Solution:
If $a-$ is the first term and $d-$ is the common difference of the arithmetic progression,
$$
\left\{\begin{array} { l }
{ \frac { a + a + 2 d } { 2 } \cdot 3 = - 141 , } \\
{ \frac { a + a + 34 d } { 2 } \cdot 35 = 35 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a+d=-47, \\
a+17 d=1
\end{array} \Rightarrow 1
|
2. $f(-x)=3(-x)^{3}-(-x)=-3 x^{3}+x=-\left(3 x^{3}-x\right)=-f(x)$ $g(-x)=f^{3}(-x)+f\left(\frac{1}{-x}\right)-8(-x)^{3}-\frac{2}{-x}=-f^{3}(x)-f\left(\frac{1}{x}\right)+8 x^{3}+\frac{2}{x}=-g(x)$
Therefore, $g$ is an odd function $\Rightarrow$ if $x_{0}$ is a root of the original equation, then $-x_{0}$ is also a root of the equation $\Rightarrow$ the sum of the roots is zero, if the roots exist
Check $\quad x=1: \quad f(1)=3 \cdot 1^{3}-1=2, \quad f\left(\frac{1}{1}\right)=f(1)=2, \quad$ substitute $\quad$ into the original equation:10=10 - correct $\Rightarrow x=1$ is a root $\Rightarrow$ the roots exist
|
Problem 8 (2nd version).
Find the sum of the roots of the equation $g^{3}(x)-g\left(\frac{1}{x}\right)=5 x^{3}+\frac{1}{x}$, where $g(x)=x^{3}+x$.
## Solution.
Consider the function $f(x)=g^{3}(x)-g\left(\frac{1}{x}\right)-5 x^{3}-\frac{1}{x}$, then the roots of the original equation are the roots of the equation $f(x)=0$.
Investigate the function $f$ for evenness:
|
0
|
Algebra
|
olympiads
| null | null |
\(\boxed{0}\)
|
2. $f(-x)=3(-x)^{3}-(-x)=-3 x^{3}+x=-\left(3 x^{3}-x\right)=-f(x)$ $g(-x)=f^{3}(-x)+f\left(\frac{1}{-x}\right)-8(-x)^{3}-\frac{2}{-x}=-f^{3}(x)-f\left(\frac{1}{x}\right)+8 x^{3}+\frac{2}{x}=-g(x)$
Therefore, $g$ is an odd function $\Rightarrow$ if $x_{0}$ is a root of the original equation, then $-x_{0}$ is also a root of the equation $\Rightarrow$ the sum of the roots is zero, if the roots exist
Check $\quad x=1: \quad f(1)=3 \cdot 1^{3}-1=2, \quad f\left(\frac{1}{1}\right)=f(1)=2, \quad$ substitute $\quad$ into the original equation:10=10 - correct $\Rightarrow x=1$ is a root $\Rightarrow$ the roots exist
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Problem 8 (2nd version).
Find the sum of the roots of the equation $g^{3}(x)-g\left(\frac{1}{x}\right)=5 x^{3}+\frac{1}{x}$, where $g(x)=x^{3}+x$.
## Solution.
Consider the function $
|
5. Garland (1 b. 6-11). A string of 100 lights was hung on a Christmas tree in a row. Then the lights started switching according to the following algorithm: all lights turned on, after a second, every second light turned off, after another second, every third light switched: if it was on, it turned off and vice versa. After another second, every fourth light switched, then every fifth light, and so on. After 100 seconds, everything was over. Find the probability that a randomly chosen light is on after this (the lights do not burn out or break).
|
Obviously, the bulb with number $n$ will remain on only if it has been switched an odd number of times, that is, if the number $n$ has an odd number of natural divisors. It is clear that only squares satisfy this condition: $n=1,4,9, \ldots, 100$. Thus, 10 bulbs out of 100 will remain on. Therefore, the probability of randomly selecting a bulb that is on is $0.1$.
|
0.1
|
Number Theory
|
olympiads
| null | null |
\(\boxed{0.1}\)
|
5. Garland (1 b. 6-11). A string of 100 lights was hung on a Christmas tree in a row. Then the lights started switching according to the following algorithm: all lights turned on, after a second, every second light turned off, after another second, every third light switched: if it was on, it turned off and vice versa. After another second, every fourth light switched, then every fifth light, and so on. After 100 seconds, everything was over. Find the probability that a randomly chosen light is on after this (the lights do not burn out or break).
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Obviously, the bulb with number $n$ will remain on only if it has been switched an odd number of times, that is, if the number $n$ has an odd number of natural divisors. It is clear t
|
15. Minimum Sum (9th grade, 4 points). There are a lot of symmetric dice. They are thrown simultaneously. With some probability $p>0$, the sum of the points can be 2022. What is the smallest sum of points that can fall when these dice are thrown with the same probability $p$?
|
The distribution of the sum rolled on the dice is symmetric. To make the sum $S$ the smallest possible, the sum of 2022 should be the largest possible. This means that the sum of 2022 is achieved when sixes are rolled on all dice. Therefore, the total number of dice is $2022: 6=337$. The smallest sum of 337 is obtained if ones are rolled on all dice.
|
337
|
Combinatorics
|
olympiads
| null | null |
\(\boxed{337}\)
|
15. Minimum Sum (9th grade, 4 points). There are a lot of symmetric dice. They are thrown simultaneously. With some probability $p>0$, the sum of the points can be 2022. What is the smallest sum of points that can fall when these dice are thrown with the same probability $p$?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The distribution of the sum rolled on the dice is symmetric. To make the sum $S$ the smallest possible, the sum of 2022 should be the largest possible. This means that the sum
|
2. Half-year grade. By the end of the half-year, Vasya Petrov had the following grades in mathematics in his journal: $4,1,2,5,2$. Before assigning the half-year grade, the math teacher told Vasya:
- Vasya, you can choose the method to determine your half-year grade. I offer two options. Method A: the arithmetic mean of the current grades, rounded to the nearest whole number. Method B: the median of the current grades.
The best method for Vasya is the one that gives him the highest grade for the half-year.
a) (from 6th grade. 1 point). Which method is the best for Vasya?
b) (from 6th grade. 2 points). Then the teacher thought and added:
- Keep in mind, Vasya, that if you manage to choose the best method for yourself, I will add two more fives to your journal, and then calculate the half-year grade.
Prove that under these conditions, Method A is not the best for Vasya.
|
# Solution.
a) The average of the current grades is 2.8 (rounded to 3), and the median of the grades is 2. It is better to choose Method A.
b) Suppose Method A is the best. Therefore, the teacher should give two fives. However, the average after rounding remains 3, while the median rises to 4. Thus, Method A now gives less than Method B, so Method A is not the best.
|
3
|
Algebra
|
olympiads
| null | null |
\(\boxed{3}\)
|
2. Half-year grade. By the end of the half-year, Vasya Petrov had the following grades in mathematics in his journal: $4,1,2,5,2$. Before assigning the half-year grade, the math teacher told Vasya:
- Vasya, you can choose the method to determine your half-year grade. I offer two options. Method A: the arithmetic mean of the current grades, rounded to the nearest whole number. Method B: the median of the current grades.
The best method for Vasya is the one that gives him the highest grade for the half-year.
a) (from 6th grade. 1 point). Which method is the best for Vasya?
b) (from 6th grade. 2 points). Then the teacher thought and added:
- Keep in mind, Vasya, that if you manage to choose the best method for yourself, I will add two more fives to your journal, and then calculate the half-year grade.
Prove that under these conditions, Method A is not the best for Vasya.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
# Solution.
a) The average of the current grades is 2.8 (rounded to 3), and the median of the grades is 2. It is better to choose Method A.
b) Suppose Method A is the best. Therefore,
|
2. Half-year grade. By the end of the half-year, Vasya Petrov had the following grades in mathematics in his journal: $4,1,2,5,2$. Before assigning the half-year grade, the math teacher told Vasya:
- Vasya, you can choose the method to determine your half-year grade. I offer two options. Method A: the arithmetic mean of the current grades, rounded to the nearest whole number. Method B: the median of the current grades.
The best method for Vasya is the one that gives him the highest grade for the half-year.
a) (from 6th grade. 1 point). Which method is the best for Vasya?
b) (from 6th grade. 2 points). Then the teacher thought and added:
- Keep in mind, Vasya, that if you manage to choose the best method for yourself, I will add two more fives to your journal, and then calculate the half-year grade.
Prove that under these conditions, Method A is not the best for Vasya.
|
# Solution.
a) The average of the current grades is 2.8 (rounded to 3), and the median of the grades is 2. It is better to choose Method A.
b) Suppose Method A is the best. Therefore, the teacher should give two fives. However, the average after rounding remains 3, while the median rises to 4. Thus, Method A now gives less than Method B, so Method A is not the best.
|
3
|
Algebra
|
olympiads
| null | null |
\(\boxed{3}\)
|
2. Half-year grade. By the end of the half-year, Vasya Petrov had the following grades in mathematics in his journal: $4,1,2,5,2$. Before assigning the half-year grade, the math teacher told Vasya:
- Vasya, you can choose the method to determine your half-year grade. I offer two options. Method A: the arithmetic mean of the current grades, rounded to the nearest whole number. Method B: the median of the current grades.
The best method for Vasya is the one that gives him the highest grade for the half-year.
a) (from 6th grade. 1 point). Which method is the best for Vasya?
b) (from 6th grade. 2 points). Then the teacher thought and added:
- Keep in mind, Vasya, that if you manage to choose the best method for yourself, I will add two more fives to your journal, and then calculate the half-year grade.
Prove that under these conditions, Method A is not the best for Vasya.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
# Solution.
a) The average of the current grades is 2.8 (rounded to 3), and the median of the grades is 2. It is better to choose Method A.
b) Suppose Method A is the best. Therefore,
|
2. Half-year grade. By the end of the half-year, Vasya Petrov had the following grades in mathematics in his journal: $4,1,2,5,2$. Before assigning the half-year grade, the math teacher told Vasya:
- Vasya, you can choose the method to determine your half-year grade. I offer two options. Method A: the arithmetic mean of the current grades, rounded to the nearest whole number. Method B: the median of the current grades.
The best method for Vasya is the one that gives him the highest grade for the half-year.
a) (from 6th grade. 1 point). Which method is the best for Vasya?
b) (from 6th grade. 2 points). Then the teacher thought and added:
- Keep in mind, Vasya, that if you manage to choose the best method for yourself, I will add two more fives to your journal, and then calculate the half-year grade.
Prove that under these conditions, Method A is not the best for Vasya.
|
# Solution.
a) The average of the current grades is 2.8 (rounded to 3), and the median of the grades is 2. It is better to choose Method A.
b) Suppose Method A is the best. Therefore, the teacher should give two fives. However, the average after rounding remains 3, while the median rises to 4. Thus, Method A now gives less than Method B, so Method A is not the best.
|
3
|
Algebra
|
olympiads
| null | null |
\(\boxed{3}\)
|
2. Half-year grade. By the end of the half-year, Vasya Petrov had the following grades in mathematics in his journal: $4,1,2,5,2$. Before assigning the half-year grade, the math teacher told Vasya:
- Vasya, you can choose the method to determine your half-year grade. I offer two options. Method A: the arithmetic mean of the current grades, rounded to the nearest whole number. Method B: the median of the current grades.
The best method for Vasya is the one that gives him the highest grade for the half-year.
a) (from 6th grade. 1 point). Which method is the best for Vasya?
b) (from 6th grade. 2 points). Then the teacher thought and added:
- Keep in mind, Vasya, that if you manage to choose the best method for yourself, I will add two more fives to your journal, and then calculate the half-year grade.
Prove that under these conditions, Method A is not the best for Vasya.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
# Solution.
a) The average of the current grades is 2.8 (rounded to 3), and the median of the grades is 2. It is better to choose Method A.
b) Suppose Method A is the best. Therefore,
|
2. Half-year grade. By the end of the half-year, Vasya Petrov had the following grades in mathematics in his journal: $4,1,2,5,2$. Before assigning the half-year grade, the math teacher told Vasya:
- Vasya, you can choose the method to determine your half-year grade. I offer two options. Method A: the arithmetic mean of the current grades, rounded to the nearest whole number. Method B: the median of the current grades.
The best method for Vasya is the one that gives him the highest grade for the half-year.
a) (from 6th grade. 1 point). Which method is the best for Vasya?
b) (from 6th grade. 2 points). Then the teacher thought and added:
- Keep in mind, Vasya, that if you manage to choose the best method for yourself, I will add two more fives to your journal, and then calculate the half-year grade.
Prove that under these conditions, Method A is not the best for Vasya.
|
# Solution.
a) The average of the current grades is 2.8 (rounded to 3), and the median of the grades is 2. It is better to choose Method A.
b) Suppose Method A is the best. Therefore, the teacher should give two fives. However, the average after rounding remains 3, while the median rises to 4. Thus, Method A now gives less than Method B, so Method A is not the best.
|
3
|
Algebra
|
olympiads
| null | null |
\(\boxed{3}\)
|
2. Half-year grade. By the end of the half-year, Vasya Petrov had the following grades in mathematics in his journal: $4,1,2,5,2$. Before assigning the half-year grade, the math teacher told Vasya:
- Vasya, you can choose the method to determine your half-year grade. I offer two options. Method A: the arithmetic mean of the current grades, rounded to the nearest whole number. Method B: the median of the current grades.
The best method for Vasya is the one that gives him the highest grade for the half-year.
a) (from 6th grade. 1 point). Which method is the best for Vasya?
b) (from 6th grade. 2 points). Then the teacher thought and added:
- Keep in mind, Vasya, that if you manage to choose the best method for yourself, I will add two more fives to your journal, and then calculate the half-year grade.
Prove that under these conditions, Method A is not the best for Vasya.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
# Solution.
a) The average of the current grades is 2.8 (rounded to 3), and the median of the grades is 2. It is better to choose Method A.
b) Suppose Method A is the best. Therefore,
|
2. Ninth-grader Gavriil decided to weigh a basketball, but he only had 400 g weights, a light ruler with the markings at the ends worn off, a pencil, and many weightless threads at his disposal. Gavriil suspended the ball from one end of the ruler and the weight from the other, and balanced the ruler on the pencil. Then he attached a second weight to the first, and to restore balance, he had to move the pencil 9 cm. When a third weight was attached to the first two, and the pencil was moved another 5 cm, balance was restored again. Calculate the mass of the ball, as Gavriil did.
|
Let the distances from the pencil to the ball and to the weight be $l_{1}$ and $l_{2}$ respectively at the first equilibrium. Denote the magnitude of the first shift by $x$, and the total shift over two times by $y$. Then the three conditions of lever equilibrium will be:
$$
\begin{gathered}
M l_{1}=m l_{2} \\
M\left(l_{1}+x\right)=2 m\left(l_{2}-x\right) \\
M\left(l_{1}+y\right)=3 m\left(l_{2}-y\right)
\end{gathered}
$$
Subtracting the first equation from the second and third, we get:
$$
\begin{gathered}
M x=m l_{2}-2 m x \\
M y=2 m l_{2}-3 m y
\end{gathered}
$$
From this,
$$
M=\frac{3 y-4 x}{2 x-y} m=600
$$
|
600
|
Algebra
|
olympiads
| null | null |
\(\boxed{600}\)
|
2. Ninth-grader Gavriil decided to weigh a basketball, but he only had 400 g weights, a light ruler with the markings at the ends worn off, a pencil, and many weightless threads at his disposal. Gavriil suspended the ball from one end of the ruler and the weight from the other, and balanced the ruler on the pencil. Then he attached a second weight to the first, and to restore balance, he had to move the pencil 9 cm. When a third weight was attached to the first two, and the pencil was moved another 5 cm, balance was restored again. Calculate the mass of the ball, as Gavriil did.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let the distances from the pencil to the ball and to the weight be $l_{1}$ and $l_{2}$ respectively at the first equilibrium. Denote the magnitude of the first shift by $x$, and the total shift over two times by $y$. Then the three conditions of lever equilibrium will be:
$$
\begin{gathered}
M l_{1}=m l_{2}
|
1. 2. A car with a load traveled from one city to another at a speed of 60 km/h, and returned empty at a speed of 90 km/h. Find the average speed of the car for the entire route. Give your answer in kilometers per hour, rounding to the nearest whole number if necessary.
$\{72\}$
|
The average speed will be the total distance divided by the total time: $\frac{2 S}{\frac{S}{V_{1}}+\frac{S}{V_{2}}}=\frac{2 V_{1} \cdot V_{2}}{V_{1}+V_{2}}=\frac{2 \cdot 60 \cdot 90}{60+90}=72(\mathrm{km} / \mathrm{h})$.
|
72
|
Algebra
|
olympiads
| null | null |
\(\boxed{72}\)
|
1. 2. A car with a load traveled from one city to another at a speed of 60 km/h, and returned empty at a speed of 90 km/h. Find the average speed of the car for the entire route. Give your answer in kilometers per hour, rounding to the nearest whole number if necessary.
$\{72\}$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The average speed will be the total distance divided by the total time: $\frac{2 S}{\frac{S}{V_{1}}+\frac{S}{V
|
4.1. From cities $A$ and $Б$, which are 240 km apart, two cars set off towards each other simultaneously at speeds of 60 km/h and 80 km/h. At what distance from point $C$, located halfway between $A$ and $Б$, will the cars meet? Give the answer in kilometers, rounding to the nearest hundredth if necessary.
## $\{17.14\}$
|
Time until meeting $=240 /(60+80)=12 / 7$ hours. In this time, the first car will travel $60 \cdot 12 / 7=720 / 7$ km. The required distance $=$ $120-60 \cdot 12 / 7=120 \cdot(1-6 / 7)=120 / 7 \approx 17.1428 \ldots$ (km).
|
17.14
|
Algebra
|
olympiads
| null | null |
\(\boxed{17.14}\)
|
4.1. From cities $A$ and $Б$, which are 240 km apart, two cars set off towards each other simultaneously at speeds of 60 km/h and 80 km/h. At what distance from point $C$, located halfway between $A$ and $Б$, will the cars meet? Give the answer in kilometers, rounding to the nearest hundredth if necessary.
## $\{17.14\}$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Time until meeting $=240 /(60+80)=12 / 7$ hours. In this time, the first car will travel $60 \cdot 12 / 7=720 /
|
12.1. Two motorcyclists are moving along straight tracks (each on their own) at constant speeds. At 11:14, the distance between them was 40 km, at 11:46 - 30 km, at 12:10 - 30 km. Determine the magnitude of the relative speed of one motorcyclist relative to the other (in kilometers per hour). Provide the answer as a whole number, rounding it if necessary.
|
We solve the problem in a coordinate system associated with the first motorcyclist, denoted as $A$. The positions of the second motorcyclist: at the first moment of time $-B$; at the second moment $-C$; at the third $-D$. Then $A B=$ $40, A C=30, A D=30$.
Since $B C: C D=\frac{46-14}{70-46}=\frac{32}{24}=4: 3=A B: A D$, then $A C$ is the bisector of $\triangle A B D$. By the formula for the length of the bisector, $30^{2}=40 \cdot 30-4 x \cdot 3 x$ (here $B C=4 x ; C D=3 x$). Therefore, $x=5, C D=15$ km. Since the distance $C D$ is covered in 24 minutes, the required relative speed is $\frac{15 \cdot 60}{24}=\frac{75}{2}=37.5$ km/h.
|
37.5
|
Algebra
|
olympiads
| null | null |
\(\boxed{37.5}\)
|
12.1. Two motorcyclists are moving along straight tracks (each on their own) at constant speeds. At 11:14, the distance between them was 40 km, at 11:46 - 30 km, at 12:10 - 30 km. Determine the magnitude of the relative speed of one motorcyclist relative to the other (in kilometers per hour). Provide the answer as a whole number, rounding it if necessary.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
We solve the problem in a coordinate system associated with the first motorcyclist, denoted as $A$. The positions of the second motorcyclist: at the first moment of time $-B$; at the second moment $-C$; at the third $-D$. Then $A B=$ $40, A C=30, A D=30$.
Since $B C: C D=\frac{46-14}{70-46}=\frac{32}{24}=4: 3=A B: A D
|
4. Gavrila placed 7 smaller boxes into a large box. After that, Glafira placed 7 small boxes into some of these seven boxes, and left others empty. Then Gavrila placed 7 boxes into some of the empty boxes, and left others empty. Glafira repeated this operation and so on. At some point, there were 34 non-empty boxes. How many empty boxes were there at this moment?
|
Instructions. Filling one box increases the number of empty boxes by 7-1=6, and the number of non-empty boxes by 1. Therefore, after filling $n$ boxes (regardless of the stage), the number of boxes will be: empty $-1+6 n$; non-empty $-n$. Thus, $n=34$, and the number of non-empty boxes will be $1+6 \cdot 34=205$.
|
205
|
Logic and Puzzles
|
olympiads
| null | null |
\(\boxed{205}\)
|
4. Gavrila placed 7 smaller boxes into a large box. After that, Glafira placed 7 small boxes into some of these seven boxes, and left others empty. Then Gavrila placed 7 boxes into some of the empty boxes, and left others empty. Glafira repeated this operation and so on. At some point, there were 34 non-empty boxes. How many empty boxes were there at this moment?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Instructions. Filling one box increases the number of empty boxes by 7-1=6, and the number of non-empty boxes by 1. Therefore, after filling $n$ boxes (regar
|
1. Gavriil decided to weigh a football, but he only had weights of 150 g, a long light ruler with the markings at the ends worn off, a pencil, and many threads at his disposal. He suspended the ball from one end of the ruler and the weight from the other, and balanced the ruler on the pencil. Then he attached a second weight to the first, and to restore balance, he had to move the pencil 6 cm. When a third weight was attached to the first two, and the pencil was moved another 4 cm, balance was restored again. Calculate the mass of the ball, as Gavriil did.
|
Let the distances from the pencil to the ball and to the weight be $l_{1}$ and $l_{2}$ respectively at the first equilibrium. Denote the magnitude of the first shift by $x$, and the total shift over two times by $y$. Then the three conditions of lever equilibrium will be:
$$
\begin{gathered}
M l_{1}=m l_{2} \\
M\left(l_{1}+x\right)=2 m\left(l_{2}-x\right) \\
M\left(l_{1}+y\right)=3 m\left(l_{2}-y\right)
\end{gathered}
$$
Subtracting the first equation from the second and third, we get:
$$
\begin{gathered}
M x=m l_{2}-2 m x \\
M y=2 m l_{2}-3 m y
\end{gathered}
$$
From this,
$$
M=\frac{3 y-4 x}{2 x-y} m=600 \text { g. }
$$
In the variant with the football, the formula is the same, the answer is 450 g.
|
600
|
Algebra
|
olympiads
| null | null |
\(\boxed{600}\)
|
1. Gavriil decided to weigh a football, but he only had weights of 150 g, a long light ruler with the markings at the ends worn off, a pencil, and many threads at his disposal. He suspended the ball from one end of the ruler and the weight from the other, and balanced the ruler on the pencil. Then he attached a second weight to the first, and to restore balance, he had to move the pencil 6 cm. When a third weight was attached to the first two, and the pencil was moved another 4 cm, balance was restored again. Calculate the mass of the ball, as Gavriil did.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let the distances from the pencil to the ball and to the weight be $l_{1}$ and $l_{2}$ respectively at the first equilibrium. Denote the magnitude of the first shift by $x$, and the total shift over two times by $y$. Then the three conditions of lever equilibrium will be:
$$
\begin{gathered}
M l_{1}=m l_{2} \\
M\left(l_{1}+x\right)=2 m\left(l_{2}-x\right)
|
. (8 points)
The fast-food network "Wings and Legs" offers summer jobs to schoolchildren. The salary is 25000 rubles per month. Those who work well receive a monthly bonus of 5000 rubles.
How much will a schoolchild who works well at "Wings and Legs" earn per month (receive after tax) after the income tax is deducted?
In your answer, provide only the number without units of measurement!
|
The total earnings will be 25000 rubles + 5000 rubles $=30000$ rubles
Income tax $13 \%-3900$ rubles
The net payment will be 30000 rubles - 3900 rubles $=26100$ rubles
## Correct answer: 26100
## 2nd Option
|
26100
|
Algebra
|
olympiads
| null | null |
\(\boxed{26100}\)
|
. (8 points)
The fast-food network "Wings and Legs" offers summer jobs to schoolchildren. The salary is 25000 rubles per month. Those who work well receive a monthly bonus of 5000 rubles.
How much will a schoolchild who works well at "Wings and Legs" earn per month (receive after tax) after the income tax is deducted?
In your answer, provide only the number without units of measurement!
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The total earnings will be 25000 rubles + 5000 rubles $=30000$ rubles
Income tax $13 \%-3900$ rubles
Th
|
. (8 points)
Announcement: "Have free time and want to earn money? Write now and earn up to 2500 rubles a day working as a courier with the service 'Food.There-Here!'. Delivery of food from stores, cafes, and restaurants.
How much will a school student working as a courier with the service 'Food.There-Here!' earn in a month (4 weeks), working 4 days a week with a minimum load for 1250 rubles a day after the income tax deduction?
In your answer, provide only the number without units of measurement! #
|
# Solution:
The total earnings will be (1250 rubles * 4 days) * 4 weeks = 20000 rubles
Income tax 13% - 2600 rubles
The amount of earnings (net pay) will be 20000 rubles - 2600 rubles = 17400 rubles
Correct answer: 17400
|
17400
|
Algebra
|
olympiads
| null | null |
\(\boxed{17400}\)
|
. (8 points)
Announcement: "Have free time and want to earn money? Write now and earn up to 2500 rubles a day working as a courier with the service 'Food.There-Here!'. Delivery of food from stores, cafes, and restaurants.
How much will a school student working as a courier with the service 'Food.There-Here!' earn in a month (4 weeks), working 4 days a week with a minimum load for 1250 rubles a day after the income tax deduction?
In your answer, provide only the number without units of measurement! #
The following text is the beginning part of the answer, which you can refer to for solving the problem:
# Solution:
The total earnings will be (1250 rubles * 4 days) * 4 weeks = 20000 rubles
Income tax 13% - 2600 r
|
# Problem 6. (10 points)
Vasily is planning to graduate from college in a year. Only 270 out of 300 third-year students successfully pass their exams and complete their bachelor's degree. If Vasily ends up among the 30 expelled students, he will have to work with a monthly salary of 25,000 rubles. It is also known that every fifth graduate gets a job with a salary of 60,000 rubles per month, every tenth graduate earns 80,000 rubles per month, every twentieth graduate cannot find a job in their field and has an average salary of 25,000 rubles per month, while the salary of all others is 40,000 rubles. When Vasily finished school, he could have chosen not to go to college and instead work as a real estate assistant, like his friend Fyodor did. Fyodor's salary increases by 3,000 rubles each year. What is Vasily's expected salary? Whose salary will be higher in a year and by how much - Vasily's expected salary or Fyodor's actual salary, if Fyodor started working with a salary of 25,000 rubles at the same time Vasily enrolled in college? Note: Bachelor's degree education lasts 4 years.
|
# Solution:
In 4 years after graduating from school, Fedor will earn $25000 + 3000 * 4 = 37000$ rubles (2 points).
The expected salary of Vasily is the expected value of the salary Vasily can earn under all possible scenarios (2 points). It will be 270/300 * $(1 / 5 * 60000 + 1 / 10 * 80$ $000 + 1 / 20 * 25000 + (1 - 1 / 5 - 1 / 10 - 1 / 20) * 40000) + 30 / 300 * 25000 = 45025$ rubles (6 points).
|
45025
|
Algebra
|
olympiads
| null | null |
\(\boxed{45025}\)
|
# Problem 6. (10 points)
Vasily is planning to graduate from college in a year. Only 270 out of 300 third-year students successfully pass their exams and complete their bachelor's degree. If Vasily ends up among the 30 expelled students, he will have to work with a monthly salary of 25,000 rubles. It is also known that every fifth graduate gets a job with a salary of 60,000 rubles per month, every tenth graduate earns 80,000 rubles per month, every twentieth graduate cannot find a job in their field and has an average salary of 25,000 rubles per month, while the salary of all others is 40,000 rubles. When Vasily finished school, he could have chosen not to go to college and instead work as a real estate assistant, like his friend Fyodor did. Fyodor's salary increases by 3,000 rubles each year. What is Vasily's expected salary? Whose salary will be higher in a year and by how much - Vasily's expected salary or Fyodor's actual salary, if Fyodor started working with a salary of 25,000 rubles at the same time Vasily enrolled in college? Note: Bachelor's degree education lasts 4 years.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
# Solution:
In 4 years after graduating from school, Fedor will earn $25000 + 3000 * 4 = 37000$ rubles (2 points).
The expected salary of Vasily is the expected value of the salary Vasily can earn u
|
2. A stationery store is running a promotion: there is a sticker on each notebook, and for every 5 stickers, a customer can get another notebook (also with a sticker). Fifth-grader Katya thinks she needs to buy as many notebooks as possible before the new semester. Each notebook costs 4 rubles, and Katya has 150 rubles. How many notebooks will Katya get?
|
Katya buys 37 notebooks for 148 rubles.
2) For 35 stickers, Katya receives 7 more notebooks, after which she has notebooks and 9 stickers.
3) For 5 stickers, Katya receives a notebook, after which she has 45 notebooks and 5 stickers.
4) For 5 stickers, Katya receives the last 46th notebook.
|
46
|
Number Theory
|
olympiads
| null | null |
\(\boxed{46}\)
|
2. A stationery store is running a promotion: there is a sticker on each notebook, and for every 5 stickers, a customer can get another notebook (also with a sticker). Fifth-grader Katya thinks she needs to buy as many notebooks as possible before the new semester. Each notebook costs 4 rubles, and Katya has 150 rubles. How many notebooks will Katya get?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Katya buys 37 notebooks for 148 rubles.
2) For 35 stickers, Katya receives 7 more notebooks, after which she has notebooks and 9 stickers.
3) For
|
5. Vanya decided to give Masha a bouquet of an odd number of flowers for her birthday, consisting of yellow and red tulips, so that the number of flowers of one color differs from the number of flowers of the other by exactly one. Yellow tulips cost 50 rubles each, and red ones cost 31 rubles. What is the largest number of tulips he can buy for Masha's birthday, spending no more than 600 rubles?
|
A bouquet with one more red tulip than yellow ones is cheaper than a bouquet with the same total number of flowers but one more yellow tulip. Therefore, Vanya should buy a bouquet with one more red tulip. The remaining flowers can be paired into red and yellow tulips, with each pair costing 81 rubles. Let's find the number of pairs of different colored tulips in the bouquet: $600: 81=7$ (remainder 33). This means Vanya needs a bouquet with $7 * 2+1=15$ tulips.
|
15
|
Algebra
|
olympiads
| null | null |
\(\boxed{15}\)
|
5. Vanya decided to give Masha a bouquet of an odd number of flowers for her birthday, consisting of yellow and red tulips, so that the number of flowers of one color differs from the number of flowers of the other by exactly one. Yellow tulips cost 50 rubles each, and red ones cost 31 rubles. What is the largest number of tulips he can buy for Masha's birthday, spending no more than 600 rubles?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
A bouquet with one more red tulip than yellow ones is cheaper than a bouquet with the same total number of flowers but one more yellow tulip. Therefore, Vanya should buy a bouquet with one more red tulip. The remaining flowers can b
|
Problem 7. (6 points)
Sergei, being a student, worked part-time in a student cafe after classes for a year. Sergei's salary was 9000 rubles per month. In the same year, Sergei paid for his medical treatment at a healthcare facility in the amount of 100000 rubles and purchased medications on a doctor's prescription for 20000 rubles (eligible for deduction). The following year, Sergei decided to apply for a social tax deduction. What amount of the paid personal income tax is subject to refund to Sergei from the budget? (Provide the answer as a whole number, without spaces and units of measurement.)
## Answer: 14040.
## Comment:
|
the amount of the social tax deduction for medical treatment will be: $100000+20000=$ 120000 rubles. The possible tax amount eligible for refund under this deduction will be $120000 \times 13\% = 15600$ rubles. However, in the past year, Sergey paid income tax (NDFL) in the amount of $13\% \times (9000 \times 12) = 14040$ rubles. Therefore, the amount of personal income tax paid, eligible for refund to Sergey from the budget, will be 14040 rubles.
|
14040
|
Algebra
|
olympiads
| null | null |
\(\boxed{14040}\)
|
Problem 7. (6 points)
Sergei, being a student, worked part-time in a student cafe after classes for a year. Sergei's salary was 9000 rubles per month. In the same year, Sergei paid for his medical treatment at a healthcare facility in the amount of 100000 rubles and purchased medications on a doctor's prescription for 20000 rubles (eligible for deduction). The following year, Sergei decided to apply for a social tax deduction. What amount of the paid personal income tax is subject to refund to Sergei from the budget? (Provide the answer as a whole number, without spaces and units of measurement.)
## Answer: 14040.
## Comment:
The following text is the beginning part of the answer, which you can refer to for solving the problem:
the amount of the social tax deduction for medical treatment will be: $100000+20000=$ 120000 rubles. The possible tax amount eligible for refund under this deduction will be $120000 \times 13\% = 15600$ rubles. However, in the
|
Problem 10. (4 points)
To buy new headphones costing 275 rubles, Katya decided to save money on sports activities. Until now, she has been buying a single-visit ticket to the swimming pool, including a visit to the sauna for 250 rubles, to warm up. However, summer has arrived, and the need to visit the sauna has disappeared. Visiting only the swimming pool costs 200 rubles more than visiting the sauna. How many times does Katya need to visit the swimming pool without the sauna to save enough to buy the headphones? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 11.
## Comment:
|
one visit to the sauna costs 25 rubles, the price of one visit to the swimming pool is 225 rubles. Katya needs to visit the swimming pool 11 times without going to the sauna in order to save up for buying headphones.
|
11
|
Algebra
|
olympiads
| null | null |
\(\boxed{11}\)
|
Problem 10. (4 points)
To buy new headphones costing 275 rubles, Katya decided to save money on sports activities. Until now, she has been buying a single-visit ticket to the swimming pool, including a visit to the sauna for 250 rubles, to warm up. However, summer has arrived, and the need to visit the sauna has disappeared. Visiting only the swimming pool costs 200 rubles more than visiting the sauna. How many times does Katya need to visit the swimming pool without the sauna to save enough to buy the headphones? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 11.
## Comment:
The following text is the beginning part of the answer, which you can refer to for solving the problem:
one visit to the sauna costs 25 rubles, the price of one visit to the swimming pool is 225 rubles. Katya nee
|
Problem 18. (4 points)
By producing and selling 4000 items at a price of 6250 rubles each, a budding businessman earned 2 million rubles in profit. Variable costs for one item amounted to 3750 rubles. By what percentage should the businessman reduce the production volume to make his revenue equal to the cost? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 20.
## Comment:
|
fixed costs $=-2$ million, gap $+4000 \times 6250-3750 \times 4000$ million $=8$ million.
$6.25 \times Q=3750 Q+8 \text{ million}$.
$Q=3200$ units.
Can be taken out of production $=4000-3200=800$ units, that is, $20\%$.
|
20
|
Algebra
|
olympiads
| null | null |
\(\boxed{20}\)
|
Problem 18. (4 points)
By producing and selling 4000 items at a price of 6250 rubles each, a budding businessman earned 2 million rubles in profit. Variable costs for one item amounted to 3750 rubles. By what percentage should the businessman reduce the production volume to make his revenue equal to the cost? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 20.
## Comment:
The following text is the beginning part of the answer, which you can refer to for solving the problem:
fixed costs $=-2$ million, gap $+4000 \times 6250-3750 \times 4000$ million $=8$ million.
$6.25 \times Q=3750
|
1. In the "Sweet Tooth" store, they are thinking of what promotion to announce before March 8. Manager Vasya suggests reducing the price of a box of candies by $20 \%$ and hopes to sell twice as much of the product as usual. Deputy Director Kolya says that it would be more profitable to increase the price of the same box of candies by a third and announce a promotion "buy two, get one free," in which case sales will remain the same (not counting the free items). In whose version of the promotion will the revenue be higher? In the answer, write how much more revenue there will be if the usual revenue from selling boxes of candies is 10000?
|
Let the box of candies cost $\mathrm{X}$ rubles.
In Vasya's option, the revenue will be $(10000 / \mathrm{X}) * 2 * 0.8 * \mathrm{X}=16000$ rubles.
In Kolya's option: $(10000 / X) * 4 / 3 * X=13333.(3)$.
Thus, the revenue will be higher in Vasya's option, and it will be 6000 rubles more than usual.
|
6000
|
Algebra
|
olympiads
| null | null |
\(\boxed{6000}\)
|
1. In the "Sweet Tooth" store, they are thinking of what promotion to announce before March 8. Manager Vasya suggests reducing the price of a box of candies by $20 \%$ and hopes to sell twice as much of the product as usual. Deputy Director Kolya says that it would be more profitable to increase the price of the same box of candies by a third and announce a promotion "buy two, get one free," in which case sales will remain the same (not counting the free items). In whose version of the promotion will the revenue be higher? In the answer, write how much more revenue there will be if the usual revenue from selling boxes of candies is 10000?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let the box of candies cost $\mathrm{X}$ rubles.
In Vasya's option, the revenue will be $(10000 / \mathrm{X}) * 2 * 0.8 * \mathrm{X}=16000$ rubles.
I
|
7. Alexey plans to buy one of two car brands: "A" for 900 thousand rubles or "B" for 600 thousand rubles. On average, Alexey drives 15 thousand km per year. The cost of gasoline is 40 rubles per liter. The cars consume the same type of gasoline. The car is planned to be used for 5 years, after which Alexey will be able to sell the car of brand "A" for 500 thousand rubles, and the car of brand "B" for 350 thousand rubles.
| Car Brand | Fuel Consumption (l/100km) | Annual Insurance Cost (rubles) | Average Annual Maintenance Cost (rubles) |
| :--- | :--- | :--- | :--- |
| "A" | 9 | 35000 | 25000 |
| "B" | 10 | 32000 | 20000 |
Using the data in the table, answer the question: how much more expensive will the purchase and ownership of the more expensive car be for Alexey?
|
Use of car brand "A":
$900000+(15000 / 100) * 9 * 5 * 40+35000 * 5+25000 * 5-500000=970000$
Use of car brand "B":
$600000+(15000 / 100) * 10 * 5 * 40+32000 * 5+20000 * 5-350000=810000$
Difference: $970000-810000=160000$
|
160000
|
Algebra
|
olympiads
| null | null |
\(\boxed{160000}\)
|
7. Alexey plans to buy one of two car brands: "A" for 900 thousand rubles or "B" for 600 thousand rubles. On average, Alexey drives 15 thousand km per year. The cost of gasoline is 40 rubles per liter. The cars consume the same type of gasoline. The car is planned to be used for 5 years, after which Alexey will be able to sell the car of brand "A" for 500 thousand rubles, and the car of brand "B" for 350 thousand rubles.
| Car Brand | Fuel Consumption (l/100km) | Annual Insurance Cost (rubles) | Average Annual Maintenance Cost (rubles) |
| :--- | :--- | :--- | :--- |
| "A" | 9 | 35000 | 25000 |
| "B" | 10 | 32000 | 20000 |
Using the data in the table, answer the question: how much more expensive will the purchase and ownership of the more expensive car be for Alexey?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Use of car brand "A":
$900000+(15000 / 100) * 9 * 5 * 40+35000 * 5+25000 * 5-500000=970000$
Use of car brand "
|
Problem 9. (12 points)
Ivan, a full-time student, started working at his long-desired job on March 1 of the current year. In the first two months, during the probationary period, Ivan's salary was 20000 rubles. After the probationary period, the salary increased to 25000 rubles per month. In December, for exceeding the plan, Ivan was awarded a bonus of 10000 rubles. In addition, while studying in a full-time budget-funded graduate program, Ivan received a scholarship of 2800 rubles per month throughout the year. What amount of personal income tax should Ivan pay to the budget? (Provide the answer as a whole number, without spaces and units of measurement)
|
## Comment
Solution: Personal Income Tax from salary $=(20000 \times 2+25000 \times 8+10000) \times 13\% = 32500$ rubles. The scholarship is not subject to Personal Income Tax.
|
32500
|
Algebra
|
olympiads
| null | null |
\(\boxed{32500}\)
|
Problem 9. (12 points)
Ivan, a full-time student, started working at his long-desired job on March 1 of the current year. In the first two months, during the probationary period, Ivan's salary was 20000 rubles. After the probationary period, the salary increased to 25000 rubles per month. In December, for exceeding the plan, Ivan was awarded a bonus of 10000 rubles. In addition, while studying in a full-time budget-funded graduate program, Ivan received a scholarship of 2800 rubles per month throughout the year. What amount of personal income tax should Ivan pay to the budget? (Provide the answer as a whole number, without spaces and units of measurement)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
## Comment
Solution: Personal Income Tax from salary $=(20000 \times 2+25000 \times 8+1
|
# Problem 6. (8 points)
Vasily is planning to graduate from the institute in a year. Only 270 out of 300 third-year students successfully pass their exams and complete their bachelor's degree. If Vasily ends up among the 30 expelled students, he will have to work with a monthly salary of 25,000 rubles. It is also known that every fifth graduate gets a job with a salary of 60,000 rubles per month, every tenth graduate earns 80,000 rubles per month, every twentieth graduate cannot find a job in their field and has an average salary of 25,000 rubles per month, while the salary of all others is 40,000 rubles.
When Vasily finished school, he could have chosen not to go to the institute but to work as a real estate assistant, as his friend Fyodor did. Fyodor's salary increases by 3,000 rubles each year. What is Vasily's expected salary? Whose salary will be higher in a year and by how much - Vasily's expected salary or Fyodor's actual salary, if Fyodor started working with a salary of 25,000 rubles at the same time Vasily enrolled in the institute?
Note: Bachelor's education lasts 4 years.
|
Four years after graduating from school, Fedor will earn $25000 + 3000 * 4 = 37000$ rubles (2 points)
The expected salary of Vasily is the expected value of the salary Vasily can earn under all possible scenarios (2 points). It will be 270/300*(1/5*60 000 + 1/10*80 000 + 1/20*25 000 + (1 - 1/5 - 1/10 - 1/20) * 40000) + 30/300 * 25000 = 45025 rubles (4 points).
|
45025
|
Algebra
|
olympiads
| null | null |
\(\boxed{45025}\)
|
# Problem 6. (8 points)
Vasily is planning to graduate from the institute in a year. Only 270 out of 300 third-year students successfully pass their exams and complete their bachelor's degree. If Vasily ends up among the 30 expelled students, he will have to work with a monthly salary of 25,000 rubles. It is also known that every fifth graduate gets a job with a salary of 60,000 rubles per month, every tenth graduate earns 80,000 rubles per month, every twentieth graduate cannot find a job in their field and has an average salary of 25,000 rubles per month, while the salary of all others is 40,000 rubles.
When Vasily finished school, he could have chosen not to go to the institute but to work as a real estate assistant, as his friend Fyodor did. Fyodor's salary increases by 3,000 rubles each year. What is Vasily's expected salary? Whose salary will be higher in a year and by how much - Vasily's expected salary or Fyodor's actual salary, if Fyodor started working with a salary of 25,000 rubles at the same time Vasily enrolled in the institute?
Note: Bachelor's education lasts 4 years.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Four years after graduating from school, Fedor will earn $25000 + 3000 * 4 = 37000$ rubles (2 points)
The expected salary of Vasily is the expected value of the salary Vasily can ea
|
7. Vanya decided to give Masha a bouquet of an odd number of flowers for her birthday, consisting of yellow and red tulips, so that the number of flowers of one color differs from the number of flowers of the other by exactly one. Yellow tulips cost 50 rubles each, and red ones cost 31 rubles. What is the largest number of tulips he can buy for Masha's birthday, spending no more than 600 rubles?
|
A bouquet with one more red tulip than yellow ones is cheaper than a bouquet with the same total number of flowers but one more yellow tulip. Therefore, Vanya should buy a bouquet with one more red tulip. The remaining flowers can be paired into red and yellow tulips, with each pair costing 81 rubles. Let's find the number of pairs of different colored tulips in the bouquet: $600: 81=7$ (remainder 33). Therefore, Vanya needs a bouquet with $7 * 2+1=$ 15 tulips.
|
15
|
Algebra
|
olympiads
| null | null |
\(\boxed{15}\)
|
7. Vanya decided to give Masha a bouquet of an odd number of flowers for her birthday, consisting of yellow and red tulips, so that the number of flowers of one color differs from the number of flowers of the other by exactly one. Yellow tulips cost 50 rubles each, and red ones cost 31 rubles. What is the largest number of tulips he can buy for Masha's birthday, spending no more than 600 rubles?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
A bouquet with one more red tulip than yellow ones is cheaper than a bouquet with the same total number of flowers but one more yellow tulip. Therefore, Vanya should buy a bouquet with one more red tulip. The remaining flowers can b
|
Problem 7. (8 points)
The earned salary of the citizen was 23,000 rubles per month from January to June inclusive, and 25,000 rubles from July to December. In August, the citizen, participating in a poetry competition, won a prize and was awarded an e-book worth 10,000 rubles. What amount of personal income tax needs to be paid to the budget? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 39540.
## Comment:
|
Personal Income Tax from salary $=(23000 \times 6+25000 \times 6) \times 13\%=37440$ rubles.
Personal Income Tax from winnings $=(10000-4000) \times 35\%=2100$ rubles.
Total Personal Income Tax = 37440 rubles +2100 rubles$=39540$ rubles.
|
39540
|
Algebra
|
olympiads
| null | null |
\(\boxed{39540}\)
|
Problem 7. (8 points)
The earned salary of the citizen was 23,000 rubles per month from January to June inclusive, and 25,000 rubles from July to December. In August, the citizen, participating in a poetry competition, won a prize and was awarded an e-book worth 10,000 rubles. What amount of personal income tax needs to be paid to the budget? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 39540.
## Comment:
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Personal Income Tax from salary $=(23000 \times 6+25000 \times 6) \times 13\%=37440$ rubles.
Personal Income Tax from w
|
# Task 14. (1 point)
Calculate the amount of personal income tax (NDFL) paid by Sergey for the past year, if he is a Russian resident and during this period had a stable income of 30000 rub./month and a one-time vacation bonus of 20000 rub. In the past year, Sergey sold a car he inherited two years ago for 250000 rub. and bought a land plot for building a residential house for 300000 rub. Sergey claimed all the tax deductions he was entitled to. (Provide the answer without spaces and units of measurement.)
|
## Comment:
Solution: tax base $=30000 \times 12+20000+250000=630000$ rubles. The amount of the tax deduction $=250000+300000=550000$ rubles. The amount of personal income tax $=13 \% \times(630000-$ $550000)=10400$ rubles.
|
10400
|
Other
|
olympiads
| null | null |
\(\boxed{10400}\)
|
# Task 14. (1 point)
Calculate the amount of personal income tax (NDFL) paid by Sergey for the past year, if he is a Russian resident and during this period had a stable income of 30000 rub./month and a one-time vacation bonus of 20000 rub. In the past year, Sergey sold a car he inherited two years ago for 250000 rub. and bought a land plot for building a residential house for 300000 rub. Sergey claimed all the tax deductions he was entitled to. (Provide the answer without spaces and units of measurement.)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
## Comment:
Solution: tax base $=30000 \times 12+20000+250000=630000$ rubles. The amount of the tax deduction $
|
Problem 20. (6 points)
Ivan Sergeyevich decided to raise quails. In a year, he sold 100 kg of poultry meat at a price of 500 rubles per kg, and also 20000 eggs at a price of 50 rubles per dozen. The expenses for the year amounted to 100000 rubles. What profit did Ivan Sergeyevich receive for this year? (Provide the answer as a whole number, without spaces and units of measurement.)
|
Comment:
Solution: revenue $=100 \times 500 + 50 \times 20000 / 10 = 150000$ rubles. Profit $=$ revenue costs $=150000-100000=50000$ rubles.
|
50000
|
Algebra
|
olympiads
| null | null |
\(\boxed{50000}\)
|
Problem 20. (6 points)
Ivan Sergeyevich decided to raise quails. In a year, he sold 100 kg of poultry meat at a price of 500 rubles per kg, and also 20000 eggs at a price of 50 rubles per dozen. The expenses for the year amounted to 100000 rubles. What profit did Ivan Sergeyevich receive for this year? (Provide the answer as a whole number, without spaces and units of measurement.)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Comment:
Solution: revenue $=100 \times 500 + 50 \times 20000 / 10 =
|
Problem 21. (8 points)
Dmitry's parents decided to buy him a laptop for his birthday. They calculated that they could save the required amount in two ways. In the first case, they need to save one-tenth of their salary for six months. In the second case, they need to save half of their salary for one month, and then deposit it in the bank for ten months at $3\%$ per month (calculated using simple interest). In the first case, the money will be just enough for the laptop, while in the second case, after buying the laptop, there will be some money left, which will be enough to buy a computer desk for 2875 rubles. What is the mother's salary, if the father's salary is 30% higher? (Provide the answer as a whole number, without spaces or units of measurement.)
|
## Comment:
Solution:
Mom's salary is $x$, then dad's salary is $1.3x$. We set up the equation:
$(x + 1.3x) / 10 \times 6 = (x + 1.3x) / 2 \times (1 + 0.03 \times 10) - 2875$
$1.38x = 1.495x - 2875$
$x = 25000$
|
25000
|
Algebra
|
olympiads
| null | null |
\(\boxed{25000}\)
|
Problem 21. (8 points)
Dmitry's parents decided to buy him a laptop for his birthday. They calculated that they could save the required amount in two ways. In the first case, they need to save one-tenth of their salary for six months. In the second case, they need to save half of their salary for one month, and then deposit it in the bank for ten months at $3\%$ per month (calculated using simple interest). In the first case, the money will be just enough for the laptop, while in the second case, after buying the laptop, there will be some money left, which will be enough to buy a computer desk for 2875 rubles. What is the mother's salary, if the father's salary is 30% higher? (Provide the answer as a whole number, without spaces or units of measurement.)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
## Comment:
Solution:
Mom's salary is $x$, then dad's salary is $1.3x$. We set up the equation:
$(x + 1.
|
Problem 7. (8 points)
The earned salary of the citizen was 23,000 rubles per month from January to June inclusive, and 25,000 rubles from July to December. In August, the citizen, participating in a poetry contest, won a prize and was awarded an e-book worth 10,000 rubles. What amount of personal income tax needs to be paid to the budget? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 39540.
## Comment:
|
Personal Income Tax from salary $=(23000 \times 6+25000 \times 6) \times 13\%=37440$ rubles.
Personal Income Tax from winnings $=(10000-4000) \times 35\%=2100$ rubles.
Total Personal Income Tax = 37440 rubles +2100 rubles$=39540$ rubles.
|
39540
|
Algebra
|
olympiads
| null | null |
\(\boxed{39540}\)
|
Problem 7. (8 points)
The earned salary of the citizen was 23,000 rubles per month from January to June inclusive, and 25,000 rubles from July to December. In August, the citizen, participating in a poetry contest, won a prize and was awarded an e-book worth 10,000 rubles. What amount of personal income tax needs to be paid to the budget? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 39540.
## Comment:
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Personal Income Tax from salary $=(23000 \times 6+25000 \times 6) \times 13\%=37440$ rubles.
Personal Income Tax from w
|
3. Once, a god sent a little cheese to two ravens. The first raven received 100 g, from which part was taken by a fox. The piece of the second raven turned out to be twice as large as that of the first, but she managed to eat only half as much as the first raven. The portion of cheese that the fox got from the second raven turned out to be three times larger than from the first. How much cheese did the fox get in total?
|
Let the first crow eat $x$ grams of cheese. Then the fox received $100-x$ grams of cheese from the first crow. The second crow ate $\frac{x}{2}$ grams of cheese. From the second crow, the fox received $200-\frac{x}{2}$ grams of cheese. This was three times more, so: $200-\frac{x}{2}=3(100-x)$. Solution: $x=40$. The fox ate 240 grams.
|
240
|
Logic and Puzzles
|
olympiads
| null | null |
\(\boxed{240}\)
|
3. Once, a god sent a little cheese to two ravens. The first raven received 100 g, from which part was taken by a fox. The piece of the second raven turned out to be twice as large as that of the first, but she managed to eat only half as much as the first raven. The portion of cheese that the fox got from the second raven turned out to be three times larger than from the first. How much cheese did the fox get in total?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let the first crow eat $x$ grams of cheese. Then the fox received $100-x$ grams of cheese from the first crow. The second crow ate $\frac{x}{2}$ grams of cheese. From t
|
5. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same moment, $B$ meets $D$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will $C$ and $D$ meet for the first time?
|
Since $A$ with $C$ and $B$ with $D$ meet every 7 minutes, their approach speeds are equal: $V_{A}+V_{C}=V_{B}+V_{D}$. In other words, the speeds of separation of $A, B$ and $C, D$ are equal: $V_{A}-V_{B}=V_{D}-V_{C}$. Therefore, since $A$ and $B$ meet for the first time at the 53rd minute, $C$ and $D$ will also meet for the first time at the 53rd minute.
Criteria. If 53 minutes is replaced with 46 - no more than 3 points. If the solution assumes that any of the speeds are equal - 1 point. Only the answer without explanation - 0 points.
|
53
|
Algebra
|
olympiads
| null | null |
\(\boxed{53}\)
|
5. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same moment, $B$ meets $D$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will $C$ and $D$ meet for the first time?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since $A$ with $C$ and $B$ with $D$ meet every 7 minutes, their approach speeds are equal: $V_{A}+V_{C}=V_{B}+V_{D}$. In other words, the speeds of separation of $A, B$ and $C, D$ are equal: $V_{A}-V_{B}=V_{D}-V_{C}$. Therefore, since $A$ and $B$ meet for the first time
|
2. Marina needs to buy a notebook, a pen, a ruler, and a pencil to participate in the Olympiad. If she buys a notebook, a pencil, and a ruler, she will spend 47 tugriks. If she buys a notebook, a ruler, and a pen, she will spend 58 tugriks. If she buys only a pen and a pencil, she will spend 15 tugriks. How much money will she need for the entire set?
|
If Marina buys all three sets from the condition at once, she will spend $47+58+$ $15=120$ tugriks, and she will buy each item twice, so the full set of school supplies costs $120 / 2=60$ tugriks.
Criteria. Only the answer without explanation - 1 point. If in the solution they try to determine the cost of the pen and pencil (although it is not stated in the condition that the cost must be an integer) - 0 points.
|
60
|
Algebra
|
olympiads
| null | null |
\(\boxed{60}\)
|
2. Marina needs to buy a notebook, a pen, a ruler, and a pencil to participate in the Olympiad. If she buys a notebook, a pencil, and a ruler, she will spend 47 tugriks. If she buys a notebook, a ruler, and a pen, she will spend 58 tugriks. If she buys only a pen and a pencil, she will spend 15 tugriks. How much money will she need for the entire set?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If Marina buys all three sets from the condition at once, she will spend $47+58+$ $15=120$ tugriks, and she will buy each item twice, so the full set of school supplies costs $120 / 2=60$ tugriks.
Criteria.
|
5. Egor wrote a number on the board and encrypted it according to the rules of letter puzzles (different letters correspond to different digits, the same letters correspond to the same digits). The result was the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible by 25?
|
The number must be divisible by 25, so “$\lambda$A” equals 25, 50, or 75 (00 cannot be, as the letters are different). If “LA” equals 50, then for the other letters (“G”, “V”, “T”, “E”, “M”) there are $A_{8}^{5}$ options; otherwise, for the other letters there are $7 \cdot A_{7}^{4}$ options. In total, $8!/ 6 + 2 \cdot 7 \cdot 7!/ 6 = 18480$ ways.
Criteria. It is explicitly stated that the option 00 is not suitable for the letters “LA”, and all 3 cases are demonstrated - 3 points. If the 2 variants are further analyzed without errors, then another 4 points. If the answer is given as an expression not fully calculated - do not deduct points.
|
18480
|
Logic and Puzzles
|
olympiads
| null | null |
\(\boxed{18480}\)
|
5. Egor wrote a number on the board and encrypted it according to the rules of letter puzzles (different letters correspond to different digits, the same letters correspond to the same digits). The result was the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible by 25?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The number must be divisible by 25, so “$\lambda$A” equals 25, 50, or 75 (00 cannot be, as the letters are different). If “LA” equals 50, then for the other letters (“G”, “V”, “T”, “E”, “M”) there are $A_{8}^{5}$ options; otherwise, for the other letters there are $7 \cdot A_{7}^{4}$ options. In total, $8!/ 6 + 2 \cdot 7 \
|
1. In one move, you can either add one of its digits to the number or subtract one of its digits from the number (for example, from the number 142 you can get $142+2=144, 142-4=138$ and several other numbers).
a) Can you get the number 2021 from the number 2020 in several moves?
b) Can you get the number 2021 from the number 1000 in several moves?
|
a) Yes, for example, like this: $20 \mathbf{2 0} \rightarrow 20 \mathbf{1 8} \rightarrow \mathbf{2 0 1 9} \rightarrow 2021$.
b) Yes. For example, by adding the first digit (one), we can reach the number 2000; by adding the first digit (two), we can reach 2020; then see part a.
Criteria. Part a) 3 points, b) 4 points. In part (b), 1 point is given for progressing to 1999 (for example, for a "solution" of the form "from 1000, we add one at a time and there is the result 2021").
|
2021
|
Number Theory
|
olympiads
| null | null |
\(\boxed{2021}\)
|
1. In one move, you can either add one of its digits to the number or subtract one of its digits from the number (for example, from the number 142 you can get $142+2=144, 142-4=138$ and several other numbers).
a) Can you get the number 2021 from the number 2020 in several moves?
b) Can you get the number 2021 from the number 1000 in several moves?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
a) Yes, for example, like this: $20 \mathbf{2 0} \rightarrow 20 \mathbf{1 8} \rightarrow \mathbf{2 0 1 9} \rightarrow 2021$.
b) Yes. For example, by adding the first digit (one), we can reach the number 2000; by adding the first digit (two)
|
2. Once, a god sent a little cheese to two ravens. The first raven received 100 g, from which a part was taken by a fox. The piece of the second raven turned out to be twice as large as that of the first, but she managed to eat only half as much as the first raven. The portion of cheese that the fox got from the second raven turned out to be three times larger than from the first. How much cheese did the fox get in total?
|
Let the first crow eat $x$ grams of cheese. Then the fox got $100-x$ grams of cheese from the first crow. The second crow ate ${ }_{2}^{x}$ grams of cheese. From the second crow, the fox received $200-\frac{x}{2}$ grams of cheese. This was three times more, so: $200-\frac{x}{2}=3(100-x)$. Solution: $x=40$. The fox ate 240 grams.
|
240
|
Algebra
|
olympiads
| null | null |
\(\boxed{240}\)
|
2. Once, a god sent a little cheese to two ravens. The first raven received 100 g, from which a part was taken by a fox. The piece of the second raven turned out to be twice as large as that of the first, but she managed to eat only half as much as the first raven. The portion of cheese that the fox got from the second raven turned out to be three times larger than from the first. How much cheese did the fox get in total?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let the first crow eat $x$ grams of cheese. Then the fox got $100-x$ grams of cheese from the first crow. The second crow ate ${ }_{2}^{x}$ grams of cheese. From the
|
5. In a row, the squares of the first 2022 natural numbers are written: $1,4,9, \ldots, 4088484$. For each written number, except the first and the last, the arithmetic mean of its left and right neighbors was calculated and written below it (for example, under the number 4, $\frac{1+9}{2}=5$ was written). For the resulting string of 2020 numbers, the same operation was performed. This process continued until a string with only two numbers remained. What are these two numbers?
|
Let's look at an arbitrary number in the row $x^{2}$. Under it, it will be written
$$
\frac{(x-1)^{2}+(x+1)^{2}}{2}=\frac{x^{2}-2 x+1+x^{2}+2 x+1}{2}=\frac{2 x^{2}+2}{2}=x^{2}+1
$$
Thus, each time the number increases by one. Initially, there are 2022 numbers, and each time their count decreases by 2, so the operation will be performed 1010 times, and two central numbers will remain, increased by 1010: $1011^{2}+1010=1023131$ and $1012^{2}+1010=$ 1025154.
Criteria. Proven in general that the number increases by 1 each time - 3 points. Solved correctly without a proper proof of the increase by 1 - 4 points. Error in counting the remaining numbers -2 points.
|
10231311025154
|
Algebra
|
olympiads
| null | null |
\(\boxed{10231311025154}\)
|
5. In a row, the squares of the first 2022 natural numbers are written: $1,4,9, \ldots, 4088484$. For each written number, except the first and the last, the arithmetic mean of its left and right neighbors was calculated and written below it (for example, under the number 4, $\frac{1+9}{2}=5$ was written). For the resulting string of 2020 numbers, the same operation was performed. This process continued until a string with only two numbers remained. What are these two numbers?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's look at an arbitrary number in the row $x^{2}$. Under it, it will be written
$$
\frac{(x-1)^{2}+(x+1)^{2}}{2}=\frac{x^{2}-2 x+1+x^{2}+2 x+1}{2}=\frac{2 x^{2}+2}{2}=x^{2}+1
$$
Thus, each time the number increases by one. Initially, there are 2022 numbers, and each time their count decreases by 2, so the operation will be perf
|
6. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same moment, $B$ meets $D$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will $C$ and $D$ meet for the first time?
|
Since $A$ with $C$ and $B$ with $D$ meet every 7 minutes, their approach speeds are equal: $V_{A}+V_{C}=V_{B}+V_{D}$. In other words, the speeds of separation of $A, B$ and $C, D$ are equal: $V_{A}-V_{B}=V_{D}-V_{C}$. Therefore, since $A$ and $B$ meet for the first time at the 53rd minute, $C$ and $D$ will also meet for the first time at the 53rd minute.
Criteria. If 53 minutes is replaced with 46 - no more than 3 points. If the solution assumes that any of the speeds are equal - 1 point. Only the answer without explanation - 0 points.
|
53
|
Algebra
|
olympiads
| null | null |
\(\boxed{53}\)
|
6. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same moment, $B$ meets $D$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will $C$ and $D$ meet for the first time?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since $A$ with $C$ and $B$ with $D$ meet every 7 minutes, their approach speeds are equal: $V_{A}+V_{C}=V_{B}+V_{D}$. In other words, the speeds of separation of $A, B$ and $C, D$ are equal: $V_{A}-V_{B}=V_{D}-V_{C}$. Therefore, since $A$ and $B$ meet for the first time
|
3. How many ways are there to cut a $10 \times 10$ square into several rectangles along the grid lines such that the sum of their perimeters is 398? Ways that can be matched by rotation or flipping are considered different.
|
ways.
If the entire square is cut into 100 unit squares, the sum of the perimeters will be $4 \times 100=400$. Therefore, we need to reduce this sum by 2, which is achieved by keeping one internal partition intact (in other words, the square is cut into 98 squares and 1 domino). There are a total of 180 internal partitions - 9 in each of the 10 rows and 9 in each of the 10 columns.
|
180
|
Combinatorics
|
olympiads
| null | null |
\(\boxed{180}\)
|
3. How many ways are there to cut a $10 \times 10$ square into several rectangles along the grid lines such that the sum of their perimeters is 398? Ways that can be matched by rotation or flipping are considered different.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
ways.
If the entire square is cut into 100 unit squares, the sum of the perimeters will be $4 \times 100=400$. Therefore, we need to reduce this sum by 2, which is achieved by keeping one int
|
5. In several packages, there are 20 candies, and there are no two packages with the same number of candies and no empty packages. Some packages may be inside other packages (then it is considered that a candy lying in the inner package also lies in the outer package). However, it is forbidden to place a package inside another package that already contains a package. What is the maximum possible number of packages?
|
Example: ((6)(2)) ((3)(4)) ((1)4) (there are other examples).
There cannot be more than 8 packages. Indeed, then the sum of the number of candies in the packages (or rather, the number of incidences of candies to packages) is not less than $1+2+\ldots+9=45$. But there are 20 candies, so at least one of them lies in at least three packages, which is not allowed.
|
8
|
Combinatorics
|
olympiads
| null | null |
\(\boxed{8}\)
|
5. In several packages, there are 20 candies, and there are no two packages with the same number of candies and no empty packages. Some packages may be inside other packages (then it is considered that a candy lying in the inner package also lies in the outer package). However, it is forbidden to place a package inside another package that already contains a package. What is the maximum possible number of packages?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Example: ((6)(2)) ((3)(4)) ((1)4) (there are other examples).
There cannot be more than 8 packages. Indeed, then the sum of the number of candies in the packages (or rather, the num
|
4. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same moment, $B$ meets $D$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will all the cars meet for the first time?
|
$A$ and $C$ meet every 7 minutes, while $A$ and $B$ meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is $7 \cdot 53 = 371$ minutes. At the same time, $B$ and $D$ also meet every 7 minutes, so on the 371st minute, car $D$ will be at the same point as the other three cars.
Criteria. If 53 minutes is replaced with 46 - no more than 3 points. If the solution assumes that any of the speeds are equal - 1 point. Only the answer without explanation - 0 points.
|
371
|
Logic and Puzzles
|
olympiads
| null | null |
\(\boxed{371}\)
|
4. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same moment, $B$ meets $D$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will all the cars meet for the first time?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
$A$ and $C$ meet every 7 minutes, while $A$ and $B$ meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is $7 \cdot 53 = 371$ minutes. At the same time, $B$ and $D$ also meet every 7 minutes, so on the 3
|
5. In a row, the squares of the first 2022 natural numbers are written: $1, 4, 9, \ldots, 4088484$. For each written number, except the first and the last, the arithmetic mean of its left and right neighbors was calculated and written below it (for example, under the number 4, $\left.\frac{1+9}{2}=5\right)$ was written. For the resulting string of 2020 numbers, the same operation was performed. This process continued until a string with only two numbers remained. What are these two numbers?
|
Let's look at an arbitrary number in the row $x^{2}$. Under it, it will be written
$$
\frac{(x-1)^{2}+(x+1)^{2}}{2}=\frac{x^{2}-2 x+1+x^{2}+2 x+1}{2}=\frac{2 x^{2}+2}{2}=x^{2}+1
$$
Thus, each time the number increases by one. Initially, there are 2022 numbers, and each time their count decreases by 2, so the operation will be performed 1010 times, and two central numbers will remain, increased by 1010: $1011^{2}+1010=1023131$ and $1012^{2}+1010=$ 1025154.
Criteria. Proven in general that the number increases by 1 each time - 3 points. Solved correctly without a proper proof of the increase by 1 - 4 points. Error in counting the remaining numbers -2 points.
|
10231311025154
|
Algebra
|
olympiads
| null | null |
\(\boxed{10231311025154}\)
|
5. In a row, the squares of the first 2022 natural numbers are written: $1, 4, 9, \ldots, 4088484$. For each written number, except the first and the last, the arithmetic mean of its left and right neighbors was calculated and written below it (for example, under the number 4, $\left.\frac{1+9}{2}=5\right)$ was written. For the resulting string of 2020 numbers, the same operation was performed. This process continued until a string with only two numbers remained. What are these two numbers?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's look at an arbitrary number in the row $x^{2}$. Under it, it will be written
$$
\frac{(x-1)^{2}+(x+1)^{2}}{2}=\frac{x^{2}-2 x+1+x^{2}+2 x+1}{2}=\frac{2 x^{2}+2}{2}=x^{2}+1
$$
Thus, each time the number increases by one. Initially, there are 2022 numbers, and each time their count decreases by 2, so the operation will be perf
|
5. In a spring math camp, between 50 and 70 children arrived. On Pi Day (March 14), they decided to give each other squares if they were just acquaintances, and circles if they were friends. Andrey calculated that each boy received 3 circles and 8 squares, while each girl received 2 squares and 9 circles. And Katya noticed that the total number of circles and squares given was the same. How many children arrived at the camp?
(P. D. Mulyenko)
|
Let the number of boys be $m$, and the number of girls be $-d$. Then $3 m + 9 d = 8 m + 2 d$ (the number of circles equals the number of squares). Transforming, we get $5 m = 7 d$, which means the number of boys and girls are in the ratio $7: 5$. Therefore, the total number of children is divisible by 12. Between 50 and 70, only the number 60 fits.
Criteria. Incorrectly understood condition (for example, that an equal number of gifts were given to boys and girls) 0 points.
Correct answer without explanation -2 points. Correct answer with verification -3 points.
|
60
|
Algebra
|
olympiads
| null | null |
\(\boxed{60}\)
|
5. In a spring math camp, between 50 and 70 children arrived. On Pi Day (March 14), they decided to give each other squares if they were just acquaintances, and circles if they were friends. Andrey calculated that each boy received 3 circles and 8 squares, while each girl received 2 squares and 9 circles. And Katya noticed that the total number of circles and squares given was the same. How many children arrived at the camp?
(P. D. Mulyenko)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let the number of boys be $m$, and the number of girls be $-d$. Then $3 m + 9 d = 8 m + 2 d$ (the number of circles equals the number of squares). Transforming, we get $5 m = 7 d$, which means the number of boys and girls are in the ratio $7: 5$. Therefore, the total number of childr
|
4. The brothers found a treasure of gold and silver. They divided it so that each received 100 kg. The eldest received the most gold - 25 kg - and one eighth of all the silver. How much gold was in the treasure?
|
The elder brother received 75 kg of silver, which is one-eighth of the total amount; therefore, the total mass of silver is 600 kg.
2) The others received more silver than the elder brother, i.e., each received more than 75 kg. If there are at least eight brothers, then in total they would receive more than 600 kg; therefore, there are no more than seven brothers.
3) But there must be at least seven of them, since the mass of the silver is 600 kg, which means the total mass of the treasure is more than 600 kg. Therefore, there are seven brothers.
4) The total mass of the treasure is 700 kg, so the mass of the gold is $700-600=100$ kg.
Criteria. The same as in problem 5.6.
|
100
|
Logic and Puzzles
|
olympiads
| null | null |
\(\boxed{100}\)
|
4. The brothers found a treasure of gold and silver. They divided it so that each received 100 kg. The eldest received the most gold - 25 kg - and one eighth of all the silver. How much gold was in the treasure?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The elder brother received 75 kg of silver, which is one-eighth of the total amount; therefore, the total mass of silver is 600 kg.
2) The others received more silver than the elder brother, i.e., each received more than 75 kg. If there are at least eight brothers, then in total they would receive more than 600 kg; therefore, there are no
|
2. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same moment, $B$ meets $D$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will all the cars meet for the first time?
|
$A$ and $C$ meet every 7 minutes, while $A$ and $B$ meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is $7 \cdot 53 = 371$ minutes. At the same time, $B$ and $D$ also meet every 7 minutes, so on the 371st minute, car $D$ will be at the same point as the other three cars.
Criteria. If 53 minutes is replaced with 46 - no more than 3 points. If the solution assumes that any of the speeds are equal - 1 point. Only the answer without explanation - 0 points.
|
371
|
Algebra
|
olympiads
| null | null |
\(\boxed{371}\)
|
2. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same moment, $B$ meets $D$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will all the cars meet for the first time?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
$A$ and $C$ meet every 7 minutes, while $A$ and $B$ meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is $7 \cdot 53 = 371$ minutes. At the same time, $B$ and $D$ also meet every 7 minutes, so on the 3
|
2. Marina needs to buy a notebook, a pen, a ruler, a pencil, and an eraser to participate in the Olympiad. If she buys a notebook, a pencil, and an eraser, she will spend 47 tugriks. If she buys a notebook, a ruler, and a pen, she will spend 58 tugriks. How much money will she need for the entire set, if the notebook costs 15 tugriks?
|
If Marina buys two sets from the condition, she will spend $47+58=105$ tugriks, but she will buy an extra notebook, so the full set of school supplies costs $105-15=90$ tugriks.
Criteria. Only the answer without explanation - 1 point. If in the solution they try to determine the cost of the pen and pencil (although it is not stated in the condition that the cost must necessarily be an integer) - 0 points.
|
90
|
Algebra
|
olympiads
| null | null |
\(\boxed{90}\)
|
2. Marina needs to buy a notebook, a pen, a ruler, a pencil, and an eraser to participate in the Olympiad. If she buys a notebook, a pencil, and an eraser, she will spend 47 tugriks. If she buys a notebook, a ruler, and a pen, she will spend 58 tugriks. How much money will she need for the entire set, if the notebook costs 15 tugriks?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If Marina buys two sets from the condition, she will spend $47+58=105$ tugriks, but she will buy an extra notebook, so the full set of school supplies costs $105-15=90$ tugriks.
Criteria. Only the answer
|
5. Egor wrote a number on the board and encrypted it according to the rules of letter puzzles (different letters correspond to different digits, the same letters correspond to the same digits). He got the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible by 5?
|
The number must be divisible by 5, so the letter "A" is equal to 0 or 5. If it is equal to 0, then for the other letters ("G", "V", "T", "E", "M", "L") there are $A_{9}^{6}=9!/ 3$! options; if "A" is equal to 5, then for the other letters there are $8 \cdot A_{8}^{5}=8$!/3! options, since "G" cannot be zero. In total, $9!/ 6+8 \cdot 8!/ 6=114240$ ways.
Criteria. It is explicitly stated that for the letter "A" there are 2 options: 5 or $0-2$ points. If the two cases are further analyzed, that's another 5 points. If the answer is given as an expression not fully calculated to the end - do not deduct points.
|
114240
|
Logic and Puzzles
|
olympiads
| null | null |
\(\boxed{114240}\)
|
5. Egor wrote a number on the board and encrypted it according to the rules of letter puzzles (different letters correspond to different digits, the same letters correspond to the same digits). He got the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible by 5?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The number must be divisible by 5, so the letter "A" is equal to 0 or 5. If it is equal to 0, then for the other letters ("G", "V", "T", "E", "M", "L") there are $A_{9}^{6}=9!/ 3$! options; if "A" is equal to 5, then for the other letters there are $8 \cdot A_{8}^{5}=8$!/3! options, since "G" cannot be ze
|
7. Three cars $A, B$ and $C$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ drives counterclockwise. All cars move at constant (but pairwise distinct) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will all three cars meet for the first time?
|
$A$ and $C$ meet every 7 minutes, and $A$ and $B$ meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is $7 \cdot 53=371$ minutes.
Criteria. If 53 minutes is replaced with 46 - 3 points. Solved by trial with lengths and speeds -1 point. Only the answer without explanation - 0 points.
|
371
|
Algebra
|
olympiads
| null | null |
\(\boxed{371}\)
|
7. Three cars $A, B$ and $C$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ drives counterclockwise. All cars move at constant (but pairwise distinct) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will all three cars meet for the first time?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
$A$ and $C$ meet every 7 minutes, and $A$ and $B$ meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is $7 \cdot 5
|
6. Find all real solutions of the system of equations
$$
\left\{\begin{array}{l}
\frac{1}{x}=\frac{32}{y^{5}}+\frac{48}{y^{3}}+\frac{17}{y}-15 \\
\frac{1}{y}=\frac{32}{z^{5}}+\frac{48}{z^{3}}+\frac{17}{z}-15 \\
\frac{1}{z}=\frac{32}{x^{5}}+\frac{48}{x^{3}}+\frac{17}{x}-15
\end{array}\right.
$$
(A. B. Vladimirov)
|
Let $F(t)=32 t^{5}+48 t^{3}+17 t-15$. Then the system has the form $F\left(\frac{1}{y}\right)=\frac{1}{x}, F\left(\frac{1}{z}\right)=\frac{1}{y}$, $F\left(\frac{1}{x}\right)=\frac{1}{z}$. From this, it follows that $F\left(F\left(F\left(\frac{1}{x}\right)\right)\right)=\frac{1}{x}$. Note that $F(0.5)=0.5$ and the function $F(t)-t$ is strictly increasing. Therefore, if $t>0.5$, then $t < F(F(F(t)))$. But this means that $\frac{1}{x}=0.5$, and the original system has a unique solution $x=y=z=2$.
Criteria. 1 point is given for finding the solution. 3 points for the solution under the assumption $x=y=z$. If the uniqueness of the solution to $F(t)=t$ is not proven, 2 points are deducted.
|
2
|
Algebra
|
olympiads
| null | null |
\(\boxed{2}\)
|
6. Find all real solutions of the system of equations
$$
\left\{\begin{array}{l}
\frac{1}{x}=\frac{32}{y^{5}}+\frac{48}{y^{3}}+\frac{17}{y}-15 \\
\frac{1}{y}=\frac{32}{z^{5}}+\frac{48}{z^{3}}+\frac{17}{z}-15 \\
\frac{1}{z}=\frac{32}{x^{5}}+\frac{48}{x^{3}}+\frac{17}{x}-15
\end{array}\right.
$$
(A. B. Vladimirov)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $F(t)=32 t^{5}+48 t^{3}+17 t-15$. Then the system has the form $F\left(\frac{1}{y}\right)=\frac{1}{x}, F\left(\frac{1}{z}\right)=\frac{1}{y}$, $F\left(\frac{1}{x}\right)=\frac{1}{z}$. From this, it follows that $F\left(F\left(F\left(\frac{1}{x}\right)\right)\right)=\frac{1}{x}$. Note that $F(0.5)=0.5$ and the function $F(t)-t$ is strictly i
|
2. It is known that on Monday, the painter painted twice as slowly as on Tuesday, Wednesday, and Thursday, and on Friday - twice as fast as on these three days, but worked 6 hours instead of 8. On Friday, he painted 300 meters more of the fence than on Monday. How many meters of the fence did the painter paint from Monday to Friday?
|
Let's take $100 \%$ of the fence length that the painter painted on Tuesday, Wednesday, and Thursday. Then Monday accounts for $50 \%$, and Friday for $-150 \%$. Therefore, 300 meters of the fence correspond to $150 \% - 50 \% = 100 \%$. For the entire week (from Monday to Friday), it is $500 \%$, i.e., 1500 meters.
Criteria. According to general rules.
|
1500
|
Algebra
|
olympiads
| null | null |
\(\boxed{1500}\)
|
2. It is known that on Monday, the painter painted twice as slowly as on Tuesday, Wednesday, and Thursday, and on Friday - twice as fast as on these three days, but worked 6 hours instead of 8. On Friday, he painted 300 meters more of the fence than on Monday. How many meters of the fence did the painter paint from Monday to Friday?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's take $100 \%$ of the fence length that the painter painted on Tuesday, Wednesday, and Thursday. Then Monday accounts for $50 \%$, and Friday for $-150 \%$. Therefore, 300 m
|
3. Find the number of five-digit numbers in which all digits are different, the first digit is divisible by 2, and the sum of the first and last digits is divisible by 3.
|
The first digit can be 2, 4, 6, or 8. If the first is 2, the last can be $1, 4, 7$; if the first is 4, the last can be $2, 5, 8$; if the first is 6, the last can be $0, 3, (6$ does not fit), 9; if the first is 8, the last can be $1, 4, 7$. In total, there are $3+3+3+3=13$ options for the first and last digits. For each of these options, there are $8 \cdot 7 \cdot 6=336$ ways to choose the two middle digits. In total, $336 \cdot 12=4032$ ways.
Criteria. The same as in 5.4.
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4032
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Combinatorics
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olympiads
| null | null |
\(\boxed{4032}\)
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3. Find the number of five-digit numbers in which all digits are different, the first digit is divisible by 2, and the sum of the first and last digits is divisible by 3.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The first digit can be 2, 4, 6, or 8. If the first is 2, the last can be $1, 4, 7$; if the first is 4, the last can be $2, 5, 8$; if the first is 6, the last can be $0, 3, (6$ does not fit), 9; if the first is 8, the last can be $1, 4, 7$
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4. At the beginning of the year, the US dollar was worth 80 European cents. An expert predicted that over the course of the year, the euro's exchange rate against the ruble would increase by 8% (i.e., 1 euro would be able to buy 8% more rubles than at the beginning of the year), and the US dollar's exchange rate against the ruble would fall by 10%. If the forecast comes true, how many US cents will the euro be worth at the end of the year?
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Let's choose the amount in rubles that will correspond to 1 euro (i.e., 100 European cents) at the end of the year. At the beginning of the year, the same amount corresponded to 108 European cents or $108 \cdot 100 / 80=135$ American cents. Consequently, at the end of the year, it will correspond to $135 \cdot 100 / 90=\mathbf{1 5 0}$ American cents.
Criteria. Arithmetic errors with the correct approach - 2 points are deducted.
If there is at least one "logical" error (the wrong percentage is taken, the wrong ratio is considered, etc.), the problem is considered unsolved. If the solution is completed and everything is correct except for one such error - 2 points can be given; in all other cases, 0.
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150
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Algebra
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olympiads
| null | null |
\(\boxed{150}\)
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4. At the beginning of the year, the US dollar was worth 80 European cents. An expert predicted that over the course of the year, the euro's exchange rate against the ruble would increase by 8% (i.e., 1 euro would be able to buy 8% more rubles than at the beginning of the year), and the US dollar's exchange rate against the ruble would fall by 10%. If the forecast comes true, how many US cents will the euro be worth at the end of the year?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's choose the amount in rubles that will correspond to 1 euro (i.e., 100 European cents) at the end of the year. At the beginning of the year, the same amount corresponded to 108 European cents or $108 \cdot 100 / 80=135$ American cents. Consequently, at the end of the year, it will correspond to $135 \cdot 100 / 90=\mathbf{1 5 0}$ American cents.
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