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4. Kolya has 440 identical cubes with a side length of 1 cm. Kolya assembled a rectangular parallelepiped from them, with all edges having a length of at least 5 cm. Find the total length of all the edges of this parallelepiped
|
Answer: 96.
Solution: $440=2^{3} \times 5 \times 11$. Since all edges have a length of at least 5, their lengths are 8, 5, and 11. Each edge is included 4 times, so the total length is ( $8+5+11)^{*} 4=96$.
|
96
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Let's call a number "marvelous" if it has exactly 3 different odd natural divisors (and an arbitrary number of even divisors). How many "marvelous" two-digit numbers exist?
|
Answer: 7.
Solution: Such numbers have the form $2^{k} \times p^{2}$, where $p-$ is an odd prime number. Clearly, $p$ does not exceed 7, since the result must be a two-digit number. If $p=3$, then $k=0,1,2,3$; If $p=5$, then $k=0,1$; if $p=7$, then $k=0$. In total, there are 7 options.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Solve the inequality
$$
\frac{2^{2+\sqrt{x-1}}-24}{2^{1+\sqrt{x-1}}-8}>1 .
$$
In your answer, specify the sum of all integer values of $x$ that satisfy the given inequality and belong to the interval $(-70 ; 34)$.
|
Solution. Let $t=2^{1+\sqrt{x-1}}$, we get
$$
\frac{2 t-24}{t-8} \geqslant 1 \Longleftrightarrow \frac{t-16}{t-8} \geqslant 0 \Longleftrightarrow t \in(-\infty ; 8) \cup[16 ;+\infty)
$$
From this, either $2^{1+\sqrt{x-1}}<2^{3}, \sqrt{x-1}<2, x \in[1 ; 5)$, or $2^{1+\sqrt{x-1}} \geqslant 2^{4}, \sqrt{x-1} \geqslant 3, x \in[10 ;+\infty)$. Therefore, the solution to the inequality is $x \in[1 ; 5) \cup(10 ;+\infty)$. The desired sum is
$$
1+2+3+4+(10+11+12+\ldots+33)=10+\frac{10+33}{2} \cdot 24=526
$$
Answer: 526.
|
526
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Inside a right triangle $ABC$ with hypotenuse $AC$, a point $M$ is taken such that the areas of triangles $ABM$ and $BCM$ are one-third and one-fourth of the area of triangle $ABC$, respectively. Find $BM$, if $AM=60$ and $CM=70$. If the answer is not an integer, round it to the nearest integer.
|
Solution. Denoting $A B=c, B C=a$, we get
$$
\left\{\begin{array}{l}
\left(c-\frac{c}{4}\right)^{2}+\left(\frac{a}{3}\right)^{2}=60^{2} \\
\left(\frac{c}{4}\right)^{2}+\left(a-\frac{a}{3}\right)^{2}=70^{2}
\end{array}\right.
$$
We solve the system and find $B M=\left(\frac{a}{3}\right)^{2}+\left(\frac{c}{4}\right)^{2}=\frac{100^{2}}{7}$. Therefore, $B M=\frac{100}{\sqrt{7}}$. We write the nearest integer in the answer.
Answer: 38.
|
38
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Find all $a$ for which the system
$$
\left\{\begin{array}{l}
x^{2}+4 y^{2}=1 \\
x+2 y=a
\end{array}\right.
$$
has a unique solution. If necessary, round it to two decimal places. If there are no solutions, put 0 in the answer.
|
Solution. Since $2 y=a-x$, from the first equation we get:
$$
x^{2}+(a-x)^{2}=1 \Longleftrightarrow 2 x^{2}-2 a x+a^{2}-1=0
$$
This equation has a unique solution when $\frac{D}{4}=2-a^{2}=0$. Therefore, $a= \pm \sqrt{2}$, and the smallest value is $-\sqrt{2} \approx-1.414214 \ldots$
Answer: $-1.41$.
I-A. The sum of 28218 natural numbers is $2016 \cdot 15$, and their product is $-\left(2016^{2}+15\right)$. Find all possible sets of such numbers. In the answer, indicate the sum of the largest and smallest numbers from all the found sets. If such numbers do not exist, then in the answer, indicate the number 0.
Solution. From the factorization $2016^{2}+15=3 \cdot 19 \cdot 113 \cdot 631$, we conclude that in the problem
$$
\left\{\begin{array}{l}
a_{1}+\ldots+a_{28218}=2016 \cdot 15 \\
a_{1} \cdot \ldots \cdot a_{28218}=2016^{2}+15=3 \cdot 19 \cdot 113 \cdot 631
\end{array}\right.
$$
the numbers $a_{k}$ not equal to 1 can be no more than 4 (i.e., no more than the number of prime factors), so the problem is equivalent to the following
$$
\left\{\begin{array}{l}
a_{1}+a_{2}+a_{3}+a_{4}=2026 \\
a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4}=3 \cdot 19 \cdot 113 \cdot 631 \\
a_{5}=a_{6}=\ldots=a_{28218}=1
\end{array}\right.
$$
We solve the last problem: The set $a_{1}, a_{2}, a_{3}, a_{4}$ consists of the numbers $1,19,113,3 \cdot 631$.
It remains to calculate $(3 \cdot 631)+1=1894$.
Answer: 1894.
|
1894
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. The sum of 1265 natural numbers is $2016+33$, and their product is $-\left(2016^{2}+33\right)$. Find all possible sets of such numbers. In your answer, indicate the sum of the largest and smallest numbers from all the found sets. If such numbers do not exist, then in your answer, indicate the number 0.
Hint: $2016^{2}+33=3 \cdot 41 \cdot 173 \cdot 191$.
|
Answer 574.
II-A. Solve the equation
$$
(1+\cos x+\cos 2 x+\cos 3 x+\cos 4 x)^{4}+(\sin x+\sin 2 x+\sin 3 x+\sin 4 x)^{4}=\frac{3}{4}+\frac{\cos 8 x}{4}
$$
Find the sum of all roots on the interval $A$, rounding to the nearest integer if necessary. If there are no roots or infinitely many roots on this interval, write the digit 0 in the answer.
Solution. 1st method of solving. The following is true:
$$
\begin{aligned}
& (1+\cos 4 x)+(\cos x+\cos 3 x)+\cos 2 x=\cos 2 x(2 \cos 2 x+2 \cos 2 x+1) \\
& \sin 2 x+\sin 4 x+(\sin x+\sin 3 x)=\sin 2 x(1+2 \cos 2 x+2 \cos x)
\end{aligned}
$$
Therefore,
$$
\begin{aligned}
& (2 \cos 2 x+2 \cos 2 x+1)^{4}\left(\cos ^{4} 2 x+\sin ^{4} 2 x\right)=\frac{3}{4}+\frac{1-2 \sin ^{2} 4 x}{4} \Longleftrightarrow \\
& \left(4 \cos ^{2} x+2 \cos x-1\right)^{4}\left(\cos ^{4} 2 x+\sin ^{4} 2 x\right)=1-\frac{\sin ^{2} 4 x}{2} \Longleftrightarrow \\
& \left(4 \cos ^{2} x+2 \cos x-1\right)^{4}\left(1-\frac{\sin ^{2} 4 x}{2}\right)=1-\frac{\sin ^{2} 4 x}{2} \Longleftrightarrow \\
& \left(4 \cos ^{2} x+2 \cos x-1\right)^{4}=1 \Longleftrightarrow \\
& 4 \cos ^{2} x+2 \cos x-1= \pm 1
\end{aligned}
$$
Solving the quadratic equations (in terms of the variable $\cos x$) $4 \cos ^{2} x+2 \cos x-2=0$ and $4 \cos ^{2} x+2 \cos x=0$, we arrive at $\cos x=1 / 2 ;-1 ; 0 ;-1 / 2$.
Answer: $x= \pm \frac{\pi}{3}+2 \pi n, x= \pm \frac{2 \pi}{3}+2 \pi n, x=\frac{\pi}{2}+\pi n, x=\pi+2 \pi n, n \in \mathbb{Z}$.
2nd method of solving. It is easy to see that there are no solutions on the set $x=2 \pi n, n \in \mathbb{Z}$. Consider all other values $x \in \mathbb{R}$. The equation is equivalent to:
$$
\begin{aligned}
& \left(\frac{\sin \frac{5 x}{2} \cos 2 x}{\sin \frac{x}{2}}\right)^{4}+\left(\frac{\sin \frac{5 x}{2} \sin 2 x}{\sin \frac{x}{2}}\right)^{4}=1-\frac{\sin ^{2} 4 x}{2} \Longleftrightarrow \\
& \left(\frac{\sin \frac{5 x}{2}}{\sin \frac{x}{2}}\right)^{4}\left(\cos ^{4} 2 x+\sin ^{4} 2 x\right)=1-\frac{\sin ^{2} 4 x}{2} \Longleftrightarrow \\
& \left(\frac{\sin \frac{5 x}{2}}{\sin \frac{x}{2}}\right)^{4}\left(1-\frac{\sin ^{2} 4 x}{2}\right)=1-\frac{\sin ^{2} 4 x}{2} \Longleftrightarrow \\
& \left(\frac{\sin \frac{5 x}{2}}{\sin \frac{x}{2}}\right)^{4}=1
\end{aligned}
$$
Solve the equation
$\sin \frac{5 x}{2}= \pm \sin \frac{x}{2} \Longleftrightarrow\left[\begin{array}{l}\frac{5 x}{2}= \pm \frac{x}{2}+2 \pi n, \\ \frac{5 x}{2}=\pi \pm \frac{x}{2}+2 \pi m,\end{array} \Longleftrightarrow\right.$
$$
\left[\begin{array} { l }
{ 3 x = 2 \pi n , } \\
{ 2 x = 2 \pi n , } \\
{ 3 x = \pi + 2 \pi n , } \\
{ 2 x = \pi + 2 \pi n }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=\frac{2 \pi n}{3} \\
x=\pi n \\
x=\frac{\pi}{3}+\frac{2 \pi n}{3} \\
x=\frac{\pi}{2}+\pi n
\end{array}\right.\right.
$$
The original equation on the set $x \neq 2 \pi n, n \in \mathbb{Z}$ is equivalent to
$$
\left[\begin{array}{l}
x=\frac{2 \pi m}{3}, m \in \mathbb{Z}, m \neq 3 s, s \in \mathbb{Z} \\
x=\frac{\pi}{3}+\frac{2 \pi n}{3}, n \in \mathbb{Z} \\
x=\frac{\pi}{2}+\pi n, n \in \mathbb{Z}
\end{array}\right.
$$
Answer: On the set $A=[2 \pi m ; 2 \pi m+\pi]$. Answer $\frac{5 \pi}{2}+8 \pi m, m \in \mathbb{Z}$.
|
574
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Equation:
$$
\cos \frac{a \pi x}{x^{2}+4}-\cos \frac{\pi\left(x^{2}-4 a x+4\right)}{4 x^{2}+16}=-\sqrt{2-\sqrt{2}}
$$
inequality $f(a) \leqslant 5$. Answer: $a \in\left[-\frac{21}{2} ; \frac{21}{2}\right]$, the length of the interval is 21.
V-A. Let $a_{n}$ be the number of permutations $\left(k_{1}, k_{2}, \ldots, k_{n}\right)$ of the numbers $(1,2, \ldots, n)$ such that the following two conditions are satisfied:
- $k_{1}=1 ;$
- For any index $i=1,2, \ldots, n-1$, the inequality $\left|k_{i}-k_{i+1}\right| \leqslant 2$ holds.
What is the number $a_{N}$?
|
Solution. We will prove in several steps:
Step 1. Prove that there is a recurrence relation: $a_{n}=a_{n-1}+a_{n-3}+1$. The following are the possible beginnings of permutations:
- In the sequence $(1,2, \ldots, n)$. Discard the first one, and decrease the remaining numbers by 1. What remains satisfies the conditions for permutations at $n-1$. Therefore, there are $a_{n-1}$ such permutations.
- In the sequence $(1,3,2,4, \ldots, n)$. Discard the first three terms of the permutation $(1,3,2)$. Decrease the remaining numbers by three. What remains satisfies the conditions for permutations at $n-3$. Therefore, there are $a_{n-3}$ such permutations.
- There is also one permutation of the form $(1,3,5,7, \ldots, 6,4,2)$. In the middle, there is a transition from the last odd number not exceeding $n$ to the last even number not exceeding $n$. Thus, we have
$$
a_{n}=a_{n-1}+a_{n-3}+1
$$
Step 2. Calculate the initial elements of the sequence
$$
\begin{aligned}
& a_{1}=1, \quad a_{2}=1, \quad a_{3}=2, \quad a_{4}=4, \quad a_{5}=6, \quad a_{6}=9, \quad a_{7}=14, \quad a_{8}=21, \quad a_{9}=31, \\
& a_{10}=46, \quad a_{11}=68, \quad a_{12}=100, \quad a_{13}=147, \quad a_{14}=216, \quad a_{15}=317, \\
& a_{16}=465, \quad a_{17}=682, \quad a_{18}=1000, \quad a_{19}=1466, \quad a_{20}=2149, \quad a_{21}=3150, \\
& a_{22}=4617, \quad a_{23}=6767, \quad a_{24}=9918, \quad a_{25}=14536, \quad a_{26}=21304, \quad a_{27}=31223, \\
& a_{28}=45760, \quad a_{29}=67065, \quad a_{30}=98289, \quad a_{31}=144050, \quad a_{32}=211116, \quad a_{33}=309406, \\
& a_{34}=453457, \quad a_{35}=664574, \quad a_{36}=973981, \quad a_{37}=1427439, \quad a_{38}=2092014, \\
& a_{39}=3065996, \quad a_{40}=4493436, \quad a_{41}=6585451, \quad a_{42}=9651448, \quad a_{43}=14144885, \\
& a_{44}=20730337, \quad a_{45}=30381786, \quad a_{46}=44526672, \quad a_{47}=65257010
\end{aligned}
$$
Answer: $a_{41}=6585451$.
|
6585451
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Ivan Semenovich leaves for work at the same time every day, drives at the same speed, and arrives exactly at 9:00. One day he overslept and left 40 minutes later than usual. To avoid being late, Ivan Semenovich drove at a speed 60% faster than usual and arrived at 8:35. By what percentage should he have increased his usual speed to arrive exactly at 9:00?
|
Answer: By $30 \%$.
Solution: By increasing the speed by $60 \%$, i.e., by 1.6 times, Ivan Semenovich reduced the time by 1.6 times and gained 40+25=65 minutes. Denoting the usual travel time as $T$, we get $\frac{T}{1.6}=T-65$, from which $T=\frac{520}{3}$. To arrive in $T-40=\frac{400}{3}$, the speed needed to be increased by $\frac{520}{3} \div \frac{400}{3}=1.3$ times, i.e., by $30 \%$.
|
30
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The English club is attended by 20 gentlemen. Some of them are acquainted (acquaintances are mutual, i.e., if A knows B, then B knows A). It is known that there are no three gentlemen in the club who are pairwise acquainted.
One day, the gentlemen came to the club, and each pair of acquaintances shook hands with each other (once). What is the maximum number of handshakes that could have been made?
|
Answer: 100 handshakes.
Solution: Choose a gentleman with the maximum number of acquaintances (if there are several, choose any one). Suppose he has $n$ acquaintances. These acquaintances cannot be pairwise acquainted with each other. Consider the remaining $(20-n-1)$ gentlemen, each of whom has no more than $n$ acquaintances, so the number of handshakes they make does not exceed $(20-n-1) \cdot n$. Therefore, the total number does not exceed $(20-n) \cdot n$. This number is maximal when $n=10$. It can be shown that 100 handshakes are possible - by dividing the gentlemen into 2 groups of 10 people each and ensuring that each person in the first group is acquainted with each person in the second (while there are no acquaintances within the groups).
|
100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Ms. Olga Ivanovna, the class teacher of 5B, is staging a "Mathematical Ballet." She wants to arrange the boys and girls so that there are exactly 2 boys 5 meters away from each girl. What is the maximum number of girls that can participate in the ballet, given that 5 boys are participating?
|
Answer: 20 girls.
Solution: Let's select and fix two arbitrary boys - $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$. Suppose they are 5 m away from some girl - G. Then $\mathrm{M}_{1}, \mathrm{M}_{2}$, and G form an isosceles triangle with the legs being 5 m. Given the fixed positions of $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$, there can be no more than two such triangles (a diagram would be helpful here!). Therefore, for any pair of boys, there can be no more than two girls. The total number of pairs of boys is $C_{5}^{2}=10$, from which we obtain the upper bound.
It is not difficult to show that this bound is achievable - simply by placing the boys in a row with an interval of 1 m and constructing all possible isosceles triangles (their vertices will obviously be different).
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Given a polynomial $P(x)$, not identically zero. It is known that for all $x$ the identity $(x-2020) \cdot P(x+1)=(x+2021) \cdot P(x)$ holds. How many roots does the equation $P(x)=0$ have? Answer: 4042
|
Solution:
First, we prove two auxiliary lemmas:
Lemma 1. If $x_{0} \neq 2020$ is a root of $P(x)$, then $x_{0}+1$ is also a root.
Proof: Substitute $\left(x_{0}-2020\right) \cdot P\left(x_{0}+1\right)=\left(x_{0}+2021\right) \cdot P\left(x_{0}\right)=0$ but $\left(x_{0}-2020\right) \neq 0$.
Lemma 2. If $x_{0} \neq -2021$ is a root of $P(x)$, then $x_{0}-1$ is also a root.
Proof: Substitute $0=\left(x_{0}-2020\right) \cdot P\left(x_{0}+1\right)=\left(x_{0}+2021\right) \cdot P\left(x_{0}\right)$ but $\left(x_{0}+2021\right) \neq 0$.
Substitute $x=2020$ into the identity, we get $0=4041 P(2020)$, hence 2020 is a root.
By sequentially applying Lemma 2, we obtain that $x=2019, 2018, \ldots, -2021$ are roots.
We will now prove that there are no other roots.
1) Suppose that the root $x_{0}$ is not an integer, then by Lemma 1, $x_{0}+1, x_{0}+2, \ldots, x_{0}+n, \ldots$ are also roots. But a polynomial (not identically zero) cannot have infinitely many roots - contradiction.
2) Suppose that $x_{0}$ is an integer and does not belong to the interval $[-2021 ; 2020]$. If $x_{0}>2020$, then (by Lemma 1) $x_{0}+1, x_{0}+2, \ldots, x_{0}+n, \ldots$ are also roots. If $x_{0}<-2021$, then (by Lemma 2) $x_{0}-1, x_{0}-2, \ldots, x_{0}-n, \ldots$ are also roots. In either case, a non-zero polynomial cannot have infinitely many roots - contradiction.
Thus, the roots are the integers from -2021 to 2020 inclusive, their number is 4042.
|
4042
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. 2-1. Find the minimum value of the function
$$
f(x)=x^{2}+(x-2)^{2}+(x-4)^{2}+\ldots+(x-100)^{2}
$$
If the result is a non-integer, round it to the nearest integer and write it as the answer.
|
Solution. Knowledge of arithmetic progression is required. It turns out to be a quadratic function
$f(x)=51 x^{2}-2(2+4+6+\ldots+100) x+(2^{2}+4^{2}+6^{2}+\ldots+100^{2})=51 x^{2}-2 \cdot 50 \cdot 51 x+4 \cdot(1^{2}+2^{2}+3^{2}+\ldots+50^{2})$.
The minimum is achieved at the point $x_{0}=50$. In this case,
$f(50)=50^{2}+48^{2}+46^{2}+\ldots+2^{2}+0^{2}+2^{2}+4^{2}+\ldots+50^{2}=2 \cdot(2^{2}+4^{2}+\ldots+50^{2})=8 \cdot(1^{2}+2^{2}+3^{2}+\ldots+25^{2})$.
Now we need the formula
$$
1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}
$$
(it is known, but in principle, it can be derived). Then $f(50)=8 \cdot \frac{25 \cdot 26 \cdot 51}{6}=17 \cdot 26 \cdot 100=44200$.
Answer: $44'200$.
|
44200
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2-2. Find the minimum value of the function
$$
f(x)=x^{2}+(x-2)^{2}+(x-4)^{2}+\ldots+(x-102)^{2}
$$
If the result is a non-integer, round it to the nearest integer and write it as the answer.
|
Solution. $f(x)=52 x^{2}-2(2+4+\ldots+102) x+2^{2}+4^{2}+\ldots+102^{2}=52 x^{2}-2 \cdot 51 \cdot 52 x+4\left(1^{2}+2^{2}+3^{2}+\ldots+51^{2}\right)$.
The minimum of the function $f$ is achieved at the point $x_{0}=51 . f(51)=51^{2}+49^{2}+\ldots 1^{2}+1^{2}+3^{2}+\ldots+49^{2}+51^{2}=$ $2\left(1^{2}+3^{2}+\ldots 51^{2}\right)$. Since $1^{2}+3^{2}+\ldots+(2 n-1)^{2}=\frac{n\left(4 n^{2}-1\right)}{3}$ (a less well-known formula, but it can also be derived), then $f(51)=\frac{26\left(52^{2}-1\right)}{3}=46852$.
Answer: 46852 .
|
46852
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2-3. Find the minimum value of the function
$$
f(x)=(x-1)^{2}+(x-3)^{2}+\ldots+(x-101)^{2}
$$
If the result is a non-integer, round it to the nearest integer and write it in the answer.
|
Solution. $f(x)=51 x^{2}-2(1+3+\ldots+101) x+\left(1^{2}+3^{2}+\ldots+101^{2}\right)=51 x^{2}-2 \cdot 51^{2} x+\left(1^{2}+3^{2}+\ldots+101^{2}\right)$.
The minimum of the function $f$ is achieved at the point $x_{0}=51$. Since $1^{2}+3^{2}+\ldots+(2 n-1)^{2}=\frac{n\left(4 n^{2}-1\right)}{3}$, then $f(51)=51^{3}-2 \cdot 51^{3}+51^{3}+\frac{51^{3}}{3}-\frac{51}{3}=\frac{51^{3}}{3}-\frac{51}{3}=44200$.
Answer: 44200.
|
44200
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2-4. Find the minimum value of the function
$$
f(x)=x^{2}+(x-2)^{2}+(x-4)^{2}+\ldots+(x-104)^{2}
$$
If the result is a non-integer, round it to the nearest integer and write it in the answer.
|
Solution. $f(x)=53 x^{2}-2(2+4+\ldots 104) x+4\left(1^{2}+2^{2}+\ldots+52^{2}\right)=53 x^{2}-2 \cdot 52 \cdot 53 x+4 \cdot \frac{52 \cdot 53 \cdot 105}{6}$ (since $\left.1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}\right)$.
The minimum of the function $f$ is achieved at the point $x_{0}=52 . f(52)=53 \cdot 52^{2}-2 \cdot 53 \cdot 52^{2}+2 \cdot 52 \cdot 53 \cdot 35=$ $52 \cdot 53(70-52)=49608$.
Answer: 49608
|
49608
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. Determine how many roots of the equation
$$
4 \sin 2 x + 3 \cos 2 x - 2 \sin x - 4 \cos x + 1 = 0
$$
are located on the interval $\left[10^{2014!} \pi ; 10^{2014!+2015} \pi\right]$. In your answer, write the sum of all digits of the found number.
|
Solution. Let $t=\sin x+2 \cos x$. Then $t^{2}=\sin ^{2} x+2 \sin 2 x+4 \cos ^{2} x=2 \sin 2 x+\frac{3}{2} \cos 2 x+\frac{5}{2}$. The original equation is equivalent to $t^{2}-t-2=0$. From which
$$
\left[\begin{array}{c}
\operatorname{sin} x + 2 \operatorname{cos} x = -1, \\
\operatorname{sin} x + 2 \operatorname{cos} x = 2
\end{array} \Longleftrightarrow \left[\begin{array}{c}
\sin (x+\varphi)=-\frac{1}{\sqrt{5}} \\
\sin (x+\varphi)=\frac{2}{\sqrt{5}}
\end{array}\right.\right.
$$
where $\varphi=\operatorname{arctg} 2$. Therefore, the original equation has exactly four roots on any interval of the form $[2 k \pi ; 2(k+1) \pi], k \in \mathbb{Z}$. Let the number of roots be denoted by $N$. Since there are four different roots in one period (the minimum period of the functions involved in the equation is $2 \pi$), then
$$
\begin{aligned}
& N=4 \cdot \frac{10^{2014!+2015} \pi-10^{2014!} \pi}{2 \pi}=2 \cdot 10^{2014!}\left(10^{2015}-1\right)= \\
&=2 \cdot 10^{2014!} \cdot \underbrace{9 \ldots 9}_{2015}=1 \underbrace{9 \ldots 9}_{2014} 8 \cdot 10^{2014!}=1 \underbrace{9 \ldots 9}_{2014} 8 \underbrace{0 \ldots 0}_{2014!}
\end{aligned}
$$
The sum of all digits of this number is $1+8+9 \cdot 2014=9 \cdot 2015=18135$.
## Answer: 18135
|
18135
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 4. Find the smallest natural $m$, for which there exists such a natural $n$ that the sets of the last 2014 digits in the decimal representation of the numbers $a=2015^{3 m+1}$ and $b=2015^{6 n+2}$ are the same, and $a<b$.
|
Solution. The coincidence of the last 2014 digits of the two specified powers means divisibility by \(10^{2014}\) of the difference
\[
2015^{6 n+2}-2015^{3 m+1}=2015^{3 m+1} \cdot\left(2015^{6 n-3 m+1}-1\right)
\]
since the first factor in the obtained decomposition is not divisible by 2, and the second is not divisible by 5, the divisibility of the obtained product by \(10^{2014}\) means two divisibilities at once
\[
2015^{3 m+1} \vdots 5^{2014}, \quad 2015^{6 n-3 m+1}-1 \vdots 2^{2014}
\]
The first divisibility means the inequality \(3 m+1 \geqslant 2014\), or \(m \geqslant m_{0}=671\). Let's check that the number \(m=m_{0}\) is realizable, i.e., for it, for some \(n=n_{0}\), the second divisibility also holds.
Indeed, let \(N=2^{2014}\) and note that for the odd \(K=2015\), the number
\[
K^{N}-1=\left(K^{N / 2}+1\right)\left(K^{N / 2}-1\right)=\ldots=\left(K^{N / 2}+1\right)\left(K^{N / 4}+1\right) \ldots\left(K^{1}+1\right)\left(K^{1}-1\right)
\]
is divisible by \(N\) (in the last representation of this number, all expressions in parentheses are even, and the number of parentheses is 2014). Therefore, \(\left(2015^{N}-1\right) \vdots 2^{2014}\), and the equality \(6 n_{0}-3 m_{0}+1=N\) holds for
\[
n_{0}=\left(\frac{N-1}{3}+m_{0}\right) / 2
\]
where \(n_{0} \in \mathbb{N}\), since the number
\[
N-1=2^{2014}-1=(3-1)^{2014}-1
\]
is divisible by three, and the quotient \((N-1) / 3\) is odd.
Answer: \(m=671\).
|
671
|
Number Theory
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math-word-problem
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Yes
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Yes
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olympiads
| false
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7. It is known that for some natural numbers $a, b$, the number $N=\frac{a^{2}+b^{2}}{a b-1}$ is also natural. Find all possible values of $N$.
---
The provided text has been translated into English while preserving the original formatting and structure.
|
Answer: 5.
Solution: It is not difficult to find the solution $a=2, b=1, N=5$. Let's prove that no other $N$ is possible.
Let $\left(a_{0}, b_{0}\right)$ be a solution corresponding to the smallest value of $a^{2}+b^{2}$. Without loss of generality, we can assume that $a_{0}>b_{0}$. If $b_{0}=1$, then $N=\frac{a^{2}+1}{a-1}$, from which $a=2$ or $a=3$, and in both cases $N=5$.
Now let $b_{0}>1$. Consider the quadratic equation $x^{2}-b_{0} N x+\left(b_{0}^{2}+N\right)=0$, then $a_{0}$ is its root. By Vieta's theorem, the second root is $a_{1}=b_{0} N-a_{0}$ and it is also positive and integer. From the minimality of $a^{2}+b^{2}$, it follows that $a_{1}>a_{0}$ (they cannot be equal). Then $\left(a_{1}-1\right)\left(a_{0}-1\right) \geqslant b_{0}\left(b_{0}+1\right)$, but on the other hand, $\left(a_{1}-1\right)\left(a_{0}-1\right)=a_{1} a_{0}-\left(a_{1}+a_{0}\right)+1=b_{0}^{2}+N-b_{0} N+1$. Therefore, $b_{0}^{2}+N-b_{0} N+1 \geqslant b_{0}^{2}+b$, which is impossible for $b_{0}>1$.
|
5
|
Number Theory
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math-word-problem
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Yes
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Yes
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olympiads
| false
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8. In a trapezoid, the diagonals intersect at a right angle, and one of them is equal to the midline. Determine the angle this diagonal forms with the bases of the trapezoid.
|
Answer: $60^{\circ}$.
Solution: We perform a parallel translation of the diagonal - we will get a right triangle, in which the leg is half the hypotenuse.
|
60
|
Geometry
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math-word-problem
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Yes
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Yes
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olympiads
| false
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2. In the wallet of the merchant Hans, there are 20 silver coins worth 2 crowns each, 15 silver coins worth 3 crowns each, and 3 gold ducats (1 ducat equals 5 crowns). In how many ways can Hans pay a sum of 10 ducats? Coins of the same denomination are indistinguishable.
|
Answer: 26
Solution. If the merchant paid $x$ coins at 2 crowns each, $y$ coins at 3 crowns each, and $z$ ducats (i.e., $z$ times 5 crowns), we get the system: $2 x+3 y+5 z=50, x \in[0 ; 20], y \in[0 ; 15], z \in[0 ; 3]$.
a) When $z=0$, we get the equation $2 x+3 y=50$, which has suitable solutions under the condition for $y=4,6,8,10,12,14$ - a total of 6 solutions.
b) When $z=1$, we get the equation $2 x+3 y=45$, which has suitable solutions under the condition for $x=0,3,6,9,12,15,18$ - a total of 7 solutions.
When $z=2$, there are 7 solutions, and when $z=3$, there are 6 solutions. These two cases could be analyzed similarly to the previous one. However, this is not necessary, using the fact that the amount to be paid is exactly half of all the merchant's money. Therefore, the number of ways using 2 ducats is equal to the number of ways using 1 ducat (since 2 ducats will remain in the wallet); similarly, the number of ways using 3 ducats is equal to the number of ways without ducats.
Thus, the total number of ways: $6+7+7+6=26$.
Answer to variant 172: 17 ways.
Answer to variant 173: 26 ways.
Answer to variant 174: 22 ways.
|
26
|
Combinatorics
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math-word-problem
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Yes
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Yes
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olympiads
| false
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1. According to the condition, these numbers are written only with the digits $0,1,2,6,8$. Then, three-digit numbers that are multiples of 4 can end with exactly 10 variants: $00,08,12,16,20,28,60,68,80,88$. In each of these 10 variants, the first place can be occupied by any of the 4 digits $1,2,6,8$. Additionally, the last 8 variants give two-digit numbers, and the second one gives a single-digit number. Therefore, the total number of such numbers is: $10 \cdot 4+8+1=49$.
|
# Answer: 49.
Answer to option: 1-2: 53.
## Solutions for option 1-2.
According to the condition, these numbers are written only with the digits $0,2,3,4,5,7$. Then, three-digit numbers that are multiples of 4 can end with exactly 9 variants: $00,04,20,24,32,40,44,52,72$. In this case, the first place in each of these 9 variants can be occupied by any of the 5 digits $2,3,4,5,7$. Additionally, the last 7 variants give two-digit numbers, and the second one gives a single-digit number. Therefore, the total number of such numbers is: $9 \cdot 5 + 7 + 1 = 53$.
|
49
|
Combinatorics
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math-word-problem
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Yes
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Yes
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olympiads
| false
|
3. A parallelepiped is inscribed in a sphere of radius $\sqrt{3}$, and the volume of the parallelepiped is 8. Find the surface area of the parallelepiped.
|
3. $S_{\text {full }}=24$
The translation is provided while maintaining the original text's formatting and structure.
|
24
|
Geometry
|
math-word-problem
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Yes
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Yes
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olympiads
| false
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6. Ms. Olga Ivanovna, the class teacher of 5B, is staging a "Mathematical Ballet." She wants to arrange the boys and girls so that exactly 2 boys are 5 meters away from each girl. What is the maximum number of girls that can participate in the ballet if it is known that 5 boys are participating?
|
Answer: 20 girls.
Solution: Let's select and fix two arbitrary boys - $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$. Suppose they are 5 m away from some girl - G. Then $\mathrm{M}_{1}, \mathrm{M}_{2}$, and G form an isosceles triangle with the legs being 5 m. Given the fixed positions of $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$, there can be no more than two such triangles (a diagram would be helpful here!). Therefore, for any pair of boys, there can be no more than two girls. The total number of pairs of boys is $C_{5}^{2}=10$, from which we obtain the upper bound.
It is not difficult to show that this bound is achievable - simply by placing the boys in a row with an interval of 1 m and constructing all possible isosceles triangles (their vertices will obviously be different).
|
20
|
Geometry
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math-word-problem
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Yes
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Yes
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olympiads
| false
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1.1 Misha noticed that a tram passed by him in 3 seconds, and a tunnel 100 meters long in 13 seconds. Find the speed of the tram (in meters per second), assuming that it remains the same throughout the entire observation period.
|
Solution. Let the speed of the tram (in meters per second) be $v$, and the length of the tram (in meters) be $l$. If $t_{1}$ and $t_{2}$ are the times it takes for the tram to pass by Mishka and through a tunnel of length $a$ respectively, then
$$
\left\{\begin{array} { l }
{ l = v \cdot t _ { 1 } } \\
{ a + l = v \cdot t _ { 2 } }
\end{array} \Longrightarrow \left\{\begin{array}{l}
v=\frac{a}{t_{2}-t_{1}} \\
l=\frac{a t_{1}}{t_{2}-t_{1}}
\end{array}\right.\right.
$$
Since $t_{1}=3, t_{2}=13$ and $a=100$, then $v=10$.
Answer: 10. (A)
|
10
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Algebra
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math-word-problem
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Yes
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Yes
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olympiads
| false
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1.2 MishΠ° noticed that the tram passed by him in 4 seconds, and a tunnel 64 meters long in 12 seconds. Find the length of the tram (in meters), assuming that its speed remains constant throughout the entire observation period.
|
Answer: 32. (E)
|
32
|
Algebra
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math-word-problem
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Yes
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Yes
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olympiads
| false
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1.3 MishΠ° noticed that the tram passed by him in 2 seconds, and a tunnel 96 meters long - in 10 seconds. Find the speed of the tram (in meters per second), assuming that it remains the same throughout the entire observation period.
|
Answer: 12. (B)
|
12
|
Algebra
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math-word-problem
|
Yes
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Yes
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olympiads
| false
|
2.1 Find $f(2013)$, if for any real $x$ and $y$ the equality holds
$$
f(x-y)=f(x)+f(y)-2xy
$$
|
Solution. Substitute $x=y=0$. We get $f(0)=2 f(0)+0$, from which we obtain that $f(0)=0$. Substitute $x=y$. We get $0=f(0)=f(x)+f(x)-2 x^{2}$. Hence, $f(x)=x^{2}$.
Answer: 4052169. (C)
|
4052169
|
Algebra
|
math-word-problem
|
Yes
|
Yes
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olympiads
| false
|
3-1. Calculate the sum
$$
S=\frac{2014}{2 \cdot 5}+\frac{2014}{5 \cdot 8}+\frac{2014}{8 \cdot 11}+\ldots+\frac{2014}{2012 \cdot 2015}
$$
In your answer, specify the remainder from dividing by 5 the even number closest to the obtained value of $S$.
|
Solution. Since
$$
\begin{aligned}
& \frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots+\frac{1}{2012 \cdot 2015}= \\
& =\frac{1}{3} \cdot\left(\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\ldots\left(\frac{1}{2012}-\frac{1}{2015}\right)\right)= \\
& \quad=\frac{1}{3} \cdot\left(\frac{1}{2}-\frac{1}{2015}\right)=\frac{1}{6} \cdot \frac{2013}{2015}
\end{aligned}
$$
the desired sum is
$$
\frac{2014 \cdot 2013}{6 \cdot 2015}=335.33 \ldots
$$
The desired natural number is 336.
Answer: 1. (V)
Options.
|
336
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3-2. Calculate the sum
$$
S=\frac{2013}{2 \cdot 6}+\frac{2013}{6 \cdot 10}+\frac{2013}{10 \cdot 14}+\ldots+\frac{2013}{2010 \cdot 2014}
$$
In your answer, specify the remainder when the nearest even number to the obtained value of $S$ is divided by 5.
|
Answer: 2. (C)
Options.
$$
\begin{array}{|l|l|l|l|l|l|l|l|l|}
\hline \mathbf{A} & 0 & \mathbf{B} & 1 & \mathbf{C} & 2 & \mathbf{D} & 3 & \mathbf{E} \\
4 & \mathbf{F} \\
\hline
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3-4. Calculate the sum
$$
S=\frac{2015}{3 \cdot 8}+\frac{2015}{8 \cdot 13}+\frac{2015}{13 \cdot 18}+\ldots+\frac{2015}{2008 \cdot 2013}
$$
In your answer, specify the remainder when dividing by 5 the natural number closest to the obtained value of $S$.
|
Answer: 4. ( ( $\mathbf{2}$ )
Options.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
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olympiads
| false
|
4-2. A circle touches the sides of an angle at points $A$ and $B$. The distance from a point $C$ lying on the circle to the line $A B$ is 6. Find the sum of the distances from point $C$ to the sides of the angle, given that one of these distances is 5 more than the other.
|
Answer: 13. (C)
Options.
|
13
|
Geometry
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math-word-problem
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Yes
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Yes
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olympiads
| false
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4-3. A circle touches the sides of an angle at points $A$ and $B$. The distance from a point $C$ lying on the circle to the line $A B$ is 6. Find the sum of the distances from point $C$ to the sides of the angle, given that one of these distances is nine times smaller than the other.
|
Answer: 20. (E)
Answer options.
|
20
|
Geometry
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math-word-problem
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Yes
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Yes
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olympiads
| false
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4-4. A circle touches the sides of an angle at points $A$ and $B$. The distance from a point $C$ lying on the circle to the line $A B$ is 8. Find the sum of the distances from point $C$ to the sides of the angle, given that one of these distances is 30 less than the other.
|
Answer: 34. (B)
Answer options.
|
34
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Geometry
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math-word-problem
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Yes
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Yes
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olympiads
| false
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5-1. Solve the inequality
$$
\sqrt{3 x-7}-\sqrt{3 x^{2}-13 x+13} \geqslant 3 x^{2}-16 x+20
$$
In your answer, specify the sum of all integer values of $x$ that satisfy the inequality.
|
Solution. As a result of the substitution $v=\sqrt{3 x-7}, u=\sqrt{3 x^{2}-13 x+13}$, we obtain the equivalent inequality $u \leqslant v$. Therefore, $x$ satisfies the inequality $3 x^{2}-13 x+13 \leqslant 3 x-7 \Longleftrightarrow 2 \leqslant x \leqslant 10 / 3$. Out of the two integer values $x=2$ and $x=3$, only $x=3$ falls within the domain of valid values.
Answer: (B) 3.
Answer options.
|
3
|
Inequalities
|
math-word-problem
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Yes
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Yes
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olympiads
| false
|
5-2. Solve the inequality
$$
\sqrt{6 x-13}-\sqrt{3 x^{2}-13 x+13} \geqslant 3 x^{2}-19 x+26
$$
In your answer, specify the sum of all integer values of $x$ that satisfy the inequality.
|
Solution. The inequality is satisfied only by the following integer values: $x=3, x=4$.
Answer: (D) 7.
Answer choices.
$$
\begin{array}{|l|l|l|l|l|l|l|l|l|}
\hline \mathbf{A} & 2 & \mathbf{B} & 3 & \mathbf{C} & 5 & \mathbf{D} & 7 & \mathbf{E} \\
\hline
\end{array}
$$
|
7
|
Inequalities
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math-word-problem
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Yes
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Yes
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olympiads
| false
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5-3. Solve the inequality
$$
\sqrt{5 x-11}-\sqrt{5 x^{2}-21 x+21} \geqslant 5 x^{2}-26 x+32
$$
In your answer, specify the sum of all integer values of $x$ that satisfy the inequality.
|
Solution. The inequality is satisfied by only one integer value: $x=3$.
Answer: (B) 3.
Answer choices.
$$
\begin{array}{|l|l|l|l|l|l|l|l|l|}
\hline \mathbf{A} & 2 & \mathbf{B} & 3 & \mathbf{C} & 5 & \mathbf{D} & 7 & \mathbf{E} \\
\hline
\end{array}
$$
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
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olympiads
| false
|
5-4. Solve the inequality
$$
\sqrt{10 x-21}-\sqrt{5 x^{2}-21 x+21} \geqslant 5 x^{2}-31 x+42
$$
In your answer, indicate the sum of all integer values of $x$ that satisfy the inequality.
|
Solution. The inequality is satisfied only by the following integer values: $x=3, x=4$.
Answer: (D) 7.
Answer choices.
|
7
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
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olympiads
| false
|
6-2. From three mathematicians and ten economists, a committee of seven people needs to be formed. At the same time, the committee must include at least one mathematician. In how many ways can the committee be formed?
|
Solution. Direct calculation gives: $C_{13}^{7}-C_{10}^{7}=1596$.
Answer: 1596.
|
1596
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6-3. From twelve students and three teachers, a school committee consisting of nine people needs to be formed. At the same time, at least one teacher must be included in it. In how many ways can the committee be formed?
|
Solution. Direct calculation gives: $C_{15}^{9}-C_{12}^{9}=4785$.
Answer: 4785.
|
4785
|
Combinatorics
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math-word-problem
|
Yes
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Yes
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olympiads
| false
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6-4. From eleven students and three teachers, a school committee consisting of eight people needs to be formed. At the same time, at least one teacher must be included in it. In how many ways can the committee be formed?
|
Solution. Direct calculation gives: $C_{14}^{8}-C_{11}^{8}=2838$.
Answer: 2838.
|
2838
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8-2. Find all common points of the graphs
$$
y=8 \cos ^{2} \pi x \cdot \cos 2 \pi x \cdot \cos 4 \pi x \quad \text { and } \quad y=\cos 6 \pi x
$$
with abscissas belonging to the interval $[-1 ; 0]$. In your answer, specify the sum of the abscissas of the found points.
|
Solution. This is equivalent to $\sin 9 \pi x=\sin (-5 \pi x), x \neq n, n \in \mathbb{Z}$. From this, $x=k / 7$ or $x=1 / 4+l / 2, k, l \in \mathbb{Z}$.
The roots of the equation in the interval $[-1 ; 0]$ are: $-1 / 7,-2 / 7,-3 / 7,-4 / 7,-5 / 7,-6 / 7$ and $-1 / 4,-3 / 4$.
Answer: -4 .
|
-4
|
Algebra
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math-word-problem
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Yes
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Yes
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olympiads
| false
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10-1. At the base of the pyramid $S A B C D$ lies a trapezoid $A B C D$ with bases $B C$ and $A D$. Points $P_{1}$, $P_{2}, P_{3}$ belong to the side $B C$, such that $B P_{1}<B P_{2}<B P_{3}<B C$. Points $Q_{1}, Q_{2}, Q_{3}$ belong to the side $A D$, such that $A Q_{1}<A Q_{2}<A Q_{3}<A D$. Denote the points of intersection of $B Q_{1}$ with $A P_{1}$, $P_{2} Q_{1}$ with $P_{1} Q_{2}$, $P_{3} Q_{2}$ with $P_{2} Q_{3}$, and $C Q_{3}$ with $P_{3} D$ as $R_{1}, R_{2}, R_{3}$, and $R_{4}$, respectively. It is known that the sum of the volumes of the pyramids $S R_{1} P_{1} R_{2} Q_{1}$ and $S R_{3} P_{3} R_{4} Q_{3}$ is 78. Find the minimum value of the quantity
$$
V_{S A B R_{1}}^{2}+V_{S R_{2} P_{2} R_{3} Q_{2}}^{2}+V_{S C D R_{4}}^{2}
$$
In your answer, specify the integer closest to the found value.
|
Solution. From the properties of the trapezoid, it follows that the triangles (see Fig. 2), shaded with the same color, have the same area. From this, the equality of the sums of the areas marked
Fig. 2:
with the same color on Fig. 3 follows. The equality of the sums of the areas takes the form
$$
S_{A B R_{1}}+S_{R_{2} P_{2} R_{3} Q_{2}}+S_{C D R_{4}}=S_{R_{1} P_{1} R_{2} Q_{1}}+S_{R_{3} P_{3} R_{4} Q_{3}}
$$
Fig. 3:
From which
$$
V_{S A B R_{1}}+V_{S R_{2} P_{2} R_{3} Q_{2}}+V_{S C D R_{4}}=V_{S R_{1} P_{1} R_{2} Q_{1}}+V_{S R_{3} P_{3} R_{4} Q_{3}}=78
$$
Let $a_{1}=V_{S A B R_{1}}, a_{2}=V_{S R_{2} P_{2} R_{3} Q_{2}}, a_{3}=V_{S C D R_{4}}$. From the problem statement: $a_{1}+a_{2}+a_{3}=78$, and we need to find $a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \rightarrow \min$, under the condition that $a_{1}, a_{2}, a_{3}$ are non-negative. The inequality holds
$$
\left(\frac{a_{1}}{3}+\frac{a_{2}}{3}+\frac{a_{3}}{3}\right)^{2} \leqslant \frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{3}
$$
For school students, this inequality is justified by applying the inequality $2 a b \leqslant a^{2}+b^{2}$ three times. We obtain
$$
a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \geqslant \frac{\left(a_{1}+a_{2}+a_{3}\right)^{2}}{3}=2028
$$
Equality is achieved when $a_{1}=a_{2}=a_{3}=26$.
Answer: 2028.
|
2028
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Let $\Sigma(n)$ denote the sum of the digits of the number $n$. Find the smallest three-digit $n$ such that $\Sigma(n)=\Sigma(2 n)=\Sigma(3 n)=\ldots=\Sigma\left(n^{2}\right)$
|
Answer: 999.
Solution: Let the desired number be $\overline{a b c}$. Note that this number is not less than 101 (since 100 does not work). Therefore, $101 \cdot \overline{a b c}=\overline{a b c 00}+\overline{a b c}$ also has the same sum of digits. But the last digits of this number are obviously $b$ and $c$, so the sum of the remaining digits must be equal to $a$. Therefore, $\Sigma(\overline{a b c}+a)=a$. If $a<9$, then $\overline{a b c}+a-$ is a three-digit number, the first digit of which is not less than $a$, which leads to a contradiction, since the second and third digits cannot be zeros. Thus, $a=9$ and $\overline{a b c}+a \leqslant 999+9=1008$. Therefore, $\overline{a b c}+a=\overline{100 d}$. But $\Sigma(\overline{100 d})=a=9$, so $d=8$, from which $\overline{a b c}=999$.
|
999
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Compare the numbers
$$
\left(1+\frac{1}{1755}\right)\left(1+\frac{1}{1756}\right) \ldots\left(1+\frac{1}{2015}\right) \text { and } \sqrt{\frac{8}{7}}
$$
Indicate in the answer Β«1Β» if the first number is greater; Β«2Β», if the second number is greater; Β«0Β», if the numbers are equal.
|
Answer: The first one is greater.
Solution: Convert each parenthesis on the left to a common denominator, then use the inequality $\frac{n \cdot n}{(n-1)(n+1)}>1$. Then we get that the square of the first number is greater than $\frac{2015}{1755}=\frac{31}{27}>\frac{8}{7}$.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The test consists of 5 sections, each containing the same number of questions. Pavel answered 32 questions correctly. The percentage of his correct answers turned out to be more than 70 but less than 77. How many questions were in the test?
ANSWER: 45.
|
Solution: from the condition $0.7<32 / x<0.77$ it follows that $41<x<46$, but $x$ is a multiple of 5, so $x=45$.
|
45
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. How many different right-angled triangles exist, one of the legs of which is equal to $\sqrt{1001}$, and the other leg and the hypotenuse are expressed as natural numbers?
ANSWER: 4.
|
Solution: Let's write down the Pythagorean theorem: $a^{2}+1001=b^{2}$. From this, we get $(b-a)(b+a)=1001=7 \times 11 \times 13$. We can represent 1001 as the product of two factors $1 \times 1001=7 \times 143=11 \times 91=13 \times 77$ - the first factor must be smaller - there are only 4 options
Lomonosov Moscow State University
## School Olympiad "Conquer Sparrow Hills" in Mathematics
Final Round Tasks for 2015/2016 Academic Year for 7-8 Grades
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Find the largest natural number that cannot be represented as the sum of two composite numbers.
OTBET: 11
|
Solution: Even numbers greater than 8 can be represented as the sum of two even numbers greater than 2. And odd numbers greater than 12 can be represented as the sum of 9 and an even composite number. By direct verification, we are convinced that 11 cannot be represented in this way.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. How many five-digit numbers of the form $\overline{a b 16 c}$ are divisible by 16? ( $a, b, c-$ are arbitrary digits, not necessarily different).
ANSWER: 90.
|
Solution: Note that the first digit does not affect divisibility, hence $a=1, \ldots, 9$. On the other hand, divisibility by 8 implies that $c=0$ or 8. If $c=0$, then $\mathrm{b}$ must be even, and if $\mathrm{c}=8$ - odd. In both cases, we get 5 options, from which the total number is $9 *(5+5)=90$.
Lomonosov Moscow State University
## School Olympiad "Conquer Vorobyovy Gory" in Mathematics
Final stage tasks for the 2015/2016 academic year for 7-8 grades
|
90
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Philatelist Andrey decided to distribute all his stamps equally into 3 envelopes, but it turned out that one stamp was extra. When he distributed them equally into 5 envelopes, 3 stamps were extra; finally, when he distributed them equally into 7 envelopes, 5 stamps remained. How many stamps does Andrey have in total, if it is known that recently he bought an additional album for them, which can hold 150 stamps, as such an old album was no longer sufficient?
ANSWER 208.
|
Solution. If the desired number is $x$, then the number $x+2$ must be divisible by 3, 5, and 7, i.e., it has the form $3 \cdot 5 \cdot 7 \cdot p$. Therefore, $x=105 p-2$. Since by the condition $150<x \leq 300$, then $p=2$. Therefore, $x=208$.
|
208
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Five runners ran a relay. If the first runner ran twice as fast, they would have spent $5 \%$ less time. If the second runner ran twice as fast, they would have spent $10 \%$ less time. If the third runner ran twice as fast, they would have spent $12 \%$ less time. If the fourth runner ran twice as fast, they would have spent $15 \%$ less time. By what percentage less time would they have spent if the fifth runner ran twice as fast?
OTBET: by $8 \%$
|
Solution: If each ran twice as fast, they would run 50% faster. This means that if the 5th runner ran faster, the time would decrease by $50-5-10-12-15=8 \%$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find the smallest natural number $N$ such that the number $99 N$ consists only of threes.
ANSWER: 3367.
|
Solution. The number 33N must consist of all ones. A number is divisible by 33 if it is divisible by 3 and by 11. A number consisting of all ones is divisible by 3 if the number of ones is a multiple of 3, and it is divisible by 11 if the number of ones is a multiple of 2. The smallest such number is 111111, so 33N = 111111, from which \( N=3367 \).
Lomonosov Moscow State University
## School Olympiad "Conquer Sparrow Hills" in Mathematics
Final Round Tasks for 2015/2016 Academic Year for 7-8 Grades
|
3367
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Find all values of $a$, for each of which the system
$$
\left\{\begin{array}{l}
|y|+|y-x| \leqslant a-|x-1| \\
(y-4)(y+3) \geqslant(4-x)(3+x)
\end{array}\right.
$$
has exactly two solutions.
|
# Problem 5.
Answer: when $a=7$.
Solution. By the triangle inequality:
$$
|y|+|y-x|+|x-1| \geqslant|y-(y-x)-(x-1)|=1
$$
therefore, the first inequality can have solutions only when $a \geqslant 1$. In this case, it is equivalent to the system of inequalities
$$
-\frac{a-1}{2} \leqslant y \leqslant \frac{a+1}{2}, \quad-\frac{a-1}{2} \leqslant x \leqslant \frac{a+1}{2}, \quad x-\frac{a+1}{2} \leqslant y \leqslant x+\frac{a-1}{2}
$$
which defines a hexagon on the coordinate plane if $a>1$, and a triangle if $a=1$.
The second inequality of the system is equivalent to the inequality $(x-0.5)^{2}+(y-0.5)^{2} \geqslant 49 / 2$ and thus defines the complement to the entire plane of the interior of a circle with center $(1 / 2,1 / 2)$ and radius $7 / \sqrt{2}$.
If $a=1$, the system has no solutions.
Since for $a>1$ the vertices $\left(-\frac{a-1}{2},-\frac{a-1}{2}\right)$ and $\left(\frac{a+1}{2}, \frac{a+1}{2}\right)$ of the larger diagonal of the hexagon are the farthest from the point $(1 / 2,1 / 2)$ among all points of the hexagon, the intersection of the considered sets will consist of exactly two points if and only if the circle of the second set passes through these vertices. The point $(1 / 2,1 / 2)$ is the midpoint of the larger diagonal, and the length of this diagonal is $a \sqrt{2}$. Therefore, $a \sqrt{2}=14 / \sqrt{2}$, i.e., $a=7$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. (9) On a plane, there are 9 points arranged in a $3 \times 3$ grid, as shown in the figure.
a) Lines are drawn through all possible pairs of points. How many different lines are obtained?
b) How many different triangles exist with vertices at these points?
|
Answer: a) 20 ; b) 76.
Solution: a) The number of pairs of points will be $C_{9}^{2}=36$, but lines coincide when three points lie on the same line. There are $8=3$ horizontals +3 verticals +2 diagonals such cases. Therefore, subtract $36-8 \times 2=20$.
b) The number of triplets of points is $C_{9}^{3}=84$, but not all form a triangle, i.e., we need to subtract 8 triplets lying on the same line. $84-8=76$.
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. On the Island of Knights and Liars, knights always tell the truth, while liars always lie. In a school on this island, both knights and liars study in the same class. One day, the teacher asked four children: Anu, Banu, Vanu, and Danu, who among them had completed their homework. They answered:
- Anu: Banu, Vanu, and Danu completed the homework.
- Banu: Anu, Vanu, and Danu did not complete the homework.
- Vanu: Don't believe them, sir! Anu and Banu are liars!
- Danu: No, sir, Anu, Banu, and Vanu are knights!
How many knights are among these children?
|
Answer: 1.
Solution: If Vanu is a knight, then all the others are liars.
Let Vanu be a liar. Then Danu is also a liar (since he says Vanu is a knight). And at least one of Anu and Banu must be a knight. Both of them cannot be knights, as they contradict each other. In any case, only one of the children is a knight.
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In triangle $\triangle A B C$, the sides $A B=5$ and $A C=6$ are known. What should the side $B C$ be so that the angle $\angle A C B$ is as large as possible? Provide the length of side $B C$, rounded to the nearest integer.
|
Answer: 3
Solution: Construct $AC=6$. Then the geometric locus of points $B$ will be a circle of radius 5 centered at point $A$. The angle $\angle ACB$ will be the largest when $CB$ is tangent to the circle (see figure). Then $CB \perp AB$ and by the Pythagorean theorem we get $BC=\sqrt{AC^{2}-AB^{2}}=\sqrt{11} \approx 3$.
## 2013/2014 Academic Year CRITERIA FOR DETERMINING WINNERS AND PRIZE WINNERS ${ }^{1}$
## of the school students' competition "CONQUER SPARROW MOUNTAINS!" IN MATHEMATICS
ELIMINATION STAGE
WINNER:
From 95 points inclusive and above.
PRIZE WINNER:
From
91 points to
94
points inclusive.
FINAL STAGE
WINNER (Diploma I degree):
From 90 points inclusive and above.
PRIZE WINNER (Diploma II degree):
From 75 points to 89 points inclusive.
PRIZE WINNER (Diploma III degree):
from 60 points to 74 points inclusive.[^0]
[^0]: ${ }^{1}$ Approved at the meeting of the jury of the school students' competition "Conquer Sparrow Mountains!" in mathematics
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Natural numbers $m, n$ are such that the fraction $\frac{m}{n}$ is irreducible, while the fraction $\frac{4 m+3 n}{5 m+2 n}$ is reducible. By which natural numbers can it be reduced?
|
# Answer: 7
Let the numbers $4 m+3 n$ and $5 m+2 n$ have a common divisor $d>1$. Then there exist natural numbers $a$ and $b$ such that the system holds:
$$
\left\{\begin{array}{l}
4 m+3 n=a d \\
5 m+2 n=b d
\end{array}\right.
$$
From this system, we can express $m$ and $n$: $7 n=d(5 a-4 b), 7 m=d(3 b-2 a)$. If 7 does not divide $d$, then $m$ and $n$ are divisible by $d>1$, which contradicts the irreducibility of the fraction $\frac{m}{n}$. Therefore, $d=7$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Find the maximum value of the expression
$$
\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\ldots+\left(x_{2010}-x_{2011}\right)^{2}+\left(x_{2011}-x_{1}\right)^{2}
$$
for $x_{1}, \ldots, x_{2011} \in[0 ; 1]$.
#
|
# Answer: 2010.
Let's prove this using mathematical induction for $2n+1$ numbers $x_{1}, \ldots, x_{2n+1} \in [0; 1]$. Specifically, we will show that the maximum value of the expression
$$
\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\ldots+\left(x_{2n}-x_{2n+1}\right)^{2}+\left(x_{2n+1}-x_{1}\right)^{2}
$$
for $x_{1}, \ldots, x_{2n+1} \in [0; 1]$ is $2n$.
First, note that for $x_{1}, \ldots, x_{2n+1} \in [0; 1]$, the following inequality always holds:
$$
\begin{aligned}
\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\ldots+\left(x_{2n}-x_{2n+1}\right)^{2}+\left(x_{2n+1}-x_{1}\right)^{2} \\
& \leqslant \left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\ldots+\left|x_{2n}-x_{2n+1}\right|+\left|x_{2n+1}-x_{1}\right|
\end{aligned}
$$
1) Let's check that for any numbers $x, y, z$ such that $x, y, z \in [0; 1]$, the inequality
$$
|x-y|+|y-z|+|z-x| \leqslant 2
$$
holds, and there exist numbers $x, y, z \in [0; 1]$ such that $|x-y|+|y-z|+|z-x|=2$. Indeed, without loss of generality, assume that $0 \leqslant x \leqslant y \leqslant z$. Then
$$
|x-y|+|y-z|+|z-x|=y-x+z-y+z-x=2(z-x) \leqslant 2
$$
Equality is achieved, for example, if $x=y=0, z=1$.
2) Assume that for $2n-1$ numbers, the maximum value of the expression
$$
\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\ldots+\left|x_{2n-2}-x_{2n-1}\right|+\left|x_{2n-1}-x_{1}\right|
$$
is $2n-2$ for $x_{1}, \ldots, x_{2n-1} \in [0; 1]$.
3) Consider the expression for $2n+1$ numbers:
$$
\begin{gathered}
\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\ldots+\left|x_{2n-2}-x_{2n-1}\right|+\left|x_{2n-1}-x_{2n}\right|+\left|x_{2n}-x_{2n+1}\right|+\left|x_{2n+1}-x_{1}\right|= \\
=\underbrace{\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\ldots+\left|x_{2n-2}-x_{2n-1}\right|+\left|x_{2n-1}-x_{1}\right|}_{\leqslant 2n-2}-\left|x_{2n-1}-x_{1}\right|+\left|x_{2n-1}-x_{2n}\right|+ \\
+\left|x_{2n}-x_{2n+1}\right|+\left|x_{2n+1}-x_{1}\right| \leqslant 2n-2-\left|x_{2n-1}-x_{1}\right|+\underbrace{\left|x_{2n-1}-x_{2n}\right|+\left|x_{2n}-x_{2n+1}\right|+\left|x_{2n+1}-x_{2n-1}\right|}_{\leqslant 2}- \\
\quad-\left|x_{2n+1}-x_{2n-1}\right|+\left|x_{2n+1}-x_{1}\right|=2n+\left|x_{2n+1}-x_{1}\right|-\left|x_{2n-1}-x_{1}\right|-\left|x_{2n+1}-x_{2n-1}\right| \leqslant 2n
\end{gathered}
$$
since $\left|x_{2n+1}-x_{1}\right| \leqslant \left|x_{2n-1}-x_{1}\right|+\left|x_{2n+1}-x_{2n-1}\right|$.
The value $2n$ of the expression
$$
\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\ldots+\left|x_{2n-2}-x_{2n-1}\right|+\left|x_{2n-1}-x_{2n}\right|+\left|x_{2n}-x_{2n+1}\right|+\left|x_{2n+1}-x_{1}\right|
$$
is achieved, for example, when $x_{1}=x_{3}=\ldots=x_{2n-1}=x_{2n+1}=1, x_{2}=x_{4}=\ldots=x_{2n}=0$.
In particular,
$$
\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\ldots+\left(x_{2n}-x_{2n+1}\right)^{2}+\left(x_{2n+1}-x_{1}\right)^{2}=2010
$$
when $x_{1}=x_{3}=\ldots=x_{2011}=x_{2n+1}=1, x_{2}=x_{4}=\ldots=x_{2010}=0$.
|
2010
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10. There are 12 pencils of pairwise different lengths. In how many ways can they be placed in a box in 2 layers of 6 pencils each, so that in each layer the pencils are arranged in increasing order of length (from left to right), and each pencil in the upper layer lies strictly above a pencil in the lower layer and is shorter than it?
|
Answer: 132
Let $0 \leqslant m \leqslant n \leqslant 6$. Denote by $K(m, n)$ the set of all arrangements of $m+n$ pencils of different lengths with the conditions: 1) in the bottom row, there are $n$ pencils, starting from the right edge without gaps in decreasing order of length; 2) in the top row, there are also $m$ pencils; 3) each pencil in the top row is shorter than the pencil directly below it. Let $k(m, n)$ denote the number of such arrangements. In each arrangement from $K(m, n)$, the shortest pencil lies either as the leftmost in the bottom row (then $m<n$), or as the leftmost in the top row. Removing it, we get a correct arrangement from the set $K(m, n-1)$ in the first case, and a correct arrangement from the set $K(m-1, n)$ in the second case. Conversely, adding a short pencil to the bottom layer of any arrangement from $K(m, n-1)$ or to the top layer of any arrangement from $K(m-1, n)$ (only if $m-1<n$) will result in a correct arrangement from the set $K(m, n)$. From this, we get that $k(m, n)=k(m-1, n)+k(m, n-1)$, if $m<n$, and $k(m, n)=k(m-1, n)$, if $m=n$. In this case, $k(0, n)=1$ for all $n$.
Consider a grid board of size $7 \times 7$. Number the rows from bottom to top with numbers $0,1, \ldots, 6$, and the columns from left to right with numbers $0,1, \ldots, 6$. In the cell located in the row with number $m$ and in the column with number $n$, we will place the number $k(m, n)$. Using the above equality, fill in the table, moving from the bottom left corner to the top right. We get the table:
| | | | | | | 132 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | | | | | 42 | 132 |
| | | | | 14 | 42 | 90 |
| | | | 5 | 14 | 28 | 48 |
| | | 2 | 5 | 9 | 14 | 20 |
| | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
We get that $k(6,6)=132$.
|
132
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Six natural numbers (possibly repeating) are written on the faces of a cube, such that the numbers on adjacent faces differ by more than 1. What is the smallest possible value of the sum of these six numbers?
|
Answer: 18.
Solution: Consider three faces that share a common vertex. The numbers on them differ pairwise by 2, so the smallest possible sum would be for $1+3+5=9$. The same can be said about the remaining three faces.
|
18
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The Slytherin faculty has 30 students. Some of them are friends (friendship is mutual), but there are no 3 people who are all friends with each other. On New Year's, everyone sent cards to all their friends. What is the maximum number of cards that could have been sent?
|
Answer: 450.
Solution: Let's find the person with the most friends. Suppose there are no fewer than 15, and denote their number as $15+a$. We will divide the students into two groups: the first group will consist of these $15+a$ students. According to the condition, they cannot be friends with each other, so each of them has no more than 15-a friends. The second group will consist of the remaining 15-a, each of whom has no more than $15+a$ friends. Thus, each group will send no more than 225- $a^{2}$ cards. Therefore, in total, no more than 450-2 $a^{2}$ cards will be sent, which does not exceed 450.
Note that this value is achievable. Divide the students into two groups of 15 people each, and let each representative of one group be friends with all representatives of the other group.
|
450
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Given a sequence of natural numbers $a_{n}$, the terms of which satisfy the relations $a_{n+1}=k \cdot \frac{a_{n}}{a_{n-1}}$ (for $n \geq 2$). All terms of the sequence are integers.
It is known that $a_{1}=1$, and $a_{2018}=2020$. Find the smallest natural $k$ for which this is possible.
|
Answer: 2020
Solution: Let a2=x. Then all terms of the sequence will have the form $x^{m} k^{n}$.
The powers of $k$ will repeat with a period of 6: $0,0,1,2,2,1,0,0, \ldots$
The powers of $x$ will also repeat with a period of 6: $0,1,1,0,-1,-1,0,1, \ldots$
Since 2018 gives a remainder of 2 when divided by 6, then $a_{2018}=a_{2}=x=2020$. For all terms of the sequence to be integers, it is necessary for $k$ to be a multiple of $x$, the smallest such $k$ is 2020.
|
2020
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Points $A_{1}, \ldots, A_{12}$ are the vertices of a regular 12-gon. How many different 11-segment open broken lines without self-intersections with vertices at these points exist? Broken lines that can be transformed into each other by rotation are considered the same.
|
Answer: 1024.
Solution: The first vertex of the broken line can be chosen in 12 ways. Each subsequent vertex (except the last one) can be chosen in two ways - it must be adjacent to the already marked vertices to avoid self-intersections. The last vertex is chosen uniquely. We get $12^{*} 2^{10}$ ways. Considering 12 possible rotations, we get that each broken line will be counted 12 times, so this number must be divided by 12.
Remark: Here, it was implied in the condition that the broken line has a starting and ending point. If, however, we consider broken lines as geometric objects, i.e., without a distinguished "head" and "tail," this significantly complicates the problem.
|
1024
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Find the smallest three-digit number with the property that if a number, which is 1 greater, is appended to it on the right, then the result (a six-digit number) will be a perfect square. Answer: 183
|
Solution: Let the required number be \(a\), then \(1000a + a + 1 = n^2\). We can write it as: \(1001a = (n - 1)(n + 1)\). Factorize \(1001 = 7 \times 11 \times 13\), so the product \((n - 1)(n + 1)\) must be divisible by 7, 11, and 13. Moreover, for the square to be a six-digit number, \(n\) must be in the interval \([317; 999]\).
Consider the following cases:
a) \(n - 1\) is divisible by 143, \(n + 1\) is divisible by 7, then we find \(n = 573\);
b) \(n - 1\) is divisible by 7, \(n + 1\) is divisible by 143, then \(n = 428\);
c) \(n - 1\) is divisible by 77, \(n + 1\) is divisible by 13, then \(n = 155\) - does not fit;
d) \(n - 1\) is divisible by 13, \(n + 1\) is divisible by 77, then \(n = 846\);
e) \(n - 1\) is divisible by 91, \(n + 1\) is divisible by 11, then \(n = 274\) - does not fit;
f) \(n - 1\) is divisible by 11, \(n + 1\) is divisible by 91, then \(n = 727\).
The smallest \(n = 428, n^2 = 428^2 = 183184\).
|
183
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14. Let's call an integer "extraordinary" if it has exactly one even divisor other than 2. How many extraordinary numbers exist in the interval $[1 ; 75]$?
|
Answer: 12
Solution: This number should be equal to a prime multiplied by 2. There are 12 such numbers:
$\begin{array}{llllllllllll}2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & 23 & 29 & 31 & 37\end{array}$
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
15. A right-angled triangle was cut along a straight line into two parts and these parts were assembled into a square (see fig). What is the length of the shorter leg if the longer leg is 10?
#
|
# Answer 5.
Solution: The short leg is equal to the side of the square, and the long leg is twice the side of the square.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
17. In a regular 1000-gon, all diagonals are drawn. What is the maximum number of diagonals that can be selected such that among any three of the selected diagonals, at least two have the same length?
|
Answer: 2000
Solution: For the condition of the problem to be met, it is necessary that the lengths of the diagonals take no more than two different values. The diagonals connecting diametrically opposite vertices are 500. Any other diagonal can be rotated to coincide with a diagonal of the corresponding length, i.e., there are 1000 of them. Therefore, 2000 can be chosen while satisfying the condition.
|
2000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
18. In how many ways can the number 1024 be factored into three natural factors such that the first factor is divisible by the second, and the second is divisible by the third?
|
Answer: 14
Solution: Note that the factors have the form $2^{a} \times 2^{b} \times 2^{c}$, where $\mathrm{a}+\mathrm{b}+\mathrm{c}=10$ and $a \geq b \geq c$. Obviously, $c$ is less than 4, otherwise the sum would be greater than 10. Let's consider the cases:
$c=0)$ Then $b=0, \ldots, 5, a=10-b-6$ options
$c=1)$ Then $b=1, . .4, a=9-b-4$ options
$c=2) b=2,3,4, a=8-b-3$ options
$c=3) b=3, a=4-1$ option.
In total, $6+4+3+1=14$ options.
In a trapezoid, the diagonals of which intersect at a right angle, it is known that the midline is 6.5 and one of the diagonals is 12. Find the second diagonal.
## Answer 5
Solution: Parallel translate one of the diagonals so that it forms a right triangle with the other. Then in this triangle, one leg is 12, and the hypotenuse is 13, so the remaining leg is 5.
|
14
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
19. Find the last two digits of the sum
$$
1^{2}+2^{2}+\ldots+50^{2}-51^{2}-\ldots-100^{2}+101^{2}+\ldots 150^{2}-151^{2}-\ldots 200^{2}+\ldots-2000^{2}+2001^{2}+\ldots+2017^{2}
$$
(i.e., 50 numbers with a plus sign, 50 with a minus sign, and so on.)
|
# Answer: 85
Solution: Note that the numbers $\mathrm{n}^{2}$ and $(\mathrm{n}+50)^{2}$ give the same remainder when divided by 100. Therefore, in each hundred, the sum will end in two zeros. The last digits of the squares from 2001 to 2017 are:
| 01 | 04 | 09 | 16 | 25 | 36 | 49 | 64 | 81 | 00 | 21 | 44 | 69 | 96 | 25 | 56 | 89 , which in sum |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
gives 685.
|
85
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
20. Find the sum of the fourth powers of the real roots of the equation
$$
x^{4}-1000 x^{2}+2017=0
$$
|
# Answer: 1991932
Solution: The roots of the equation are of the form $\pm \sqrt{t_{1}}, \pm \sqrt{t_{2}}$, where $t_{1,2}$ are the roots of the equation $t^{2}-1000 t+2017=0$. Therefore, the sum of the fourth powers is $2\left(t_{1}^{2}+t_{2}^{2}\right)=2\left(t_{1}+t_{2}\right)^{2}-4 t_{1} t_{2}=2 \cdot 1000^{2}-4 \cdot 2017=1991932$ - here we used Vieta's formulas.
## Variant 3-a
|
1991932
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. We will call a number "remarkable" if it has exactly 4 distinct natural divisors, and among them, there are two such that neither is a multiple of the other. How many "remarkable" two-digit numbers exist?
|
Answer 36.
Solution: Such numbers must have the form $p_{1} \cdot p_{2}$, where $p_{1}, p_{2}$ are prime numbers. Note that the smaller of these prime numbers cannot be greater than 7, otherwise the product will be at least 121. It is sufficient to check $p_{1}=2,3,5,7$. For 2, we get the second factor: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, or 47 - 14 options, for 3 we get: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 - 10 options, for 5: 3, 5, 7, 11, 13, 17, 19 - 7 options, and for 7: 3, 5, 7, 11, 13 - 5 options. In total, $14+10+7+5=36$ options.
|
36
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Arrange the smallest square area using square tiles of sizes $1 \times 1$, $2 \times 2$, and $3 \times 3$, such that the number of tiles of each size is the same.
|
Answer:
Solution. Let $n$ be the number of squares of each type. Then $n + 4n + 9n = 14n$ must be a perfect square. The smallest $n$ for which this is possible is 14. The figure shows an example of how to construct a 14x14 square (other arrangements are possible, as long as there are 14 squares of each type).
|
14
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The company conducted a survey among its employees - which social networks they use: VKontakte or Odnoklassniki. Some employees said they use VKontakte, some - Odnoklassniki, some said they use both social networks, and 40 employees said they do not use social networks. Among all those who use social networks, 75% use VKontakte, and 65% use both networks. The proportion of employees who use Odnoklassniki from the total number of all employees is 5/6. How many employees work in the company
|
Answer: 540
Solution: Since 75% of social media users use VKontakte, it follows that only 25% use Odnoklassniki. Additionally, 65% use both networks, so in total, Odnoklassniki is used by $65+25=90\%$ of social media users. These $90\%$ constitute $5 / 6$ of the company's employees, so $100\%$ constitutes $10 / 9 * 5 / 6 = 50 / 54$ of all employees. Therefore, those who do not use social media constitute 1-50/54 = 4/54, and there are 40 such people. Thus, the total number of employees is 540.
|
540
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In a regular 2017-gon, all diagonals are drawn. Petya randomly selects some $\mathrm{N}$ diagonals. What is the smallest $N$ such that among the selected diagonals, there are guaranteed to be two of the same length?
|
Answer: 1008.
Solution: Let's choose an arbitrary vertex and consider all the diagonals emanating from it. There are 2014 of them, and by length, they are divided into 1007 pairs. Clearly, by rotating the polygon, any of its diagonals can be aligned with one of these. Therefore, there are only 1007 different sizes of diagonals. Thus, by choosing 1008, Petya is guaranteed to get at least two of the same.
|
1008
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In how many ways can the number 10000 be factored into three natural factors, none of which is divisible by 10? Factorizations that differ only in the order of the factors are considered the same.
|
Answer: 6.
Solution: Each of the factors should only include powers of 2 and 5 (they cannot be included simultaneously, as it would be a multiple of 10). There can be two factors that are powers of two, in which case the third factor is 625. Or conversely, two factors are powers of five, and the third factor is 16. In each case, there are 3 variants.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. How many zeros does the number
$$
(1 \underbrace{000 \ldots 001}_{2017 \text { zeros }})^{2017}-1 ?
$$
end with?
|
Answer 2018
Solution: Factorize:
$(1 \underbrace{000 \ldots 001}_{2017 \text { zeros }})^{2017}-1=(1 \underbrace{000 \ldots 00}_{2017 \text { zeros }} 1-1) \times(1 \underbrace{000 \ldots 00}_{2017 \text { zeros }} 1^{2016}+1 \underbrace{000 \ldots 00}_{2017 \text { zeros }} 1^{2015}+$
$\ldots .+1 \underbrace{000 \ldots 00}_{2017 \text { zeros }} 1+1$ ). The first bracket ends with 2018 zeros, while the second is not divisible by 10.
|
2018
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Vovochka approached a slot machine, on the screen of which the number 0 was lit. The game rules stated: Β«The screen shows the number of points. If you throw a 1 ruble coin, the number of points will increase by 1. If you throw a 2 ruble coin, the number of points will double. If you score 50 points, the machine will give a prize. If you get a number greater than 50, all the points will be lost.Β» What is the minimum amount of rubles Vovochka can spend to get the prize?
|
# Answer: 11 rubles.
Solution: Let's try to solve it from the end - how to get from 50 to 1 with the least amount of rubles, if you can only divide by 2 and subtract 1. We get: 50 >25->24->12->6->3->2->1. That is, it will take 4 two-ruble and 3 one-ruble coins. Obviously, if you use 3 two-ruble coins and fewer than 5 one-ruble coins (which corresponds to multiplying by 8), you cannot get a number greater than 40. Fewer one-ruble coins will also not suffice - this can be shown by enumeration.
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Petya is coming up with a password for his smartphone. The password consists of 4 decimal digits. Petya wants the password not to contain the digit 7, and at the same time, the password should have at least two (or more) identical digits. In how many ways can Petya do this?
|
Answer 3537.
Solution: The total number of passwords not containing the digit 7 is $9^{4}=6561$. Of these, 9 98x7x6=3024 consist of different digits. Therefore, 6561-3024=3537 passwords contain identical digits.
|
3537
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In the computer center, there are 200 computers, some of which (in pairs) are connected by cables, a total of 345 cables are used. We will call a "cluster" a set of computers such that a signal from any computer in this set can reach all other computers via cables (possibly through intermediate computers). Initially, all computers formed one cluster. But one night, a malicious hacker cut several cables, resulting in 8 clusters. Find the maximum possible number of cables that were cut.
|
Answer: 153.
Solution: Let's try to imagine the problem this way: an evil hacker has cut all the wires. What is the minimum number of wires the admin needs to restore to end up with 8 clusters? Obviously, by adding a wire, the admin can reduce the number of clusters by one. This means that from 200 clusters, 8 can be obtained by restoring 192 wires. Therefore, the hacker could have cut a maximum of 153 wires.
|
153
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In trapezoid $A B C D$ with bases $A D / / B C$
diagonals intersect at point $E$. It is known that
the areas $S(\triangle A D E)=12$ and $S(\triangle B C E)=3$. Find
the area of the trapezoid.
|
Answer: 27
Solution: Triangles ADE and CBE are similar, their areas are in the ratio of the square of the similarity coefficient. Therefore, this coefficient is 2. This means that point E divides the diagonals in the ratio 1:2. Therefore, the areas of triangles ABE and CDE are twice the area of BCE and are equal to 6.
We get $S(ABCD)=12+3+6+6=27$.
|
27
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find the smallest natural number ending in the digit 2 that doubles when this digit is moved to the beginning.
|
Answer: 105263157894736842
Solution: Let's write the number in the form ***...** 2 and gradually restore the "asterisks" by multiplying by 2:
$* * * . . . * * 2 \times 2=* * * . . * * 4$
$* * * . . . * 42 \times 2=* * * . . . * 84$
$* * * \ldots * 842 \times 2=* * * \ldots * 684$
$* * * . . * 6842 \times 2=* * * \ldots * 3684$
***...*36842 $2=* * * \ldots * 73684$
***... $* 736842 \times 2=* * * . . . * 473684$
***...*4736842 x $2={ }^{* * *} \ldots . . * 9473684$
...
$105263157894736842 \times 2=2105263157$
89473684
|
105263157894736842
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find the smallest $n \geq 2017$, such that $1^{\mathrm{n}}+2^{\mathrm{n}}+3^{\mathrm{n}}+4^{\mathrm{n}}$ is not divisible by 10.
|
Answer: 2020.
Solution: Powers of 1 always end in 1. The last digit of powers of 2 changes with a period of 4:
$2,4,8,6$. Powers of 3 also follow the pattern $3,9,7,1$. Powers of 4 change with a period of 2: 4,6,4,6. Therefore, the last digit will repeat with a period of 4. Checking for $n=1,2,3$, we find that the last digit is 0, and for $n=4$ it is 4. Therefore, the smallest N>2016 will be 2020.
|
2020
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the Empire of Westeros, there were 1000 cities and 2017 roads (each road connected some two cities). From any city, you could travel to any other. One day, an evil wizard enchanted $N$ roads, making it impossible to travel on them. As a result, 7 kingdoms formed, such that within each kingdom, you could travel from any city to any other by roads, but you could not travel from one kingdom to another by roads. What is the largest $N$ for which this is possible?
|
Answer: 1024.
Solution: Suppose the evil wizard enchanted all 2017 roads. This would result in 1000 kingdoms (each consisting of one city). Now, imagine that the good wizard disenchants the roads so that there are 7 kingdoms. He must disenchant at least 993 roads, as each road can reduce the number of kingdoms by no more than 1. Therefore, the evil wizard could not have enchanted more than 2017-993=1024 roads.
For graders: Partial credit can be given for a correct answer obtained for some specific case - without a general proof.
|
1024
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Jack Sparrow needed to distribute 150 piastres into 10 purses. After placing a certain number of piastres in the first purse, he put more in each subsequent purse than in the previous one. As a result, it turned out that the number of piastres in the first purse was not less than half the number of piastres in the last purse. How many piastres are in the 6th purse?
|
Answer: 16
Solution: Let there be $\mathrm{x}$ piastres in the first purse. Then in the second there are no less than $\mathrm{x}+1$, in the third - no less than $\mathrm{x}+2$... in the 10th - no less than $\mathrm{x}+9$. Thus, on the one hand, $x+$ $x+1+\cdots+x+9=10 x+45 \leq 150$, from which $x \leq 10$. On the other hand, $x \geq(x+$ 9 )/2, from which $x \geq 9$. Therefore, in the first purse, there are 9 or 10 piastres. But 9 cannot be, because then in the last one there will be no more than 18 and the sum will not reach 150.
So, in the 1st - 10, in the 2nd - no less than 11, in the 3rd - no less than 12, in the 4th - no less than 13, in the 5th - no less than 14, and in the 6th - no less than 15. But with 15, the sum is less than 150. Therefore, 16. And it cannot be more, because then in the last purse there will be 21.
For graders: A partial score can be given for a correct example of distribution (trial and error) without proof of uniqueness.
Find the smallest natural $N$ such that the decimal representation of the number $N \times 999$ consists entirely of sevens (the " $x$ " symbol denotes multiplication of numbers).
Answer 778556334111889667445223
Solution: $N \times 999=77 \ldots 7$, then $N$ is divisible by 7, denote $n=N / 7$. We get $999 n=$ $1000 n-n=11 \ldots 1$, so $1000 \mathrm{n}-111 \ldots 1=$ n. Write it as a subtraction in a column and repeat the found digits of $N$ with a shift of 3 to the left
********000
$* * * * * * 889$
$* * * * 889000$
1111111111
$* * * 777889$
777889000
111111111
6667778889
Notice that the digits repeat every 3, so we get $n=111222333444555666777889$.
Therefore, $N=7 n=778556334111889667445223$.
For graders: An alternative approach is to take the number 111... 1 and start dividing it by 999 in a column until it divides evenly. I think the score should not be reduced if at the very end the number $\mathrm{n}$ was incorrectly multiplied by 7.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The Mad Hatter's clock is fast by 15 minutes per hour, while the March Hare's clock is slow by 10 minutes per hour. One day, they set their clocks by the Dormouse's clock (which is stopped and always shows 12:00) and agreed to meet at 5 o'clock in the evening for their traditional five o'clock tea. How long will the Mad Hatter wait for the March Hare if each arrives exactly at 17:00 by their own clocks?
|
Answer: 2 hours.
Solution: The Mad Hatter's clock runs at a speed of $5 / 4$ of normal, so it will take 4 hours of normal time to pass 5 hours. Similarly, the March Hare's clock runs at a speed of $5 / 6$ of normal, so it will take 6 hours of normal time to pass 5 hours.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. At the international StarCraft championship, 100 participants gathered. The game is played in a knockout format, meaning in each match, two players compete, the loser is eliminated from the tournament, and the winner remains. Find the maximum possible number of participants who won exactly two games.
|
Answer: 49
Solution: each participant (except the winner) lost one game to someone. There are 99 such participants, which means no more than 49 participants could have won 2 games (someone must lose 2 games to them).
We will show that there could have been 49. Let's say β3 won against β1 and β2, β5 - against β3 and β4, ... β99 - against β97 and β98, and β100 won against β99. Then all participants with odd numbers (except the first) won exactly 2 games.
|
49
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. We will call a number "remarkable" if it has exactly 4 distinct natural divisors, and among them, there are two such that neither is a multiple of the other. How many "remarkable" two-digit numbers exist?
|
Answer 36.
Solution: Such numbers must have the form $p_{1} \cdot p_{2}$, where $p_{1}, p_{2}$ are prime numbers. Note that the smaller of these prime numbers cannot be greater than 7, because otherwise the product will be at least 121. It is sufficient
to check $p_{1}=2,3,5,7$. For 2, we get the second factor: $3,5,7,11,13,17,19$, $23,29,31,37,41,43$ or 47 - 14 options, for 3 we get: 3, 5, 7, 11, 13, 17, 19, 23, 29, $31-10$ options, for $5-: 3, \quad 5, \quad 7,11,13,17,19,-7$ options and for $7: 3,5, \quad 7,11,13,-5$ options. In total $14+10+7+5=36$ options.
|
36
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. On graph paper, a right-angled triangle with legs equal to 7 cells was drawn (see fig.). Then all the grid lines inside the triangle were outlined. What is the maximum number of triangles that can be found in this drawing?
#
|
# Answer: 28 triangles
Solution: One of the sides of the triangle must be inclined, i.e., lie on the segment BC. If we fix some diagonal segment, the remaining vertex is uniquely determined. That is, we need to choose 2 points out of 8, which can be done in $7 \times 8 \backslash 2=28$ ways.
|
28
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find all three-digit numbers $\overline{\Pi B \Gamma}$, consisting of distinct digits Π, B, and Π, for which the equality $\overline{\Pi B \Gamma}=(Π+B+\Gamma) \times(Π+B+\Gamma+1)$ holds.
|
Answer: 156.
Solution: Note that $P+B+\Gamma \geq 3$ and $\leq 24$ (since the digits are different). Moreover, the numbers $\overline{P B \Gamma}$ and $(P + B + \Gamma)$ should give the same remainder when divided by 9. This is only possible when $P+B+\Gamma$ is a multiple of 3. Note that $P+B+\Gamma=9$ does not work, since $(P + B + \Gamma) \times (P+B+\Gamma+1)=90$ is a two-digit number. By trying $12,15,18,21,24$, we get $\overline{P B \Gamma}=$ $156=12 \times 13$.
|
156
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$. We will call a point "equidistant" if there exist two vertices of the cube for which this point is the midpoint of the segment. How many "equidistant" points are there in the cube
|
Answer: 19.
Solution: Midpoints of 12 edges, centers of 6 faces, and the center of the cube.
|
19
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-x-3=0$. Find $\left(x_{1}^{5}-20\right) \cdot\left(3 x_{2}^{4}-2 x_{2}-35\right)$.
|
Answer: -1063.
Solution: If $x$ is one of the roots of the equation, then $x^{2}=x+3$. Squaring, we get $x^{4}=$ $x^{2}+6 x+9=7 x+12$. Multiplying by $x$, we get $x^{5}=7 x^{2}+12 x=19 x+21$. Substituting the roots of the equation for $x$, we get: $\left(x_{1}^{5}-20\right) \cdot\left(3 x_{2}^{4}-2 x_{2}-35\right)=\left(19 x_{1}+21-20\right) \cdot\left(21 x_{2}+36-\right.$ $\left.2 x_{2}-35\right)=\left(19 x_{1}+1\right)\left(19 x_{2}+1\right)=361 x_{1} x_{2}+19\left(x_{1}+x_{2}\right)+1$. By Vieta's theorem $x_{1} x_{2}=$ $-3, x_{1}+x_{2}=1$. Substituting, we get $361 *(-3)+19+1=-1063$.
## Variant 1-b
|
-1063
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. We will call a number "marvelous" if it has exactly 3 different odd natural divisors (and an arbitrary number of even divisors). How many "marvelous" two-digit numbers exist?
|
Answer: 7.
Solution: Such numbers have the form $2^{k} \times p^{2}$, where $p-$ is an odd prime number. Clearly, $p$ does not exceed 7, since the result must be a two-digit number. If $p=3$, then $k=0,1,2,3$; If $p=5$, then $\mathrm{k}=0,1$; if $\mathrm{p}=7$, then $k=0$. In total, there are 7 options.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. On graph paper, a stepped right-angled triangle with legs equal to 6 cells (see fig.) was drawn. Then all the grid lines inside the triangle were outlined. What is the maximum number of rectangles that can be found in this drawing?
|
Answer: 126
Solution: For each cell, find the number of rectangles in which this cell is the top right corner. This is not difficult to do, you can simply multiply the cell number horizontally and vertically (if starting from the lower left corner and numbering from one).
| | | | | | | |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 6 | | | | | | |
| 5 | 10 | | | | | |
| 4 | 8 | 12 | | | | |
| 3 | 6 | 9 | 12 | | | |
| 2 | 4 | 6 | 8 | 10 | | |
| 1 | 2 | 3 | 4 | 5 | 6 | |
| | | | | | | |
| | | | | | | |
Summing the numbers by columns: $1+\ldots+6+2(1+\ldots 5)+3(1+\ldots 4)+4(1+\ldots 3)+5(1+2)+6=21+$ $30+30+24+15+6=126$.
|
126
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-x-4=0$. Find $\left(x_{1}^{5}-20 x_{1}\right) \cdot\left(x_{2}^{4}+16\right)$.
|
Answer 76
Solution: If $x$ is one of the roots of the equation, then $x^{2}=x+4$. Squaring, we get $x^{4}=$ $x^{2}+8 x+16=9 x+20$. Multiplying by $x$, we get $x^{5}=9 x^{2}+20 x=29 x+36$. Substituting $x$ with the roots of the equation, we get: $\left(x_{1}^{5}-20 x_{1}\right) \cdot\left(x_{2}^{4}+16\right) .=\left(29 x_{1}+36-20 x_{1}\right) \cdot\left(9 x_{2}+16\right)=$ $\left(9 x_{1}+16\right)\left(9 x_{2}+16\right)=81 x_{1} x_{2}+144\left(x_{1}+x_{2}\right)+256$. By Vieta's theorem $x_{1} x_{2}=-4, x_{1}+$ $x_{2}=1$. Substituting, we get $81 *(-4)+144+256=76$
All problems were scored out of 15 points.
|
76
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Find $\sqrt{\frac{x}{63}-32} \times \sqrt{\frac{y}{63}-32}$, given that $\frac{1}{x}+\frac{1}{y}=\frac{1}{2016}$.
ANSWER: 32.
|
Solution: Let's make the substitution: $a=x / 63, b=y / 63$. Then we can rewrite the condition as: find $\sqrt{(a-32) \cdot(b-32)}$, given that $\frac{1}{a}+\frac{1}{b}=\frac{1}{32}$, i.e., $a b=32(a+b)$. Transforming: $\sqrt{(a-32) \cdot(b-32)}=\sqrt{a b-32(a+b)+1024}=32$.
|
32
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Find the smallest $n>2016$, such that $1^{n}+2^{n}+3^{n}+4^{n}$ is not divisible by 10.
|
Answer: 2020.
Solution: Powers of 1 always end in 1. The last digit of powers of 2 changes with a period of 4: 2,4,8,6. Powers of 3 also change with a period of 4: 3,9,7,1. Powers of 4 change with a period of 2: 4,6,4,6. That is, the last digit will repeat with a period of 4. Checking for $n=1,2,3$ we get that the last digit is 0, and for $n=4-4$. Therefore, the smallest N>2016 will be 2020.
|
2020
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Masha has 2 kg of "Swallow" candies, 3 kg of "Truffle" candies, 4 kg of "Bird's Milk" candies, and 5 kg of "Citron" candies. What is the maximum number of New Year's gifts she can make if each gift must contain 3 different types of candies, 100 grams of each?
|
Answer: 45
Solution: Even if Masha puts "Citron" in all the gifts, she will still have 2+3+4 = 9 kg of candies left, and in each gift, she must put at least 200g (and if she doesn't put Citron in all of them, then more than 200g). This means there can be no more than 45 gifts. 45 gifts can be made if she creates 5 gifts of the type Swallow+Truffle+Citron, 15 gifts of the type Swallow+Bird's Milk + Citron, and
25 gifts of the type Truffle+Bird's Milk + Citron.
Comment for graders: Here, half a point can be given for the correct selection without proving that it is optimal.
|
45
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. In the Empire of Westeros, there were 1000 cities and 2017 roads (each road connected some two cities). From any city, you could travel to any other. One day, an evil wizard enchanted $N$ roads, making it impossible to travel on them. As a result, 7 kingdoms formed, such that within each kingdom, you could travel from any city to any other by roads, but you could not travel from one kingdom to another by roads. What is the largest $N$ for which this is possible?
|
# Answer 1024.
Solution: Suppose the evil wizard enchanted all 2017 roads. This would result in 1000 kingdoms (each consisting of one city). Now, imagine that the good wizard disenchants the roads so that there are 7 kingdoms. He must disenchant at least 993 roads, as each road can reduce the number of kingdoms by no more than 1. Therefore, the evil wizard could not have enchanted more than 2017-993=1024 roads.
For graders: Partial credit can be given for a correct answer obtained for a specific case - without a general proof.
## Variant 3-b
|
1024
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Six numbers are given, the arithmetic mean of which is equal to some number A. Petrov calculated the arithmetic mean of the first four numbers - it turned out to be A+10. Vasechkin calculated the arithmetic mean of the last four numbers - it is A - 7. In which direction and by how much does the arithmetic mean of the first, second, fifth, and sixth of these numbers differ from A?
|
Answer: 3 less.
Solution - similar to option v3a.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In a regular $1000-$gon, all diagonals are drawn. What is the maximum number of diagonals that can be selected such that among any three of the selected diagonals, at least two have the same length?
|
Answer: 2000
Solution: For the condition of the problem to be met, it is necessary that the lengths of the diagonals take no more than two different values. The diagonals connecting diametrically opposite vertices are 500. Any other diagonal can be rotated to coincide with a diagonal of the corresponding length, i.e., there are 1000 of them. Therefore, 2000 can be chosen while meeting the condition.
|
2000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. If you open the cold water tap, the bathtub will fill up in 10 minutes, if you open the hot water tap, it will take 15 minutes. If you pull the plug, the bathtub will completely drain in 12 minutes. How long will it take to fill the bathtub if you open both taps and pull the plug?
#
|
# Answer: 12 min.
Solution: Let's take one bathtub as a unit of volume. The cold water tap flows at a rate of $1 / 10$ of a bathtub per minute, and the hot water tap flows at a rate of $1 / 15$ of a bathtub per minute. Water drains from the drain hole at a rate of $1 / 12$ of a bathtub per minute. Therefore, the total rate is $1 / 10 + 1 / 15 - 1 / 12 = 1 / 12$, so the bathtub will be filled in 12 minutes.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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