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6. Petya formed all possible natural numbers that can be formed from the digits $2,0,1$, 8 (each digit can be used no more than once). Find their sum. Answer: 78331
|
Solution: First, consider the units place. Each of the digits 1, 2, 8 appears once in this place for single-digit numbers, twice for two-digit numbers, four times for three-digit numbers, and four times for four-digit numbers - a total of 11 times.
In the tens place, each of them appears 3 times for two-digit numbers, 4 times for three-digit numbers, and 4 times for four-digit numbers - also 11 times.
In the hundreds place, each appears 6 times in three-digit numbers and 4 times in single-digit numbers. In the thousands place, each appears 6 times.
In total, we get $11 \times 11 + 11 \times 11 \times 10 + 11 \times 10 \times 100 + 11 \times 1000 \times 6 = 78331$.
|
78331
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (5-7,8,9) There are 2014 boxes on the table, some of which contain candies, while the others are empty.
On the first box, it is written: “All boxes are empty.”
On the second - “At least 2013 boxes are empty.”
On the third - “At least 2012 boxes are empty.”
...
On the 2014th - “At least one box is empty.”
It is known that the inscriptions on the empty boxes are false, while those on the boxes with candies are true. Determine how many boxes contain candies.
|
Answer: 1007.
Solution: Suppose that $N$ boxes are empty, then $2014-N$ boxes contain candies. Note that on the box with number $k$, it is written that there are at least $2015-k$ empty boxes. Therefore, the inscriptions on the boxes with numbers $1,2, \ldots, 2014-N$ are false. Consequently, $N=2014-N$, from which $N=1007$.
|
1007
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (5-7) Sergei collects toy trains. He has several sets, each with a different number of cars. If all the sets are combined into one train, there will be 112 cars. If you take the three smallest sets, there will be 25 cars in them, and in the three largest - 50 cars. How many sets does Sergei have? How many cars are in the largest set?
|
Answer: 9 sets. 18 or 19 cars.
Solution: Let $a_{1}, a_{2}, \ldots, a_{n}$ be the number of cars in the sets, ordered in ascending order. Note that $a_{3} \geqslant 9$, otherwise the total length of the three smallest sets would be less than $a_{1}+a_{2}+a_{3} \leqslant 7+8+9=2450$.
Thus, the remaining $112-50-25=37$ cars form sets whose lengths are in the range from 10 to 15. If we assume that the number of these sets is 1 or 2, then their total length is no more than 30. And if we assume that there are 4 or more, then their total length will be more than 40. Therefore, the number of such sets is 3, and thus the total number of sets is 9.
Since $a_{7} \leqslant 16$, then $a_{8}+a_{9} \geqslant 50-16=34$. This is only possible when $a_{9} \geqslant 18$. The case $a_{9}>19$ is impossible, because then $a_{7}+a_{8}<50-19=31$, so $a_{7} \leqslant 14$, therefore, $a_{4}+a_{5}+a_{6} \leqslant 11+12+13=36<37$. It can be shown that both cases are possible by explicitly indicating the lengths: $\{7,8,10,11,12,14,15,17,18\}$ and $\{7,8,10,11,12,14,15,16,19\}$.
|
9
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. $(5-7,8)$ Nезнayka and Ponchik have the same amounts of money, composed of coins worth $1, 3, 5$, and 7 ferthings.
Nезнayka has as many 1-ferthing coins as Ponchik has 3-ferthing coins;
3-ferthing coins - as many as Ponchik has 5-ferthing coins; 5-ferthing coins - as many as Ponchik has 7-ferthing coins; and 7-ferthing coins - as many as Ponchik has 1-ferthing coins;
Determine how many 7-ferthing coins Nезнayka has, given that each has 20 coins.
|
Answer: 5 coins.
Solution: Let $x, y, z, t$ be the number of 1, 3, 5, and 7-ferting coins that Nезнайка has. It is known that $x+y+z+t=20$ and $x+$ $3 y+5 z+7 t=3 x+5 y+7 z+t$. From the last equation, it follows that $6 t=2(x+y+z)$. Substituting $x+y+z=20-t$, we get the equation $6 t=2(20-t)$, the solution of which is $t=5$.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. $(5-7,8)$ The phone PIN code consists of 4 digits (and can start with zero, for example, 0951). Petya calls "lucky" those PIN codes where the sum of the outer digits equals the sum of the middle digits, for example $1357: 1+7=3+5$. In his phone, he uses only "lucky" PIN codes. Petya says that even if he forgets one digit (but remembers its position), he can easily restore it. And if he forgets two digits (but remembers their positions), he will have to try only a small number of PIN codes.
a) How many PIN codes will Petya have to try in the worst case?
b) How many "lucky" PIN codes exist in total?
|
Answer: a) $10 ;$ b) 670 .
Solution: a) Obviously, in the worst case, Petya tries all possible values for one digit and reconstructs the PIN code from them. Thus, no more than 10 combinations need to be tried. Using the example of the PIN code 0099, we can see that if you forget the two middle digits, you need to try exactly 10 combinations. b) The sum of a pair of digits can take values from 0 to 18. It is easy to notice that the sum equal to 0 can be obtained in only one way, equal to 1 in two ways, ..., equal to 10 in 11 ways, ..., equal to 18 in one way. In total, the number of lucky PIN codes is $N=1^{2}+2^{2}+\ldots+10^{2}+11^{2}+10^{2}+\ldots+1^{2}=670$.
|
670
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (5-7,8) There are 10 segments, the length of each of which is expressed as an integer not exceeding some $N$. a) Let $N=100$. Provide an example of a set of 10 segments such that no three of them can form a triangle. b) Find the maximum $N$ for which it can be guaranteed that there will be three segments that can form a triangle.
|
Answer: a) $1,1,2,3,5,8,13,21,34,55 ;$ b) $N=54$.
Solution: a) If each new segment is chosen to be equal to the sum of the two largest of the remaining ones, then it is impossible to form a triangle with its participation.
b) From the previous point, it is clear that for $N=55$ such a sequence can be constructed.
We will show that for $N=54$ it is already impossible to construct such a sequence. Let's order the lengths of the segments in ascending order: $a_{1} \leqslant a_{2} \leqslant \ldots \leqslant a_{10} \leqslant 54$. If a triangle cannot be formed from segments of lengths $a_{i}, a_{i+1}, a_{i+2}$, this means that the triangle inequality is violated, i.e., $a_{i+2} \geqslant a_{i}+a_{i+1}$. Thus, since it is obvious that $a_{1}, a_{2} \geqslant 1$, by sequentially applying this inequality for $i=1,2, \ldots, 8$, we get: $a_{3} \geqslant a_{1}+a_{2} \geqslant 2, a_{4} \geqslant a_{2}+a_{3} \geqslant 3, \ldots a_{10} \geqslant a_{8}+a_{9} \geqslant 55$, which leads to a contradiction.
|
54
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. $(8,9)$ What is the maximum possible area of quadrilateral $A B C D$, the sides of which are $A B=1, B C=8, C D=7$ and $D A=4$?
|
Answer: 18.
Solution: Note that $1^{2}+8^{2}=7^{2}+4^{2}=65$. With fixed lengths of $A B$ and $B C$, the area of $\triangle A B C$ will be maximized if $\angle A B C=$ $90^{\circ}$. In this case, $A C=\sqrt{65}$, and thus $\angle B C D=$ $90^{\circ}$ as well, and the area of $\triangle B C D$ is also maximized.
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10. $(8,9)$ Find the smallest possible value of $\left|2015 m^{5}-2014 n^{4}\right|$, where $m$ and $n$ are natural numbers.
|
Answer: 0.
Solution: Consider numbers of the form $m=2014^{a} \cdot 2015^{b}$ and $n=2014^{c} \cdot 2015^{d}$. Then $\left|2015 m^{5}-2014 n^{4}\right|=\left|2014^{5 a} \cdot 2015^{5 b+1}-2014^{4 c+1} \cdot 2015^{4 d}\right|$. This value equals 0 in the case $5 a=4 c+1,5 b+1=4 d$. It is not difficult to find such numbers, for example, $a=c=1, b=3, d=4$.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. (9) Integers $a, b$ and $c$ are such that $a \cdot\left(1-\frac{\sqrt{7}}{2}\right)^{2}+b \cdot\left(1-\frac{\sqrt{7}}{2}\right)+c=5$. What is the minimum value of $|a+b+c|$ under this condition?
|
Answer: 2.
Solution: If $a=0$, then $b=0$ and $c=5$, hence $|a+b+c|=5$. If $a \neq 0$, then consider the quadratic function $f(x)=a x^{2}+b x+c$. Notice that $f(x)-5$ has roots $1 \pm \frac{\sqrt{7}}{2}$. Therefore, $f(x)=$ $5+k \cdot\left(4 x^{2}-8 x-3\right), k \neq 0$, i.e., $|a+b+c|=|5-7 k|$, The minimum value of 2 is achieved when $k=1$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
15. (9) Find $q$, for which $x^{2}+x+q=0$ has two distinct real roots satisfying the relation $x_{1}^{4}+2 x_{1} x_{2}^{2}-x_{2}=19$.
|
Answer: $q=-3$.
Solution: If $x$ is a root, then $x^{2}=-x-q$, hence $x^{4}=x^{2}+$ $2 q x+q^{2}=(2 q-1) x+q^{2}-q$. Then $x_{1}^{4}+2 x_{1} x_{2}^{2}-x_{2}=(2 q-1) x_{1}+q^{2}-q+$ $2 q x_{2}-x_{2}=(2 q-1)\left(x_{1}+x_{2}\right)+q^{2}-q$. Using Vieta's theorem
$x_{1}+x_{2}=-1$, we get $q^{2}-3 q+1=19$. Solving, we obtain $q_{1}=-3$, $q_{2}=6$, but for $q=6$ the original equation has no roots.
## Lomonosov Moscow State University
## "Conquer Sparrow Hills" Olympiad Variant 3-1
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. From the condition about 15 kg, it follows that the container can contain either only items weighing 5 and 10 kg, or only items weighing 2, 3, and 10 kg. Let $x, y, z, u$ denote the number of items weighing $2, 3, 5$, and 10 kg, respectively.
In the first case, the conditions of the problem give the system $\left\{\begin{array}{l}u=z+5, \\ 5 z+10 u=100,\end{array}\right.$ which has no solutions in natural numbers.
In the second case, for natural numbers $x, y, u$, we obtain the system of equations
$$
\left\{\begin{array} { l }
{ u = x + y + 5 , } \\
{ 2 x + 3 y + 1 0 u = 1 0 0 , }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ u = x + y + 5 , } \\
{ 1 2 x = 5 0 - 1 3 y , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=y=2 \\
u=9
\end{array}\right.\right.\right.
$$
|
Answer: 2 items weighing 2 and 3 kg, 9 items weighing 10 kg.
Answer to variant 4-2: 2 weighing 3 and 5 kg, 12 weighing 7 kg.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The robotics club accepts only those who know mathematics, physics, or programming. It is known that 8 members of the club know physics, 7 know mathematics, and 11 know programming. Additionally, it is known that at least two know both physics and mathematics, at least three know both mathematics and programming, and at least four know both physics and programming. What is the maximum number of club members possible under these conditions?
|
Answer 19.
2. From the sequence of natural numbers $1,2,3, \ldots$, all perfect squares (squares of integers) have been removed. What number will be in the 2018th position among the remaining numbers?
Answer: 2063.
|
19
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. The sequence $a_{n}$ is defined as follows:
$$
a_{1}=1, a_{n+1}=a_{n}+\frac{2 a_{n}}{n}, \text { for } n \geq 1 . \text { Find } a_{200}
$$
|
Answer: 20100.
## Variant 3a (Chelyabinsk)
|
20100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. The sequence $a_{n}$ is defined as follows:
$a_{1}=1, a_{n+1}=a_{n}+\frac{2 a_{n}}{n}$, for $n \geq 1$. Find $a_{999}$.
|
Answer: 499500.
## School Olympiad "Conquer Sparrow Hills"
## Tasks for 7-8 Grades
## Variant 1b (Zheleznovodsk)
|
499500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Thirteen millionaires arrived at the economic forum and settled in the "Super Luxury+" hotel. The hotel has rooms of 3 different types: 6-star, 7-star, and 8-star. The millionaires need to be accommodated in such a way that all three types of rooms are used (i.e., at least one person must be placed in a 6-star room, at least one in a 7-star room, and at least one in an 8-star room). Additionally, a richer millionaire cannot be placed in a room with fewer stars than a less wealthy one.
In how many ways can they be accommodated (all millionaires have different wealth levels)?
|
Answer: $C_{12}^{2}=66$.
|
66
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. The sequence $a_{n}$ is defined as follows:
$a_{1}=2, a_{n+1}=a_{n}+\frac{2 a_{n}}{n}$, for $n \geq 1$. Find $a_{100}$.
|
Answer: 10100.
## Variant 2b (Saratov)
|
10100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A school coach decided to reward 12 students who ran the distance in the best time. Each of them should be awarded a "gold", "silver", or "bronze" medal. All three types of medals must be used (at least once), and the one who finished earlier cannot be awarded a less valuable medal than the one who finished later.
In how many ways can the coach distribute the medals (all runners have different times)?
|
Answer: $C_{11}^{2}=55$.
|
55
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Petrov lists the odd numbers: $1,3,5, \ldots, 2013$, while Vasechkin lists the even numbers $2,4, \ldots, 2012$. Each of them calculated the sum of all digits of all their numbers and told the result to the excellent student Masha. Masha subtracted the result of Vasechkin from the result of Petrov. What did she get?
|
Answer: 1007.
Solution: Let's break down the numbers of Petrov and Vasechkin into pairs as follows: $(2,3),(4,5), \ldots,(98,99),(100,101), \ldots$ (2012,2013), with 1 left unpaired for Petrov. Notice that in each pair, the sum of the digits of the second number is 1 greater than that of the first (since they differ only in the last digit). There will be a total of $\frac{2012}{2}=1006$ such pairs. Therefore, the difference in the sums of the digits will be 1006, and with the 1 left unpaired for Petrov, it becomes -1007.
|
1007
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Dima went to school in the morning, but after walking exactly half the distance, he realized he had forgotten his mobile phone at home. Dima estimated (he had an A in mental arithmetic) that if he continued walking at the same speed, he would arrive at school 3 minutes before the first bell, but if he ran home for the phone and then ran to school, he would arrive 3 minutes after the bell. Dima decided to run home, but he got out of breath while running (he had a C in physical education), and walked from home to school at his usual speed. As a result, he was 15 minutes late for the first class! How many times faster is the speed at which he runs compared to the speed at which he walks?
|
Answer: 2.
Solution: Let $x$ be the time it takes for Dima to walk from home to school, $y$ be the time it takes for Dima to run from home to school, and $T$ be the remaining time until the bell rings (at the moment Dima noticed the loss). Then the conditions of the problem can be written as $\left\{\begin{array}{l}\frac{x}{2}=T-3 \\ \frac{y}{2}+y=T+3 \\ \frac{y}{2}+x=T+15\end{array}\right.$ Solving this, we get that $x=24, y=12, T=15$, from which it follows that he runs twice as fast as he walks.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Famous skater Tony Hawk is riding a skateboard (segment $A B$) in a ramp, which is a semicircle with diameter $P Q$. Point $M$ is the midpoint of the skateboard, $C$ is the foot of the perpendicular dropped from point $A$ to the diameter $P Q$. What values can the angle $\angle A C M$ take if it is known that the angular measure of the arc $A B$ is $24^{\circ}$?
|
Answer: $12^{\circ}$.
Solution: Extend the line $A C$ to intersect the circle at point $D$ (see figure). The chord $A D$ is perpendicular to the diameter $P Q$, therefore, it is bisected by it. Thus, $C M$ is the midline of triangle $A B D$, so $C M \| B D$ and, therefore, $\angle A C M=\angle A D B$. The angle $\angle A D B$ is inscribed, subtends the arc $A B$, and thus is equal to half of it.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Place the numbers $1,2,3,4,5,6,7,8$ and 9 in the nine cells of the figure shown in the diagram, so that the sum of the numbers in each column, starting from the second, is 1 more than in the previous one. It is sufficient to find at least one such arrangement. In your answer, indicate the number in the first column.
|
Answer: 7.
Solution: For now, we will not pay attention to the order of numbers in one column.
The sum of the given numbers is 45. Let $x$ be the number in the bottom-left cell. Then $5x + 10 = 45$, from which $x = 7$. Therefore, the sum of the numbers in the second column is $8 = 5 + 3 = 6 + 2$. If the second column contains 3 and 5, then the third column must contain 1 and 8, the fourth column must contain 6 and 4, and the last column must contain 2 and 9. If the second column contains 6 and 2, then the third column can contain 1 and 8 or 4 and 5. It can be shown that if the third column contains 1 and 8, it is impossible to select numbers for the fourth column. Therefore, the third column must contain 4 and 5, then the fourth column must contain 1 and 9, and the last column must contain 3 and 8. This results in 2 arrangements without considering the order of the numbers.
Notice that in each column (except the first), the numbers can be swapped, which gives 16 options for each arrangement. In the end, we get 32 options considering the order of the numbers in the columns.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The board for playing "Battleship" is $10 \times 10$. What is the maximum number of ships of size $1 \times 4$ that can be placed on it?
|
Answer: 24.
Solution: It is easy to show by trial that 24 ships can be placed. Let's color the board in 4 colors as shown in the figure. Note that each ship of size $1 \times 4$ contains one cell of each color, and there will be 24 yellow cells.
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Find the number of pairs of natural numbers $(x, y), 1 \leqslant x, y \leqslant 1000$, such that $x^{2}+y^{2}$ is divisible by 5.
|
Answer: 360000.
Solution: We have 200 numbers for each of the remainders $0,1,2,3,5$ when divided by 5. There are two cases: a) the numbers $x, y$ are both divisible by 5; b) or one number gives a remainder of 1 or 4, and the other gives a remainder of 2 or 3 when divided by 5. In the first case, we get $200 \times 200=40000$ options, in the second case $2 \times 400 \times 400=320000$, in total $40000+320000=$ 360000.
|
360000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. 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
|
1. Solution. From the formulas for the sum of a geometric progression, we know
$$
\begin{aligned}
& \frac{b_{1}}{1-q}=60 \\
& \frac{b_{1}^{2}}{1-q^{2}}=1200
\end{aligned}
$$
By dividing the second equation by the first, we get $\frac{b_{1}}{1+q}=20$, which is the answer.
Remark. We could also find $b_{1}=30, q=1 / 2$.
|
Answer: 20 (option $1-2:-30)$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. However, if the task is to restate it as a translation, here it is:
Answer: 20 (option $1-2:-30)$.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Answer: 1 (in all options).
|
Solution. (Option 6-1). Let's represent $k$ as $k=10 a+b$, where $a, b$ are integers and $1 \leqslant b \leqslant 9$. Then $k^{2}+6 k=10\left(10 a^{2}+2 a b+6 a\right)+b^{2}+6 b$. The expression $b^{2}+6 b$ ends in 6 only if $b=2$. But in this case $k^{2}+6 k=$ $100\left(a^{2}+a\right)+16$, which means the second-to-last digit is 1.
In option $6-2$, we need to consider that $k^{2}+2 k-8=(k-2)^{2}+6(k-2)$.
|
1
|
Other
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
2. Thirteen millionaires arrived at the economic forum and settled in the "Super Luxury+" hotel. The hotel has rooms of 3 different types: 6-star, 7-star, and 8-star. The millionaires need to be accommodated in such a way that all three types of rooms are used (i.e., at least one person must be placed in a 6-star room, at least one in a 7-star room, and at least one in an 8-star room). Additionally, a richer millionaire cannot be placed in a room with fewer stars than a less wealthy one.
In how many ways can they be accommodated (all millionaires have different wealth levels)?
|
Answer: $C_{12}^{2}=66$.
|
66
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. A school coach decided to reward 12 students who ran the distance in the best time. Each of them must be awarded a "gold", "silver", or "bronze" medal. All three types of medals must be used, and the one who finished earlier cannot be awarded a less valuable medal than the one who finished later.
In how many ways can the coach distribute the medals (all runners have different times)?
|
Answer: $C_{11}^{2}=55$.
|
55
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Petrov and Vasechkin were repairing a fence. Each had to nail a certain number of boards (the same amount). Petrov nailed two nails into some boards and three nails into others. Vasechkin nailed three nails into some boards and five nails into the rest. Find out how many boards each of them nailed, given that Petrov nailed 87 nails, and Vasechkin nailed 94.
|
Answer: 30.
Solution: If Petrov had nailed 2 nails into each board, he would have nailed 43 boards and had one extra nail. If he had nailed 3 nails into each board, he would have nailed 29 boards. Therefore, the desired number lies between 29 and 43 (inclusive). Similarly, if Vasechkin had nailed 3 nails into each board, he would have nailed 31 boards and had 1 extra nail, and if he had nailed 5 nails into each board, he would have nailed 18 boards and had 4 extra nails. This means there were 29, 30, or 31 boards. Note that Vasechkin nailed an odd number of nails into each board, so the number of boards must be even - 30.
|
30
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. 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.
Thus, the sum cannot be less than 18. We will show that 18 can be achieved - place the number 1 on the top and bottom faces of the cube, 3 on the right and left faces, and 5 on the front and back faces.
|
18
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In the Slytherin faculty at Hogwarts, there are 30 students. Some are friends (friendship is mutual, i.e., if A is friends with B, then B is also friends with A), but there are no 3 people who are pairwise friends with each other. On New Year's, each student 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. 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
|
5. 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
|
6. Pete came up with all the numbers that can be formed using the digits 2, 0, 1, 8 (each digit can be used no more than once). Find their sum.
|
Answer: 78331
Solution: First, consider the units place. Each of the digits 1, 2, 8 appears once in this place for single-digit numbers, twice for two-digit numbers, four times for three-digit numbers, and four times for four-digit numbers - a total of 11 times.
In the tens place, each of them appears 3 times for two-digit numbers, 4 times for three-digit numbers, and 4 times for four-digit numbers - also 11 times.
In the hundreds place, each appears 6 times in three-digit numbers and 4 times in single-digit numbers.
In the thousands place, each appears 6 times.
In total, we get $11 \times 11 + 11 \times 11 \times 10 + 11 \times 10 \times 100 + 11 \times 1000 \times 6 = 78331$.
|
78331
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. 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 $a_2=x$. 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,...
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, $k$ must be a multiple of $x$, the smallest such $k$ is 2020.
|
2020
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Solve the inequality
$$
(2+\sqrt{3})^{x}+2<3(\sqrt{2-\sqrt{3}})^{2 x}
$$
In your answer, specify the sum of all integer values of $x$ that satisfy the given inequality and belong to the interval $(-20 ; 53)$.
|
Answer: The solution to the inequality is $(-\infty ; 0)$. The desired sum $-19-18-\cdots-2-1=-190$. We write -190 as the answer.
|
-190
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The area of triangle $\triangle A B C$ is 10 cm $^{2}$. What is the smallest value in centimeters that the circumference of the circle circumscribed around triangle $\triangle A B C$ can take, given that the midpoints of the heights of this triangle lie on the same line? If the answer is not an integer, round it to the nearest integer.
|
Answer: The smallest value is $2 \pi \sqrt{10}$ cm. In the answer, we write 20, since $2 \pi \sqrt{10} \approx 19.8691 \ldots$
|
20
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Find the set of pairs of real numbers $(x, y)$ satisfying the conditions:
$$
\left\{\begin{array}{l}
3^{-x} y^{4}-2 y^{2}+3^{x} \leq 0 \\
27^{x}+y^{4}-3^{x}-1=0
\end{array}\right.
$$
Calculate the values of the expression $x_{k}^{3}+y_{k}^{3}$ for each solution $\left(x_{k}, y_{k}\right)$ of the system and find the minimum among them. In the answer, specify the found minimum value, rounding it to two decimal places if necessary. If the original system has no solutions, then put the digit 0 in the answer field.
|
Answer: The solution to the system is $(x, y)=(0, \pm 1)$. Answer -1.
## 1 Creative problems (7 items)
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Let $n$ be the time in minutes it takes for the slower boy to complete a lap. Then $n>5, n \in \mathbb{N}$. The speed at which the faster boy catches up to the slower one is
$$
\frac{1}{5}-\frac{1}{n}=\frac{n-5}{5 n}\left(\frac{\text { lap }}{\text { min }}\right)
$$
therefore, the time between meetings is
$$
\frac{5 n}{n-5}=\frac{5(n-5)+25}{n-5}=5+\frac{25}{n-5}(\text { min })
$$
and since this is an integer, $(n-5)$ must divide 25. The natural divisors of $25$ are $1, 5,$ and $25$. Thus, $n=6$ or $n=10$ or $n=30$, which correspond to meeting times of 30 minutes, 10 minutes, and 6 minutes, respectively. Since the time between meetings is at least 12 minutes, the only value that satisfies the given condition is $n=6$.
|
Answer: 6 min.
Answer to option 172: 12 min.
Answer to option $173: 56$ min.
Answer to option $174: 4$ min.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The total number of students in the school must be divisible by 7 and 4, which means it must be divisible by 28. Since the number is no more than 40, there were 28 students in total. The number of prize winners is $(1 / 7 + 1 / 4 + 1 / 4) \cdot 28 = 18$ people. Therefore, 10 students did not receive any medals.
|
Answer: $x=10$.
Answer to the variant: $4-2: x=18$.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. First method: Since $x_{1}^{3}-2015 x_{1}+2016=0$ and $x_{2}^{3}-2015 x_{2}+2016=0$, then $x_{1}^{3}-x_{2}^{3}=2015 \cdot\left(x_{1}-x_{2}\right)$. Therefore, $x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=2015$.
Second method: By Vieta's theorem (but then it needs to be justified that there are three distinct roots): $x_{1}+x_{2}+x_{3}=0, x_{1} x_{2} x_{3}=-2016$. Therefore,
$$
\begin{aligned}
x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=\left(-x_{3}\right)^{2}+\frac{2016}{x_{3}} & \\
=\frac{x_{3}^{3}+2016}{x_{3}} & =\frac{2015 x_{3}}{x_{3}}=2015
\end{aligned}
$$
|
Answer: 2015.
Answer to option: 4-2: 2016.
|
2015
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The right side of the inequality is zero when $|x| \leqslant 10$ (for other $x$ it is undefined). Denoting $\alpha=\frac{\pi x}{4}$, we get:
$$
\begin{aligned}
& \frac{\sin 2 \alpha-\cos 2 \alpha+1}{\sin 2 \alpha+\cos 2 \alpha-1} \geqslant 0 \Longleftrightarrow \frac{2 \sin \alpha \cos \alpha+2 \sin ^{2} \alpha}{2 \sin \alpha \cos \alpha-2 \sin ^{2} \alpha} \geqslant 0 \Longleftrightarrow \\
& \Longleftrightarrow\left\{\begin{array} { l }
{ \frac { \cos \alpha + \sin \alpha } { \cos \alpha - \sin \alpha } \geqslant 0 , } \\
{ \sin \alpha \neq 0 }
\end{array} \Longleftrightarrow \left\{\begin{array}{l}
\frac{1+\operatorname{tg} \alpha}{1-\operatorname{tg} \alpha} \geqslant 0 \\
\operatorname{tg} \alpha \neq 0
\end{array}\right.\right.
\end{aligned}
$$
Therefore, $x \in[-1+4 k ; 4 k) \cup(4 k ; 4 k+1)$. On the interval $[-10,10]$, the integers are $-9,-5,-1,3,7$, of which the numbers $-9,-5,-1,3$ are within the specified interval, and their sum is -12.
|
Answer: -12.
Answer to the option: $4-2: 12$.
|
-12
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. If there are $x$ black pieces, then there are $32-x$ white pieces. Since each black piece (pentagonal) borders only white pieces, the number of black and white piece borders will be $5 x$. On the other hand, such borders are $3 \cdot(32-x)$. From the equation $5 x=3 \cdot(32-x)$ we get $x=12$.
|
Answer: 12.
Answer to option: 5-2: 20.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. After substituting $x=\cos \alpha, y=\sin \alpha$, the second equation is satisfied, and the first equation can be written as $\cos ^{15} 3 \alpha+\sin ^{16} 3 \alpha=1$. From the chain
$$
1=\cos ^{15} 3 \alpha+\sin ^{16} 3 \alpha \leqslant \cos ^{2} 3 \alpha+\sin ^{2} 3 \alpha=1
$$
it follows that $\cos ^{15} 3 \alpha=\cos ^{2} 3 \alpha$ and $\sin ^{16} 3 \alpha=\sin ^{2} 3 \alpha$. Therefore, either $3 \alpha=\pi / 2+\pi k$ or $3 \alpha=2 \pi k$, which means $\alpha=\frac{\pi}{6}+\frac{\pi k}{3}$ or $\alpha=\frac{2 \pi n}{3}$. This results in 9 pairs of solutions $(x, y)$:
$$
\left(\frac{\sqrt{3}}{2} ; \pm \frac{1}{2}\right) ;\left(-\frac{\sqrt{3}}{2} ; \pm \frac{1}{2}\right) ;\left(-\frac{1}{2} ; \pm \frac{\sqrt{3}}{2}\right) ;(0 ; \pm 1) ;(1 ; 0)
$$
|
Answer: 9.
Answer to option: $6-2$ : 9.
The equation $\cos 3 x=0$ is equivalent to
$$
3 x=\pi / 2+2 \pi k, k \in \mathbb{Z}, \quad 3 x=-\pi / 2+2 \pi n, n \in \mathbb{Z}
$$
Answers and solutions to option $\mathbf{7}-1$
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Let $\alpha=\pi x$.
$$
\begin{aligned}
& \cos (8 \alpha)+2 \cos (4 \alpha)-\cos (2 \alpha)+2 \sin (\alpha)+3=0 \Longleftrightarrow \\
& 2 \cos ^{2} 4 \alpha-1+2 \cos 4 \alpha-1+2 \sin ^{2} \alpha+2 \sin \alpha+3=0 \Longleftrightarrow \\
& 2\left(\cos 4 \alpha+\frac{1}{2}\right)^{2}+2\left(\sin \alpha+\frac{1}{2}\right)^{2}=0
\end{aligned}
$$
From which
$$
\left\{\begin{array} { l }
{ \operatorname { cos } 4 \alpha = - 1 / 2 } \\
{ \operatorname { sin } \alpha = - 1 / 2 }
\end{array} \Longleftrightarrow \left[\begin{array}{l}
\alpha=-\frac{\pi}{6}+2 \pi m, m \in \mathbb{Z} \\
\alpha=-\frac{5 \pi}{6}+2 \pi n, n \in \mathbb{Z}
\end{array}\right.\right.
$$
Thus $x=-1 / 6+2 m, x=-5 / 6+2 n, m, n \in \mathbb{Z}$. The sum of the first two positive roots: $1 \frac{1}{6}+1 \frac{5}{6}=3$. The next two positive roots are 4 greater, i.e., equal to 7, and so on. The desired sum is
$$
3+7+11+\cdots+(3+4 \cdot 49)=\frac{6+4 \cdot 49}{2} \cdot 50=5050
$$
|
Answer: 5050
Answer to option: 7-2: 4950.
Solutions for option 17-1 and answers to all options
|
5050
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Let $n$ be the time in minutes it takes for the slower boy to complete a lap. Then $n>5, n \in \mathbb{N}$. The speed at which the faster boy catches up to the slower one is
$$
\frac{1}{5}-\frac{1}{n}=\frac{n-5}{5 n}\left(\frac{\text { lap }}{\text { min }}\right)
$$
therefore, the time between meetings is
$$
\left.\frac{5 n}{n-5}=\frac{5(n-5)+25}{n-5}=5+\frac{25}{n-5} \text { (min }\right)
$$
and since this is an integer, $(n-5)$ must divide 25. The natural divisors of $25$ are $1, 5,$ and $25$. Thus, $n=6$ or $n=10$ or $n=30$, which correspond to meeting times of 30 minutes, 10 minutes, and 6 minutes, respectively. Since the time between meetings is at least 12 minutes, the only value that satisfies the given condition is $n=6$.
|
Answer: 6 min.
Answer to option 17 - $2: 12$ min.
Answer to option 17 - $\mathbf{\text { : }}$ : 56 min.
Answer to option $17-4: 4$ min.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Drop perpendiculars $D D_{1}, D D_{2}, D D_{3}$ from point $D$ to the planes $S B C$, $S A C$, and $S A B$ respectively. Let $D D_{1}=x, D D_{2}=y, D D_{3}=z$. According to the condition, we form the system of equations
$$
\left\{\begin{array}{l}
y^{2}+z^{2}=5 \\
x^{2}+z^{2}=13 \\
x^{2}+y^{2}=10
\end{array}\right.
$$
Fig. 6:
From here, we find $x=3, y=1, z=2$. Let the lengths of the edges $S A, S B$, and $S C$ be $a, b$, and $c$ respectively. Since points $A, B, C$, and $D$ lie in the same plane, the relation $\frac{3}{a}+\frac{1}{b}+\frac{2}{c}=1$ holds.
Using the inequality between the arithmetic mean and the geometric mean for three variables, we get:
$$
\begin{aligned}
& \frac{\frac{3}{a}+\frac{1}{b}+\frac{2}{c}}{3} \geqslant \sqrt[3]{\frac{3}{a} \cdot \frac{1}{b} \cdot \frac{2}{c}}=\sqrt[3]{\frac{6}{a b c}} \Longleftrightarrow \\
& \Longleftrightarrow 1=\left(\frac{3}{a}+\frac{1}{b}+\frac{2}{c}\right)^{3} \geqslant \frac{6 \cdot 27}{a b c} \Longleftrightarrow a b c \geqslant 6 \cdot 27
\end{aligned}
$$
with equality holding when $\frac{3}{a}=\frac{1}{b}=\frac{2}{c}=\frac{1}{3}$. The volume of the pyramid $V=\frac{a b c}{6}$, so $V \geqslant 27$. Equality holds when $a=9, b=3, c=6$.
|
Answer: 27.
Answer to option 17-2: 108.
Answer to option $17-3: 27$.
Answer to option $17-4: 108$.
[^0]: ${ }^{1}$ This equality can be proven by expressing $B C^{2}$ from two triangles $B A C$ and $B D C$ using the planimetric cosine theorem.
|
27
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. A natural number $N$ ends in ...70, and it has exactly 72 natural divisors (including 1 and itself). How many natural divisors does the number 80N have? ANSWER: 324.
|
Solution: Each divisor of the number $\mathrm{N}$ can be represented in the form $2^{a} \cdot 5^{b} \cdot q$, where q is not divisible by 2 and 5, and the numbers $a, b$ are 0 or 1. Therefore, there are only 4 different combinations for the pair a and b, which means there are $72 / 4=18$ possible values for q. These correspond to divisors of the number $80 \mathrm{~N}$ in the form $2^{c} \cdot 5^{d} \cdot q$, where c=0,...,5 and $d=0,1,2$. There are 18 for each q, i.e., a total of $18 * 18=324$.
|
324
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Solve the inequality
$$
\frac{4+\sqrt{-3-x}}{3-|x+4|} \leqslant 1
$$
In your answer, specify the sum of all integer values of $x$ that satisfy the given inequality and do not exceed 12 in absolute value.
|
Solution. Transferring 1 to the left side and bringing to a common denominator:
$$
\frac{1+\sqrt{-3-x}+|x+4|}{3-|x+4|} \leqslant 0
$$
For $x>-3$, the expression under the square root is negative, meaning there are no solutions. For $x \leqslant-3$, the numerator is positive, and thus,
$$
3-|x+4|3 \Longleftrightarrow x>-1
$$
or $x<-7$.
Therefore, the solution to the inequality is: $x<-7$. The required integer roots according to the condition are: $-12, -11, \ldots$, -8. Their sum is -50.
Answer: -50.
|
-50
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Find the side $B C$ of the quadrilateral $A B C D$, if $\angle B A C=\alpha, \angle A C D=\beta, \angle B C A+\angle C A D=\frac{\pi}{2}$ and $A D=a$. In the answer, write the result rounded to two decimal places.
$$
\alpha=\arcsin \frac{5}{13}, \beta=\arcsin \frac{12}{13}, a=24
$$
|
Solution. Since the sum of angles $\angle B A C+\angle A C D=\arcsin \frac{5}{13}+\arcsin \frac{12}{13}=\frac{\pi}{2}$, then $\angle B A D+\angle B C D=$ $\pi$. Therefore, a circle can be circumscribed around quadrilateral $A B C D$. Next, by the Law of Sines $\frac{A D}{\sin \beta}=\frac{B C}{\sin \alpha} \Rightarrow B C=\frac{24 \cdot 5}{12}=10$.
Answer: 10.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The number 2015 can be represented as the sum of consecutive integers in various ways, for example, $2015=1007+1008$ or $2015=$ $401+402+403+404+405$. In how many ways can this be done?
|
Answer: 16.
Solution: The sum of $k$ numbers starting from $n$ is $S(k, n)=\frac{1}{2}(2 n+k-1) \cdot k$. That is, we need to solve the equation $(2 n+k-1) \cdot k=4030$ in integers. Obviously, $k$ can be any divisor of $4030=2 \times 5 \times 13 \times 31$. Note that each of the prime factors $(2,5,13$ and 31$)$ can be in the power of 0 or 1 - in total 16 options.
|
16
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. Given triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$, the areas of which are 1 and 2025, respectively. It is known that rays $A B$ and $A^{\prime} B^{\prime}$ are parallel and point in opposite directions (see figure). The same is true for pairs $B C$ and $B^{\prime} C^{\prime}, C A$ and $C^{\prime} A^{\prime} . A^{\prime \prime}, B^{\prime \prime}$ and $C^{\prime \prime}$ are the midpoints of segments $A A^{\prime}, B B^{\prime}$ and $C C^{\prime}$. Find the area of triangle $A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$.
|
Answer: 484.
Solution: Obviously, triangles $A B C A^{\prime} B^{\prime} C^{\prime}$ are similar with a coefficient of 45.
Three trapezoids are formed, in which $A^{\prime \prime} B^{\prime \prime}, B^{\prime \prime} C^{\prime \prime}$ and $C^{\prime \prime} A^{\prime \prime}$ are segments connecting the midpoints of the diagonals (i.e., equal to half the difference of the bases). Therefore, triangle $A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$ is also similar to $\triangle A B C$ and its coefficient is $(45-1) / 2=22$. Consequently, its area is $22^{2}=484$ times larger than the area of the small triangle.
|
484
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A school coach decided to reward 12 students who ran the distance in the best time. Each of them must be awarded a "gold", "silver", or "bronze" medal. All three types of medals must be used, and the one who finished earlier cannot be awarded a less valuable medal than the one who finished later.
How many ways can the coach distribute the medals (all runners have different times)?
|
Answer: $C_{11}^{2}=55$.
Solution: see 9th grade.
|
55
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. On a line, there are 16 points $A_{1}, \ldots A_{16}$, spaced 1 cm apart. Misha constructs circles according to the following rules:
a) The circles do not intersect or touch.
b) Inside each circle, there is at least one of the specified points $\mathrm{A}_{1}, \ldots \mathrm{A}_{16}$.
c) None of these points lie on the circle.
d) Different circles contain different sets of points. For example, if one circle contains points $A_{1}$ and $A_{2}$ inside and the rest outside, then it is not possible to construct another circle that contains only $A_{1}$ and $A_{2}$ inside.
What is the maximum number of circles Misha can construct according to these rules?
|
Answer: 31.
Solution: see 9th grade.
## Tasks for 7-8 grades
## Option 3b (Moscow)
|
31
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Each worker on the construction site has at least one construction specialty. 10 people have the specialty of a mason, 9 - the specialty of a painter, 8 - the specialty of a plasterer. It is known that at least four people have the specialties of a mason and a plasterer simultaneously, at least five people - the specialties of a mason and a painter, and at least three people - the specialties of a painter and a plasterer. What is the maximum number of workers that can be on the construction site under these conditions?
|
# Answer 18.
Solution: According to the principle of inclusion-exclusion, the total number of workers is K+M+S - KM-KS-MS + KMS = 10+9+8-4-5-3 + KMS = 15 + KMS. Note that the number of workers proficient in all three specialties cannot exceed 3.
|
18
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Natural numbers, for which the sum of the digits equals 5, were arranged in ascending order. What number is in the $122-$nd position?
|
Answer: 40001.
Solution: Let's calculate the number of such numbers for different digit counts.
Let $n$ be the number of digits. Subtract 1 from the most significant digit, we get a number (which can now start with zero), the sum of whose digits is 4. Represent this as follows - there are 4 balls, between which $n-1$ partitions are placed. This can be chosen in $C_{n+3}^{4}$ ways.
Let's construct a table:
| Number of digits | Number of ways |
| :--- | :---: |
| 1 | $C_{4}^{4}=1$ |
| 2 | $C_{5}^{4}=5$ |
| 3 | $C_{6}^{4}=15$ |
| 4 | $C_{7}^{4}=35$ |
| 5 | $C_{8}^{4}=70$ |
| TOTAL | $\mathbf{1 2 6}$ |
Obviously, the 126th number - the largest 5-digit number - will be 50000.
Thus, the 125th number is 41000.
the 124th is 40100
the 123rd is 40010
the 122nd is 40001
|
40001
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. On a line, there are 16 points $A_{1}, \ldots A_{16}$, spaced $1 \mathrm{~cm}$ apart. Misha constructs circles according to the following rules:
a) The circles do not intersect or touch.
b) Inside each circle, there is at least one of the specified points $\mathrm{A}_{1}, \ldots \mathrm{A}_{16}$.
c) None of these points lie on the circle.
d) Different circles contain different sets of points. For example, if one circle contains points $\mathrm{A}_{1}$ and $\mathrm{A}_{2}$ inside and the rest outside, then it is not possible to construct another circle that contains only $\mathrm{A}_{1}$ and $\mathrm{A}_{2}$ inside.
What is the maximum number of circles Misha can construct according to these rules?
Answer: 31.
|
Solution: We can represent such a system of circles as a tree with 16 leaves.
In such a tree, there cannot be more than 31 nodes. It is easy to construct a binary tree in which there are exactly 31 nodes.
|
31
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. The sequence $a_{n}$ is defined as follows:
$a_{1}=2, a_{n+1}=a_{n}+\frac{2 a_{n}}{n}$, for $n \geq 1$. Find $a_{999}$.
|
Answer: 999000.
Solution: Consider the beginning of the sequence: $a_{1}=2, a_{2}=6, a_{3}=12, a_{4}=20, \ldots$
We can notice a pattern - the difference between consecutive terms forms an arithmetic progression: $4,6,8, \ldots$.
From this, we get the formula for the $\mathrm{n}$-th term $a_{n}=n \cdot(n+1)$, which can be proven by induction.
Thus, $a_{999}=999 \cdot 1000=999000$.
|
999000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The robotics club accepts only those who know mathematics, physics, or programming. It is known that 8 members of the club know physics, 7 know mathematics, and 11 know programming. Additionally, it is known that at least two know both physics and mathematics, at least three know both mathematics and programming, and at least four know both physics and programming. What is the maximum number of club members possible under these conditions?
|
Answer 19.
3 Natural numbers, the sum of whose digits is 5, were arranged in ascending order. What number is in the $111-$th place?
Answer: 23000.
|
23000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. On a line, there are 15 points $A_{1}, \ldots A_{15}$, spaced 1 cm apart. Petya constructs circles according to the following rules:
a) The circles do not intersect or touch.
b) Inside each circle, there is at least one of the specified points $\mathrm{A}_{1}, \ldots \mathrm{A}_{15}$.
c) None of these points lie on the circle.
d) Different circles contain different sets of points. For example, if one circle contains points $\mathrm{A}_{1}$ and $\mathrm{A}_{2}$ inside and the rest outside, then it is not possible to construct another circle that contains only $\mathrm{A}_{1}$ and $\mathrm{A}_{2}$ inside.
What is the maximum number of circles Petya can construct according to these rules?
|
Answer: 29.
## Option 2b (Saratov)
|
29
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The digits of numbers from 20 to 99 occupy the first $80 \cdot 2=160$ places in this sequence. There are $2021-160=1861$ places left. The digits of numbers from 100 to 719 occupy the next $(719-99) \cdot 3=1860$ places. Therefore, the 2021st place is the first digit of the number 720, which is 7.
|
Answer: 7. Answer to option: $1-2: 0$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The equation is transformed into the form
$$
t^{\alpha}=\alpha(t+1), \quad \alpha=\lg 2 \in(0,1), \quad t=x^{2}-3>0
$$
where:
1) the left side of the equation is a concave up function (since $\alpha \in(0 ; 1)$), defined for $t \geqslant 0$
2) the right side of the equation is a linear function with a positive slope
3) at $t=0$, the value of the left side is less than the value of the right side; at $t=1$, the value of the left side is, on the contrary, greater than the value of the right side, since
$$
1^{\alpha}=\lg 10>\lg 4=\alpha \cdot(1+1)
$$
Therefore, the graphs intersect at two points (one between 0 and 1, the other to the right of 1), each of which generates two roots of the original equation.
|
Answer: 4. Answer to option: $1-2: 4$.
保留源文本的换行和格式，直接输出翻译结果。
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Let there be $n$ voters, and $k$ votes were cast for a given candidate. Then $0.50332 \leqslant$ $\frac{k}{n} \leqslant 0.50333$, and $1.00664 \leqslant \frac{2 k}{n} \leqslant 1.00666$. If we denote $m=2 k-n$, then $0.00664 \leqslant \frac{m}{n} \leqslant$ 0.00666.
- If $m=1$, then $150.1<\frac{1}{0.00666} \leqslant n \leqslant \frac{1}{0.00664}<150.7$ - no integer solutions.
- If $m=2$, then $300<\frac{2}{0.00666} \leqslant n \leqslant \frac{2}{0.00664}<302$. But if $n=301$ and $m=2$, then the corresponding value of $k$ does not exist.
- If $m=3$, then $450<\frac{3}{0.00666} \leqslant n \leqslant \frac{3}{0.00664}<452$. Therefore, $n=451, k=227$, and all inequalities are satisfied.
|
Answer: 451. Answer to the option: $5-2: 341$.
#
|
451
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 3. We have
$$
\begin{aligned}
& 42=\frac{1}{\sqrt{x+2}+\sqrt{x+3}}+\frac{1}{\sqrt{x+3}+\sqrt{x+4}}+\ldots+\frac{1}{\sqrt{x+2017}+\sqrt{x+2018}} \equiv \\
& \equiv(\sqrt{x+3}-\sqrt{x+2})+(\sqrt{x+4}-\sqrt{x+3})+\ldots(\sqrt{x+2018}-\sqrt{x+2017}) \equiv \\
& \equiv \sqrt{x+2018}-\sqrt{x+2} \Leftrightarrow \\
& \Leftrightarrow 42(\sqrt{x+2018}+\sqrt{x+2})=2018-2 \Leftrightarrow x=7
\end{aligned}
$$
|
Answer: $x=7$. Answer to option $5-2: x=6$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. 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
|
4. 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 "1" if the first number is greater; "2", if the second number is greater; "0", if the numbers are equal.
|
Answer: The first is greater
Solution: (Solution: on the left, bring each parenthesis to a common denominator, use the inequality $\frac{n \cdot n}{(n-1)(n+1)}>1$ and we get that the square of the first number is greater than $\frac{2015}{1755}=\frac{31}{27}>\frac{8}{7$.)
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. Calculate (without using a calculator) $\sqrt[3]{9+4 \sqrt{5}}+\sqrt[3]{9-4 \sqrt{5}}$, given that this number is an integer.
|
Answer: 3.
Solution: Note that $2<\sqrt[3]{9+4 \sqrt{5}}<3$ and $0<\sqrt[3]{9-4 \sqrt{5}}<1$. Therefore, $2<\sqrt[3]{9+4 \sqrt{5}}+\sqrt[3]{9-4 \sqrt{5}}<4$, hence this number is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Igor Gorshkov has all seven books about Harry Potter. In how many ways can Igor arrange these seven volumes on three different bookshelves, so that each shelf has at least one book? (Arrangements that differ by the order of books on the shelf are considered different).
|
Answer: $C_{6}^{2} \times 7!=75600$.
Solution: First, we can arrange the books in any order, which gives 7! options. Place two dividers into 6 gaps between the books - this can be done in $C_{6}^{2}$ ways. The dividers will divide the books into three parts, which need to be placed on the 1st, 2nd, and 3rd shelves. In total, we get $C_{6}^{2} \times 7!=75600$ options.
|
75600
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Find the sum of all natural numbers that have exactly four natural divisors, three of which (of the divisors) are less than 15, and the fourth is not less than 15.
|
Answer: $\left((2+3+5+7+11+13)^{2}-\left(2^{2}+3^{2}+5^{2}+7^{2}+11^{2}+13^{2}\right)\right) / 2-$ $6-10-14+27=649$.
Solution: The numbers $N$ specified have exactly 4 divisors either if $N=$ $p^{3}$, or if $N=p q$, where $p$ and $q$ are primes. In the first case, only 27 fits. In the second case, we need to consider the pairwise products of all prime numbers less than 15. These numbers are $p_{i}=2,3,5,7,11,13$. The sum of their pairwise products can be calculated as $S_{p q}=2 \cdot(3+5+7+$ $11+13)+3 \cdot(5+7+11+13)+5 \cdot(7+11+13)+7 \cdot(11+13)+11 \cdot 13=652$. From the pairwise products, we need to exclude 6, 10, and 14, as they are less than 15. In the end, $S=27+S_{p q}-6-10-14=649$.
|
649
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Famous skater Tony Hawk is riding a skateboard (segment $A B$) in a ramp, which is a semicircle with diameter $P Q$. Point $M$ is the midpoint of the skateboard, $C$ is the foot of the perpendicular dropped from point $A$ to the diameter $P Q$. What values can the angle $\angle A C M$ take if it is known that the angular measure of the arc $A B$ is $24^{\circ}$?
|
Answer: $12^{\circ}$.
Solution: Extend the line $A C$ to intersect the circle at point $D$ (see figure). The chord $A D$ is perpendicular to the diameter $P Q$, therefore, it is bisected by it. Thus, $C M$ is the midline of triangle $A B D$, so $C M \| B D$ and, therefore, $\angle A C M=\angle A D B$. The angle $\angle A D B$ is an inscribed angle, subtending the arc $A B$, hence it is equal to half of the arc.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. In the school Spartakiad, teams from classes $8^{\mathrm{A}}$, $8^{\mathrm{E}}$, and $8^{\mathrm{B}}$ participated. In each of the competitions, one of these teams took 1st place, another took 2nd place, and another took 3rd place. After the Spartakiad, points were tallied: $x$ points were awarded for 1st place, $y$ for 2nd place, and $z$ for 3rd place $\left(x>y>z>0\right.$ - integers). In the end, team $8^{\mathrm{A}}$ received 22 points, and teams $8^{\text {B }}$ and $8^{\text {E }}$ each received 9 points. How many competitions were there in total, and which team took 2nd place in the grenade throwing competition, given that the team $8^{\mathrm{B}}$ took 1st place in the vault over the "buck"?
|
Answer: 5 competitions, $8^{\text{B}}$.
Solution: See the solution to problem 7 for 8th grade.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. A right-angled triangle is called Pythagorean if the lengths of all its sides are natural numbers. Find the greatest integer that divides the product of the lengths of the sides of any Pythagorean triangle.
|
# Problem 5.
Answer: 60.
Solution. Since a triangle with sides $3,4,5$ is a right triangle, the desired number cannot be greater than 60. We will show that the product of the lengths of the sides of any Pythagorean triangle is divisible by 60, i.e., by 3, by 4, and by 5.
The remainders when the square of an integer is divided by 3 and 4 can only be 0 and 1. Therefore, if $a$ and $b$ are not divisible by 3 and 4, then $a^{2}+b^{2}$, when divided by 3 and 4, gives a remainder of 2, and thus cannot be a perfect square.
The remainders when an integer is divided by 5 can only be 0, 1, and 4. Therefore, if $a$ and $b$ are not divisible by 5 and $a^{2}+b^{2}$ is a perfect square, then one of the numbers - $a^{2}$ or $b^{2}$ - has a remainder of 1, and the other has a remainder of 4. Consequently, $a^{2}+b^{2}$ is divisible by 5.
|
60
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Given three numbers $a, b, c$. It is known that the arithmetic mean of the numbers $a$ and $b$ is 5 more than the arithmetic mean of all three numbers. And the arithmetic mean of the numbers $a$ and $c$ is 8 less than the arithmetic mean of all three numbers. By how much does the arithmetic mean of the numbers $b$ and $c$ differ from the arithmetic mean of all three numbers?
|
Answer: 3 more.
Solution: The arithmetic mean of the numbers $\frac{a+b}{2}, \frac{b+c}{2}, \frac{a+c}{2}$, obviously, coincides with the arithmetic mean of the numbers $a, b, c$. Therefore, if one $\frac{a+b}{2}$ is 5 more, and $\frac{a+c}{2}$ is 8 less, then, to get the same result, $\frac{b+c}{2}$ must be 3 more.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Let's 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, 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
|
9. Assemble a square of the smallest area from squares of size $1 \times 1$, $2 \times 2$, and $3 \times 3$, such that the number of squares 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
|
10. 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 the total percentage of users of Odnoklassniki is $65 + 25 = 90\%$ 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
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. 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
|
12. 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
|
2. 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
|
4. A pirate schooner has taken a merchant ship by boarding. Ten pirates did not participate in the fight, while the rest lost either a hand, a leg, or both a hand and a leg in the battle. $54\%$ of the participants in the battle lost a hand, and $34\%$ lost both a hand and a leg. It is known that $2/3$ of all the pirates on the schooner lost a leg. How many pirates were on the schooner?
|
Answer: 60 pirates.
Solution - similar to option v3a.
|
60
|
Logic and Puzzles
|
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 satisfying the condition.
|
2000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. 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.
|
14
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. 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: Let's parallel translate one of the diagonals so that it forms a right-angled triangle with the other. Then, in this triangle, one leg is 12, and the hypotenuse is 13, so the remaining leg is 5.
## Option $2-$ -
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A rectangle $A D E C$ is described around a right triangle $A B C$ with legs $A B=5$ and $B C=6$, as shown in the figure. What is the area of $A D E C$?
|
# Answer 30.
Solution: The area of the rectangle is twice the area of the triangle.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. 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
|
6. 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 $x$ 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 one side $x+$ $x+1+\cdots+x+9=10 x+45 \leq 150$, From which $x \leq 10$. On the other side $x \geq(x+$ 9 )/2, from which $x \geq 9$. Therefore, in the first 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 be 150.
So in the 1st there are 10, in the second - 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 $6-$th - no less than 15. But with 15 the sum is less than 150. So it must be 16. And it cannot be more because then in the last purse there will be 21.
For graders: Partial credit can be given for a correct example of distribution (trial and error) without proof of uniqueness.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Find the smallest natural number $N$, such that the decimal representation of the number $N \times 999$ consists entirely of sevens.
|
Answer 778556334111889667445223
Solution: $N \times 999=77 \ldots 7$, then $N$ is a multiple of 7, denote $n=N / 7$. We get $999 n=$ $1000 n-n=11 \ldots 1$, so $1000 n-111 \ldots 1=n$. Write it as a column subtraction and repeat the found digits of $\mathrm{N}$ with a shift of 3 to the left $* * * * * * * 000$
1111111111
$* * * * * * 889$
$* * * * 889000$
1111111111
$* * * 777889$
777889000
111111111
6667778889
Notice that the digits repeat every 3, so we get
$n=111222333444555666777889$.
Therefore, $\mathrm{N}=7 \mathrm{n}=778556334111889667445223$.
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 you shouldn't deduct points if at the very end the number n was incorrectly multiplied by 7.
## Variant $2-\mathrm{b}$
|
778556334111889667445223
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. How many natural numbers from 1 to 2017 have exactly three distinct natural divisors?
|
Answer: 14.
Solution: Only squares of prime numbers have exactly three divisors. Note that $47^{2}>2017$, so it is sufficient to consider the squares of prime numbers from 2 to 43. There are 14 of them.
|
14
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. 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 will be $9^{4}=6561$. Among these, 9x8x7x6=3024 consist of different digits. Therefore, the number of passwords containing identical digits is 6561-3024=3537 passwords.
|
3537
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. 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
|
5. In trapezoid $A B C D$ with bases $A D / / B C$, the 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: Triangle ADE and CBE are similar, their areas are in the ratio of the square of the similarity coefficient. Therefore, this coefficient is
|
27
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. 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:
```
\(* * \ldots * * 2 \times 2=* * * . . . * * 4\)
\(* * * . . * 42 \times 2=* * * \ldots * 84\)
\(* * * . . .842 \times 2=* * * \ldots * 684\)
***... \(* 6842 \times 2=* * * \ldots * 3684\)
\(* * * . . * 36842 \times 2=* * * \ldots * 73684\)
\(* * * . . * 736842 \times 2=* * * . . . * 473684\)
\(* * * . . * 4736842 \times 2=* * * \ldots * 9473684\)
\(\cdots\)
\(105263157894736842 \times 2=210526315789473684\)
```
|
105263157894736842
|
Number Theory
|
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 5:00 PM 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, so no more than 49 participants could have won 2 games (someone must lose 2 games to them).
We can show that there could be 49 such participants. Let's say №3 won against №1 and №2, №5 won against №3 and №4, ... №99 won against №97 and №98, and №100 won against №99. Then all participants with odd numbers (except the first) won exactly 2 games.
|
49
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The robot moves along straight segments, making turns of 90 degrees to the right or left every minute (neglect the time for turning). The robot travels 10 meters per minute. What is the minimum distance from the starting position that the robot can be after 9 minutes, if it did not turn during the first minute?
#
|
# Answer: $10 \text{~m}$
Solution: We can consider a coordinate grid with nodes spaced 10m apart. Clearly, the robot travels along the nodes of this grid. It cannot return to the initial position in 9 minutes. However, it can be at a distance of 10 m - it is easy to construct an example.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A box of sugar has the shape of a rectangular parallelepiped. It contains 280 sugar cubes, each of which is a $1 \times 1 \times 1$ cm cube. Find the surface area of the box, given that the length of each of its sides is less than 10 cm.
|
# Answer: 262
Solution: $\mathbf{2 8 0}=\mathbf{2}^{3} \mathbf{x} \mathbf{x} \mathbf{x}$. From the condition that the lengths of the sides are integers less than 10, it follows that its edges are equal to 5, 7, and 8. The surface area is $2 *\left(5^{*} 7+5 * 8+7 * 8\right)=262$
|
262
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. 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 no less than 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 $5: 3, \quad 5,7,11,13,17,19, \quad-7$ 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
|
7. Find all three-digit numbers $\overline{\Pi B \Gamma}$, consisting of different digits $\Pi, B$, and $\Gamma$, for which the equality $\overline{\Pi B \Gamma}=(\Pi+B+\Gamma) \times(\Pi+B+\Gamma+1)$ holds.
|
Answer: 156.
Solution: Note that П+В $\Gamma \geq 3$ and $\leq 24$ (since the digits are different). Moreover, the numbers
$\overline{\Pi В \Gamma}$ and $(П+B+\Gamma)$ must give the same remainder when divided by 9. This is only possible when П+В $\Gamma) \times(П+B+\Gamma+1)=90$-a two-digit number. By trying $12,15,18,21,24$, we get $\overline{\Pi В \Gamma}=$ $156=12 \times 13$.
## Variant 1-b
|
156
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. At the international table tennis championship, 200 participants gathered. The game is played in a knockout format, i.e., in each match, two players participate, the loser is eliminated from the championship, and the winner remains. Find the maximum possible number of participants who won at least three matches.
|
Answer: 66.
Solution: Each participant (except the winner) lost one game to someone. There are 199 such participants, so no more than 66 participants could have won 3 games (someone must lose 3 games to them).
We will show that there could be 66 such participants. Let №4 win against №1,2,3; №7 - against №4,5,6,... №199 - against №196,197,198, and №200 win against №199. Then all participants with numbers giving a remainder of 1 when divided by 3 (except the first) won exactly 3 games.
|
66
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The turtle crawled out of its house and moved in a straight line at a constant speed of 5 m/hour. After an hour, it turned $90^{\circ}$ (right or left) and continued moving, then crawled for another hour, then turned $90^{\circ}$ (right or left) again... and so on. It crawled for 11 hours, turning $90^{\circ}$ at the end of each hour. What is the shortest distance from the house it could have been?
|
Answer 5 m.
Solution: We can consider a coordinate grid with nodes spaced 5 m apart. It is clear that the turtle crawls along the nodes of this grid. In 11 hours, it cannot return to the initial position. However, it can be 5 m away - it is easy to construct an example.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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