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Problem 9.1. Find the largest five-digit number, the product of whose digits is 120. | Answer: 85311.
Solution. The largest single-digit divisor of the number $120-8$, so the desired number definitely starts with this digit. The product of all the remaining digits is 15.
The largest single-digit divisor of the number $15-5$, so the digit in the thousands place will be this digit. The product of the las... | 85311 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.2. During the first half of the year, lazy Pasha forced himself to solve math problems. Every day he solved no more than 10 problems, and if on any day he solved more than 7 problems, then for the next two days he solved no more than 5 problems per day. What is the maximum number of problems Pasha could solve... | Answer: 52.
Solution. Suppose Pasha solved at least 8 tasks (but no more than 10) in one of the first five days, then in the next two days he solved no more than 5 tasks per day. Thus, in these three days, he solved no more than $20(10+5+5)$ tasks. If he solved 7 tasks each day, it would be more.
It turns out that in... | 52 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.3. Given a convex quadrilateral $ABCD$, $X$ is the midpoint of diagonal $AC$. It turns out that $CD \parallel BX$. Find $AD$, if it is known that $BX=3, BC=7, CD=6$.
 | Answer: 14.
Solution. Double the median $B X$ of triangle $A B C$, to get point $M$. Quadrilateral $A B C M$ is a parallelogram (Fig. 1).
Notice that $B C D M$ is also a parallelogram, since segments $B M$ and $C D$ are equal in length (both 6) and parallel. This means that point $M$ lies on segment $A D$, since $A M... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. A white checkered $8 \times 8$ table is given. In it, 20 cells are painted black. What is the minimum number of pairs of adjacent white cells that could remain? | Answer: 34.
Solution. We will call a pair of cells that are adjacent by side simply a pair. Let's first count the total number of pairs. There are 7 pairs in each row and column, so there are a total of $7 \cdot 8 \cdot 2 = 112$ pairs.
We will call cells that touch the border of the table boundary cells, and those th... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. The sides of the square $A B C D$ are parallel to the coordinate axes, with $A B$ lying on the y-axis, and the square is positioned as shown in the figure. The parabola defined by the equation
$$
y=\frac{1}{5} x^{2}+a x+b
$$
passes through points $B$ and $C$. Additionally, the vertex of this parabola (po... | Answer: 20.
Solution. Note that the parabola is symmetric with respect to the vertical axis passing through its vertex, point $E$. Since points $B$ and $C$ are on the same horizontal line, they are symmetric with respect to this axis. This means that this axis passes through the midpoint of $B C$, and therefore, throu... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. 73 children are standing in a circle. A mean Father Frost walks around the circle clockwise and distributes candies. At first, he gave one candy to the first child, then skipped 1 child, gave one candy to the next child, then skipped 2 children, gave one candy to the next child, then skipped 3 children, an... | Answer: 36.
Solution. Let's number the children clockwise from 0 to 72. Initially, Santa Claus gives a candy to the child with number 1.
Consider the sequence of numbers $a_{n}=1+2+3+\ldots+n$, where $n=1,2,3, \ldots, 2020$. Notice that the $n$-th candy is given to the child whose number is the remainder of the divis... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 10.2. How many roots does the equation
$$
\overbrace{f(f(\ldots f}^{10 \text { times } f}(x) \ldots))+\frac{1}{2}=0
$$
have, where $f(x)=|x|-1$? | Answer: 20.
Solution. Let
$$
f_{n}(x)=\overbrace{f(f(\ldots(f}^{n \text { times } f}(x) \ldots))
$$
We will solve the equation $f_{10}(x)=-\frac{1}{2}$ graphically.
We will use the fact that the graph of $y=f_{k}(x)$ can be obtained from the graph of $y=f_{k-1}(x)$, based on the relation $f_{k}(x)=\left|f_{k-1}(x)\... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.3. Anton wrote three natural numbers $a, b$, and $c$ on the board. Ira drew three rectangles $a \times b, a \times c$, and $b \times c$ on the board. It turned out that the difference in the areas of one pair of rectangles is 1, and the difference in the areas of another pair of rectangles is 49. What can $a... | Answer: 16.
Solution. Without loss of generality, we assume that $1=ac-ab=a(c-b)$, then $a=1$, $c=b+1$. Thus, the numbers written on the board were $1, b, b+1$.
Notice that either $b(b+1)-b \cdot 1=49$, or $b(b+1)-(b+1) \cdot 1=49$.
In the first case, we get $b^{2}=49, b=7$, and $a+b+c=1+b+(b+1)=16$.
In the second ... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.4. An isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ is such that $\angle ADC = 2 \angle CAD = 82^{\circ}$. Inside the trapezoid, a point $T$ is chosen such that $CT = CD, AT = TD$. Find $\angle TCD$. Give your answer in degrees.
.
![](https://cdn.mathpix.com/cropped/2024_05_06_2befe97065... | 38 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. On an island, there are two tribes: knights and liars. Knights always tell the truth, while liars always lie. One day, 80 people sat around a round table, and each of them stated: "Among the 11 people sitting after me in a clockwise direction, there are at least 9 liars." How many knights are sitting at t... | Answer: 20.
Solution. First, we prove that among 12 consecutive people, there are no more than 3 knights. Suppose this is not the case. Consider the first knight in this group. Among the 11 people sitting after him in a clockwise direction, there are at least 3 knights, which contradicts the problem's condition.
Next... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. Given a right triangle $ABC$ with legs $AB=42$ and $BC=56$. A circle passing through point $B$ intersects side $AB$ at point $P$, side $BC$ at point $Q$, and side $AC$ at points $K$ and $L$. It is known that $PK=KQ$ and $QL: PL=3: 4$. Find $PQ^2$.
. From this similarity and the cyclic nature of the pentagon
 into a bag, and the second bandit chooses who gets this bag; then this action is repeated several times. The division ends when
- either all the money is gone,
- or someone gets 11 ba... | Answer: 146.
Solution. First, we will show that the first bandit can get at least 146 coins. His strategy will be to put 14 coins in each bag. First, note that he will always be able to do this: since $21 \cdot 14=294$, he will do this at least 21 times, and when the coins start to run out, the process will certainly ... | 146 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.1. The teacher wrote a two-digit number on the board. Each of the three boys made two statements.
- Andrey: “the number ends with the digit 6” and “the number is divisible by 7”.
- Borya: “the number is greater than 26” and “the number ends with the digit 8”.
- Sasha: “the number is divisible by 13” and “th... | Answer: 91.
Solution. Let's consider Sasha's statements. If his second statement that the number is less than 27 is true, then Borya's first statement is definitely false, so the number must end in 8. The only two-digit number that meets these conditions is 18, but then none of Andrei's statements are true. Contradict... | 91 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.2. Vera has a set of weights of different masses, each weighing an integer number of grams. It is known that the lightest weight in the set weighs 71 times less than all the other weights combined. It is also known that the two lightest weights in the set together weigh 34 times less than all the other weigh... | Answer: 35.
Solution. All weights in the solution are expressed in grams.
Let the lightest weight be $m$, then the other weights are $71 m$, and the total weight is $72 \mathrm{~m}$. Let also the two lightest weights together weigh $n$, then the other weights weigh $34 n$, and the total weight is $35 n$, which is div... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.3. On the coordinate plane, all points $(x, y)$ such that $x$ and $y$ are integers satisfying the inequalities $0 \leqslant x \leqslant 2$ and $0 \leqslant y \leqslant 26$ are marked. How many lines exist that pass through exactly 3 of the marked points? | Answer: 365.
Solution. All marked points are located on three vertical lines $x=0, x=1, x=2$. Let's call these lines the left, middle, and right lines, respectively.
Consider any three marked points lying on one line. If at least two of them lie on a vertical line, then there are more than 3 marked points on such a l... | 365 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.4. On the side $AC$ of triangle $ABC$, points $M$ and $N$ are marked ($M$ lies on the segment $AN$). It is known that $AB = AN$, $BC = MC$. The circumcircles of triangles $ABM$ and $CBN$ intersect at points $B$ and $K$. How many degrees does the angle $AKC$ measure if $\angle ABC = 68^\circ$?
 \cos \left(\frac{\angle A-\angle B}{2}\right) \\
2-\sqrt{2} & =\cos \angle A+\cos \angle B=2 \cos \left(\frac{\angle A+\angle B}{2}\... | 168 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.7. Natural numbers $a$ and $b$ are such that $a^{a}$ is divisible by $b^{b}$, but $a$ is not divisible by $b$. Find the smallest possible value of the number $a+b$, given that the number $b$ is coprime with 210. | Answer: 374.
Solution. Obviously, $b \neq 1$. Let $p$ be a prime divisor of the number $b$; then $p \geqslant 11$, since $b$ is coprime with $210=2 \cdot 3 \cdot 5 \cdot 7$. Since $a^{a}$ is divisible by $b^{b}$, which is divisible by $p$, then $a$ is also divisible by $p$. From this, it immediately follows that the n... | 374 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.8. Inside the tetrahedron $ABCD$, points $X$ and $Y$ are given. The distances from point $X$ to the faces $ABC, ABD, ACD, BCD$ are $14, 11, 29, 8$ respectively. And the distances from point $Y$ to the faces $ABC, ABD, ACD, BCD$ are $15, 13, 25, 11$ respectively. Find the radius of the inscribed sphere of the... | Answer: 17.
Solution. Consider a point $Z$ lying on the ray $XY$ such that $XY: YZ = 1: 2$. We will prove that this point is the center of the inscribed sphere of the tetrahedron.
Drop perpendiculars $X_{\alpha}, Y_{\alpha}, Z_{\alpha}$ from points $X, Y, Z$ to the plane $\alpha$ - obviously, they will lie in the sam... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Misha and Masha had the same multi-digit integer in their notebooks, ending in 9876. Masha placed a plus sign between the third and fourth digits from the right, while Misha placed a plus sign between the fourth and fifth digits from the right. To the surprise of the schoolchildren, both resulting sums turned out to... | Solution. Let the written number have the form $\overline{x 9876}$, where $x$ is also some natural number. Then Misha got the sum $x+9876$, and Masha got the sum $10 x+9+876$. From the equality $x+9876=10 x+9+876$ we find that $x$ $=999$.
Answer: 9999876 and there is no other number.
## CONDITION | 9999876 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. On the first day, Masha collected $25 \%$ fewer berries than Vanya, and on the second day, $20 \%$ more than Vanya. Over the two days, Masha collected $10 \%$ more berries than Vanya. What is the smallest number of berries they could have collected together? | # Solution
Masha collected $3 / 4$ on the first day and $-6 / 5$ of the number of berries collected by Vanya over these days. Let Vanya collect $4 x$ berries on the first day and $5 y$ on the second day, then Masha collected $3 x$ and $6 y$ berries respectively. According to the condition, $3 x + 6 y = 11 / 10 (4 x + ... | 189 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task No. 1.1
# Condition:
The figure shows 4 circles.
Find the sum of the numbers that are in exactly two circles.
# | # Answer: 22
## Exact match of the answer -1 point
## Solution.
The gray areas on the diagram represent the regions that are included in exactly two circles. The number 10 is in the orange and blue circles, the number 2 is in the blue and brown circles, the number 1 is in the green and brown circles, and the number ... | 59 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. (a) (2 points) A natural number $n$ is less than 120. What is the largest remainder that the number 209 can give when divided by $n$?
(b) (2 points) A natural number $n$ is less than 90. What is the largest remainder that the number 209 can give when divided by $n$? | Answer: (a) 104. (b) 69.
Solution. Let $209=n k+r$, where $k-$ is the quotient, and $r$ is the remainder of the division. Since $rn k+r=209=n k+r>r k+r=r(k+1)$, hence
$$
k+1>\frac{209}{n} \quad \text { and } \quad r\frac{209}{119}$, i.e., $k \geqslant 1$. Then $r\frac{209}{89}$, i.e., $k \geqslant 2$. Then $r<\frac{2... | 104 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.3. On the board, natural numbers $a, b, c, d$ are written. It is known that among the six sums
$$
a+b, \quad b+c, c+d, d+a, a+c, b+d
$$
three are equal to 23, and the other three are equal to 34.
(a) (1 point) What is the value of $a+b+c+d$?
(b) (3 points) What is the smallest of the numbers $a, b, c, d$... | Answer: (a) 57. (b) 6.
Solution. (a) Let's add all 6 sums $a+b, b+c, c+d, d+a, a+c, b+d$. Since three of them are equal to 23 and the other three are equal to 34, we get $23 \cdot 3 + 34 \cdot 3$. On the other hand, we get $3(a+b+c+d)$. Therefore,
$$
a+b+c+d=\frac{23 \cdot 3 + 34 \cdot 3}{3}=57
$$
(b) Suppose that a... | 57 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.4. Given a parallelogram $A B C D$, point $M$ is the midpoint of side $B C$. On side $A D$, there is a point $K$ such that $B K=B M$ and the quadrilateral $K B M D$ is cyclic.
(a) (2 points) What is the length of segment $M D$, if $A D=17$?
(b) (2 points) How many degrees does the angle $K M D$ measure, if... | Answer: (a) 8.5. (b) 48.
Solution. (a) Note that KBMD is a cyclic trapezoid, so it is isosceles, i.e., $M D=K B$. Therefore,
$$
M D=K B=\frac{B C}{2}=\frac{A D}{2}=8.5
$$
Fig. 7: to the so... | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.5. One winter day, 43 children were throwing snowballs at each other. Each of them threw exactly one snowball at someone else. It is known that:
- the first threw a snowball at the one who threw a snowball at the second,
- the second threw a snowball at the one who threw a snowball at the third,
- the forty... | # Answer: 24.
Solution. First, note that not only did each throw exactly one snowball, but each was hit by exactly one snowball. Indeed, from the phrase "the first threw a snowball at the one who threw a snowball at the second," it follows that someone threw a snowball at the second; similarly, it is established that ... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. A pair of natural numbers ( $a, p$ ) is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 13)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 20. | Answer: (a) any of the numbers $14, 26, 182$. (b) 24.
Solution. Since $a^{3}+p^{3}=(a+p)\left(a^{2}-a p+p^{2}\right)$, and $a^{2}-p^{2}=(a+p)(a-p)$, the condition of divisibility is equivalent to $a^{2}-a p+p^{2}=a(a-p)+p^{2}$ being divisible by $a-p$. Note that $a(a-p)$ is divisible by $a-p$, so $p^{2}$ must be divis... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Variant 10.4.2. Given a parallelogram $A B C D$, point $M$ is the midpoint of side $B C$. On side $A D$, there is a point $K$ such that $B K = B M$ and quadrilateral $K B M D$ is cyclic.
(a) (2 points) What is the length of segment $M D$, if $A D = 19$?
(b) (2 points) How many degrees does angle $K M D$ measure, if $... | Answer: (a) 9.5. (b) 42. | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 10.6.1. A pair of natural numbers $(a, p)$ is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 13)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 20. | Answer: (a) any of the numbers $14,26,182$. (b) 24. | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Variant 10.6.2. A pair of natural numbers $(a, p)$ is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 17)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 18. | Answer: (a) any of the numbers $18,34,306$. (b) 21. | 21 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Variant 10.6.3. A pair of natural numbers ( $a, p$ ) is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 19)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 24. | Answer: (a) any of the numbers $20,38,380$. (b) 27. | 27 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Variant 10.6.4. A pair of natural numbers $(a, p)$ is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 11)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 16. | Answer: (a) any of the numbers $12, 22, 132$. (b) 18. | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. Several consecutive natural numbers are written on the board. Exactly $52 \%$ of them are even. How many even numbers are written on the board | # Answer: 13.
Solution. Since the natural numbers written down are consecutive, even and odd numbers alternate. According to the condition, there are more even numbers, which means the sequence starts and ends with even numbers.
First method. Let $n$ be the number of even numbers, then the number of odd numbers is $(... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.4. In a square table of size $100 \times 100$, some cells are shaded. Each shaded cell is the only shaded cell either in its column or in its row. What is the maximum number of cells that can be shaded? | Answer: 198.
Solution. Example. We will color all the cells of one row and all the cells of one column, except for their common cell. In this case, the condition of the problem is satisfied, and exactly 198 cells are colored.
Estimate. We will prove that no more than 198 cells could have been colored in the required ... | 198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.6. Twenty-five coins are arranged into piles as follows. First, they are arbitrarily divided into two groups. Then any of the existing groups is again divided into two groups, and so on until each group consists of one coin. Each time a group is divided into two, the product of the number of coins in the two resultin... | Answer: 300.
Solution. First method. Let's represent the coins as points and connect each pair of points with a segment. We will get $\frac{25(25-1)}{2}$ $=300$ segments. With each division of one group of coins into two, we will erase all segments connecting points corresponding to coins that end up in different grou... | 300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.5. To a natural number $N$, the largest divisor of $N$ less than $N$ was added, and the result was a power of ten. Find all such $N$. (N. Agakhanov) | Answer: 75.
Solution. Let $m$ be the greatest divisor of the number $N$, less than $N$. Then $n=mp$, where $p$ is the smallest prime divisor of the number $N$. We have $N+m=10^{k}$, that is, $m(p+1)=10^{k}$. The number on the right side is not divisible by 3, so $p>2$. From this, it follows that $N$ is an odd number, ... | 75 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. To a natural number $N$, the largest divisor of $N$ less than $N$ was added, and the result was a power of ten. Find all such $N$.
(N. Agakhanov) | Answer: 75.
Solution. Let $m$ be the greatest divisor of the number $N$, less than $N$. Then $n=m p$, where $p$ is the smallest prime divisor of the number $N$. We have $N+m=10^{k}$, that is, $m(p+1)=10^{k}$. The number on the right side is not divisible by 3, so $p>2$. From this, it follows that $N$ is an odd number,... | 75 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. First, we will calculate the number of ways to choose three points out of the 60 available. The first point can be chosen in 60 ways, the second in 59 ways, and the third in 58 ways; as a result, we get $205320=60 \cdot 59 \cdot 58$ options. Since the selections of vertices ABC, ACB, CBA, CAB, BAC, and BCA all give ... | Answer: 34190
Recommendations for evaluating solutions: to determine the number of triplets of points, one can use the combination formula $C_{60}^{3}=\frac{60!}{3!57!}=34220$. | 34190 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that the quadratic function $f(x)=x^{2}+a x+b$ has zeros $x_{1}$, $x_{2}$ and $f(2014)=f(2016)$. Find $\frac{x_{1}+x_{2}}{2}$. | # Answer: 2015
## First Solution.
Since the graph of the quadratic function $f(x)$ is symmetric with respect to the line $x=-a / 2$ and $f(2014)=f(2016), f\left(x_{1}\right)=f\left(x_{2}\right)=0$, then
$-\frac{a}{2}=\frac{x_{1}+x_{2}}{2}=\frac{2014+2016}{2}=2015$.
## Second Solution.
Since $f(2014)=f(2016)$, then... | 2015 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Three bees collect nectar from 88 garden flowers. Each flower was visited by at least one bee. Each bee visited exactly 54 flowers. We will call a flower sweet if it was visited by all three bees, and bitter if it was visited by exactly one. Which of these 88 flowers are more: sweet or bitter, and by how many? | Solution. Let among the 88 garden flowers, $s$ be sweet and $g$ be bitter. Then the number of flowers visited by two bees is exactly $88-s-g$. On one hand, the total number of bee landings on the flowers is $3 \cdot 54=162$ (each bee visited 54 flowers), and on the other hand, it is $3s + 2(88-s-g) + g = s - g + 176$. ... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. The bisectors $\mathrm{AD}$ and $\mathrm{BE}$ of triangle $\mathrm{ABC}$ intersect at point I. It turns out that the area of triangle ABI is equal to the area of quadrilateral CDIE. Find the greatest possible value of angle ACB. | Answer: $60^{\circ}$.
Solution. Let $\mathrm{S}(\mathrm{CDIE})=\mathrm{S}_{1}, \mathrm{~S}(\mathrm{ABI})=\mathrm{S}_{2}$, $S(B D I)=S_{3}, S(A I E)=S_{4}$ (see figure). Since the ratio of the areas of triangles with a common height is equal to the ratio of the bases, and the angle bisector divides the opposite side in... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. In the cells of an $8 \times 8$ board, the numbers 1 and -1 are placed (one number per cell). Consider all possible placements of a four-cell figure
$\square$ on the board (the figure can be rotated, but its cells must not go beyond the board's boundaries). A placement is called unsuccessful if the sum of the numbe... | Answer: 36. Solution: We will show that in each "cross" of five cells on the board, there will be at least one unsuccessful placement. Suppose the opposite; let the numbers in the outer cells of the cross be \(a, b, c, d\), and in the central cell be \(e\); denote the sum of all these five numbers by \(S\). Then, accor... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. On the day when Dima's brother and sister were congratulating him on his birthday, Dima said: "Look how interesting, I am now twice as old as my brother and three times as old as my sister!" - "And your average age is 11 years," - added Dad. How old did Dima turn? | Answer: 18 years.
Solution. The first method. According to the problem, we can form an equation. Let Dima's age be $x$ years, then his sister's age is $x / 3$, and his brother's age is $-x / 2 ;(x+x / 3+x / 2): 3=11$. After solving this equation, we get that $x=18$. Dima is 18 years old.
It will be useful to provide ... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. How many three-digit numbers exist that are 5 times the product of their digits? | Answer. One number is 175.
Solution. First method. The digits that make up the number do not include the digit 0, otherwise the condition of the problem cannot be met. The given three-digit number is obtained by multiplying 5 by the product of its digits, so it is divisible by 5. Therefore, its notation ends with the ... | 175 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In a circle, a diameter $A B$ and a chord $C D$ parallel to it were drawn such that the distance between them is half the radius of this circle (see figure). Find the angle $CAB$.
| Answer: $75^{\circ}$.
Solution. Consider triangle $A O C$, where $O$ is the center of the circle. This triangle is isosceles because $O C$ and $O A$ are radii. Therefore, by the property of isosceles triangles, angles $A$ and $C$ are equal. Draw the perpendicular $C M$ to side $A O$ and consider the right triangle $O ... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.2. In four classes of a school, there are more than 70 children, all of whom came to the grade meeting (no other children were present at the meeting).
Each girl who came was asked: "How many people from your class, including you, came to the meeting?"
Each boy who came was asked: "How many boys from your c... | # Answer:
(a) (1 point) 21 students.
(b) (3 points) 33 girls.
Solution. (a) Since all 8 numbers listed in the condition are distinct, exactly 4 of them represent the number of children in the classes, and the other 4 represent the number of boys in the classes. The number 21, being the largest of the listed numbers,... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. The numbers $1, 2, 3, \ldots, 57$ are written on a board. What is the maximum number of these numbers that can be chosen so that no two chosen numbers differ exactly by a factor of 2.5? | Answer: 48.
Solution. Consider sequences of natural numbers that satisfy the following set of conditions: in each sequence
- each number does not exceed 57
- there are at least two numbers, and they all go in ascending order;
- each subsequent number is 2.5 times the previous one.
Let's list them all.
- 2,5
- $4,10... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. On a plane, 36 points are marked, no three of which lie on the same line. Some pairs of marked points are connected by segments such that no more than 3 segments emanate from each marked point.
What is the maximum number of different closed 4-segment broken lines that can result?
The vertices of the brok... | # Answer: 54.
Solution. First, let's prove that there are no more than 54 broken lines.
Consider a "tick" structure, consisting of three points $A, B$, and $C$, as well as two segments $AB$ and $AC$ (the segment $BC$ may or may not be present; point $A$ will be called the vertex of the tick). Since from $B$ and $C$ n... | 54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2.1. Points $A, B, C, D$ are marked on a line, in that exact order. Point $M$ is the midpoint of segment $A C$, and point $N$ is the midpoint of segment $B D$. Find the length of segment $M N$, given that $A D=68$ and $B C=20$.
}{2}=44-\frac{x}{2}
$$
Now it is not difficult to calculate $M N$:
$$
M N=A D-A M-N D=68-\frac{x}{2}-\left(44-\frac{x}{2}\right)=24
$$ | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.1. Along the road connecting Masha's and Sasha's houses, there are 17 apple trees and 18 poplars. When Masha was going to visit Sasha, she took photos of all the trees. Right after the tenth apple tree, Masha's phone memory ran out, and she couldn't photograph the remaining 13 trees. The next day, when Sasha ... | Answer: 22.
Solution. Note that the tenth apple tree, counting from Masha's house, is the eighth apple tree, counting from Sasha's house. Therefore, Sasha will not pick a leaf from exactly 13 trees. We get that he will pick a total of $17+18-13=22$ leaves. | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.4. Along the road connecting Masha's and Sasha's houses, there are 17 apple trees and 20 poplars. When Masha was going to visit Sasha, she took photos of all the trees. Right after the tenth apple tree, Masha's phone memory ran out, and she couldn't photograph the remaining 13 trees. The next day, when Sasha ... | Answer: 24.
## 7th grade, problem 5 | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.1. At a physical education class, 27 seventh-graders arrived, some of them brought one ball each. Sometimes during the class, one of the seventh-graders would give their ball to another seventh-grader who didn't have one.
At the end of the class, $N$ seventh-graders said: "I received balls less frequently th... | Answer: 13.
Solution. If a seventh-grader received the ball less frequently than he gave it away, then he originally had the ball, but no longer had it at the end. Thus, at the beginning of the lesson, the students collectively had at least $N$ balls, which ultimately ended up with some of the remaining $27-N$ student... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.4. For a physical education class, 29 seventh-graders came, some of them brought one ball each. Sometimes during the class, one of the seventh-graders would give their ball to another seventh-grader who didn't have one.
At the end of the class, \( N \) seventh-graders said: “I received balls less frequently ... | Answer: 14.
## 7th grade, problem 6 | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. Let $s(n)$ denote the sum of all odd digits of the number $n$. For example, $s(4)=0$, $s(173)=11, s(1623)=4$.
Calculate the value of the sum $s(1)+s(2)+s(3)+\ldots+s(321)$. | Answer: 1727.
Solution. We will separately count the sum of odd digits by place value.
## Units place.
Among the numbers from 1 to 321, there are 32 complete tens:
- from 1 to 10;
- from 11 to 20;
- from 21 to 30;
- ...
- from 311 to 320.
In each ten, the sum of the odd digits in the units place is $1+3+5+7+9=25... | 1727 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. At an open evening at the conservatory, four quartets, five duets, and six trios were supposed to perform (a quartet consists of four musicians, a trio of three, and a duet of two; each musician is a member of only one musical group). However, one quartet and two duets unexpectedly went on tour, and a solois... | Answer: 35.
Solution. If no one was absent, then at the evening there would have been
$4 \cdot 4(4$ quartets $)+5 \cdot 2(5$ duets $)+6 \cdot 3$ ( 6 trios $)=44$ people.
But $4(1$ quartet) $+2 \cdot 2$ (2 duets) +1 (a soloist from one of the trios) $=9$ people were absent. Thus, at the evening, $44-9=35$ people perf... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Dima stood on one of the steps of the staircase and suddenly noticed that there were an equal number of steps above and below him. Then he climbed up 7 steps, and after that, he went down 15 steps. In the end, he found himself on the 8th step of the staircase (counting from the bottom). How many steps does t... | Answer: 31.
Solution. Since Dima ended up on the 8th step at the end, before that he was on the $8+15=23$ step. He got there by climbing up 7 steps, so he started from the $23-7=16$ step. Thus, we have that the 16th step is the middle of the ladder. Therefore, the ladder consists of 31 steps. | 31 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.2. Electronic clocks display time from 00.00.00 to 23.59.59. How much time during the day does the number on the display that reads the same from left to right and from right to left light up? | Answer: 96 seconds.
Solution. If the digits on the display are $a b . c d . m n$, then $a=0,1,2,0 \leq b \leq 9,0 \leq c \leq 5$, $0 \leq d \leq 9,0 \leq m \leq 5,0 \leq n \leq 9$. Therefore, if $a=n, b=m, c=d$, the symmetrical number on the display is uniquely determined by the digits $a, b$ and $c$, where $a=0,1,2,0... | 96 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.5. $A L$ and $B M$ are the angle bisectors of triangle $A B C$. The circumcircles of triangles $A L C$ and $B M C$ intersect again at point $K$, which lies on side $A B$. Find the measure of angle $A C B$. | Answer. $\angle A C B=60^{\circ}$.
Solution. Draw the segment $C K . \angle L C K=\angle L A K$ (these angles are inscribed in the circle and subtend the same arc). Similarly, $\angle M C K$ $=\angle M B K$. Since $\angle A C B=\angle L C K+\angle M C K$, the desired angle $A C B$ is one-third of the sum of the angles... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 1. Option 1
On the sheet, three rectangles A, B, and C are drawn.
A
5
B
Rectangles A and B have the same width, and rectangles B and C have the same length (width - top to bottom, lengt... | Answer: 88.
Solution: Let rectangle A have a length of $a$ cm and a width of $b$ cm. If the length is increased by 2 cm, the area will increase by $2 b$ cm $^{2}$. Therefore, $2 b=22, b=11$. The area of rectangle B is 40 cm $^{2}$ less than that of rectangle B, so the length of rectangle B is $40: 4=10$ cm. Therefore,... | 88 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 2. Option 1
There are candies in three boxes. It is known that there are 2 times fewer candies in the first box than in the second. It is also known that there are a total of 24 candies in the first and third boxes, and a total of 34 candies in the second and third boxes. How many candies are there in total in the b... | Answer: 44.
Solution: From the condition, it follows that in the second box there are $34-24=10$ more candies than in the first. And this difference is equal to the number of candies in the first box. Therefore, there are 10 candies in the first box, 20 in the second, and 14 in the third.
## Variant 2
In three boxes... | 44 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 5. Variant 1
Given nine cards with the numbers $5,5,6,6,6,7,8,8,9$ written on them. From these cards, three three-digit numbers $A$, B, V were formed, each of which has all three digits different. What is the smallest value that the expression A + B - V can have? | Answer: 149.
Solution. By forming the smallest sum of numbers A and B, as well as the largest number C, we get the smallest value of the expression A + B - C: $566 + 567 - 988 = 145$. However, this combination is not suitable: two numbers have the same digits. By swapping the digits 6 and 8 in the units place, we get ... | 149 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 6. Variant 1
Given the road map of the kingdom. The cities are marked with numbers, and the segments represent roads. One day, a traveling knight started his journey from one of the cities in the kingdom and managed to construct his route in such a way that he traveled each road exactly once. In which city, marked w... | Answer: 2, 5.
Solution: If a city is not the beginning or the end of the knight's journey, then every time he enters through one road, he must exit through another road. This means that the roads from such a city come in "pairs," and there is an even number of them in total. Therefore, cities 2 and 5, from which an od... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 8. Variant 1.
Each of the 10 students came up with 5 natural numbers. It turned out that each number was thought of by at least three students. What is the maximum number of different numbers that could have been thought of? | Answer: 16.
Solution: In total, the students came up with 50 numbers, with each number being counted at least 3 times. We will prove that there could not have been more than 16 different numbers. If at least 17 different numbers were thought of and each by at least three students, then a total of no less than $17 \cdo... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Twelve pencils are sharpened so that they all have different lengths. Masha wants to put the pencils in a box in two rows of 6 each, so that in each row the lengths of the pencils decrease from left to right, and each pencil in the second row lies on a longer pencil. In how many ways can she do this?
Answer: 132. | A layout of pencils that satisfies the conditions in the problem will be called correct. We will stack the pencils in a box in descending order of length. The order of placement will be recorded in the form of a broken path from the point (0,0) in a rectangular Cartesian coordinate system on a plane to the point $(6,6)... | 132 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of triples of natural numbers $a, b$, and $c$ not exceeding 2017 such that the polynomial $x^{11} + a x^{7} + b x^{3} + c$ has a rational root.
Trunov K.V. | Answer: 2031120
Solution: Since all coefficients of the polynomial are natural and the leading coefficient is 1, any rational root must be an integer.
Notice that for $x \geq 0$, $x^{11} + a x^{7} + b x^{3} + c \geq c \geq 1$. Therefore, there are no integer roots for $x \geq 0$. If $x \leq -2$, then $x^{11} + a x^{7... | 2031120 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.3. Nadya baked pies with raspberries, blueberries, and strawberries. The number of pies with raspberries was half of the total number of pies; the number of pies with blueberries was 14 less than the number of pies with raspberries. And the number of pies with strawberries was half the number of pies with raspberries... | Answer: 21 raspberry pies, 7 blueberry pies, and 14 strawberry pies.
Solution. First method. Since the number of raspberry pies is half of the total, the number of blueberry and strawberry pies together is the same as the number of raspberry pies (see Fig. 5.3). Moreover, there are 14 fewer blueberry pies than raspber... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.4. Postman Pechkin delivers mail by bicycle to four villages: Prostokvashino, Smetanino, Tvorozhnoye, and Molochnoye (see figure). He knows that the distance from Prostokvashino to Tvorozhnoye is 9 km, from Prostokvashino to Smetanino is 13 km, from Tvorozhnoye to Smetanino is 8 km, and from Tvorozhnoye to Molochnoye... | Answer: 19 km.
Solution. The distance from Prostokvashino to Smetanino with a detour to Tvorozhnoye is $9+8=17$ (km), and without the detour - 13 km. Therefore, the detour to Tvorozhnoye (from the highway and back) is 4 km. The distance from Prostokvashino to Molochnoye with a detour to Tvorozhnoye is $9+14=23$ (km), ... | 19 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.4. The teacher is going to give the children a problem of the following type. He will inform them that he has thought of a polynomial $P(x)$ of degree 2017 with integer coefficients, the leading coefficient of which is 1. Then he will tell them $k$ integers $n_{1}, n_{2}, \ldots, n_{k}$, and separately inform them o... | Answer. For $k=2017$.
Solution. First, we prove that $k>2016$. Suppose the teacher used some $k \leqslant 2016$, thinking of the polynomial $P(x)$. Consider the polynomial $Q(x)=P(x)+\left(x-n_{1}\right)\left(x-n_{2}\right) \ldots(x-n_{k})$. Note that the degree of the polynomial $Q(x)$ is also 2017, and its leading c... | 2017 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.1. Petya wrote ten natural numbers on the board, none of which are equal. It is known that among these ten numbers, three can be chosen that are divisible by 5. It is also known that among the ten numbers written, four can be chosen that are divisible by 4. Can the sum of all the numbers written on the board be less... | Answer. It can.
Solution. Example: $1,2,3,4,5,6,8,10,12,20$. In this set, three numbers $(5,10,20)$ are divisible by 5, four numbers $(4,8,12,20)$ are divisible by 4, and the total sum is 71.
Remark. It can be proven (but, of course, this is not required in the problem), that in any example satisfying the problem's c... | 71 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.4. In a company, some pairs of people are friends (if $A$ is friends with $B$, then $B$ is also friends with $A$). It turned out that for any selection of 101 people from this company, the number of pairs of friends among them is odd. Find the largest possible number of people in such a company.
(E. Bakayev, I. Bog... | Answer: 102.
Solution: In all solutions below, we consider the friendship graph, where vertices are people in the company, and two people are connected by an edge if they are friends.
Consider 102 vertices, and construct the following graph on them. Connect one vertex $x$ to three others $v_{1}, v_{2}, v_{3}$. Divide... | 102 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.5. Let $S$ be a 100-element set consisting of natural numbers not exceeding 10000. Mark all points in space where each of the coordinates belongs to the set $S$. Attach to each of the 1000000 marked points $(x, y, z)$ a ball with the number $\frac{x^{2}+y^{2}+z^{2}}{x y+y z+z x}$ written on it. What is the maximum n... | Answer. $3 \cdot C_{100}^{2}=14850$.
Solution. Let's call a triplet of natural numbers $(x, y, z)$, whose elements belong to $S$, good if
$$
x^{2}+y^{2}+z^{2}=2(x y+y z+z x)
$$
Thus, we need to find the maximum possible number of good triplets.
Let's determine when a triplet is good. Rewrite (*) as a quadratic equa... | 14850 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A company is called public if it has at least 15 shareholders. A shareholder of a company is called a minority shareholder if they own no more than $25 \%$ of the shares of that company. On the stock exchange where the shares are traded, one sixth of the companies are public. Prove that among all shareholders partic... | Solution. Let the number of firms on the exchange be N. We will call a shareholder who owns more than $25 \%$ of a firm's shares a real shareholder. The number of real shareholders in one firm cannot exceed three. (If there are four, then one of them cannot own more than $25 \%$ of the shares). Therefore, the total num... | 20 | Combinatorics | proof | Yes | Yes | olympiads | false |
1. Kolya, Seryozha, and Vanya regularly go to the cinema: Kolya goes there every 4th day, Seryozha - every 5th day, and Vanya - every 6th day. Today all the boys were at the cinema. When will all three meet at the cinema again? | 1. Answer: 60.
Let's start numbering the days beginning with tomorrow (today's day is "zero"). Kolya will go to the cinema on the days whose numbers are divisible by 4, Seryozha - on the days whose numbers are divisible by 5, Vanya - on the days whose numbers are divisible by 6. For them to all be at the cinema togeth... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Four friends came back from fishing. Every two of them counted the sum of their catches. Six numbers were obtained: $7,9,14,14,19,21$. How many fish were caught in total? | 4. 28 fish.
Note that the catch of each person is counted in exactly three sums. Therefore, if we add up all six numbers, we get the tripled total catch. Thus, the total number of fish caught is $(7+9+14+14+19+21): 3=28$ fish. | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.2. Find the value of the expression $a^{3}+12 a b+b^{3}$, given that $a+b=4$. | Answer: 64.
Solution. $a^{3}+12 a b+b^{3}=a^{3}+b^{3}+12 a b=(a+b)\left(a^{2}-a b+b^{2}\right)+12 a b=4 a^{2}-4 a b+4 b^{2}+12 a b=4(a+b)^{2}=64$. | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5. A natural number is called interesting if all its digits are different, and the sum of any two adjacent digits is a square of a natural number. Find the largest interesting number. | Answer: 6310972.
Solution. Let's mark 10 points on the plane, representing the digits from 0 to 9, and connect those points whose sum is a square of a natural number.
On this diagram, we ne... | 6310972 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.3. In an intergalactic hotel, there are 100 rooms with capacities of
$101, 102, \ldots, 200$ people. In these rooms, a total of $n$ people live. A VIP guest has arrived at the hotel, and a whole room needs to be vacated for them. To do this, the hotel manager selects one room and relocates all its occupants to anot... | Answer: 8824.
Solution: Suppose that with 8824 guests, the director cannot carry out the relocation. Let's divide the rooms into pairs by capacity: $101-200, 102-199, \ldots, 150-151$. Note that for each pair of rooms, the total number of people living in the two rooms is greater than the capacity of the larger room i... | 8824 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.6. In some cells of a $200 \times 200$ square, there is one chip - either red or blue; the other cells are empty. A chip can see another if they are in the same row or column. It is known that each chip sees exactly five chips of the other color (and possibly some chips of its own color). Find the maximum possible n... | Answer: 3800 chips.
Solution: An example containing 3800 chips can be constructed as follows. Highlight the "border" of width 5 in a $200 \times 200$ square. This border consists of four corner squares $5 \times 5$ and four rectangles $5 \times 190$. Place the chips in these four rectangles: red chips in the left and ... | 3800 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A square $10 \times 10$ is divided into unit squares. How many triangles are formed after drawing one diagonal? | Answer: 110.
Solution. The figure shows one of the obtainable triangles. All such triangles are right-angled, with the right angle vertex being any lattice node except those lying on the diagonal. There are a total of $11 \times 11$ nodes, and 11 of them are on the diagonal, so the number of triangles is $11 \times 11... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.1. Find the largest six-digit number, all digits of which are different, and each of the digits, except for the extreme ones, is either the sum or the difference of the adjacent digits. | Answer: 972538.
Solution. Let $A$ be the desired number. Let's try to find the number $A$ with the first digit being 9. We will try different options for the second digit. If the second digit is 8, then we get: $A=98176-$ it does not form a six-digit number. If the second digit is 7, then we get: $A=972538-$ a six-dig... | 972538 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.2. On the board, there are $n$ different integers, any two of which differ by at least 10. The sum of the squares of the three largest of them is less than three million. The sum of the squares of the three smallest of them is also less than three million. For what largest $n$ is this possible? | Answer. For $n=202$.
Solution. Notice immediately that $990^{2}+1000^{2}+1010^{2}=(1000-10)^{2}+1000^{2}+(1000+10)^{2}=3 \cdot 1000^{2}+2 \cdot 10^{2}$, which is greater than three million. On the other hand, $989^{2}+999^{2}+1009^{2}=(1000-11)^{2}+(1000-1)^{2}+(1000+9)^{2}=3 \cdot 1000^{2}-6 \cdot 1000+(9^{2}+1^{2}+1... | 202 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Scheherazade tells the Shah mathematical tales. A mathematical tale is an equation of the form $a x+b y=c$, where $a, b, c$ are natural numbers ranging from 1 to 17 (inclusive). A tale is sad if the corresponding equation has no solutions in integers (for example, the tale $2 x+4 y=7$ is sad, while the tale $3 x+5 y... | Answer: will be able to.
Solution. The following fairy tales will be sad:
a) a and b are even, c is odd. There are 8$\cdot$8$\cdot$9=576 such fairy tales.
b) a and b are divisible by 3, c is not divisible by 3. There are 5$\cdot$5$\cdot$12 such fairy tales. However, we need to account for the overlap with a): a and ... | 1009 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In triangle $ABC$ (angle B is obtuse), the altitude $BH$ and the bisector $AK$ are drawn. Find the angle $AKB$, if the angle $KHC$ is $45^{\circ}$. | Answer: $45^{\circ}$.
Solution. Point K is equidistant from lines AB and AC (since AK is the bisector), and from lines HB and HC (since HK is the bisector of the right angle). Therefore, point K is equidistant from lines AB and BH. This means that BK is the bisector of the angle external to angle ABH. Therefore, angle... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.3 A weirdo chose 677 different natural numbers from the list $1,2,3, \ldots, 2022$. He claims that the sum of no two of the chosen numbers is divisible by 6. Did he go too far with his claim? | Solution: Suppose the eccentric is right. From the list specified in the condition, there are exactly 377 remainders of division by 6 of each type from 0 to 5. Numbers with remainders 0 and 3 can be taken no more than one each. It is impossible to take numbers with remainders 1 and 5 simultaneously, so in the eccentric... | 676 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Dwarves Glóin, Óin, and Thráin found 70 identical precious stones and want to divide them among themselves so that each of them gets no less than 10 stones. In how many ways can the dwarves do this? | Answer: 861.
Solution: Give each dwarf 9 stones, and lay out the remaining 43 stones in a row. To distribute the remaining stones among the dwarves, it is sufficient to place two dividers in the 42 spaces between the stones. Gloin will receive the stones to the left of the first divider, Oin will receive the stones be... | 861 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Given a $101 \times 101$ grid, all cells of which are painted white. It is allowed to choose several rows and repaint all cells in these rows to black. Then, choose exactly the same number of columns and repaint all cells in these columns to the opposite color (i.e., white to black, and black to white). What is the ... | Answer: 5100.
Solution. Let $k$ rows be repainted first, then $k$ columns. After the first stage of repainting, each column will contain $k$ black and $101-k$ white cells. Since $101-k$ columns will remain untouched, the total number of black cells in these columns will be $k(101-k)$. In each of the repainted columns,... | 5100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) In an $8 \times 8$ frame with a width of 2 cells (see figure), there are a total of 48 cells. How many cells are in a $254 \times 254$ frame with a width of 2 cells?
 | Answer: 2016.
Solution. First method. Cut the frame into four identical rectangles as shown in the figure. The width of the rectangles is equal to the width of the frame, i.e., 2 cells. The length of each rectangle is 2 less than the side of the frame: $254-2=252$ cells. Then the area of one rectangle is $2 \cdot 252=... | 2016 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) Replace each letter with a digit so that the operations performed horizontally and vertically are valid.
| $a b$ | + | $c d$ | $=$ | $e f$ |
| :---: | :---: | :---: | :---: | :---: |
| $\times$ | | $:$ | | + |
| $k$ | $:$ | $m$ | $=$ | $m$ |
| $m n d$ | $:$ | $e$ | $=$ | $e m$ |
Identical letters corr... | # Solution.
| 42 | + | 18 | $=$ | 60 |
| :---: | :---: | :---: | :---: | :---: |
| $\times$ | | $:$ | | + |
| 9 | $:$ | 3 | $=$ | 3 |
| 378 | $:$ | 6 | $=$ | 63 | | 63 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. A three-digit number, all digits of which are different, will be called balanced if it is equal to the sum of all possible two-digit numbers formed from the different digits of this number. Provide an example of any balanced number. Justify your answer. | Solution: For example, the number $132=13+12+32+21+31+23$ is balanced (there are other options).
Grading criteria: Any suitable number with verification - 7 points, incorrect solution or only answer - 0 points. | 132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In the tournament, each participant was supposed to play exactly one game with each of the remaining participants, but two participants dropped out during the tournament, having played only 4 games each. In the end, the total number of games played turned out to be 62. How many participants were there in total? | Answer: 13.
Solution: Let the total number of participants be $n$. Then, excluding the two who dropped out, the remaining participants played $\frac{(n-2)(n-3)}{2}$ matches. If these two managed to play against each other, then 7 matches were played with their participation, and if they did not, then 8 matches. Thus, ... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. In a school quiz, 100 students participated. After the results were tallied, it turned out that any 66 of them together earned no less than $50 \%$ of the total prize points. What is the highest percentage of points that one participant could have earned? | Answer: $25 \%$.
Solution: Suppose participant $\mathrm{X}$ scored the highest percentage of points $-x \%$. Divide the remaining participants into three groups A, B, and C, each with 33 people. Let the total percentages of points scored by these groups be $a, b$, and $c$ respectively. Then,
$$
2(100-x)=2(a+b+c)=(a+b... | 25 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
# 3. CONDITION
There is an unlimited number of chips in six colors. What is the smallest number of chips that need to be arranged in a row so that for any two different colors, there are two adjacent chips of these colors in the row? | Solution. From the condition, it follows that for each fixed color A, a chip of this color must be paired with a chip of each of the other 5 colors. In a row, a chip has no more than two neighbors, so a chip of color A must appear at least 3 times. Similarly for each other color. Thus, there should be no less than $3 \... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. Misha suggested that Yulia move a chip from cell $A$ to cell $B$. In one step, you can move the chip to an adjacent cell by side or by corner. To make it more interesting, Misha put 30 candies in the prize fund, but said that he would take 2 candies for each horizontal or vertical move and 3 candies for ea... | Answer: 14.
Solution. From $A$ to $B$, one can get through the top or the bottom. If going through the top, the first 2 moves are diagonal (a diagonal move is more advantageous than 2 horizontal moves), and the next 5 moves are horizontal. Misha will take $2 \cdot 3 + 5 \cdot 2 = 16$ candies, and Yulia will win 14. If... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4. The diagonals of quadrilateral $A B C D$ intersect at point $K$. It turns out that $A B=B K=K D$. On segment $K C$, a point $L$ is marked such that $A K=L C$. Find $\angle B L A$, given that $\angle A B D=52^{\circ}$ and $\angle C D B=74^{\circ}$.
. We have
$$
\begin{aligned}
\angle BLA & =180^{\circ}-\angle BAL-\angle ABL=180^{\circ}-\frac{180^{\circ}-\angle ABD}{2}-\angle CDB= \\
& =90^{\circ}+\frac{1}{2} \angle ABD-\angle CDB=42^{\circ... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.6. Six princesses have a magic chest. Every minute, a dress of one of 10 colors and one of 9 styles can be taken out of it. However, within one hour, it is impossible to take out two dresses from the chest that match both in color and style. What is the minimum number of dresses the princesses will have to ta... | Answer: 46.
Solution. Note that 45 dresses would not be enough, as the chest can issue exactly 5 dresses of each of the 9 styles.
We will prove that if the princesses take out 46 dresses, there will definitely be 6 dresses of the same style, and thus of different colors (identical dresses could not have occurred over... | 46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. Along an alley, maples and larches were planted in one row, a total of 75 trees. It is known that there are no two maples between which there are exactly 5 trees. What is the maximum number of maples that could have been planted along the alley? | Answer: 39.
Solution. Let's divide all the trees into groups of 12 standing in a row. There will be 6 complete groups and 3 more trees at the end. In each group, we will divide the trees into 6 pairs: the first with the seventh, the second with the eighth, ..., the sixth with the twelfth. Note that there are exactly 5... | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.8. In trapezoid $A B C D(A D \| B C) \angle A B C=108^{\circ}$ and $\angle A D C=54^{\circ}$. On ray $B A$ beyond point $A$, point $K$ is marked such that $A K=B C$. Find the angle $D K C$, given that $\angle B K C=27^{\circ}$.
Fig. 1: to the solu... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Before the lesson, Nestor Petrovich wrote several words on the board. When the bell rang for the lesson, he noticed a mistake in the first word. If he corrects the mistake in it, the words with mistakes will make up $24 \%$, and if he erases the first word from the board altogether, the words with mistakes will make... | Answer: $28 \%$.
## Solution:
Let there be $n$ words written on the board before the lesson, of which $x$ have errors. If the error in the first word is corrected, then there will be $x-1$ words with errors out of $n$, and by the condition $x-1=0.24 n$. If a word with an error is erased, then there will be $x-1$ word... | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3. The integer 23713 has the following two properties:
(1) any two adjacent digits form a prime two-digit number,
(2) all these prime two-digit numbers are pairwise distinct.
Find the largest of all integers with properties (1) and (2). | Answer: 617371311979.
Solution. Since even digits and the digit 5 can only be in the highest place (and only one of these digits can form a single prime number), and the remaining prime two-digit numbers are $11, 13, 17, 19, 31, 37, 71, 73, 79$, 97, the maximum number of digits in the desired number is 12. In the desi... | 617371311979 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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