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7.1. Lesha wrote down a number, and then replaced the same digits with the same letters, and different digits with different letters. He ended up with the word NOVOSIBIRSK. Could the original number be divisible by 3? | Answer: Yes, it could.
Solution: For example, 10203454638.
Criteria: Any correct example without verification - 7 points. | 10203454638 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,704 |
7.2. There are broken scales that are off by no more than 500g in their readings. When Alexei weighed the melon, they showed 4kg. When he weighed the watermelon - 3 kg. When he weighed both the melon and the watermelon together - 8.5 kg. What is the actual weight of the melon and the watermelon separately? The scales c... | Answer: The melon weighs 4.5 kg, and the watermelon weighs 3.5 kg.
Solution: From the condition, it follows that the melon weighs no more than 4.5 kg. The watermelon weighs no more than 3.5 kg. Therefore, together they weigh no more than 8 kg. On the other hand, from the second condition, it follows that together they... | The\melon\weighs\4.5\,\\the\watermelon\weighs\3.5\ | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,705 |
7.3. Danil drew several lines through one point. From all the angles formed, he began to consider only angles with integer degree measures. Danil claims that among them, there are exactly 15 more angles with odd measures than with even measures. Can this be true? Answer: No, it cannot. | Solution: Let's divide all angles into pairs that are vertical to each other. In each pair, the angles are equal to each other, so the number of even angles is even, as there is an integer number of pairs. Similarly, the number of odd angles is even, so the difference between the number of even and odd angles is also e... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,706 |
7.4. The steamship "Raritet" after leaving the city moves at a constant speed for three hours, then drifts for an hour, moving with the current, then moves for three hours at the same speed, and so on. If the steamship starts its journey from city A and heads to city B, it takes 10 hours. If it starts from city B and h... | Answer: 60 hours.
Solution: Let the speed of the steamboat be $U$, and the speed of the river be $V$. When the steamboat travels from B to A, it arrives at the destination just before the fourth engine stop, meaning it travels 12 hours at a speed of $U-V$ towards A, and then 3 hours back at a speed of $V$ towards B. T... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,707 |
7.5. A square box 3 by 3 is divided into 9 cells. It is allowed to place balls in some cells (possibly a different number in different cells). What is the minimum number of balls that need to be placed in the box so that each row and each column of the box contains a different number of balls? | Answer: 8.
Solution: Add up the number of balls in all rows and columns. Since these are 6 different non-negative numbers, this sum is at least $0+1+\ldots+5=15$.
Now notice that the sum of the numbers in the rows is equal to the sum of the numbers in the columns, since these sums are equal to the total number of bal... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,708 |
10.1. Find all natural numbers $n$ such that $\frac{1}{n}=\frac{1}{p}+\frac{1}{q}+\frac{1}{p q}$ for some primes $p$ and $q$ | Answer. $n=1$.
Solution. Bringing the expression in the condition to a common denominator, we get: $n(p+q+1)=pq$. From the simplicity of $p$ and $q$, it follows that the divisors of the right-hand side can only be the numbers $1, p, q$, and $pq$, one of which must equal $p+q+1$. Since $1, p$, and $q$ are less than $p+... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,709 |
10.2. On a coordinate plane, a grasshopper jumps starting from the origin. The first jump, one centimeter long, is directed along the OX axis, each subsequent jump is 1 cm longer than the previous one, and is directed
perpendicular to the previous one in either of two directions at its choice. Can the grasshopper end u... | Answer: Yes.
Solution. The grasshopper will make 16 horizontal jumps of odd lengths $1,3,5, . ., 31$ cm. We will divide them into 4 quartets of consecutive odd numbers, in each quartet the first and last jumps will be made to the right, and the second and third - to the left. After every four such jumps, the grasshopp... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,710 |
10.4. Find all functions $f(x)$ defined on the entire real line that satisfy the equation $f(y-f(x))=1-x-y$ for any $x$ and $y$. | Answer. $f(x)=\frac{1}{2}-x$.
Solution. 1) First, let $y=f(x)$ in the condition, then $f(x)=1-x-f(0)$.
2) Substitute the obtained expression for $f(x)$ into the condition, then: $f(y-f(x))=1-y+f(x)-f(0)=1-y+1-x-f(0)-f(0)=1-x-y, \quad$ from which $\quad f(0)=\frac{1}{2}$.
Therefore, $f(x)=\frac{1}{2}-x$ is the only c... | f(x)=\frac{1}{2}-x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,711 |
10.5. Can five consecutive natural numbers be found such that if they are denoted by the letters $a, b, c, d, e$ in some order, the equality $(a+b)(b+c)(c+d)(d+e)(e+a)=(a+c)(c+e)(e+b)(b+d)(d+a)$ holds? | Answer: No.
Solution: Suppose such numbers exist, denote them as $x-2, x-1, x, x+1, x+2$ for some natural number $x \geq 3$. Notice that the ten numbers in parentheses on both sides of the equation in the problem are all possible pairwise sums of the numbers $a, b, c, d, e$, that is, the pairwise sums of the numbers $... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,712 |
11.1. Can both numbers $\cos x + \sqrt{2}$ and $\cos 2x + \sqrt{2}$ be rational for some value of $x$ | Answer. No.
Solution. Let these numbers be $a$ and $b$ respectively. Then
$b-\sqrt{2}=2(a-\sqrt{2})^{2}-1=2a^{2}+3-4a\sqrt{2}$, from which $b-2a^{2}-3=(1-4a)\sqrt{2}$. Due to the rationality of $a$ and $b$, the last equality is possible only if $1-4a=0$, from which $a=\frac{1}{4}$, but in this case $\cos x=\frac{1}{4... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,713 |
11.2. Solve the system of equations in real numbers:
$$
x^{2}+x y+y^{2}=4, x^{4}+x^{2} y^{2}+y^{4}=8
$$ | Answer. $\pm\left(\sqrt{\frac{3 \pm \sqrt{5}}{2}}, \sqrt{\frac{3 \mp \sqrt{5}}{2}}\right)$ - a total of 4 solutions.
## Solution. Consider the cases.
1) $x=y$, then $3 x^{2}=4,3 x^{4}=8$, from which $x^{2}=2$, which clearly does not satisfy both equations. No solutions.
2) $x=-y$, then $x^{2}=4,3 x^{4}=8$, from which... | \(\sqrt{\frac{3\\sqrt{5}}{2}},\sqrt{\frac{3\\sqrt{5}}{2}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,714 |
11.5. Find all natural $n$ for which all natural numbers from 1 to $n$ inclusive can be arranged in a row in such an order that the sum of the first $k$ numbers from the left will either divide the sum of all $n-k$ remaining numbers or be divisible by it for any $k$ from 1 to $n-1$. | Answer: $n=3,4,5$.
Solution. Examples for these numbers: $\{1,2,3\},\{1,4,3,2\},\{1,4,5,2,3\}$.
Let $a_{1}, a_{2}, \ldots, a_{n}$ be all numbers from 1 to $n$, written in the required order. Denote their sum by $S$, and consider the maximum $k$ such that $A_{k}=a_{1}+a_{2}+\ldots+a_{k}$ does not exceed $\frac{S}{2}$.... | 3,4,5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,715 |
7.1. A kilogram of meat with bones costs 165 rubles, a kilogram of meat without bones costs 240 rubles, and a kilogram of bones costs 40 rubles. How many grams of bones are there in a kilogram of meat with bones? | Answer: 375 grams.
Solution: Let $x$ kilos of bones be in a kilogram of meat with bones. Then $40 x + 240(1-x) = 165$, from which $x=0.375$. | 375 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,716 |
7.2. In a family, there are three brothers. It is known that Kolya was born the year after the year when there were 5 years left until the birth of the youngest of the brothers, and Vanya was born two years earlier than when the middle one turned 3 years old. Now Petya is 10 years old. How old are Vanya and Kolya? | Answer: Vanya is now 9 years old, and Kolya is 13 years old.
Solution: From the second condition about Vanya, it follows that he is 1 year younger than the middle brother, so Vanya is the youngest. Kolya is 4 years older than the youngest, so he is the oldest. Therefore, Petya is the middle one, which means Vanya is n... | Vanya\is\9\\old,\\Kolya\is\13\\old | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,717 |
7.3. From the highway, four roads sequentially lead to four villages A, B, C, D. It is known that the route by road/highway/road from A to B is 9 km, from A to C - 13 km, from B to C - 8 km, from B to D - 14 km. Find the length of the route by road/highway/road from A to D. Explain your answer. | Answer: 19 km.
Solution. Let's add the distances from $A$ to $C$ and from $B$ to $D$. Then the highway segment from the turn to $B$ to the turn to $C$ will be counted twice, while the highway segments from the turn to $A$ to the turn to $B$ and from the turn to $C$ to the turn to $D$, as well as the four roads from th... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,718 |
7.4. What can be the sum of the digits of a number divisible by 7? | Answer. Any natural number, greater than or equal to 2.
Solution. Note that the numbers 21 and 1001 are divisible by 7, and the sums of their digits are 3 and 2, respectively. Therefore, to get a sum of digits equal to an even number $n$, we need to take a number whose decimal representation consists of $\frac{n}{2}$ ... | Anynatural,greaterthanorequalto2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,719 |
7.5. On a 5 by 5 grid, several cells are marked in such a way that every 3 by 3 square of cells contains exactly one marked cell. How many cells can be marked | Answer. Any number from 1 to 4.
Solution. It is clear that there must be at least one cell. On the other hand, four corner 3x3 squares completely cover the 5x5 square, and each of them contains exactly 1 marked cell, so the total number of marked cells is no more than 4. Examples for quantities from 1 to 4 are shown i... | 1to4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,720 |
9.1. What is the maximum number of different rectangles that an 8 by 8 chessboard can be cut into? All cuts must follow the grid lines. Rectangles are considered different if they are not equal as geometric figures. | Answer: 12.
Solution. Let's list the possible sizes of different integer rectangles of minimal areas that can fit along the grid lines on an 8 by 8 board in ascending order of these areas: 1 by 1, 1 by 2, 1 by 3, 1 by 4, 2 by 2, 1 by 5, 1 by 6, 2 by 3, 1 by 7, 1 by 8, 2 by 4, 3 by 3, 2 by 5. There are already 13 recta... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,721 |
9.2. Can the points of intersection of the angle bisector of angle A, the altitude drawn from vertex B, and the median drawn from vertex C in some acute-angled triangle ABC be the vertices of a non-degenerate equilateral triangle? | Answer: No.
Solution. Let the intersection point of the angle bisector of angle A and the altitude BK from vertex B be M, the intersection of the altitude from vertex B and the median from vertex C be P, and the intersection of the angle bisector of angle A and the median from vertex C be T. Suppose these points are d... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,722 |
9.3. Let the two-digit numbers ${ }^{\overline{a b}}$ and ${ }^{\overline{c d}}$ be such that the ratio of the four-digit number $\overline{a b c d}$
to the sum $\overline{a b}+\overline{c d}$ is an integer. Find all possible values that this number can take. | Answer. All natural numbers from 11 to 90 inclusive.
Solution. Let the ratio in question be denoted by ${ }^{n}$, and the numbers $\overline{a b}$ and $\bar{c} \bar{d}$ by $A$ and $C$ respectively. Then $\overline{a b c d}=100 \cdot A+C=n(A+C)$, from which $\frac{100-n}{n-1}=\frac{C}{A}$. Note that $C \leq 99, A \geq ... | Allnaturalfrom11to90inclusive | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,723 |
9.4. It is known that the values of the quadratic trinomial $a x^{2}+b x+c$ on the interval $[-1,1]$ do not exceed 1 in absolute value. Find the maximum possible value of the sum $|a|+|b|+|c|$. Answer. 3. | Solution. Substituting the values $x=0,1,-1$ from the interval $[-1,1]$ into the polynomial $a x^{2}+b x+c$, we obtain three inequalities: $-1 \leq c \leq 1$, $-1 \leq a+b+c \leq 1$, and $-1 \leq a-b+c \leq 1$. Adding the second and third inequalities, we also get $-1 \leq a+c \leq 1$. Subtracting the second from the t... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,724 |
9.5. What is the maximum number of integers that can be written in a row so that the sum of any five consecutive ones is greater than zero, and the sum of any seven consecutive ones is less than zero? | Answer. Ten.
Solution. The sum of any seven written numbers is negative, while the sum of the five outermost numbers from this seven is positive, which means the sum of the two leftmost and the sum of the two rightmost numbers from this seven are negative. Therefore, the sum of any two adjacent numbers, with at least ... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,725 |
7.2. A rectangle is cut into several rectangles, the perimeter of each of which is a number of meters divisible by 4. Is it true that the perimeter of the original rectangle is divisible by 4? | Answer: No.
Solution: For example, a rectangle with sides 1 and 2 can be cut into squares with side 1. The perimeters of the squares are 4, while the perimeter of the rectangle is 6.
Criterion: If a correct counterexample is found but the verification is not performed - 7 points. | No | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,727 |
7.3. In a family, there are six children. Five of them are respectively 2, 6, 8, 12, and 14 years older than the youngest, and the age of each child is a prime number. How old is the youngest? | Answer: $5,7,11,13,17$ and 19
Solution: The remainders from dividing the age differences by 5 are 2, 1, 3, 2, and 4, respectively. Therefore, if the age of the youngest does not divide by 5, then the age of some other child does. Since all numbers are prime, this number is equal to 5. Only the second child fits, other... | 5,7,11,13,17,19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,728 |
7.4. On an island, there lives an odd number of people, each of whom is either a knight, who always tells the truth, or a liar, who always lies. One day, all the knights declared: “I only befriend 1 liar,” while all the liars said: “I do not befriend knights.” Who is more numerous on the island, knights or liars? | Answer: There are more knights
Solution: Each liar is friends with at least one knight. But since each knight is friends with exactly one liar, two liars cannot have a common friend who is a knight. Therefore, each liar can be matched with their friend who is a knight, which means there are at least as many knights as... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,729 |
7.5. There are 100 boxes numbered from 1 to 100. One of the boxes contains a prize, and the host knows where it is. The audience can send the host a batch of notes with questions that require a "yes" or "no" answer. The host shuffles the notes in the batch and, without reading the questions aloud, honestly answers all ... | Answer: 99.
Solution: To be able to definitively determine which of the 100 boxes contains the prize, it is necessary to have the possibility of receiving at least 100 different answers to one set of questions. Since the host's answers for different prize positions can only differ by the number of "yes" responses, it ... | 99 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,730 |
10.1. Solve the equation: $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=\sqrt{2}$. | Answer: $\frac{1}{2} \leq x \leq 1$.
Solution. First, we find the domain of the equation, which is $x \geq \frac{1}{2}$ from the inner radical. The condition $x \geq \sqrt{2 x-1}$ is also satisfied, as both sides are non-negative and after squaring, it is equivalent to the inequality $(x-1)^{2} \geq 0$.
Both sides of... | \frac{1}{2}\leqx\leq1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,731 |
10.2. Find the number of all five-digit numbers $\overline{a b c d e}$, all digits of which are distinct and $ad>e$. | Answer: 1134.
Solution. For the correct notation of a number satisfying the condition of the problem, one needs to arbitrarily select a quintet of different digits from 10 possible ones, and then arrange two of them to the left of the maximum in ascending order and the two remaining to the right of the maximum in desc... | 1134 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,732 |
10.5. Find all natural numbers $\boldsymbol{n}$ such that $\boldsymbol{n}$ is equal to the sum of three numbers, the first of which is the largest divisor of the number $\boldsymbol{n} \boldsymbol{- 1}$, different from $\boldsymbol{n}-1$, the second is the largest divisor of the number $\boldsymbol{n}-\mathbf{2}$, diff... | Answer: 58 and 66.
Solution. The maximum divisor of a number, different from the number itself, is equal to the number divided by its smallest prime divisor. 1) If $n$ is odd, then the smallest prime divisors of the numbers $n-1$ and $n-3$ are 2, and the smallest prime divisor of $n-2$ is denoted by $p$ - an odd prime... | 5866 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,733 |
10.3. Prove that for any $0 \leq x, y \leq 1$ the inequality $\frac{x}{1+y}+\frac{y}{1+x} \leq 1$ holds.
Proof 1. Replace the ones in the denominators of the fractions on the left side of the inequality with $0 \leq x \leq 1$ and $0 \leq y \leq 1$ respectively. In this case, the denominators of the fractions will not ... | Answer. $A P M=90^{\circ}$.
Solution. Mark a point T on the extension of CM beyond M such that MT=MP. In this case, segments $\mathrm{AB}$ and TP are bisected by their intersection point M, so quadrilateral ATBP is a parallelogram. In particular, segments AP and BT are equal and parallel. Consider triangles $\mathrm{B... | 90 | Inequalities | proof | Yes | Yes | olympiads | false | 9,734 |
9.1. Petya wrote 10 integers on the board (not necessarily distinct).
Then he calculated the pairwise products (that is, he multiplied each of the written numbers by each other). Among them, there were exactly 15 negative products. How many zeros were written on the board? | Answer: 2.
Solution. Let there be $A$ positive numbers and $B$ negative numbers on the board. Then $A+B \leq 10$ and $A \cdot B=15$. Since a negative product is obtained when we multiply a negative and a positive number. From this, it is easy to understand that the numbers $A$ and $B$ are 3 and 5 (1). Therefore, $A+B=... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,735 |
9.2. In a trapezoid, one lateral side is twice as large as the other, and the sum of the angles at the larger base is 120 degrees. Find the angles of the trapezoid. | Answer: 90 and 30 degrees.
Solution: Let the vertices of the trapezoid be A, B, C, D, with the larger base AD, and assume CD is twice as long as AB. Choose a point E on AD such that BE is parallel (and equal) to CD, and let M be the midpoint of segment BE. Then triangle ABM is isosceles with a 60-degree angle at verte... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,736 |
9.4. $N$ different natural numbers, not exceeding 1000, are written in a circle such that the sum of any two of them, standing one apart, is divisible by 3. Find the maximum possible value of $N$.
# | # Answer: 664.
Solution. Consider the remainders of the numbers when divided by 3. Divisibility by 3 means that in each pair of numbers standing one apart, either both numbers are divisible by 3, or one has a remainder of 1 and the other has a remainder of 2 when divided by 3. Among the numbers from 1 to 1000, 333 are... | 664 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,737 |
9.5. On a chessboard of size 8 by 8, 8 domino pieces are arbitrarily placed, each occupying two adjacent cells. Different dominoes do not share any cells. Prove that there will always be a 2 by 2 square on the board where none of the cells are covered by a domino. Is this true if 9 dominoes are placed on the board? | Answer. For 9 dominoes, this is incorrect.
Solution. On an 8 by 8 board, a 2 by 2 square can be chosen in 49 ways, and each domino has at least one common cell with a maximum of 6 2 by 2 squares. Therefore, 8 dominoes "cover" cells in a maximum of 48 2 by 2 squares, so there will be at least one 2 by 2 square where no... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,738 |
11.1. Find all natural numbers $n$ such that there exist $n$ consecutive natural numbers whose sum is equal to $n^{2}$. | Answer. All odd $n$.
Solution. Let $n$ satisfy the condition and the first of $n$ consecutive natural numbers be $x+1$, then the last is $x+n$, and their sum is $n\left(x+\frac{n+1}{2}\right)$, from which $n=2 x+1$ - odd. On the other hand, for any odd $n$ the sum of $n$ consecutive
natural numbers $\frac{n+1}{2}, \f... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,739 |
11.2. Find all solutions of the equation: $\cos ^{2} x+\cos ^{2} 2 x+\cos ^{2} 3 x=1$. | Answer. $x= \pm \frac{\pi}{2}+2 k \pi, \pm \frac{\pi}{4}+\frac{l \pi}{2}, \pm \frac{\pi}{6}$
Solution. One can, acting straightforwardly, replace
after transformations we get: $8 \cos ^{6} x-10 \cos ^{4} x+3 \cos ^{2} x=0$, from which $\cos x=0, \pm \frac{1}{\sqrt{2}}, \pm \frac{\sqrt{3}}{2}$. Therefore, $x= \pm \fra... | \\frac{\pi}{2}+2k\pi,\\frac{\pi}{4}+\frac{\pi}{2},\\frac{\pi}{6}+\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,740 |
11.3. For what smallest $n$ is the following condition satisfied: if $n$ crosses are placed in some cells of a $6 \times 6$ table in any order (no more than one per cell), then there will definitely be three cells forming a strip of length 3, either vertical or horizontal, each containing a cross? | Answer. $n=25$.
Solution. If there are no fewer than 25 crosses, then one of the rows of the table contains no fewer than 5 crosses, and no more than one empty cell. Then either the three left cells of this row, or the three right cells of it, all contain crosses and form the desired strip.
If there are fewer than 25... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,741 |
11.4. Find all natural numbers $x$ such that the product of all digits in the decimal representation of $x$ equals $x^{2}-10 x-22$ | Answer: $x=12$.
Solution: First, the product of all digits of a natural number is non-negative, so $x^{2}-10 x-22$, from which $x \geq \frac{10+\sqrt{188}}{2}$, that is, $x \geq 12$. Second, if in the product of all digits of a natural number, all digits except the first are replaced by tens, the product will not decr... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,742 |
8.1. Lesha wrote down a number, and then replaced the same digits with the same letters, and different digits with different letters. He ended up with the word NOVOSIBIRSK. Could the original number be divisible by 9? | Answer: Yes, it could.
Solution: For example, 10203454638.
Criteria: Any correct example without verification - 7 points. | 10203454638 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,743 |
8.2. In triangle $A B C$, points $M$ and $N$ are taken on sides $B C$ and $A C$ respectively. Segments $A M$ and $B N$ intersect at point $O$. Prove that the sum of angles $A M B$ and $A N B$ is greater than angle $A O B$. | Solution: Consider the quadrilateral NOMC. The sum of its angles
is
$$
360^{\circ}=\angle M O N+\angle O N C+\angle N C M+\angle C M O=\angle A O B+180^{\circ}-\angle A N B+\angle N C M+180^... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,744 |
8.4. The steamship "Raritet" after leaving the city moves at a constant speed for three hours, then drifts for an hour, moving with the current, then moves for three hours at the same speed, and so on. If the steamship starts its journey from city A and heads to city B, it takes 10 hours. If it starts from city B and h... | Answer: 60 hours.
Solution: Let the speed of the steamboat be $U$, and the speed of the river be $V$. When the steamboat travels from B to A, it arrives at the destination just before the fourth engine stop, meaning it travels 12 hours at a speed of $U-V$ towards A, and then 3 hours back at a speed of $V$ towards B. T... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,746 |
8.5. A confectioner uses various spices when making buns. Once, he baked 10 buns, and it turned out that each of them contained more than half of the entire range of spices. Prove that it is possible to choose three spices in such a way that each bun will contain at least one of these spices. | Solution: We will construct a bipartite graph where the vertices of the first part are 10 buns, and the vertices of the second part are all $n$ spices. An edge between bun $k$ and spice $m$ will mean that spice $m$ was used to prepare bun $k$.
Let's calculate the sum of the degrees of the vertices in the first part. B... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,747 |
10.1. Find all four-digit numbers $\overline{x y z t}$, where all digits $x, y, z, t$ are distinct and not equal to 0, such that the sum of all four-digit numbers obtained from $\overline{x y z t}$ by all possible permutations of the digits is 10 times the number $\overline{x x x x}$. | Answer. The number 9123 and all numbers obtained from it by permuting the last three digits, a total of 6 answers.
Solution. The number of four-digit numbers obtained from $\overline{x y z t}$ by all possible permutations of the digits is 24, in each of which each of the digits $x, y, z, t$ appears exactly 6 times in ... | 9123 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,748 |
10.2. Find all solutions of the system of equations in real numbers:
$$
\left(\begin{array}{l}
x^{3}-x+1=y^{2} \\
y^{3}-y+1=x^{2}
\end{array}\right.
$$ | Answer.. ( $\pm 1, \pm 1)$ - a total of 4 solutions.
Solution. Move 1 to the right in each equation and multiply the equations, obtaining the equality $\left(x^{2}-1\right)\left(y^{2}-1\right) x y=\left(x^{2}-1\right)\left(y^{2}-1\right)$, from which either $x^{2}-1=0$, or $y^{2}-1=0$, or $x y=1$. In the first two cas... | (\1,\1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,749 |
10.3. Find all natural numbers $n$ such that a square of size $n$ by $n$ cells can be cut along the grid lines into a single-cell square and four rectangles, all nine side lengths of which are pairwise distinct. | Answer: All $n \geq 11$.
Solution. We will show that the cutting scheme must look like the one shown in the left figure (note that in the solution, this proof is not necessary; it is sufficient to simply indicate the scheme itself! (
, and then the central cell was marked in each of the resulting squares. Prove that an equal number of white and black cells were marked. | Solution: Let's take an arbitrary square with an odd side. Notice that in it, either there is one more black cell than white, or vice versa. Moreover, if there are more black cells, then the central cell is black, and if there are more white cells, then the central cell is white. Indeed, let's divide all rows into pair... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,757 |
8.1. Lada, Lera, and Lara were solving problems. It turned out that Lada solved as many more problems than Lera as Lera solved more than Lara. Could it be that they solved a total of 2015 problems together? | Answer: No.
Solution: Let Lara solve x problems, and Lera solve x + a problems. Then Lada solved $\mathrm{x}+2 \mathrm{a}$ problems, and all together the girls solved $3 x+3 a=2015$ problems. However, 2015 is not divisible by three, leading to a contradiction.
Criteria: Only answer - 0 points. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,758 |
8.2. A number was decreased by $1 \%$, then the remainder by $2 \%$, then the remainder by $3 \%$, and so on, until finally the remainder was decreased by $30 \%$. Another number was first decreased by $30 \%$, then the remainder by $29 \%$, and so on, until the remainder was decreased by $1 \%$. The results turned out... | Answer: they were equal.
Solution: If the first number was equal to a, then the final number is $0.99 * 0.98 * \ldots * 0.70 * a$. If the second number is $b$, then the final number is $0.70 * 0.71 * \ldots * 0.99 * b$. Since according to the condition $0.99 * 0.98 * \ldots * 0.70 * a = 0.70 * 0.71 * \ldots * 0.99 * b... | b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,759 |
8.3. In a parallelogram, the diagonals were drawn, and then the bisectors of all the angles formed by them were drawn to the intersection with the sides of the parallelogram. These points were named $A, B, C$ and $D$ respectively. Prove that $ABCD$ is a rhombus. | Solution: Note that the bisectors of adjacent angles form a right angle, and the bisectors of vertical angles form a straight line. Thus, the diagonals of quadrilateral $A B C D$ are perpendicular. In addition, a parallelogram has a center of symmetry - the point of intersection of the diagonals $O$. Under this symmetr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,760 |
8.4. Let's call the average magic of a set of numbers the ratio of their sum to their product. Initially, several (more than one) different natural numbers are written on the board. After the smallest of them is erased, the average magic of all the numbers written on the board increases threefold. Find what the numbers... | Answer: 4 and $12 ; 4,5,7$.
Solution 1: Let the original sum be $s+a$, the product be $ap$, and the smallest number be $a$. Then we get and simplify the following relationship:
$$
\begin{gathered}
3(s+a) / a p=s / p \\
3 s / a+3=s \\
1 / a+1 / s=1 / 3
\end{gathered}
$$
On one hand, this expression can be estimated a... | 412;4,5,7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,761 |
8.5. Dasha has 4 coins, one of which is counterfeit and differs in weight from the genuine ones. You are allowed to take two groups of coins and ask Dasha which one is lighter. If there is a lighter one, Dasha will point it out. If the groups turn out to be equal in weight, Dasha will point to one of the groups arbitra... | Solution: Let's number the coins. We will conduct three trials: 1 and 2 against 3 and 4, 1 and 3 against 2 and 4, 1 and 4 against 2 and 3. It is clear that each time Dasha will point to the lighter group, because all the coins participate in the weighing.
Without loss of generality, let's assume that the group 3 and 4... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,762 |
11.2. Can the bisectors of two adjacent exterior angles of a triangle (adjacent to one of its sides) intersect on its circumscribed circle? | Answer: They cannot.
Solution. Let's take the adjacent external angles of triangle $\mathrm{ABC}$, adjacent to its side $\mathrm{BC}$, and denote the intersection point of their bisectors as P. Denoting the measures of the angles of the triangle at vertices B and C by these letters themselves, we get that the measures... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,764 |
11.4. Find all natural numbers $n$ that can be represented as $n=\frac{x+\frac{1}{x}}{y+\frac{1}{y}}$, for some natural numbers $x$ and $y$. | Answer. $n=1$
Solution. Transform the equality in the condition to the form $n=\frac{\left(x^{2}+1\right) y}{\left(y^{2}+1\right) x}$. Note that the numbers $x^{2}+1$ and $x$, as well as the numbers $y^{2}+1$ and $y$ are coprime, so $x$ in the denominator can only cancel out with $y$ in the numerator, meaning $y$ is d... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,765 |
11.5. a) A square of size 1 by 1 is divided into 25 not necessarily identical rectangles, each of which has the same perimeter \( \boldsymbol{p} \). Find the minimum and maximum possible value of \( \boldsymbol{p} \). b) Can a unit square be divided into 30 not necessarily identical rectangles with a perimeter of 2? | Answer. a) The minimum value of $p$ is $0.8=4/5$, the maximum value of $p$ is $2.08=2+2/25$. b) Yes, it is possible, the method is indicated in the solution.
Solution. a) One of the rectangles in the partition must have an area of at least $1/25$. Let the sides of this rectangle be $x$ and $y$. By the inequality of th... | ) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,766 |
8.1 Roman wants to buy a football club, a yacht, and a small mansion. If he buys only the football club, he will have 2 billion left. If he buys only the yacht, he will have 3 billion left. If he buys only the mansion, he will have 6 billion left. Will Roman be able to buy the club, the yacht, and the mansion all at on... | Answer: will not be able to.
Solution: Let the mansion cost $x$ billion, then the yacht costs $x+3$ billion, and the club $x+4$, while Roman has $-x+6$ billion in total. Let's compare this amount with the total cost of the club, yacht, and mansion:
$$
\begin{gathered}
x+x+3+x+4>x+6 \\
3 x+7>x+6 \\
2 x+1>0
\end{gather... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,767 |
8.2 Sergey arranged several (more than two) pairwise distinct real numbers in a circle so that each number turned out to be equal to the product of its neighbors. How many numbers could Sergey have arranged? | Answer: 6.
Solution: Let's denote the two adjacent numbers as $a$ and $b$. Then, next to them stands $b / a$, followed by $1 / a, 1 / b, a / b$, and again $a$. Thus, it is impossible to arrange more than 6 numbers.
If 3 numbers can be arranged, then $a=1 / a$, which means $a$ is 1 or -1. In the first case, $b$ and $b... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,768 |
8.3 There are twenty-one visually identical coins, one of which weighs 9.9 g, two others weigh 9.8 g each, and all the rest weigh 10 g. Can at least one 10-gram coin be identified in two weighings using a balance scale without weights? | Answer: Yes.
Solution: Let's denote the coins by numbers from 1 to 20. Weigh coins 1 and 2 against coins 3 and 4. Among these four coins, there is at least one genuine coin.
In the case of balance, each pan either has two genuine coins or one genuine and one 9.8 g coin. These cases can be distinguished by the second ... | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,769 |
8.4 Given a triangle \(ABC\) and points \(D\) and \(E\) such that angles \(ADB\) and \(CEB\) are right angles. Prove that the length of segment \(DE\) is not greater than the semiperimeter of triangle \(ABC\). | Solution: Mark the midpoints of $AB$ and $BC$ - points $K$ and $L$. Notice that triangles $ADB$ and $CEB$ are right triangles, so the medians in them are equal to half the hypotenuse: $DK = AB / 2, EL = CB / 2$. Additionally, $KL = AC / 2$ as the midline. From this, we get that $DE \leq DK + KL + LE = AB / 2 + AC / 2 +... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,770 |
8.5 Mrs. Hudson and Dr. Watson each thought of a natural number. Each of them calculated the sum of all divisors of their number (including 1 and the number itself) and the sum of the reciprocals of the divisors. When Sherlock Holmes learned that both the first and the second sums of his friends were the same, he immed... | Solution: Let Mrs. Hudson have guessed the number $n$. We will prove that the first sum of Mrs. Hudson is $n$ times larger than the second. Let $a$ be a divisor of $n$ that is included in the first sum. Then the second sum includes the term $1 / a$ respectively. After increasing the second sum by $n$ times, the term $1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 9,771 |
8.1. A number is called good if any two adjacent digits in its notation differ by at least 4. Vera wrote some good number, and then replaced identical digits with identical letters, and different ones with different letters. Could she have ended up with the word NOVOSIBIRSK? | Answer: For example, the number 82729161593 could work ( $\mathrm{H}=8, \mathrm{O}=2, \mathrm{~B}=7, \mathrm{C}=9$, I = 1, B = 6, $\mathrm{P}=5, \mathrm{~K}=3$).
Criterion: any valid example without verification - 7 points. | 82729161593 | Other | math-word-problem | Yes | Yes | olympiads | false | 9,772 |
8.2. Matvey left from Totma, and at the same time, Petr ran towards him from Kaluga along the same road. Under identical conditions, the boys' speeds are in the ratio of $2: 3$ and are constant. At some point, they encounter a rough road (possibly at different times) that continues until the moment they meet. On the ro... | Answer: The distances traveled off-road by Matvei and Petr are in the ratio of $1: 3$.
Solution: If the off-road terrain on Matvei's side were more or equal to that on Petr's side, Petr's speed would always be higher, and the meeting would not have occurred halfway. Therefore, Petr walked on off-road terrain for a lon... | 1:3 | Other | math-word-problem | Yes | Yes | olympiads | false | 9,773 |
8.3. Find the angle $D A C$, given that $A B=B C$ and $A C=C D$, and the lines on which points $A, B, C, D$ lie are parallel, with the distances between adjacent lines being equal. Point $A$ is to the left of $B$, $C$ is to the left of $B$, and $D$ is to the right of $C$ (see figure).
 + 8 + 9, k \geq 5, n \geq 21 \), in the second case - \( n = 4k + 3 = 4(k-3) + 6 + 9, k \geq 4, n \geq 19 \). On the other hand, the three smallest compos... | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,778 |
9.3. In an acute-angled triangle $A B C$, a point $H$ is chosen such that the radii of the circumcircles of triangles $A H B, B H C$, and $C H A$ are equal. Prove that $H$ is the orthocenter of triangle $A B C$. | Solution. Let the midpoints of segments $A H, B H$, and $C H$ be denoted by $P, Q$, and $R$ respectively, and the centers of the circumcircles of triangles $A H B, B H C$, and $C H A$ by $O_{1}, O_{2}, O_{3}$ respectively. From the condition, it follows that the quadrilaterals $\mathrm{AO}_{1} \mathrm{HO}_{3}, \mathrm{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,779 |
9.5. In the cells of an 8 by 8 board, tokens are placed such that for each token, the row or column of the board in which it lies contains no more than three tokens. What is the maximum possible number of tokens on the board? | Answer: 30.
Solution: By swapping the verticals and horizontals, we assume that only the left $x$ verticals and the bottom $y$ horizontals contain more than 3 chips. From the condition, it follows that there are no chips at all in the lower left rectangle at the intersection of these verticals and horizontals. Each ve... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,781 |
7.1. Prove that if $a+\frac{1}{a}$ is an integer, then $a^{2}+\frac{1}{a^{2}}$ is also an integer. | Solution: It is clear that $\left(a+\frac{1}{a}\right)^{2}$ is an integer, but $\left(a+\frac{1}{a}\right)^{2}=a^{2}+\frac{1}{a^{2}}+2$.
That is, $\left(a+\frac{1}{a}\right)^{2}-2=a^{2}+\frac{1}{a^{2}}-$ is an integer. This completes the proof. | proof | Algebra | proof | Yes | Yes | olympiads | false | 9,782 |
7.2. Can a plane be painted in 2016 colors in such a way that among the vertices of any triangle, there will be at least two different colors? | Solution: We will prove that all points of the same color lie on one straight line. Since there are more than 2016 lines on the plane, this will prove that it is impossible to do so, and there will inevitably be a triangle with three vertices of the same color.
Suppose there are points of the same color that do not li... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,783 |
7.3. Given a triangle $A B C$, side $A B$ is divided into 4 equal segments $A B_{1}=B_{1} B_{2}=B_{2} B_{3}=B_{3} B$, and side $A C$ into 5 equal segments $A C_{1}=C_{1} C_{2}=C_{2} C_{3}=C_{3} C_{4}=C_{4} C$. How many times larger is the area of triangle $A B C$ compared to the sum of the areas of triangles $C_{1} B_{... | Answer: 2 times.
Solution: Let the area of $A B_{1} C_{1}$ be $S$. Then the area of $B_{1} C_{1} C_{2}$ is also $S$, since $B_{1} C_{1}$ is the median in $A B_{1} C_{2}$. Similarly, the area of $B_{1} B_{2} C_{2}$ is $2S$, as $C_{2} B_{1}$ is the median in $A B_{2} C_{2}$. The area of $B_{2} C_{2} C_{3}$ is half the a... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,784 |
7.4. Masha and Misha set out to meet each other simultaneously, each from their own house, and met one kilometer from Masha's house. Another time, they again set out to meet each other simultaneously, each from their own house, but Masha walked twice as fast, and Misha walked twice as slow as the previous time. This ti... | # Answer: 3 km.
Solution: We will prove that they spent the same amount of time on the first and second occasions. Suppose this is not the case, and they spent less time on the second occasion. Then, on the first occasion, Masha walked 1 km, and on the second occasion, less than 2 km (her speed was twice as high, but ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,785 |
7.5. Given the number 1836549, you can take two adjacent non-zero digits and swap their places, after which you subtract 1 from each of them. What is the smallest number that can result from these operations? | Answer: 1010101
Solution: The digits in the number alternate in parity: odd, even, etc. Note that with the described operation, even and odd numbers swap places, and then 1 is subtracted from them, thereby not disrupting the order of parity. Thus, it is impossible to obtain a number less than 1010101 (in each place, t... | 1010101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,786 |
11.1. Find all positive integer solutions of the equation $(n+2)!-(n+1)!-(n)!=n^{2}+n^{4}$. Answer. $n=3$. | Solution. Rewrite the equation as $n!=\left(n^{*}\left(n^{2}+1\right)\right) /(n+2)$. Transforming the right side, we get $n!=n^{2}-2 n+5-10:(n+2)$. The last fraction will be an integer for $n=3$ and $n=8$, but the latter number is not a solution (substitute and check!)
Grading criteria. Acquiring extraneous solutions... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,787 |
11.2. Which of the numbers is greater, $2^{\sqrt{\log _{3}}}$ or $3^{\sqrt{\log _{2}}}$? | Answer. $3^{\sqrt{\log _{2} 3}}$ is greater than $2^{\sqrt{\log _{3} 2}}$.
Solution. Two is less than three, so $\log _{3} 2 < 1$ and $\sqrt{\log _{2} 3} > 1$, therefore, $3^{\sqrt{\log _{2} 3}} > 3^{1} = 3 > 2 > 2^{\sqrt{\log _{3} 2}}$, which is what we needed to prove.
Grading criteria. Approximate calculations on ... | 3^{\sqrt{\log_{2}3}}>2^{\sqrt{\log_{3}2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,788 |
11.3. A square with a side of 4 cm is divided by three parallel horizontal and three parallel vertical lines into 16 smaller squares with a side of 1 cm. The sides of these smaller squares, including those located on the boundary of the larger square, will be called unit segments. In how many ways can an orientation be... | Answer. $C_{20}^{10} \cdot C_{20}^{10}$.
Solution. In total, there will be 20 horizontal and 20 vertical segments. If we consider them to be located along the coordinate axes OX and OY, respectively, and denote the number of positively oriented horizontal and vertical segments by $x$ and $y$ respectively, then the coo... | C_{20}^{10}\cdotC_{20}^{10} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,789 |
11.4. Let $\mathrm{O}$ be the point of intersection of the diagonals of a convex quadrilateral $A B C D$, and let $P, Q, R$, $S$ be the points of intersection of the medians of triangles $A O B, B O C, C O D$ and $D O A$ respectively. Find the ratio of the areas of quadrilaterals $P Q R S$ and $A B C D$. | Answer: $\frac{2}{9}$
Solution: Let the midpoints of sides $AB, BC, CD,$ and $DA$ of quadrilateral $ABCD$ be denoted by $X, Y, Z,$ and $T$ respectively. By the property of medians, points $P, Q, R, S$ divide segments $OX, OY, OZ, OT$ in the ratio 2:1, counting from $O$. Therefore, the area of quadrilateral $PQRS$ is
... | \frac{2}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,790 |
11.5. The alphabet consists of $n$ letters. A word composed of these letters is called allowed if all adjacent letters in it are different and it is impossible to obtain a word of the form $a b a b$ by erasing letters from it, where the letters $a$ and $b$ are different. What is the maximum length that an allowed word ... | Answer: $2 n-1$.
Solution. We will prove by mathematical induction. For $n=1,2$ - it is obvious. In an arbitrary word $w$ of $n+1$ letters, consider the leftmost letter, call it $a$. If it does not appear again in $w$, the remaining part of $w$ is a word of $n$ letters and, by the induction hypothesis, has a length no... | 2n-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,791 |
7.1 Cut a $5 \times 5$ square into rectangles of $1 \times 3$ and $1 \times 4$.
# | # Solution:
| 1 | 1 | 1 | 1 | 2 |
| :--- | :--- | :--- | :--- | :--- |
| 4 | 5 | 6 | 7 | 2 |
| 4 | 5 | 6 | 7 | 2 |
| 4 | 5 | 6 | 7 | 2 |
| 4 | 3 | 3 | 3 | 3 |
Other solutions are possible! | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,792 |
7.2 To buy an apartment, you need to take either exactly 9 small, 6 medium, and one large loan, or exactly 3 small, 2 medium, and 3 large loans. How many only large loans would be required to buy an apartment? | Answer: 4 large loans.
Solution: Let's denote small, medium, and large loans by the letters m, c, and b, respectively. Rewrite the condition using these notations:
$$
9 \mathrm{M}+6 \mathrm{c}+\sigma=3 \mathrm{~m}+2 \mathrm{c}+3 b
$$
Simplifying, we get $6 \mathrm{~m}+4 \mathrm{c}=2$ b or $3 \mathrm{~m}+2 \mathrm{c}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,793 |
7.3 Among 9 coins, there are 4 counterfeit ones. All genuine coins weigh the same, while the counterfeit ones differ in weight from each other and from the genuine ones. Using only a balance scale without weights, find at least one genuine coin in four weighings. | Solution: Let's denote the coins with numbers from 1 to 9. We will weigh the coins in pairs: 1 and 2, 3 and 4, 5 and 6, 7 and 8. If in one of these weighings the scales balance, then both coins involved in it are genuine. Otherwise, there are no more than four genuine coins in the first four weighings, which means the ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,794 |
7.4 Diagonals $A C$ and $B D$ of a convex quadrilateral $A B C D$ intersect at point $O$. It is known that the perimeter of triangle $A B C$ is equal to the perimeter of triangle $A B D$. Additionally, the perimeter of triangle $A C D$ is equal to the perimeter of triangle $B C D$. Prove that $A O=O B$. | Solution: Let's write the equality of perimeters, add two equalities, and combine like terms:
$$
\begin{gathered}
A B + B C + A C = B D + A B + A D \\
A C + C D + A D = B C + C D + B D \\
A B + B C + A C + A C + C D + A D = B C + C D + B D + B D + A B + A D \\
A C + A C = B D + B D \\
A C = B D
\end{gathered}
$$
Usin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,795 |
7.5 In the cells of a 7 x $\square 7$ square, pluses and minuses are arranged. It is allowed to change all the signs to their opposites in any row or any column. Prove that by such actions, one can achieve that in each row and in each column, there are more pluses than minuses. | Solution: Consider any arrangement of pluses and minuses. If there is a column (or row) in which there are fewer pluses than minuses, then we will change all the signs in it (them) to the opposite. In this way, the number of pluses will increase. By doing this and so on, we will increase the number of pluses in the tab... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,796 |
9.1. It is known that the sum of the digits of number A is 59, and the sum of the digits of number B is 77. What is the minimum sum of the digits that the number A+B can have? | Answer. 1.
Solution. It is sufficient to consider A=9999995, B=999999990000005, then $\mathrm{A}+\mathrm{B}=1000000000000000$, the sum of the digits is 1 - the minimum possible.
Grading Criteria. Correct answer and example: 7 points. Presence of arithmetic errors: minus 1-2 points. Any other answer: 0 points. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,797 |
9.2. On an island, 20 people live, some of whom are knights who always tell the truth, and the rest are liars who always lie. Each islander knows for sure who among the others is a knight and who is a liar. When asked by a visitor how many knights live on the island, the first islander answered: "None," the second: "No... | Answer: 10.
Solution: If the first islander were a knight, he would have lied in his answer, which cannot be the case. Therefore, the first is a liar, and there are no more than 19 knights on the island. This means the twentieth islander told the truth, so he is a knight, and there is at least one knight on the island... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,798 |
9.3. Find the value of the expression $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}$, given that $\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}=5$ and $x+y+z=2$. | Answer: 7.
Solution. Transform: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{x+y+z}{y+z}+\frac{y+x+z}{x+z}+\frac{z+x+y}{x+y}-3=$ $=(x+y+z)\left(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}\right)-3=2 \cdot 5-3=7$.
Grading Criteria. Presence of arithmetic errors: minus 1-2 points. Correct answer calculated on some ex... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,799 |
8.1. The numbers from 1 to 2017 are arranged in some order in a circle. Can it happen that the sum of any three consecutive numbers is divisible by $2$? | Solution: Suppose the sum of any three consecutive numbers is even. Consider the quartet $a b c d$. Since $a+b+c$ and $b+c+d$ are even, $a$ and $d$ have the same parity. Therefore, any numbers that are 2, 5, 8, 11, ... places apart have the same parity. Now, let's fix any triplet of consecutive numbers. There are two p... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,800 |
8.2. In triangle $ABC$, angle $A$ is twice as small as angle $C$, and point $D$ on side $AC$ is the foot of the perpendicular from $B$. Prove that the difference between the segments into which $D$ divides $AC$ is equal to one of the sides of triangle $ABC$. | Solution: Reflect point $C$ with respect to $D$, obtaining point $K$. In triangle $B K C$, segment $B D$ is both a median and an altitude, so $B K = B C$ and angle $B K C$ equals angle $B C K$. However, angle $B K C$ is an exterior angle to triangle $A B K$, so angle $A B K$ equals angle $B A K$, hence $A K = B K$.
Th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,801 |
8.3. There is a steamship route between the cities of Dzerzhinsk and Lviv. Every midnight, a steamship departs from Dzerzhinsk, arriving exactly eight days later in Lviv. How many steamships will the steamship "Raritet" meet on its way to Dzerzhinsk if it departs from Lviv exactly at midnight and spends the same eight ... | Answer: 17.
Solution: As "Raritet" departs from Lviv, a steamer arrives there, which left Dzerzhinsk 8 days ago. By the time "Raritet" arrives at its final destination, 8 days have passed since the initial moment, and at this moment, the last steamer departs from Dzerzhinsk, which "Raritet" meets on its way. Thus, "Ra... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,802 |
8.4. On a square board with side $N$, where $N$ is some natural number, a chess bishop stands in the bottom-left corner. Alexei and Danil take turns moving this bishop, with Alexei moving first, and players are not allowed to move to cells the bishop has already visited (the bishop moves diagonally to any distance). Th... | Answer: $N$ odd - Danil wins, $N$ even - Alexey wins.
Solution: If $N$ is odd, let's describe Danil's game strategy and prove that he wins.
1) If Alexey ends up on the main diagonal with his move, we move to any cell on the same diagonal. This is always possible because initially, there is an even number of unvisited... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,803 |
8.5. Egor, Nikita, and Innokentiy took turns playing chess with each other (two play, one watches). After each game, the loser gave up their place at the board to the spectator (there were no draws). In the end, it turned out that Egor participated in 13 games, and Nikita in 27. How many games did Innokentiy play? | Answer: 14.
Solution: On the one hand, there were no fewer than 27 games. On the other hand, a player cannot skip two games in a row, meaning each player participates in at least every other game. Therefore, if there were at least 28 games, Egor would have participated in at least 14, which contradicts the condition. ... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,804 |
10.1. Find all numbers $a$ and $b$ for which the equality $|a x+b y|+|b x+a y|=|x|+|y|$ holds for all values of the variables $x$ and $\cdot y$. | Answer. $a= \pm 1, b=0$ or $a=0, b= \pm 1$, a total of four pairs of values.
Solution. Substitute $x=1, y=0$ into the formula from the condition, we get $|a|+|b|=1$. Then substitute $x=1, y=1$, from which $|a+b|=1$. Finally, by setting $x=1, y=-1$, we have $|a-b|=1$. From the last two equations, we get $a+b= \pm(a-b)$... | =\1,b=0or=0,b=\1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,805 |
10.2. Find all pairs of natural numbers $x$ and $y$ such that their least common multiple is equal to $1 + 2x + 3y$. | Answer: $x=4, y=9$ or $x=10, y=3$.
Solution. Let's first assume $x \leq y$. Note that $y$ cannot be divisible by $x$, otherwise the least common multiple of $x$ and $y$ would be $y$, which is less than $1+2x+3y$. In particular, $x>1$.
Next, the least common multiple of $x$ and $y$ is divisible by $x$ and $y$, so $1+2... | 4,9or10,3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,806 |
10.3. Find the number of different ways to place 8 rooks in the cells of an 8x8 chessboard such that every cell of the board is under attack by at least one of them. The rooks can attack each other, a rook attacks all cells in the row and column it is in, including the cell it occupies. | Answer: $2 \cdot 8^{8}-8$ !
Solution. We will prove that the placement of 8 rooks satisfies the condition if and only if either there is a rook in each row, or there is a rook in each column, or both simultaneously. The positions of the rooks in the rows or columns are arbitrary. The sufficiency of the given condition... | 2\cdot8^{8}-8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,807 |
1. Let's call a partition any side of some unit square on a grid sheet of paper. Can several non-intersecting partitions be marked in a 7 by 7 square of cells so that the projections of all marked partitions onto the horizontal and vertical sides of the square completely cover them? Partitions do not intersect if they ... | 9.1. Answer. Yes, for example, as shown in the figure on the right.
There are many other ways, including when all the marked partitions lie inside the square, it is sufficient to provide any one of them.
Evaluation remarks. Any correct answer is worth 7 points. If the projections of the partitions do not cover just o... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,808 |
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