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Problem 2. In a football tournament, seven teams played: each team played once with each other. In the next round, teams that scored thirteen or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next rou... | Answer: 4.
Solution. In total, the teams played $\frac{7 \cdot 6}{2}=21$ games, each of which awarded 2 or 3 points. Therefore, the maximum total number of points that all teams can have is $21 \cdot 3=63$. Hence, the number of teams $n$ advancing to the next stage satisfies the inequality $n \cdot 13 \leqslant 63$, f... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. For what least natural $k$ is the expression $2017 \cdot 2018 \cdot 2019 \cdot 2020+k$ a square of a natural number? | Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2018$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In a football tournament, seven teams played: each team played once with each other. In the next round, teams that scored twelve or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next round... | Answer: 5.
Solution. In total, the teams played $\frac{7 \cdot 6}{2}=21$ games, each of which awarded 2 or 3 points. Therefore, the maximum total number of points that all teams can have is $21 \cdot 3=63$. Hence, the number of teams $n$ advancing to the next stage satisfies the inequality $n \cdot 12 \leqslant 63$, f... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. For what least natural $k$ is the expression $2019 \cdot 2020 \cdot 2021 \cdot 2022 + k$ a square of a natural number? | Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2020$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 1. What is greater: 1 or $\frac{23}{93}+\frac{41}{165}+\frac{71}{143}$? | Answer: One is greater.
## First solution.
$$
\frac{23}{93}+\frac{41}{165}+\frac{71}{143}<\frac{23}{92}+\frac{41}{164}+\frac{71}{142}=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}=1 .
$$
## Second solution.
$$
\begin{aligned}
\frac{23}{93}+\frac{41}{165}+\frac{71}{143} & =\frac{23 \cdot 165 \cdot 143+93 \cdot 41 \cdot 143+93... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In a football tournament, eight teams played: each team played once with each other. In the next round, teams that scored fifteen or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next roun... | Answer: 5.
Solution. In total, the teams played $\frac{8 \cdot 7}{2}=28$ games, in each of which 2 or 3 points were at stake. Therefore, the maximum total number of points that all teams can have is $28 \cdot 3=84$. Hence, the number of teams $n$ that advance to the next stage satisfies the inequality $n \cdot 15 \leq... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. For what least natural $k$ is the expression $2018 \cdot 2019 \cdot 2020 \cdot 2021+k$ a square of a natural number | Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2019$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In a football tournament, six teams played: each team played once with each other. In the next round, teams that scored twelve or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next round? | Answer: 3
Solution. In total, the teams played $\frac{6 \cdot 5}{2}=15$ games, each of which awarded 2 or 3 points. Therefore, the maximum total number of points that all teams can have is $15 \cdot 3=45$. Hence, the number of teams $n$ that advance to the next stage satisfies the inequality $n \cdot 12 \leqslant 45$,... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. For what least natural $k$ is the expression $2016 \cdot 2017 \cdot 2018 \cdot 2019 + k$ a square of a natural number? | Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2017$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. In the surgical department, there are 4 operating rooms: I, II, III, and IV. In the morning, they were all empty. At some point, an operation began in operating room I, then after some time in operating room II, then after some more time in operating room III, and finally in operating room IV.
All four oper... | Answer: Only the duration of the operation in Operating Room IV can be determined.
Solution. First, let's prove that the durations of the operations in Operating Rooms I, II, and III cannot be determined uniquely. Indeed, it is not difficult to verify that if the durations of the operations are 70, 39, 33, 10 or 56, 5... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. For what greatest value of the parameter $a$ will the coefficient of $x^{4}$ in the expansion of the polynomial $\left(1-2 x+a x^{2}\right)^{8}$ be equal to -1540? | Answer: -5.
Solution. Applying the polynomial formula, we get
$$
\left(1-2 x+a x^{2}\right)^{8}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot 1^{n_{1}} \cdot(-2 x)^{n_{2}} \cdot\left(a x^{2}\right)^{n_{3}}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot(-2)^{n_{... | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. For what greatest value of the parameter $a$ will the coefficient of $x^{4}$ in the expansion of the polynomial $\left(1-3 x+a x^{2}\right)^{8}$ be equal to $70$? | Answer: -4.
Solution. Applying the polynomial formula, we get
$$
\left(1-3 x+a x^{2}\right)^{8}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot 1^{n_{1}} \cdot(-3 x)^{n_{2}} \cdot\left(a x^{2}\right)^{n_{3}}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot(-3)^{n_{... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (Option 1) The decimal representation of the natural number $n$ contains sixty-three digits. Among these digits, there are twos, threes, and fours. No other digits are present. The number of twos is 22 more than the number of fours. Find the remainder when $n$ is divided by 9. | # Answer. 5.
(Option 2) The decimal representation of a natural number $n$ contains sixty-one digits. Among these digits, there are threes, fours, and fives. No other digits are present. The number of threes is 11 more than the number of fives. Find the remainder when $n$ is divided by 9.
Answer. 8.
Criteria. "干" Th... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. For what values of the parameter $a$ does the equation
$$
3^{x^{2}-2 a x+a^{2}}=a x^{2}-2 a^{2} x+a^{3}+a^{2}-4 a+4
$$
have exactly one solution? | Answer: Only when $a=1$.
Solution. Let's denote $x-a$ by $t$. Note that the number of solutions of the equation does not change with such a substitution. Then the original equation will take the form
$$
3^{t^{2}}=a t^{2}+a^{2}-4 a+4.
$$
Notice that the expressions on both sides do not change when $t$ is replaced by ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10. Points $M, N$, and $K$ are located on the lateral edges $A A_{1}, B B_{1}$, and $C C_{1}$ of the triangular prism $A B C A_{1} B_{1} C_{1}$ such that $A M: A A_{1}=1: 2, B N: B B_{1}=1: 3, C K: C C_{1}=1: 4$. Point $P$ belongs to the prism. Find the maximum possible value of the volume of the pyramid $M N K... | Answer: 4.
Solution. Suppose we have found the position of point $P$ such that the volume of pyramid $M N K P$ is maximized. Draw a plane $\alpha$ through it, parallel to the plane $M N K$, and call $M_{1}, N_{1}$, and $K_{1}$ the points of intersection of this plane with the edges $A A_{1}, B B_{1}$, and $C C_{1}$, r... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. For what values of the parameter $a$ does the equation
$$
3^{x^{2}+6 a x+9 a^{2}}=a x^{2}+6 a^{2} x+9 a^{3}+a^{2}-4 a+4
$$
have exactly one solution? | Answer: Only when $a=1$.
Solution. Let's denote $x+3a$ by $t$. Note that the number of solutions to the equation does not change with this substitution. Then the original equation will take the form
$$
3^{t^{2}}=a t^{2}+a^{2}-4 a+4
$$
Notice that the expressions on both sides do not change when $t$ is replaced by $-... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10. Points $M, N$, and $K$ are located on the lateral edges $A A_{1}, B B_{1}$, and $C C_{1}$ of the triangular prism $A B C A_{1} B_{1} C_{1}$ such that $A M: A A_{1}=2: 3, B N: B B_{1}=3: 5, C K: C C_{1}=4: 7$. Point $P$ belongs to the prism. Find the maximum possible value of the volume of the pyramid $M N K... | # Answer: 6.
Solution. Suppose we have found the position of point $P$ such that the volume of pyramid $M N K P$ is maximized. Draw a plane $\alpha$ through it, parallel to the plane $M N K$, and call $M_{1}, N_{1}$, and $K_{1}$ the points of intersection of this plane with the edges $A A_{1}, B B_{1}$, and $C C_{1}$,... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 8. For what values of the parameter $a$ does the equation
$$
5^{x^{2}-6 a x+9 a^{2}}=a x^{2}-6 a^{2} x+9 a^{3}+a^{2}-6 a+6
$$
have exactly one solution? | Answer: Only when $a=1$.
Solution. Let's denote $x-3a$ by $t$. Note that the number of solutions to the equation does not change with this substitution. Then the original equation will take the form
$$
5^{t^{2}}=a t^{2}+a^{2}-6a+6
$$
Notice that the expressions on both sides do not change when $t$ is replaced by $-t... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10. Points $M, N$, and $K$ are located on the lateral edges $A A_{1}, B B_{1}$, and $C C_{1}$ of the triangular prism $A B C A_{1} B_{1} C_{1}$ such that $A M: A A_{1}=3: 7, B N: B B_{1}=2: 5, C K: C C_{1}=4: 9$. Point $P$ belongs to the prism. Find the maximum possible value of the volume of the pyramid $M N K... | Answer: 8.
Solution. Suppose we have found the position of point $P$ such that the volume of pyramid $M N K P$ is maximized. Draw a plane $\alpha$ through it, parallel to the plane $M N K$, and call $M_{1}, N_{1}$, and $K_{1}$ the points of intersection of this plane with the edges $A A_{1}, B B_{1}$, and $C C_{1}$, r... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. For what values of the parameter $a$ does the equation
$$
5^{x^{2}+2 a x+a^{2}}=a x^{2}+2 a^{2} x+a^{3}+a^{2}-6 a+6
$$
have exactly one solution? | Answer: Only when $a=1$.
Solution. Let's denote $x+a$ by $t$. Note that the number of solutions of the equation does not change with such a substitution. Then the original equation will take the form
$$
5^{t^{2}}=a t^{2}+a^{2}-6 a+6
$$
Notice that the expressions on both sides do not change when $t$ is replaced by $... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10. Points $M, N$, and $K$ are located on the lateral edges $A A_{1}, B B_{1}$, and $C C_{1}$ of the triangular prism $A B C A_{1} B_{1} C_{1}$ such that $A M: A A_{1}=5: 6, B N: B B_{1}=6: 7, C K: C C_{1}=2: 3$. Point $P$ belongs to the prism. Find the maximum possible value of the volume of the pyramid $M N K... | Answer: 10.
Solution. Suppose we have found the position of point $P$ at which the volume of pyramid $M N K P$ is maximized. Draw a plane $\alpha$ through it, parallel to the plane $M N K$, and call $M_{1}, N_{1}$, and $K_{1}$ the points of intersection of this plane with the edges $A A_{1}, B B_{1}$, and $C C_{1}$, r... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In a right triangle $ABC$, a circle is constructed on the leg $AC$ as its diameter, which intersects the hypotenuse $AB$ at point $E$. A tangent to the circle is drawn through point $E$, which intersects the leg $CB$ at point $D$. Find the length of $DB$, if $AE=6$, and $BE=2$. | Answer: 2.
Solution. The solution is based on two simple observations. First, $\angle A E C=90^{\circ}$, since it subtends the diameter. Second, $D E$ and $D C$ are tangents to the circle from the condition, so $D E=D C$. Therefore, in the right triangle $C E B$, a point $D$ is marked on the hypotenuse $B C$ such that... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In a right triangle $P Q R$, a circle is constructed on the leg $P R$ as its diameter, which intersects the hypotenuse $P Q$ at point $T$. A tangent to the circle is drawn through point $T$, which intersects the leg $R Q$ at point $S$. Find the length of $S Q$, if $P T=15$, and $Q T=5$. | Answer: 5.
Solution. The solution is based on two simple observations. First, $\angle P T R=90^{\circ}$, since it subtends the diameter. Second, $S T$ and $S R$ are tangents to the circle from the given conditions, so $S T=S R$. Therefore, in the right triangle $R T Q$, a point $S$ is marked on the hypotenuse $Q R$ su... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In a right triangle $X Y Z$, a circle is constructed on the leg $X Z$ as its diameter, which intersects the hypotenuse $X Y$ at point $W$. A tangent to the circle is drawn through point $W$, which intersects the leg $Z Y$ at point $V$. Find the length of $V Y$, if $X W=12$, and $Y W=4$. | Answer: 4.
Solution. The solution is based on two simple observations. First, $\angle X W Z=90^{\circ}$, since it subtends the diameter. Second, $V W$ and $V Z$ are tangents to the circle from the condition, so $V W=V Z$. Therefore, in the right triangle $Z W Y$, a point $V$ is marked on the hypotenuse $Y Z$ such that... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In a right triangle $KLM$, a circle is constructed on the leg $KM$ as its diameter, which intersects the hypotenuse $KL$ at point $G$. A tangent to the circle is drawn through point $G$, intersecting the leg $ML$ at point $F$. Find the length of $FL$, if $KG=5$ and $LG=4$. | Answer: 3.
Solution. The solution is based on two simple observations. First, $\angle K G M=90^{\circ}$, since it subtends the diameter. Second, $F G$ and $F M$ are tangents to the circle from the condition, so $F G=F M$. Therefore, in the right triangle $M G L$, a point $F$ is marked on the hypotenuse $L M$ such that... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. Each knight gives one affirmative answer to four questions, while a liar gives three. In total, there were $105+45+85+65=300$ affirmative answers. If all the residents of the city were knights, the total number of affirmative answers would be 200. The 100 extra "yes" answers come from the lies of the liars. ... | Answer: in Block B, on 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Each of the two workers was assigned to process the same number of parts. The first completed the work in 8 hours. The second spent more than 2 hours on setting up the equipment and with its help finished the work 3 hours earlier than the first. It is known that the second worker processed as many parts in 1... | Answer: 4 times.
Solution: Let $x$ be the time spent on equipment setup. Then the second worker worked (on the equipment) $8-3-x=5-x$ hours, producing as much per hour as the first worker in $x+1$ hours. Therefore, $\frac{8}{5-x}=\frac{x+1}{1}$. We get $x^{2}-4 x+3=0$. But by the condition $x>2$, so $x=3$, and the req... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. The function $f$ is such that $f(2 x-3 y)-f(x+y)=-2 x+8 y$ for all $x, y$. Find all possible values of the expression $\frac{f(5 t)-f(t)}{f(4 t)-f(3 t)}$. | Answer: 4.
Solution. By substituting $y=-x$, we get that $f(5 x)=-10 x+f(0)$, i.e., $f(t)=-2 t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. The function $f$ is such that $f(2 x-3 y)-f(x+y)=-2 x+8 y$ for all $x, y$. Find all possible values of the expression $\frac{f(4 t)-f(t)}{f(3 t)-f(2 t)}$. | Answer: 3.
Solution. By substituting $y=-x$, we get that $f(5 x)=-10 x+f(0)$, i.e., $f(t)=-2 t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. The function $f$ is such that $f(x+2 y)-f(3 x-2 y)=2 y-x$ for all $x, y$. Find all possible values of the expression $\frac{f(5 t)-f(t)}{f(4 t)-f(3 t)}$. | Answer: 4.
Solution. Substituting $x=-2 y$, we get that $f(0)-f(-8 y)=-4 y$, i.e., $f(t)=\frac{1}{2} t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. The function $f$ is such that $f(x+2 y)-f(3 x-2 y)=2 y-x$ for all $x, y$. Find all possible values of the expression $\frac{f(4 t)-f(t)}{f(3 t)-f(2 t)}$. | Answer: 3.
Solution. Substituting $x=-2 y$, we get that $f(0)-f(-8 y)=-4 y$, i.e., $f(t)=\frac{1}{2} t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. A team of workers was working on pouring the rink on the large and small fields, with the area of the large field being twice the area of the small field. In the part of the team that worked on the large field, there were 4 more workers than in the part that worked on the small field. When the pouring of the la... | Answer: 10.
Solution. Let the number of workers on the smaller field be $n$, then the number of workers on the larger field is $n+4$, and the total number of people in the team is $2n+4$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivities of... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $BC=8$ and $AC=4$. Find the length of side $AB$ if the length of the vector $4 \overrightarrow{OA} - \overrightarrow{OB} - 3 \overrightarrow{OC}$ is 10. | Answer: 5.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \cd... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. For what values of the parameter $a$ does the equation
$$
\log _{2}^{2} x+(a-6) \log _{2} x+9-3 a=0
$$
have exactly two roots, one of which is four times the other? | Answer: $-2.2$
Solution. Let $t=\log _{2} x$, then the equation becomes $t^{2}+(a-6) t+(9-3 a)=0$. Notice that $3 \cdot(3-a)=9-3 a, 3+(3-a)=6-a$, from which, by the theorem converse to Vieta's theorem, the roots of this equation are -3 and $3-a$. We make the reverse substitution: $\log _{2} x=3$ or $\log _{2} x=3-a$, ... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $AB=8$ and $AC=5$. Find the length of side $BC$ if the length of the vector $\overrightarrow{OA}+3 \overrightarrow{OB}-4 \overrightarrow{OC}$ is 10. | Answer: 4.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \cd... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. A team of workers was laying linoleum in a store's warehouse and in the cash hall, with the warehouse area being 3 times larger than the cash hall area. In the part of the team working in the warehouse, there were 5 more workers than in the part working in the cash hall. When the work in the warehouse was co... | Answer: 9.
Solution. Let the number of workers in the cash hall be denoted as $n$, then the number of workers in the warehouse is $n+5$, and the total number of people in the team is $2n+5$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivity o... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $BC=5$ and $AB=4$. Find the length of side $AC$ if the length of the vector $3 \overrightarrow{OA}-4 \overrightarrow{OB}+\overrightarrow{OC}$ is 10. | Answer: 8.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \cd... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $AB=5, AC=8$, and $BC=4$. Find the length of the vector $\overrightarrow{O A}-4 \overrightarrow{O B}+3 \overrightarrow{O C}$. | Answer: 10.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \c... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. How many natural numbers $n>1$ exist, for which there are $n$ consecutive natural numbers, the sum of which is equal to 2016? | Answer: 5.
Solution. Suppose that for some $n$ there exist $n$ consecutive natural numbers $a$, $a+1, \ldots, a+(n-1)$, the sum of which is 2016. Then $n a+(n-1) n / 2=2016$, or, after algebraic transformations, $n(2 a+n-1)=4032=2^{6} \cdot 3^{2} \cdot 7$.
Note that $n$ and $2 a+n-1$ have different parity. Therefore,... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.4. On the surface of a pentagonal pyramid (see fig.), several gnomes live in pairwise distinct points, and they can live both inside the faces and on the edges or at the vertices. It turned out that on each face (including the vertices and edges that bound it) a different number of gnomes live. What is the minimum nu... | Answer: 6
Solution: There are 6 faces in total, so at least 5 gnomes live on the most "populated" one. If there are exactly 5 gnomes in total, then all of them live on one face (let's call it face $A$), so the faces where 4 and 3 gnomes live (let's call them $B$ and $C$ respectively) are adjacent to it. Then, on the e... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.4. Ten chess players over nine days played a full round-robin tournament, during which each of them played exactly one game with each other. Each day, exactly five games were played, with each chess player involved in exactly one of them. For what maximum $n \leq 9$ can it be claimed that, regardless of the schedule... | Answer. $n=5$.
Solution. By the end of the eighth day, each chess player has played 8 games, meaning they have not played one. The unplayed games divide the chess players into 5 non-intersecting pairs. By the Pigeonhole Principle, among any six chess players, there will always be two belonging to the same pair, that i... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.5. At a round table, 410 deputies sat down, each of whom was either a knight, who always tells the truth, or a liar, who always lies. Each of the deputies said: "Among my twenty neighbors to the left and twenty neighbors to the right, there are exactly 20 liars in total." It is known that at least half of the people ... | Answer: None.
Solution: Let's divide all the people sitting at the table into ten groups of 41 people each. Then, at least one group will have at least 21 liars. Otherwise, in each group, there would be a maximum of 20, i.e., no more than \(20 \times 10 = 200\) in total, which is less than half of the total number. Co... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.2. On a circular route, 12 trams run in one direction at the same speed and at equal intervals. How many trams need to be added so that at the same speed, the intervals between trams decrease by one fifth?
# | # Answer: 3.
Solution: Let's take the entire distance as 60 arbitrary units, which means the trams are currently 5 arbitrary units apart. We want this distance to be reduced by $1 / 5$, making it equal to 4 arbitrary units. For this, we need $60: 4=15$ trams, which is 3 more than the current number.
Criteria: only th... | 3 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.5. Each digit of the natural number $N$ is strictly greater than the one to its left. What is the sum of the digits of the number $9 N$? | Answer: 9.
Solution: Note that $9 N=10 N-N$. Let's perform this subtraction in a column. In the units place, there will be the difference between 10 and the last digit of the number $N$, in the tens place - the last and the second-to-last digit, decreased by 1. In all subsequent places, there will be the difference be... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. In the morning, a dandelion blooms, it flowers yellow for three days, on the fourth morning it turns white, and by the evening of the fifth day, it withers. On Monday afternoon, there were 20 yellow and 14 white dandelions on the meadow, and on Wednesday - 15 yellow and 11 white. How many white dandelions will the... | Answer. Six dandelions.
Solution. A blooming dandelion is white on the fourth and fifth day. This means that on Saturday, the dandelions that bloomed on Tuesday or Wednesday will be white. Let's determine how many there are. The dandelions that were white on Monday had already flown away by Wednesday, and 20 yellow on... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.3. Inside a semicircle of radius 12, there are a circle of radius 6 and a small semicircle, each touching the others pairwise, as shown in the figure. Find the radius of the small semicircle. | Answer: 4.
Solution. Let the radius of the small semicircle be $x$, the center of the large semicircle be $A$, the center of the circle be $B$, and the center of the small semicircle be $C$. The centers of the tangent circles and semicircle, and the corresponding points of tangency, lie on the same straight line, so $... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.4. In a semicircle with a radius of 18 cm, a semicircle with a radius of 9 cm is constructed on one of the halves of the diameter, and a circle is inscribed, touching the larger semicircle from the inside, the smaller semicircle from the outside, and the second half of the diameter. Find the radius of this circle.
... | # Answer: 8 cm.
Solution. Let O, O_1, O_2 be the centers of the large semicircle, the small semicircle, and the inscribed circle, respectively, and let P, Q, R be the points of tangency of the inscribed circle with the diameter of the large semicircle, the small semicircle, and the large semicircle, respectively. Then... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.2. The road from point A to point B first goes uphill and then downhill. A cat takes 2 hours and 12 minutes to travel from A to B, and the return trip takes 6 minutes longer. The cat's speed going uphill is 4 km/h, and downhill is 5 km/h. How many kilometers is the distance from A to B? (Provide a complete solution, ... | Solution: When a cat goes uphill, it takes 15 minutes for 1 km, and when it goes downhill, it takes 12 minutes. That is, when the direction changes, the time spent on 1 km changes by 3 minutes. Since the cat spent 6 minutes more on the return trip, the uphill section on the return trip is 2 km longer.
Let the length o... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.4. On a certain island, there live 100 people, each of whom is either a knight, who always tells the truth, or a liar, who always lies. One day, all the inhabitants of this island lined up, and the first one said: "The number of knights on this island is a divisor of the number 1." Then the second said: "The number o... | Solution: If there are no knights, then all the speakers are lying, since 0 is not a divisor of any natural number.
If there are knights, let there be $a$ of them. Then only the people with numbers $a k$ for $k=1,2, \ldots$ are telling the truth. On the other hand, since exactly $a$ people are telling the truth, $k$ c... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.3. The perimeter of triangle $ABC$ is 24 cm, and the segment connecting the point of intersection of its medians with the point of intersection of its angle bisectors is parallel to side $AC$. Find the length of $AC$. | Answer: 8 cm.
Solution. Let AK be the median from vertex A, M - the point of intersection of the medians ABC, and I - the point of intersection of its angle bisectors AA1, BB1, CC1. Draw a line through K parallel to AC, intersecting the angle bisector BB1 at point P - its midpoint. By Thales' theorem, $PI: IB1 = KM: M... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.5. What is the smallest number of colors needed to color all the cells of a 6 by 6 square so that in each row, column, and diagonal of the square, all cells have different colors? Explanation: a diagonal of the square is understood to mean all rows of at least two cells running diagonally from one edge of the square ... | Answer: In 7 colors.
Solution. Let's provide an example of coloring in 7 colors that satisfies the condition of the problem. Consider a 7 by 7 square, and color it in the required way using 7 colors with a known technique: the coloring of each subsequent row is obtained from the coloring of the previous row by a cycli... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.5. In the city, there are 9 bus stops and several buses. Any two buses have no more than one common stop. Each bus has exactly three stops. What is the maximum number of buses that can be in the city | Answer: 12.
Solution: if some stop is common for 5 routes, then no two of them have any more common stops, which means there are at least $1+5*2=11$ stops, which contradicts the condition. Therefore, each stop is the intersection of no more than 4 routes, then the total number of stops made by buses is no more than 94... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.5. In the city, there are 9 bus stops and several buses. Any two buses have no more than one common stop. Each bus has exactly three stops. What is the maximum number of buses that can be in the city? | Answer: 12.
Solution: if some stop is common for 5 routes, then no two of them have any more common stops, which means there are at least $1+5*2=11$ stops, which contradicts the condition. Therefore, each stop is the intersection of no more than 4 routes, then the total number of stops made by buses is no more than 94... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. The electronic clock on the building of the station shows the current hours and minutes in the format HH:MM from 00:00 to 23:59. How much time in one day will the clock display four different digits? | Answer: 10 hours 44 minutes.
Solution. Each possible combination of four digits burns on the clock for one minute. Consider separately the time of day from 00:00 to 19:59 and from 20:00 to 23:59. In the first case, the number of valid combinations according to the problem's conditions will be: 2 (the tens digit of the... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.2. In a fairy-tale country, every piglet either always lies or always tells the truth, and each piglet reliably knows whether every other piglet is a liar. One day, Nif-Nif, Naf-Naf, and Nuf-Nuf met for a cup of tea, and two of them made statements, but it is unknown who exactly said what. One of the three piglets sa... | # Answer: Two
Solution: If at least one of the statements is true, then the piglets mentioned in it are liars, which means there are at least two liars. At the same time, the one making this true statement must be telling the truth. Therefore, there are no more than two liars. In total, if at least one of the spoken p... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.2. On a certain island, only knights, who always tell the truth, and liars, who always lie, live. One day, 99 inhabitants of this island stood in a circle, and each of them said: "All ten people following me in a clockwise direction are liars." How many knights could there be among those standing in the circle? | Answer: 9 knights.
Solution. Note that all people cannot be liars, because then it would mean that each of them is telling the truth. Therefore, among these people, there is at least one knight. Let's number all the people so that the knight is 99th in the sequence. Then, the 10 people with numbers from 1 to 10 are li... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.3. We consider all possible tilings of an 8 by 8 chessboard with dominoes, each consisting of two adjacent squares. Determine the maximum natural number \( n \) such that for any tiling of the 8 by 8 board with dominoes, one can find some rectangle composed of \( n \) squares of the board that does not contain any do... | Answer. $n=4$.
Solution. 1) We will prove that $n \leq 4$. Consider the following tiling of an 8 by 8 chessboard with dominoes. Divide the board into 2 by 2 squares, color each of them in red and blue in a checkerboard (relative to the 4 by 4 board) pattern, and divide the red squares into pairs of horizontal dominoes... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. What is the minimum sum of digits in the decimal representation of the number $f(n)=17 n^{2}-11 n+1$, where $n$ runs through all natural numbers?
# Answer. 2. | Solution. When $n=8$, the number $f(n)$ is 1001, so the sum of its digits is 2. If $f(n)$ for some $n$ had a sum of digits equal to 1, it would have the form $100, \ldots 00$ and would either be equal to 1 or divisible by 10. The function of a real variable $f(x)$ reaches its minimum at $x=\frac{11}{34}1$, and $f(n)$ c... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. It is known that $70 \%$ of mathematicians who have moved to IT regret their change of activity. At the same time, only $7 \%$ of all people who have moved to IT regret the change. What percentage of those who have moved to IT are mathematicians, if only they regret the change of activity? | Solution. Let a total of $x$ people went into IT, and $y$ of them are mathematicians. According to the condition of the change in activity, on the one hand, $0.07 x$ people regret, and on the other - $0.7 y$. From this, we get that $0.07 x=0.7 y$, from which $y / x=0.1$, that is, $10 \%$.
Criteria. Only the answer - 1... | 10 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.3. What is the maximum number of rooks that can be placed on an 8x8 chessboard so that each rook attacks no more than one other? A rook attacks all squares on the same row and column it occupies. | Answer: 10.
Solution: It is clear that in each column and row there are no more than two rooks. Let $k$ rooks be placed while satisfying the condition. On each square where a rook is placed, write the number 0. In each of the 8 columns, perform the following operation: if there are two numbers in the column, add 1 to ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.3. Find all natural $n$, for which on a square grid of size $n$ by $n$ cells, it is possible to mark $n$ cells, each in different rows and different columns, that can be sequentially visited by a knight's move in chess, starting from some cell, without landing on the same cell twice, and returning to the starting ce... | Answer. $n=4$.
Solution. An example for $n=4$ is not difficult:
We will prove that for $n \neq 4$, the required set of cells does not exist.
Suppose that for a given $n$, it is possible t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.1. Two athletes with constant speeds run on an oval track of a sports ground, the first of them runs the track completely 5 seconds faster than the second. If they run on the track from the same starting point in the same direction, they will meet again for the first time after 30 seconds. How many seconds will it t... | Answer. In 6 seconds.
Solution. Let the length of the track be $S$ meters, and the speeds of the first and second runners be $x$ and $y$ meters per second, respectively. From the first condition: $\frac{S}{x} + 5 = \frac{S}{y}$, and from the second condition $\frac{S}{x-y} = 30$, since in this case the first runner ca... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3. In a row from left to right, all natural numbers from 1 to 37 are written in such an order that each number, starting from the second to the 37th, divides the sum of all numbers to its left: the second divides the first, the third divides the sum of the first and second, and so on, the last divides the sum of the... | Answer: 2.
Solution. If the first number is the prime number 37, then the second must be 1, and the third must be a divisor of the number $37+1=38$, that is, 2 or 19. However, 19 must be in the last position, since the number 37 minus an even number must divide the sum of all the other numbers and itself, that is, div... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. Each cell of a $5 \times 5$ table is painted in one of several colors. Lada shuffled the rows of this table so that no row remained in its original position. Then Lera shuffled the columns so that no column remained in its original position. To their surprise, the girls noticed that the resulting table was the sam... | # Answer: 7.
Solution: Let's renumber the colors and reason about numbers instead. Both columns and rows could have been cyclically permuted or divided into a pair and a triplet. If a cyclic permutation of columns was used, then all columns consist of the same set of numbers, i.e., no more than five different numbers.... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.5. For what minimum natural $n$ can $n$ distinct natural numbers $s_{1}, s_{2}, \ldots, s_{n}$ be found such that $\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \ldots\left(1-\frac{1}{s_{n}}\right)=\frac{7}{66} ?$ | Answer. $n=9$.
Solution. We can assume that $1<s_{1}<s_{2}<\ldots<s_{n}$, then for any $k=1, \ldots, n$ the inequality $s_{k} \geq k+1$ holds. Therefore, $\frac{7}{66}=\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \ldots\left(1-\frac{1}{s_{n}}\right) \geq\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\rig... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. A finite set of distinct real numbers $X$ is called good if each number in $X$ can be represented as the sum of two other distinct numbers in $X$. What is the minimum number of elements that a good set $X$ can contain?
# | # Answer: 6.
Solution. From the condition, it follows that $X$ contains no less than three numbers, which means there are non-zero numbers in it. By multiplying all numbers by minus one if necessary, we can assume that $X$ contains positive numbers. Let's choose the largest number $M$ from them. According to the condi... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.4. What is the maximum number of colors in which all cells of a 4 by 4 square can be painted so that any 2 by 2 square of cells necessarily contains at least two cells of the same color? | Answer: In 11 colors.
Solution. We will prove that the maximum number of colors under the conditions of the problem does not exceed 11. Consider in a 4x4 square five 2x2 squares: four corner ones and the central one. The corner 2x2 squares do not intersect, and the central one shares one common cell with each of the c... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.2. In a kindergarten, each child was given three cards, each of which had either "MA" or "NYA" written on it. It turned out that 20 children could form the word "MAMA" from their cards, 30 children could form the word "NYANYA," and 40 children could form the word "MANYA." How many children had all three cards the sam... | Answer: 10 children.
Solution. Let's denote the number of children who received three "MA" cards as $x$, two "MA" cards and one "NA" card as $y$, two "NA" cards and one "MA" card as $z$, and three "NA" cards as $t$. Then, the word "MAMA" can be formed by all children from the first and second groups and only them, the... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.5. For seven natural numbers $a, b, c, a+b-c, a+c-b, b+c-a, a+b+c$ it is known that all of them are different prime numbers. Find all values that the smallest of these seven numbers can take. | Answer: 3.
Solution. From the condition, it follows that $a, b, c$ are also primes. If the smallest of the seven numbers were equal to two, the last four numbers would be different even numbers, which means they could not all be prime. If all seven numbers are greater than three, due to their primality, they are not d... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2. Daniil has 6 cards with letters, from which he managed to form the word WNMWNM shown in the picture. Note that this word has a remarkable property: if you rotate it 180 degrees, you get the same word. How many words with such a property can Daniil form using all 6 cards at once?
 According to the problem, Danil has 2 cards with the letter $\mathrm{N}$, which remains the same when flipped, and 4 cards with the letter M, which turns into the letter W when flipped. Clearly, to get a word with the desired properties, we need to arrange 2 letters M and one $\math... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.2 The banker leaves home, and at that exact moment, a car from the bank arrives to take him to the bank. The banker and the car always leave at the same time, and the car always travels at a constant speed. One day, the banker left home 55 minutes earlier than usual and, for fun, started walking in the direction oppo... | Answer. The speed of the car is 12 times greater than the speed of the banker.
Solution. Indeed, the car delivered the banker to the bank 10 minutes later than usual, which means it caught up with him 5 minutes later than usual, i.e., from the moment he usually leaves his house. Therefore, the car traveled from the ba... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.1. Find the value of the expression $\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{2}{1+x y}$, given that $x \neq y$ and the sum of the first two terms of the expression is equal to the third. | Answer: 2.
Solution. Let's write the condition of the equality of the sum of the first two terms to the third one as: $\frac{1}{1+x^{2}}-\frac{1}{1+x y}=\frac{1}{1+x y}-\frac{1}{1+y^{2}} \quad$ and bring it to a common denominator:
$\frac{x(y-x)}{(1+x^{2})(1+x y)}=\frac{y(y-x)}{(1+y^{2})(1+x y)}$. Given $x \neq y$, we... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. A set for playing lotto contains 90 barrels numbered with natural numbers from 1 to 90. The barrels are somehow distributed among several bags (each bag contains more than one barrel). We will call a bag good if the number of one of the barrels in it equals the product of the numbers of the other barrels in the sa... | Answer: 8.
Solution: In each good bag, there are no fewer than three barrels. The smallest number in each good bag must be unique; otherwise, the largest number in this bag would not be less than $10 \times 11=110$, which is impossible. For the same reason, if a good bag contains a barrel with the number 1, it must al... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. Experienced sawyer Garik knows how to make cuts. In one day of continuous work, he can saw 600 nine-meter logs into identical three-meter logs (they differ from the original only in length). How much time will the experienced sawyer Garik need to saw 400 twelve-meter logs (they differ from the nine-meter logs only... | Answer: one day
Solution: to turn a 9-meter log into 3 three-meter pieces, 2 cuts are needed. Therefore, Garik makes $2 * 600=1200$ cuts per day. To turn a 12-meter log into three-meter pieces, 3 cuts are needed, and for 400 logs, $400 * 3=1200$ cuts are needed, which means it will take the same amount of time.
Crite... | 1 | Other | math-word-problem | Yes | Yes | olympiads | false |
11.1. The ore contains $21 \%$ copper, enriched - $45 \%$ copper. It is known that during the enrichment process, $60 \%$ of the mined ore goes to waste. Determine the percentage content of ore in the waste. | Answer: $5 \%$.
Solution. In the extracted 100 kg of ore, there are 21 kg of copper. From these 100 kg of enriched ore, 40 kg will be obtained, containing 18 kg of copper. Therefore, the 60 kg of waste that went to the dump contain 3 kg of copper, that is, $5 \%$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
9.1. 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 |
9.4. 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 |
9.5. 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 |
9.1. 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
9.1. 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 |
9.2. 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 |
9.3. 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 |
2. The bathtub fills up in 23 minutes from the hot water tap, and in 17 minutes from the cold water tap. Pete first opened the hot water tap. After how many minutes should he open the cold water tap so that by the time the bathtub is full, one and a half times more hot water has been added than cold water? | # 9.2. Answer. In 7 minutes.
In one minute, the hot water tap fills $\frac{1}{23}$ of the bathtub, and the cold water tap fills $\frac{1}{17}$ of the bathtub. After filling the bathtub, the hot water should make up $\frac{3}{5}$ of the bathtub, and the cold water should make up $\frac{2}{5}$ of the bathtub. Therefore,... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.5. What is the maximum number of 4-element subsets that can be selected from a set of 8 elements such that the intersection of any three of the selected subsets contains no more than one element? | Answer. Eight.
Solution. We will provide two different examples of choosing eight 4-element subsets in a set $\mathrm{X}$ of eight elements, satisfying the condition of the problem. Both examples are constructed as geometric objects.
Example 1. We will consider the elements of $\mathrm{X}$ as the vertices of a unit c... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. At the Olympiad, students from gymnasiums, lyceums, and regular schools met. Some of them stood in a circle. Gymnasium students always lie to regular school students, lyceum students lie to gymnasium students, and regular school students lie to lyceum students. In all other cases, the students tell the truth. Each... | Answer: There were no ordinary school students in the circle.
Solution: Let's assume there was an ordinary school student in the circle. Consider his left neighbor. This neighbor could not be another ordinary school student or a lyceum student, because they would tell the truth. But he also cannot be a gymnasium stude... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.5. Given a convex quadrilateral $A B C D$ with side $A D$ equal to 3. Diagonals $A C$ and $B D$ intersect at point $E$, and it is known that the areas of triangles $A B E$ and $D C E$ are both 1. Find the side $B C$, given that the area of $A B C D$ does not exceed 4. | # Answer: 3.
Solution: Triangles $A B D$ and $A C D$ have the same area, as they are composed of the common $A E D$ and equal-area $A B E$ and $D C E$. Since $A B D$ and $A C D$ have the same base $A D$, their heights to this base are equal. Therefore, $B C$ is parallel to $A D$, meaning our quadrilateral is a trapezo... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.3. Find all triples of distinct natural numbers, the least common multiple of which is equal to their sum. The least common multiple of several numbers is the smallest natural number that is divisible by each of these numbers. | Answer. All triples of the form $\{n, 2 n, 3 n\}$ for any natural number $n$.
Solution. Let the required numbers be $a<b<c$. By the condition, their least common multiple, which is equal to $a+b+c$, is divisible by $c$, so $c$ divides the sum $a+b<2 c$. Therefore, $a+b=c$. Next, $a+b+c=2 a+2 b$ is divisible by $b$, so... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. For non-negative numbers $a, b, c, d$, the following equalities are satisfied:
$\sqrt{a+b}+\sqrt{c+d}=\sqrt{a+c}+\sqrt{b+d}=\sqrt{a+d}+\sqrt{b+c}$. What is the maximum number of distinct values that can be among the numbers $a, b, c, d$? | Answer. Two.
Solution. Square the first equality, combine like terms, cancel by 2, square again, combine like terms once more, ultimately obtaining the equality $a c+b d=a b+c d$, equivalent to $(a-d)(c-b)=0$, from which either $a=d$ or $b=c$. Performing similar manipulations with the second equality, we get $a=b$ or ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.3. What is the maximum number of colors needed to color all cells of a 4 by 4 square so that for every pair of different colors, there are two cells of these colors that are either in the same row or in the same column of the square? | Answer: In 8 colors.
Solution. If the cells were painted in 9 or more colors, there would be a color in which only one cell is painted. There are only 6 cells located in the same row or column with it, so there are no more than 6 other colors forming a pair with it, as required by the condition - a contradiction. Ther... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.5. A set for playing lotto contains 90 barrels numbered with natural numbers from 1 to 90. The barrels are somehow distributed into several bags (each bag contains more than one barrel). We will call a bag good if the number of one of the barrels in it equals the product of the numbers of the other barrels in the sam... | # Answer: 8.
Solution: In each good bag, there are no fewer than three barrels. The smallest number in each good bag must be unique, otherwise the largest number in this bag would be no less than $10 \times 11=110$, which is impossible. For the same reason, if a good bag contains a barrel with the number 1, it must al... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.3. In the math test, each of the 25 students could receive one of four grades: $2, 3, 4$ or 5. It turned out that the number of students who received fours was 4 more than those who received threes. How many people received twos, if it is known that the sum of all grades for this test is 121? | # 7.3. Answer: 0.
Let's assume that the 4 students who received fours skipped school, then 21 students should have scored 105 points on the test, which is only possible if all of them received fives. Therefore, in the real situation, no one received a two.
## Comments on Evaluation.
Answer only: 1 point. Answer with... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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