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Problem 2. $n$ mushroom pickers went to the forest and brought a total of 200 mushrooms (it is possible that some of them did not bring any mushrooms home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't... | Answer: 21.
Solution. First, let's prove that when $n \leqslant 20$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $0,1, \ldots, n-2$ mushrooms, and the $n$-th collected all the rest. Since
$$
0+1+\ldots+(n-2) \leqslant 0+1+\ldots+18=171=200-29
$$
the last mushroom picker collected at least ... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. A circle is inscribed in trapezoid $ABCD$, touching the lateral side $AD$ at point $K$. Find the area of the trapezoid if $AK=16, DK=4$ and $CD=6$. | Answer: 432.
Solution. Let $L, M, N$ be the points of tangency of the inscribed circle with the sides $BC, AB, CD$ respectively; let $I$ be the center of the inscribed circle. Denote the radius of the circle by $r$. Immediately note that $DN = DK = 4$ (the first equality follows from the equality of the segments of ta... | 432 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. Points $A_{1}, B_{1}, C_{1}$ are the points of intersection of the extensions of the altitudes of an acute-angled triangle $A B C$ with the circumcircle of $A B C$. The incircle of triangle $A_{1} B_{1} C_{1}$ touches one of the sides of $A B C$, and one of the angles of triangle $A B C$ is $40^{\circ}$. Fin... | Answer: $60^{\circ}$ and $80^{\circ}$.
Solution. Without loss of generality, let the circle $\omega$, inscribed in $A_{1} B_{1} C_{1}$, touch the side $B C$. Let $H$ be the orthocenter of triangle $A B C$, $K$ be the point of tangency of $\omega$ and $B C$, and $L$ be the point of tangency of $\omega$ and $A_{1} C_{1}... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. In the decimal representation of an even number $M$, only the digits $0, 2, 4, 5, 7$, and 9 are used, and digits can repeat. It is known that the sum of the digits of the number $2M$ is 35, and the sum of the digits of the number $M / 2$ is 29. What values can the sum of the digits of the number $M$ take? Li... | # Answer: 31.
Solution. Let's denote the sum of the digits of a natural number $n$ by $S(n)$. Notice the following facts, each of which is easy to verify if you add numbers in a column.
Lemma 1. Let $n$ be a natural number. Then the number of odd digits in the number $2n$ is equal to the number of carries when adding... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. $n$ mushroom pickers went to the forest and brought a total of 338 mushrooms (it is possible that some of them did not bring any mushrooms home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't... | Answer: 27.
Solution. First, let's prove that when $n \leqslant 26$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $0,1, \ldots, n-2$ mushrooms, and the $n$-th collected all the rest. Since
$$
0+1+\ldots+(n-2) \leqslant 0+1+\ldots+24=300=338-38
$$
the last mushroom picker collected at least ... | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. A circle with radius 4 is inscribed in trapezoid $ABCD$, touching the base $AB$ at point $M$. Find the area of the trapezoid if $BM=16$ and $CD=3$. | Answer: 108.
Solution. Let $K, L, N$ be the points of tangency of the inscribed circle with the sides $AD, BC, CD$ respectively; let $I$ be the center of the inscribed circle. Immediately note that $BL = BM = 16$.
$ denote the sum of the digits of a natural number $n$. Notice the following facts, each of which is easy to verify by adding numbers in a column.
Lemma 1. Let $n$ be a natural number. Then the number of odd digits in the number $2n$ is equal to the number of carries when adding $n$ and... | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 2. $n$ mushroom pickers went to the forest and brought a total of 450 mushrooms (each brought at least one mushroom home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify your answe... | Answer: 30.
Solution. First, let's prove that when $n \leqslant 29$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $1, \ldots, n-1$ mushrooms, and the $n$-th collected all the rest. Since
$$
1+\ldots+(n-1) \leqslant 1+\ldots+28=406=450-44
$$
the last mushroom picker collected at least 44 mus... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. Points $A_{1}, B_{1}, C_{1}$ are the points of intersection of the extensions of the altitudes of an acute-angled triangle $A B C$ with the circumcircle of $A B C$. The incircle of triangle $A_{1} B_{1} C_{1}$ touches one of the sides of $A B C$, and one of the angles of triangle $A B C$ is $70^{\circ}$. Fin... | Answer: $60^{\circ}$ and $50^{\circ}$.
Solution. Without loss of generality, let the circle $\omega$, inscribed in $A_{1} B_{1} C_{1}$, touch the side $B C$. Let $H$ be the orthocenter of triangle $A B C$, $K$ be the point of tangency of $\omega$ and $B C$, and $L$ be the point of tangency of $\omega$ and $A_{1} C_{1}... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. In the decimal representation of an even number $M$, only the digits $0, 2, 4, 5, 7$, and 9 are used, and digits can repeat. It is known that the sum of the digits of the number $2M$ is 43, and the sum of the digits of the number $M / 2$ is 31. What values can the sum of the digits of the number $M$ take? Li... | # Answer: 35.
Solution. Let $S(n)$ denote the sum of the digits of a natural number $n$. Notice the following facts, each of which is easy to verify by adding numbers in a column.
Lemma 1. Let $n$ be a natural number. Then the number of odd digits in the number $2n$ is equal to the number of carries when adding $n$ a... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. $n$ mushroom pickers went to the forest and brought a total of 162 mushrooms (each brought at least one mushroom home). Boy Petya, upon learning this, declared: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify you... | Answer: 18.
Solution. First, let's prove that when $n \leqslant 17$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $1, \ldots, n-1$ mushrooms, and the $n$-th collected all the rest. Since
$$
1+\ldots+(n-1) \leqslant 1+\ldots+16=136=162-26
$$
the last mushroom picker collected at least 26 mus... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. A circle with radius 2 is inscribed in trapezoid $ABCD$, touching the base $CD$ at point $N$. Find the area of the trapezoid if $DN=1$ and $AB=12$. | Answer: 27.
Solution. Let $K, L, M$ be the points of tangency of the inscribed circle with the sides $AD, BC, AB$ respectively; let $I$ be the center of the inscribed circle. Immediately note that $DK = DN = 1$.
$ denote the sum of the digits of a natural number $n$. Notice the following facts, each of which is easy to verify by adding numbers in a column.
Lemma 1. Let $n$ be a natural number. Then the number of odd digits in the number $2n$ is equal to the number of carries when adding $n$ and... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. In a football championship, 20 teams participate, each playing against each other once. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other? | Answer: 90 games.
Solution. We will consider the games that have not been played. The condition means that the unplayed games do not form triangles. We will prove by induction on $k$ that for $2k$ teams, the maximum number of unplayed games is no more than $k^2$.
Base case: $k=1$ (the estimate is obvious).
Inductive... | 90 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. In a football championship, 16 teams participate, each playing against each other once. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other?
# | # Answer: 56 games.
Solution. We will consider the games that have not been played. The condition means that there will be no three teams that have not played against each other at all. We will prove by induction on $k$ that for $2k$ teams, the maximum number of unplayed games is no more than $k^2$.
Base case: $k=1$ ... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. In a football championship, 16 teams participate, each playing against each other once. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other? | Answer: 56 games.
Solution. We will consider the games that have not been played. The condition means that there will be no three teams that have not played with each other at all. We will prove by induction on $k$ that for $2k$ teams, the maximum number of unplayed games is no more than $k^2$.
Base case: $k=1$ (the ... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 30? | Answer: 32.
Solution. First, let's provide an example of 32 numbers whose sum is 0 and the sum of their squares is 30. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{16}=\sqrt{\frac{15}{16}}, x_{17}=x_{18}=\ldots=x_{32}=-\sqrt{\frac{15}{16}}$ will work.
Now, we will prove that fewer than 32 numbers will not suffice... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. For what values of the parameter $a$ does the equation $x^{3}-11 x^{2}+a x-8=0$ have three distinct real roots that form a geometric progression? | Answer: only 22.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}-11 x^{2}+a x-8=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a polynomial of the third degree:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=11 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2... | 22 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 42? | Answer: 44.
Solution. First, let's provide an example of 44 numbers whose sum is 0 and the sum of their squares is 42. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{22}=\sqrt{\frac{21}{22}}, x_{23}=x_{24}=\ldots=x_{44}=-\sqrt{\frac{21}{22}}$ will work.
Now, we will prove that it is impossible to achieve this with ... | 44 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. For what values of the parameter $a$ does the equation $x^{3}-14 x^{2}+a x-27=0$ have three distinct real roots that form a geometric progression? | Answer: only 42.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}-14 x^{2}+a x-27=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a cubic polynomial:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=14 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2} x_{3}=27
\e... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 36?
# | # Answer: 38.
Solution. First, let's provide an example of 38 numbers whose sum is 0 and the sum of their squares is 36. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{19}=\sqrt{\frac{18}{19}}, x_{20}=x_{21}=\ldots=x_{38}=-\sqrt{\frac{18}{19}}$ will work.
Now, we will prove that fewer than 38 numbers will not suffi... | 38 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. For what values of the parameter $a$ does the equation $x^{3}-15 x^{2}+a x-64=0$ have three distinct real roots that form a geometric progression? | Answer: only 60.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}-15 x^{2}+a x-64=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a polynomial of the third degree:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=15 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 40? | Answer: 42.
Solution. First, let's provide an example of 42 numbers whose sum is 0 and the sum of their squares is 40. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{21}=\sqrt{\frac{20}{21}}, x_{22}=x_{23}=\ldots=x_{42}=-\sqrt{\frac{20}{21}}$ will work.
Now, we will prove that fewer than 42 numbers will not suffice... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. For what values of the parameter $a$ does the equation $x^{3}+16 x^{2}+a x+64=0$ have three distinct real roots that form a geometric progression? | Answer: only 64.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}+16 x^{2}+a x+64=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a cubic polynomial:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=-16 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2} x_{3}=-64
... | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10. The sum of the surface areas of the polyhedra into which a parallelepiped is divided by sections is equal to the sum of the surface area of the parallelepiped and the areas of the internal surfaces. The sum of the areas of the internal surfaces is equal to twice the sum of the areas of the sections.
Let's ... | Answer: 194.
[^0]: ${ }^{1}$ This statement is a particular case of Cauchy's inequality. | 194 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 13 adventurers have rubies; exactly 9 have emeralds; exactly 15 have sapphires; exactly 6 have diamonds. In addition, it is known that
- if an adventurer has sapphires, then they have either emeralds or diamonds (but not both at the ... | Answer: 22.
Solution. Note that the number of adventurers who have sapphires is equal to the total number of adventurers who have emeralds or diamonds. Then, from the first condition, it follows that 9 adventurers have sapphires and emeralds, and 6 have sapphires and diamonds. That is, every adventurer who has emerald... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. In triangle $A B C$, side $A C=42$. The bisector $C L$ is divided by the point of intersection of the bisectors of the triangle in the ratio $2: 1$, counting from the vertex. Find the length of side $A B$, if the radius of the circle inscribed in triangle $A B C$ is 14. | Answer: 56.
Answer: Let $I$ be the center of the inscribed circle in triangle $ABC$ (i.e., the point of intersection of the angle bisectors). Noting that $AI$ is the angle bisector in triangle $ALC$, by the angle bisector theorem, we have: $AC: AL = CI: IL = 2: 1$, from which $AL = AC / 2 = 21$.
Next, $AC \cdot AL \c... | 56 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. The function $F$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $F(n a, n b, n c)=n \cdot F(a, b, c)$, $F(a+n, b+n, c+n)=F(a, b, c)+n$, $F(a, b, c)=F(c, b, a)$ hold. Find $F(58,59,60)$. | Answer: 59.
Solution. Note that $F(-1,0,1)=F(1,0,-1)=(-1) \cdot F(-1,0,1)$, from which $F(-1,0,1)=0$. Then $F(58,59,60)=F(-1,0,1)+59=59$.
Comment. The function $F$ cannot be uniquely determined. For example, the functions $F(a, b, c)=(a+b+c) / 3$, $F(a, b, c)=b$, and $F(a, b, c)=$ median of the numbers $\{a, b, c\}$ ... | 59 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 5 adventurers have rubies; exactly 11 have emeralds; exactly 10 have sapphires; exactly 6 have diamonds. In addition, it is known that
- if an adventurer has diamonds, then they have either emeralds or sapphires (but not both at the ... | Answer: 16.
Solution. Note that the number of adventurers who have emeralds is equal to the total number of adventurers who have rubies or diamonds. Then, from the second condition, it follows that 5 adventurers have rubies and emeralds, and 6 have emeralds and diamonds. That is, every adventurer who has diamonds must... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 3. A landscaping team worked on a large and a small football field, 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 6 more workers than in the part that worked on the small field. When the landscaping of the large field wa... | Answer: 16.
Solution. Let the number of workers on the smaller field be $n$, then the number of workers on the larger field is $n+6$, and the total number of people in the team is $2n+6$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivities of... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. The function $f$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $f(n a, n b, n c)=n \cdot f(a, b, c)$, $f(a+n, b+n, c+n)=f(a, b, c)+n$, $f(a, b, c)=f(c, b, a)$ hold. Find $f(24,25,26)$. | Answer: 25.
Solution. Note that $f(-1,0,1)=f(1,0,-1)=(-1) \cdot f(-1,0,1)$, from which $f(-1,0,1)=0$. Then $f(24,25,26)=f(-1,0,1)+25=25$.
Comment. The function $f$ cannot be uniquely determined. For example, the functions $f(a, b, c)=(a+b+c) / 3$, $f(a, b, c)=b$, and $f(a, b, c)=$ median of the numbers $\{a, b, c\}$ ... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 4 adventurers have rubies; exactly 10 have emeralds; exactly 6 have sapphires; exactly 14 have diamonds. Moreover, it is known that
- if an adventurer has rubies, then they have either emeralds or diamonds (but not both at the same t... | Answer: 18.
Solution. Note that the number of adventurers who have emeralds is equal to the total number of adventurers who have rubies or sapphires. Then, from the second condition, it follows that 4 adventurers have rubies and emeralds, and 6 have emeralds and sapphires. That is, every adventurer who has rubies must... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. The function $G$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$, the equalities $G(n a, n b, n c)=n \cdot G(a, b, c)$, $G(a+n, b+n, c+n)=G(a, b, c)+n$, $G(a, b, c)=G(c, b, a)$ hold. Find $G(89,90,91)$. | Answer: 90.
Solution. Note that $G(-1,0,1)=G(1,0,-1)=(-1) \cdot G(-1,0,1)$, from which $G(-1,0,1)=0$. Then $G(89,90,91)=G(-1,0,1)+90=90$.
Comment. The function $G$ cannot be uniquely determined. For example, the functions $G(a, b, c)=(a+b+c) / 3, G(a, b, c)=b$ and $G(a, b, c)=$ median of the numbers $\{a, b, c\}$ are... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 9 adventurers have rubies; exactly 8 have emeralds; exactly 2 have sapphires; exactly 11 have diamonds. Moreover, it is known that
- if an adventurer has diamonds, then they have either rubies or sapphires (but not both at the same t... | Answer: 17.
Solution. Note that the number of adventurers who have diamonds is equal to the total number of adventurers who have rubies or sapphires. Then, from the first condition, it follows that 9 adventurers have rubies and diamonds, and 2 have sapphires and diamonds. That is, every adventurer who has rubies must ... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. A team of lumberjacks was cutting trees on a large and a small plot, with the area of the small plot being 3 times less than that of the large plot. In the part of the team that worked on the large plot, there were 8 more lumberjacks than in the part that worked on the small plot. When the tree harvesting on... | Answer: 14.
Solution. Let the number of workers on the smaller plot be denoted as $n$, then the number of workers on the larger plot is $n+8$, and the total number of workers in the team is $2n+8$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the product... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. In triangle $A B C$, side $B C=28$. The bisector $B L$ is divided by the point of intersection of the bisectors of the triangle in the ratio $4: 3$, counting from the vertex. Find the radius of the circumscribed circle around triangle $A B C$, if the radius of the inscribed circle in it is 12. | Answer: 50.
Solution. Let $I$ be the center of the inscribed circle in triangle $ABC$ (i.e., the point of intersection of the angle bisectors). Noting that $CI$ is the angle bisector in triangle $BLC$, by the angle bisector theorem, we have: $BC: CL = BI: IL = 4: 3$, from which $CL = 3BC / 4 = 21$.
Next, $BC \cdot CL... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. The function $g$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $g(n a, n b, n c)=$ $n \cdot g(a, b, c), g(a+n, b+n, c+n)=g(a, b, c)+n, g(a, b, c)=g(c, b, a)$ hold. Find $g(14,15,16)$. | Answer: 15.
Solution. Note that $g(-1,0,1)=g(1,0,-1)=(-1) \cdot g(-1,0,1)$, from which $g(-1,0,1)=0$. Then $g(14,15,16)=g(-1,0,1)+15=15$.
Comment. The function $g$ cannot be uniquely determined. For example, the functions $g(a, b, c)=(a+b+c) / 3$, $g(a, b, c)=b$, and $g(a, b, c)=$ median of the numbers $\{a, b, c\}$ ... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 2. In the $1^{\text{st}}$ grade class, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 2. Among the ans... | Answer: 15 boys and 12 girls.
Solution. First solution. Let's denote the children who gave the answers (13,11), (17,11), (14,14) as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the answers of the girls has the same parity as $m$, and in the answers of the boys - the... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=5: 4$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $4 \cdot \overrightarrow{A B}+5 \cdot \overrightarrow{A C}$ is 2016. | Answer: 224.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=5: 4$, from which $\overrightarrow{B L}=5 / 9 \cdot \overrightarrow{B C}$. Now
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{5}{9} \cdot \overrightarrow{B C}=\overrigh... | 224 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 9. The Federation of Wrestling has assigned each participant in the competition a qualification number. It is known that in matches between wrestlers whose qualification numbers differ by more than 2, the wrestler with the lower number always wins. The tournament for 256 wrestlers is held on an Olympic system: at ... | Answer: 16
Solution. Note that a wrestler with number $k$ can only lose to a wrestler with number $k+1$ or $k+2$, so after each round, the smallest number cannot increase by more than 2 numbers. In a tournament with 256 participants, there are 8 rounds, so the number of the tournament winner does not exceed $1+2 \cdot... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 2. In $1^{st}$ grade, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 4. Among the answers received wer... | Answer: 16 boys and 14 girls.
Solution. First solution. Let's denote the children who gave the answers (15,18), (15,10), (12,13) as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the answers of girls has the same parity as $m$, while in the answers of boys, it has the... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=4: 3$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $3 \cdot \overrightarrow{A B}+4 \cdot \overrightarrow{A C}$ is 2016. | Answer: 288.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=4: 3$, from which $\overrightarrow{B L}=4 / 7 \cdot \overrightarrow{B C}$. Now
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{4}{7} \cdot \overrightarrow{B C}=\overrigh... | 288 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 9. The Federation of Wrestling has assigned each participant in the competition a qualification number. It is known that in matches between wrestlers whose qualification numbers differ by more than 2, the wrestler with the lower number always wins. The tournament for 512 wrestlers is held on an Olympic system: at ... | Answer: 18.
Solution. Note that a wrestler with number $k$ can only lose to a wrestler with number $k+1$ or $k+2$, so after each round, the smallest number cannot increase by more than 2 numbers. In a tournament with 512 participants, there are 9 rounds, so the number of the tournament winner does not exceed $1+2 \cdo... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 2. In $1^{\text {st }}$ grade, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 2. Among the answers rec... | Answer: 13 boys and 16 girls.
Solution. First solution. Let's denote the children who gave the answers (12,18), (15,15), (11,15) as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the girls' answers has the same parity as $m$, and in the boys' answers - the opposite. T... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=5: 2$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $2 \cdot \overrightarrow{A B}+5 \cdot \overrightarrow{A C}$ is 2016. | Answer: 288.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=5: 2$, from which $\overrightarrow{B L}=5 / 7 \cdot \overrightarrow{B C}$. Now,
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{4}{7} \cdot \overrightarrow{B C}=\overrig... | 288 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 2. In the $1^{\text{st}}$ grade, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 4. Among the answers r... | Answer: 14 boys and 15 girls.
Solution. First solution. Let's denote the children who gave the answers $(10,14),(13,11),(13,19)$ as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the girls' answers has the same parity as $m$, and in the boys' answers - the opposite. T... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=7: 2$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $2 \cdot \overrightarrow{A B}+7 \cdot \overrightarrow{A C}$ is 2016. | Answer: 224.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=7: 2$, from which $\overrightarrow{B L}=7 / 9 \cdot \overrightarrow{B C}$. Now,
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{7}{9} \cdot \overrightarrow{B C}=\overrig... | 224 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.3. Egor borrowed 28 rubles from Nikita, and then returned them in four payments. It turned out that Egor always returned a whole number of rubles, and the amount of each payment increased and was divisible by the previous one. What was the last amount Egor paid? | Answer: 18 rubles
Solution:
1) If Egor paid $a$ rubles the first time, then the second time - not less than $2a$, the third - not less than $4a$, the fourth - not less than $8a$, and in total - not less than $15a$. Since $15a \leq 28$, we get that $a=1$.
2) The second time he paid 2 or 3 rubles (because if 4, then he... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. On a circular route, two buses operate with the same speed and a movement interval of 21 minutes. What will be the movement interval if 3 buses operate on this route at the same constant speed? | Answer: 14.
Solution: Since the interval of movement with two buses on the route is 21 minutes, the length of the route in "minutes" is 42 minutes. Therefore, the interval of movement with three buses on the route is $42: 3=14$ minutes.
Criteria: only answer, answer with verification - 3 points. | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.4. Natasha and Inna bought the same box of tea bags. It is known that one tea bag is enough for two or three cups of tea. Natasha's box was enough for only 41 cups of tea, while Inna's was enough for 58 cups. How many tea bags were in the box? | Answer: 20.
Solution: Let there be $n$ tea bags in the box. Then the number of brewings can vary from $2n$ to $3n$. Therefore, 58 is not greater than $3n$, which means $19<n$. Additionally, 41 is not less than $2n$, so $n<21$. Since the number of tea bags must be a natural number that is less than 21 but greater than ... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. On a line, one hundred points are marked: green, blue, and red. It is known that between any two red points there is a blue one, and between any two blue points there is a green one. In addition, there are no fewer red points than blue ones, and no fewer blue points than green ones. How many points are painted blu... | Answer: 33.
Solution: Let the number of red points be $n$, then the number of blue points is not less than $n-1$ (the number of intervals between "adjacent" red points), and since by condition the number of red points is not less than the number of blue points, the number of blue points is either $n$ or $n-1$. Similar... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.4. On a cubic planet, there live cubic mice, and they live only on the faces of the cube, not on the edges or vertices. It is known that different numbers of mice live on different faces, and the number on any two adjacent faces differs by at least 2. What is the minimum number of cubic mice that can live on this pla... | Solution: We will prove that no three consecutive numbers can be the number of mice on the faces. Indeed, if there were $x, x+1$, and $x+2$ mice on some three faces, then $x$ and $x+1$ would have to be on opposite faces. But then $x+2$ mice could not be anywhere.
Consider the first 8 natural numbers. Among the first t... | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.2. Find the number of different ways to arrange all natural numbers from 1 to 9 inclusive in the cells of a 3 by 3 table, one number per cell, such that the sums of the numbers in each row and each column are equal. The table cannot be rotated or reflected. | Answer: 72 ways
Solution. Among the numbers from 1 to 9, there are 5 odd numbers. Since the sums in all rows and columns are $\frac{1}{3}(1+2+\ldots+9)=15$ - which are odd, there must be an odd number of odd numbers in each row and each column. This is only possible if one row contains three odd numbers, and the other... | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.1. Provide an example of a natural number that is a multiple of 2020, such that the sum of its digits is also a multiple of 2020. | Solution: For example, the number 20202020...2020, where the fragment 2020 repeats 505 times, fits. Such a number is obviously divisible by 2020, and the sum of its digits is $505 \times 4=2020$.
Criteria: Any correct answer with or without verification - 7 points.
Correctly conceived example, but with a calculation ... | 2020 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.4. Vera Aleksandrovna urgently needed to cut out three 20-sided polygons (not necessarily identical) from one rectangular sheet of paper. She can take this sheet and cut it along a straight line into two parts. Then take one of the resulting parts and cut it along a straight line. Then take one of the available piece... | Answer: 50 cuts.
Solution: With each cut, the total number of paper pieces increases by 1 (one piece turns into two new pieces), so after $n$ cuts, there will be $(n+1)$ pieces of paper. Let's calculate how many vertices all the pieces together can have after $n$ cuts. With each cut, the total number of vertices incre... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.5. On a 10 by 10 cell board, some 10 cells are marked. For what largest $n$ is it always possible, regardless of which cells are marked, to find a rectangle consisting of several cells, the perimeter of which will be at least $n$? The length or width of the rectangle can be equal to one cell. | Answer: $n=20$.
Solution. First, we prove that when $n=20$, it is always possible to find such a rectangle. Suppose 10 cells are colored. If there is a column or row without colored cells, then a rectangle of $1 \times 9$ with a perimeter of 20 (or even $1 \times 10$ with a perimeter of 22) can be cut from it. Now, le... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. Let's call a four-digit number $\overline{a b c d}$ curious if the sum of the two-digit numbers $\overline{a b}$ and $\overline{c d}$ equals the two-digit number $\overline{b c}$. For example, the number 1978 is curious because 19+78=97. Find the number of curious numbers. Answer. 36. | Solution. Let's form the equation $\overline{a b}+\overline{c d}=10(a+c)+b+d=\overline{b c}=10 b+c$, from which we get $10 a+9 c+d=9 b$. The difference $9(b-c)$ is divisible by 9, so the sum $10 a+d=9(b-c)$ is also divisible by 9, which is equivalent to the divisibility by 9 of the sum $a+d=9(b-c-a)$. The sum of two di... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.3. Inside an isosceles triangle $\mathrm{ABC}$ with equal sides $\mathrm{AB}=\mathrm{BC}$ and an angle of 80 degrees at vertex $\mathrm{B}$, a point $\mathrm{M}$ is taken such that the angle
$\mathrm{MAC}$ is 10 degrees, and the angle $\mathrm{MCA}$ is 30 degrees. Find the measure of angle $\mathrm{AMB}$. | Answer: 70 degrees.
Solution. Draw a perpendicular from vertex B to side AC, and denote the points of its intersection with lines AC and CM as P and T, respectively. Since angle MAC is less than angle MCA, side CM of triangle MAC is shorter than side AM, so point M is closer to C than to A, and therefore T lies on the... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.3. It is known that all krakozyabrs have horns or wings (possibly both). According to the results of the world census of krakozyabrs, it turned out that $20 \%$ of the krakozyabrs with horns also have wings, and $25 \%$ of the krakozyabrs with wings also have horns. How many krakozyabrs are left in the world, if it i... | Answer: 32.
Solution: Let $n$ be the number of krakozyabrs with both wings and horns. Then the number of horned krakozyabrs is $-5 n$, and the number of winged krakozyabrs is $-4 n$. Using the principle of inclusion-exclusion, the total number of krakozyabrs is $5 n + 4 n - n = 8 n$. There is only one integer between ... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. At the school for slackers, a competition on cheating and giving hints was organized. It is known that $75 \%$ of the students did not show up for the competition at all, and all the rest participated in at least one of the competitions. When the results were announced, it turned out that $10 \%$ of all those who ... | Answer: 200.
Solution. Let the number of students in our school be $n$ people. $\frac{n}{4}$ people attended the competitions. Taking into account $\frac{n}{40}$ people who participated in both competitions, the number of participants in the competition with hints was $\frac{3}{5}\left(\frac{n}{4}+\frac{n}{40}\right)=... | 200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.2. Ellie and Toto painted daisies in the field. On the first day, Ellie painted one fourteenth of the entire field. On the second day, she painted twice as much as on the first day, and on the third day, she painted twice as much as on the second day. Toto, in total, painted 7000 daisies. How many daisies were there ... | Solution. Let there be $n$ daisies in the field. Then Ellie painted $n / 14 + 2 n / 14 + 4 n / 14 = 7 n / 14 = n / 2$ daisies in total. This means Toto also painted half of the field, from which it follows that half of the field is 7000 daisies, and the entire field is 14000.
Criteria. Only the answer - 1 point.
The ... | 14000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.4. Anton from the village was given several zucchini, and he decided to give them to his friends. He gave half of the received zucchini to Arina, and a third (also from the received amount) to Vera. It turned out that after this, the number of zucchini Arina had became a square of some natural number, and the number ... | Solution. Let Anton receive $n$ zucchinis. Since both half and a third of $n$ are integers, $n$ is divisible by 6, that is, $n=6k$ for some natural number $k$. Then it is known that $3k$ is the square of a natural number, and $2k$ is a cube. Let $k=2^p 3^q m$. In other words, let $p$ and $q$ be the highest powers of tw... | 648 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.1. What two digits need to be appended to the right of the number 2013 so that the resulting six-digit number is divisible by 101? Find all possible solutions. | Answer: 94, the obtained number will be equal to 201394.
Solution: The remainder of dividing the number $\overline{2013 x y}$ by 101 is $\overline{x y}+7$ and this must be divisible by 101. This is greater than 0 but less than 202, so $\overline{x y}+7=101, \overline{x y}=94, x=9, y=4$.
Grading: Just the answer with ... | 94 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.4. On the cells of an 8 by 8 board, chips are placed such that for each chip, the row or column of the board in which it lies contains only one chip. What is the maximum possible number of chips on the board? | Answer: 14.
Solution: Let's match each chip to the row or column of the board in which it is the only one. If it is the only one in both, we match it to the row. From the condition, it follows that different chips are matched to different rows and columns. If not all rows and columns are matched, then their total numb... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.5. At a diplomatic reception, there are 99 persons, each of whom has heard of no fewer than $n$ other attendees. If A has heard of B, it does not automatically mean that B has heard of A. For what minimum $n$ is it guaranteed that there will be two attendees who have heard of each other? | Answer. When $n=50$.
Solution. Let's call a situation where one of the guests has heard about another a half-acquaintance. If each guest has heard about at least 50 other participants at the reception, then there are at least $99 \cdot 50$ half-acquaintances, which is more than the total number of pairs of guests at t... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.5. What is the maximum number of 2 by 2 squares that can be placed on a 7 by 7 grid of squares so that any two placed squares share no more than one common cell? The 2 by 2 squares are placed along the grid lines such that each covers exactly 4 cells. The squares do not extend beyond the boundaries of the board. | Answer: 18 squares.
Solution: First, let's provide an example of laying out 18 squares of 2 by 2: 9 of them cover the left bottom square of size 6 by 6 of the board, and the other 9 cover the right top square of size 6 by 6 of the board.
We will prove that it is impossible to lay out 19 squares correctly. Note that i... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.2. A bus with programmers left Novosibirsk for Pavlodar. When it had traveled 70 km, Pavel Viktorovich set off from Novosibirsk in a car along the same route, and caught up with the programmers in Karasuk. After that, Pavel drove another 40 km, while the bus traveled only 20 km in the same time. Find the distance fro... | Solution. Since by the time the car has traveled 40 km, the bus has traveled half that distance, its speed is exactly half the speed of the car. However, when the bus has traveled 70 km after the car's departure, the car will have traveled 140 km and will just catch up with the bus. According to the problem, this happe... | 140 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. Come up with at least one three-digit PAU number (all digits are different) such that $(П+\mathrm{A}+\mathrm{У}) \times \Pi \times \mathrm{A} \times \mathrm{Y}=300$ (it is sufficient to provide one example) | Solution: For example, PAU = 235 is suitable. There are other examples as well. Criteria: Any correct example - 7 points. | 235 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2. Students in the seventh grade send each other New Year's stickers on Telegram. It is known that exactly 26 people received at least one sticker, exactly 25 - at least two stickers, ..., exactly 1 - at least 26 stickers. How many stickers did the students in this class receive in total, if it is known that no one r... | # Answer: 351
Solution: Note that exactly 1 sticker was received by one person, as it is precisely in this case that the difference between those who received at least 1 and those who received at least 2. Similarly, one person received exactly $2, 3, \ldots, 26$ stickers, so the total number of stickers is $1+\ldots+2... | 351 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.1. In each of the four volleyball teams, there are six players, including a captain and a setter, and these are different people. In how many ways can a team of six players be formed from these four teams, such that there is at least one player from each team and there must be a pair of captain and setter from at le... | Answer: 9720.
Solution. Case 1. Three players, including the captain and the setter, are chosen from one of the teams, and one player is chosen from each of the remaining three teams. The team is chosen in four ways, the third player from it in another four ways, and the three players from the remaining teams in $6 \c... | 9720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.2. Find the number of five-digit numbers containing two digits, one of which is divisible by the other. | Answer. $89760=9 \cdot 10^{4}-2 \cdot 5!$.
Solution. We will count the number of five-digit numbers in which no digit is divisible by any other digit, and then subtract this number from the total number of five-digit numbers, which is 90000, to get the answer to the problem.
Note that a number, none of whose digits a... | 89760 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.4. There are 100 coins, 99 of which are genuine and weigh the same, and 1 is counterfeit and lighter than the others. Dmitry has a pair of balance scales without weights, which always show incorrect results (for example, if the left pan is heavier, they will show either balance or the right pan being heavier, but it ... | # Solution:
Let's number the coins from 1 to 100. Weigh the first coin against the second. If the scales show equality, then one of them is fake, and all the other coins are genuine, and we have achieved the desired result. If the scales do not balance, assume the coin numbered 1 is heavier. Then the second coin is de... | 98 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.1. How much of a $5 \%$ and a $20 \%$ salt solution in water should be taken to obtain 90 kg of a $7 \%$ solution | Answer: 78 kg of 5% solution and 12 kg of 20% solution.
Solution. Let the mass of the 5% solution be $x$ kg, the mass of the 20% solution will be $90-x$ kg, and the total mass of salt in the 5% and 20% solutions is equal to the mass of salt in 90 kg of 7% solution: $\frac{5}{100} x+\frac{20}{100}(90-x)=\frac{7}{100} 9... | 78 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.1. Paramon set out from point A to point B. At $12^{00}$, when he had walked half the way to B, Agafon ran out from A to B, and at the same time, Solomon set out from B to A. At $13^{20}$, Agafon met Solomon, and at $14^{00}$, he caught up with Paramon. At what time did Paramon and Solomon meet? | Answer: At 13 o'clock.
Solution: Let the distance between A and B be $\mathrm{S}$ km, and the speeds of Paramon, Solomon, and Agafon be $x, y, z$ km per hour, respectively. Then from the condition, we get: $\frac{S / 2}{z-x}=2, \frac{S}{y+z}=\frac{4}{3}$, from which $x+y=\frac{1}{2} S$. Therefore, Paramon and Solomon ... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.2. The median $A M$ of triangle $A B C$ divides the segment $P R$, parallel to side $A C$, with endpoints on sides $\mathrm{AB}$ and $\mathrm{BC}$, into segments of lengths 5 cm and 3 cm, starting from side $\mathrm{AB}$. What is the length of side AC? | Answer: 13 cm.
Solution. Let the ends of the segment be denoted as $\mathrm{P}$ and $\mathrm{R}$, and the point of intersection with the median $\mathrm{AM}$ as $\mathrm{Q}$, with $\mathrm{P}$ lying on side $\mathrm{AB}$ and $\mathrm{R}$ on side $\mathrm{BC}$. Draw the midline $\mathrm{MN}$ of the triangle, its length... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.5. In a row from left to right, there are $n$ coins. It is known that two of them are counterfeit, they lie next to each other, the left one weighs 9 grams, the right one 11 grams, and all the remaining ones are genuine and each weighs 10 grams. The coins are weighed on balance scales, which either show which of the ... | Answer. $n=28$.
Solution. Let $n=28$. Divide all 28 coins into three piles: the first pile contains coins numbered $11,13,15,17,19,21,23,25,27$, the second pile contains coins numbered $12,14,16,18,20,22,24,26,28$, and the third pile contains coins numbered from 1 to 10. The first weighing compares the first and secon... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.4. On the table, 28 coins of the same size but possibly different masses are arranged in a triangular shape (see figure). It is known that the total mass of any triplet of coins that touch each other pairwise is 10 g. Find the total mass of all 18 coins on the boundary of the triangle. | Answer: 60 g.
Solution 1: Take a rhombus made of 4 coins. As can be seen from the diagram, the masses of two non-touching coins in it are equal. Considering such rhombi, we get that if we color the coins in 3 colors, as shown in the diagram, then the coins of the same color will have the same mass.
Now it is easy to ... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.5. Represent the number 1000 as the sum of the maximum possible number of natural numbers, the sums of the digits of which are pairwise distinct. | Answer: 19.
Solution: Note that the smallest natural number with the sum of digits A is a99..99, where the first digit is the remainder, and the number of nines in the record is the incomplete quotient of the division of A by 9. From this, it follows that if A is less than B, then the smallest number with the sum of d... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.5. In Nastya's room, 16 people gathered, each pair of whom either are friends or enemies. Upon entering the room, each of them wrote down the number of friends who had already arrived, and upon leaving - the number of enemies remaining in the room. What can the sum of all the numbers written down be, after everyone h... | Answer: 120
Solution: Consider any pair of friends. Their "friendship" was counted exactly once, as it is included in the sum by the person who arrived later than their friend. Therefore, after everyone has arrived, the sum of the numbers on the door will be equal to the total number of friendships between people. Sim... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.2. Losharik is going to visit Sovunya along the river at a speed of 4 km/h. Every half hour, he launches paper boats that float to Sovunya at a speed of 10 km/h. With what time interval do the boats arrive at Sovunya? (Provide a complete solution, not just the answer.) | Solution: If Losyash launched the boats from one place, they would arrive every half hour. But since he is walking, the next boat has to travel a shorter distance than the previous one. In half an hour, the distance between Losyash and the last boat will be $(10-4) \cdot 0.5=3$. This means that the distance between adj... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.3. Katya wrote a four-digit number on the board that was divisible by each of its digits without a remainder (there were no zeros in the number). Then she erased the first and last digits, and the number 89 remained on the board. What could have been written on the board initially? (Find all options and show that the... | Solution: The original number was divisible by 8. Therefore, the number formed by its last three digits is also divisible by 8. From the condition, this number has the form $\overline{89 x}$. Clearly, $x=6$ fits, and other numbers divisible by 8 do not fit into this decade. The entire number is divisible by 9, which me... | 4896 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.4. Vikentiy has two jars, a red one and a blue one, and a pile of 20 pebbles. Initially, both jars are empty. A move in Vikentiy's game consists of transferring a pebble from the pile to one of the jars or returning a pebble from one of the jars to the pile. The number of pebbles in the jars determines the game posit... | Answer: 110.
Solution: The position in Vikenty's game is uniquely defined by a pair of non-negative integers $(x, y)$, and $x+y \leq 20$, where $0 \leq y \leq x \leq 20$ are the numbers of pebbles in the blue and red jars, respectively. In total, there are $21+19+17 \ldots+1=121$ positions in Vikenty's game. We will c... | 110 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.5. Find the number of different arrangements in a row of all natural numbers from 1 to 10 such that the sum of any three consecutive numbers is divisible by 3. | Answer: $4! \cdot 2 \cdot 3! \cdot 3! = 1728$
Solution. From the condition, it follows that the remainders of the numbers standing two apart when divided by 3 are equal. Therefore, the numbers standing at positions 1, 4, 7, and 10, as well as those at positions 2, 5, and 8, and at positions 3, 6, and 9, have equal rem... | 1728 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.5. All natural numbers from 1 to 100 are written in some order in a circle. For each pair of adjacent numbers, the sum is calculated. Out of the hundred resulting numbers, what is the maximum number that can be divisible by 7? | Answer: 96.
Solution. For the sum of a pair of adjacent numbers to be divisible by 7, their remainders when divided by 7 must sum to 7: 0+0, 1+6, 2+5, and 3+4. Let's call such pairs of remainders suitable pairs. Among the numbers from 1 to 100, there are 14, 15, 15, 14, 14, 14, and 14 numbers with remainders 0, 1, 2, ... | 96 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
9.2. 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 |
9.4. $N$ 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 |
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 |
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 |
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 |
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 |
10.5. Around a circle, 32 numbers $a_{1}, a_{2}, \ldots, a_{32}$ are written, each of which is either -1 or 1. In one operation, each number $a_{n}, n=1,2, \ldots, 32$ is replaced by the product $a_{n} a_{n+1}$ of it and the next number in the cycle, with indices considered cyclically, $a_{33}=a_{1}, a_{34}=a_{2}$, and... | Answer: 32.
Solution: We will prove by induction on $n$ that if $2^n$ numbers are written in a circle, the answer to the problem is $2^n$. The base case $n=1$ is straightforward: either $\{1,-1\} \rightarrow\{-1,-1\} \rightarrow\{1,1\}$, or $\{-1,-1\} \rightarrow\{1,1\}$, in any case, two operations are always suffici... | 32 | Combinatorics | proof | Yes | Yes | olympiads | false |
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