problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
4. How many three-digit numbers exist that are 5 times the product of their digits? | Answer. One number is 175.
Solution. First method. The digits that make up the number do not include the digit 0, otherwise the condition of the problem cannot be met. The given three-digit number is obtained by multiplying 5 by the product of its digits, so it is divisible by 5. Therefore, its notation ends with the ... | 175 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,212 |
5. In a circle, a diameter $A B$ and a chord $C D$ parallel to it were drawn such that the distance between them is half the radius of this circle (see figure). Find the angle $CAB$.
| Answer: $75^{\circ}$.
Solution. Consider triangle $A O C$, where $O$ is the center of the circle. This triangle is isosceles because $O C$ and $O A$ are radii. Therefore, by the property of isosceles triangles, angles $A$ and $C$ are equal. Draw the perpendicular $C M$ to side $A O$ and consider the right triangle $O ... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,213 |
6. Plot the graph of the equation $x^{2}-y^{4}=\sqrt{18 x-x^{2}-81}$, that is, represent on the coordinate plane all points with coordinates $(x ; y)$ that satisfy this equation. | Answer. See the figure.
Solution. Transform the given equation by extracting a complete square under the root: $x^{2}-y^{4}=\sqrt{-(x-9)^{2}}$. The expression on the right side is meaningful only when $x=9$. Substituting this value into the equation, we get: $9^{2}-y^{4}=0$. Factor the left side: $(3-y)(3+y)\left(9+y^... | (9,3)(9,-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,214 |
Problem 9.1. For a natural number $a$, the product $1 \cdot 2 \cdot 3 \cdot \ldots \cdot a$ is denoted as $a$ !.
(a) (2 points) Find the smallest natural number $m$ such that $m$ ! is divisible by $23 m$.
(b) (2 points) Find the smallest natural number $n$ such that $n$ ! is divisible by $33n$. | # Answer:
(a) (2 points) 24.
(b) (2 points) 12.
Solution. (a) The condition is equivalent to $(m-1)!$ being divisible by 23. Since 23 is a prime number, at least one of the numbers $1, 2, \ldots, m-1$ must be divisible by 23, so $m-1 \geqslant 23$ and $m \geqslant 24$. Clearly, $m=24$ works, since in this case $\fra... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,215 |
Problem 9.2. In four classes of a school, there are more than 70 children, all of whom came to the grade meeting (no other children were present at the meeting).
Each girl who came was asked: "How many people from your class, including you, came to the meeting?"
Each boy who came was asked: "How many boys from your c... | # Answer:
(a) (1 point) 21 students.
(b) (3 points) 33 girls.
Solution. (a) Since all 8 numbers listed in the condition are distinct, exactly 4 of them represent the number of children in the classes, and the other 4 represent the number of boys in the classes. The number 21, being the largest of the listed numbers,... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,216 |
Problem 9.3. Point $I$ is the intersection of the angle bisectors of triangle $ABC$. On side $BC$, point $X$ is marked. It turns out that $AI = BX, AC = CX, \angle ABC = 42^\circ$.
(a) (2 poi... | # Answer:
(a) (2 points) $21^{\circ}$.
(b) (2 points) $54^{\circ}$.
Solution. (a) Note that triangles XIC and AIC are congruent (IC is a common side, $X C = C A, \angle X C I = \angle A C I$ by the problem statement; see Fig. 3). Therefore, $X I = A I = B X$, and triangle $B X I$ is isosceles. Hence,
$$
\angle B I ... | 21,54 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,217 |
Problem 9.4. Foma and Yerema were traveling to Moscow on a cart at a constant speed along a straight road.
- At 12:00, Foma asked: "How many versts to Moscow?"
- Yerema answered: "82".
- At 13:00, Foma asked: "How many versts to Moscow?"
- Yerema answered: "71".
- At 15:00, Foma asked: "How many versts to Moscow?"
- Y... | Answer: 34.5 versts.
Solution. From Yeremy's answers, we understand that
- at 12:00, the distance to Moscow was between 81.5 and 88.5 versts;
- at 15:00, the distance to Moscow was between 4... | 34.5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,218 |
Problem 9.5. For real non-zero numbers $a, b, c$ it is known that
$$
a^{2}-b c=b^{2}-a c=c^{2}-a b
$$
(a) (1 point) What positive values can the expression
$$
\frac{a}{b+c}+\frac{2 b}{a+c}+\frac{4 c}{a+b} ?
$$
take? List all possible options. If the expression cannot take positive values, write 0 as the answer.
(b... | # Answer:
(a) (1 point) $7 / 2$.
(b) (3 points) -7.
Solution. First case: all numbers $a, b, c$ are equal to each other.
Then
$$
\frac{a}{b+c}+\frac{2 b}{a+c}+\frac{4 c}{a+b}=\frac{1}{2}+\frac{2}{2}+\frac{4}{2}=\frac{7}{2}
$$
Second case: among $a, b, c$ there are different numbers.
Without loss of generality, $... | \frac{7}{2},-7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,219 |
Problem 9.6. Given a triangle $ABC$, point $M$ is the midpoint of side $BC$. Let $\ell$ be the bisector of the external angle $A$ of triangle $ABC$. The line passing through $M$ and parallel to $\ell$ intersects side $AB$ at point $K$. Find the length of segment $AK$ if $AB=23$ and $AC=8$.
. Let $X$ be the point of intersection of line $\ell_{1}$ and $AB$.
Notice that $\angle AXC = \angle ACX$, since both these angles are equal to half of the external angle $A$, so $AX = AC = 8$.
![](https://cdn.mathpix.co... | 15.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,220 |
Problem 9.7. The numbers $1, 2, 3, \ldots, 57$ are written on a board. What is the maximum number of these numbers that can be chosen so that no two chosen numbers differ exactly by a factor of 2.5? | Answer: 48.
Solution. Consider sequences of natural numbers that satisfy the following set of conditions: in each sequence
- each number does not exceed 57
- there are at least two numbers, and they all go in ascending order;
- each subsequent number is 2.5 times the previous one.
Let's list them all.
- 2,5
- $4,10... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,221 |
Problem 9.8. On a plane, 36 points are marked, no three of which lie on the same line. Some pairs of marked points are connected by segments such that no more than 3 segments emanate from each marked point.
What is the maximum number of different closed 4-segment broken lines that can result?
The vertices of the brok... | # Answer: 54.
Solution. First, let's prove that there are no more than 54 broken lines.
Consider a "tick" structure, consisting of three points $A, B$, and $C$, as well as two segments $AB$ and $AC$ (the segment $BC$ may or may not be present; point $A$ will be called the vertex of the tick). Since from $B$ and $C$ n... | 54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,222 |
Task 1.1. A $4 \times 4$ table is divided into four $2 \times 2$ squares.
Vika wrote 4 ones, 4 twos, 4 threes, and 4 fours in the cells of the table such that in each column, each row, and each $2 \times 2$ square, all numbers are different.
The hooligan Andrey erased some of the numbers. Help Vika restore which cell... | Answer: $A-4, B-1, C-3, D-2$.
Solution.
In the fourth column, the number 1 is already present, so in the upper right $2 \times 2$ square, the number 1 will be in cell $A 3$. By similar reasoning, the number 3 will be in cell $D 4$.
| | 1 | 2 | 3 | 4 |
| :---: | :---: | :---: | :---: | :---: |
| $A$ | | | 1 | |
|... | A-4,B-1,C-3,D-2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,223 |
Task 1.3. A $4 \times 4$ table is divided into four $2 \times 2$ squares.
Vika wrote 4 ones, 4 twos, 4 threes, and 4 fours in the cells of the table such that in each column, each row, and each $2 \times 2$ square, all numbers are different.
The hooligan Andrey erased some of the numbers. Help Vika restore which cell... | Answer: $A-3, B-2, C-4, D-1$. | A-3,B-2,C-4,D-1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,225 |
Task 1.4. A $4 \times 4$ table is divided into four $2 \times 2$ squares.
Vika wrote 4 ones, 4 twos, 4 threes, and 4 fours in the cells of the table such that in each column, each row, and each $2 \times 2$ square, all numbers are different.
The hooligan Andrey erased some of the numbers. Help Vika restore which cell... | Answer: $A-1, B-3, C-4, D-2$.
## Grade 7, Problem 2 | A-1,B-3,C-4,D-2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,226 |
Problem 2.1. Points $A, B, C, D$ are marked on a line, in that exact order. Point $M$ is the midpoint of segment $A C$, and point $N$ is the midpoint of segment $B D$. Find the length of segment $M N$, given that $A D=68$ and $B C=20$.
}{2}=44-\frac{x}{2}
$$
Now it is not difficult to calculate $M N$:
$$
M N=A D-A M-N D=68-\frac{x}{2}-\left(44-\frac{x}{2}\right)=24
$$ | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,227 |
Problem 3.1. Pasha knows the speed of his motorboat. He calculated that it would take him 20 minutes to travel from the pier to the bridge and back. But in his calculations, he forgot that the river has a current. How many minutes will Pasha actually need for the planned route, if it is known that the speed of the curr... | Answer: 22.5.
Solution. Let $x$ be the speed of the current, then $3x$ is the speed of the boat. According to Pasha's calculations, traveling at a speed of $3x$, he would spend 10 minutes on the trip in one direction.
To cover the same distance downstream, traveling at a speed of $4x$ (which is $\frac{4}{3}$ times th... | 22.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,229 |
Problem 3.4. Pasha knows the speed of his motorboat. He calculated that it would take him 44 minutes to travel from the pier to the bridge and back. But in his calculations, he forgot that the river has a current. How many minutes will Pasha actually need for the planned route if it is known that the speed of the curre... | Answer: 49.5 .
## 7th grade, problem 4 | 49.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,230 |
Problem 4.1. Along the road connecting Masha's and Sasha's houses, there are 17 apple trees and 18 poplars. When Masha was going to visit Sasha, she took photos of all the trees. Right after the tenth apple tree, Masha's phone memory ran out, and she couldn't photograph the remaining 13 trees. The next day, when Sasha ... | Answer: 22.
Solution. Note that the tenth apple tree, counting from Masha's house, is the eighth apple tree, counting from Sasha's house. Therefore, Sasha will not pick a leaf from exactly 13 trees. We get that he will pick a total of $17+18-13=22$ leaves. | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,231 |
Problem 4.4. Along the road connecting Masha's and Sasha's houses, there are 17 apple trees and 20 poplars. When Masha was going to visit Sasha, she took photos of all the trees. Right after the tenth apple tree, Masha's phone memory ran out, and she couldn't photograph the remaining 13 trees. The next day, when Sasha ... | Answer: 24.
## 7th grade, problem 5 | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,232 |
Problem 5.1. At a physical education class, 27 seventh-graders arrived, some of them brought one ball each. Sometimes during the class, one of the seventh-graders would give their ball to another seventh-grader who didn't have one.
At the end of the class, $N$ seventh-graders said: "I received balls less frequently th... | Answer: 13.
Solution. If a seventh-grader received the ball less frequently than he gave it away, then he originally had the ball, but no longer had it at the end. Thus, at the beginning of the lesson, the students collectively had at least $N$ balls, which ultimately ended up with some of the remaining $27-N$ student... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,233 |
Problem 5.4. For a physical education class, 29 seventh-graders came, some of them brought one ball each. Sometimes during the class, one of the seventh-graders would give their ball to another seventh-grader who didn't have one.
At the end of the class, \( N \) seventh-graders said: “I received balls less frequently ... | Answer: 14.
## 7th grade, problem 6 | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,234 |
Problem 6.1. Given a quadrilateral $A B C D$, in which $A D \| B C$. The bisector of angle $A$ intersects side $C D$ at point $X$, and the extension of side $B C$ beyond point $C$ - at point $Y$. It turns out that $\angle A X C=90^{\circ}$. Find the length of segment $A B$, if it is known that $A D=16$ and $C Y=13$.
!... | Answer: 14.5.
Solution. $\quad$ Let $AB=a$. Note that $\angle BYA=\angle YAD=\angle YAB$. We get that triangle $ABY$ is isosceles $(AB=BY)$, and $BC=BY-CY=AB-CY=a-13$ (Fig. 1).
Fig. 1: to t... | 14.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,235 |
Problem 7.1. Each of the seven dwarfs thought of a natural number. They all know what the others have thought of. Snow White asked each of the dwarfs what number they thought of.
- The 1st dwarf remained silent.
- The 2nd dwarf said: “My number is the same as the number of the first dwarf.”
- The 3rd dwarf said: “My n... | Answer: 7 or 14.
Solution. Let the first dwarf have guessed the number $a_{1}$, the second $a_{2}$, the third $-a_{3}, \ldots$, the seventh $-a^{7}$.
Assume that the 7th dwarf told the truth. Then his number is exactly half of the sum of all numbers, that is, $a_{7}=\frac{46}{2}=23$. Reasoning similarly, we get that ... | 7or14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,237 |
Problem 8.1. Let $s(n)$ denote the sum of all odd digits of the number $n$. For example, $s(4)=0$, $s(173)=11, s(1623)=4$.
Calculate the value of the sum $s(1)+s(2)+s(3)+\ldots+s(321)$. | Answer: 1727.
Solution. We will separately count the sum of odd digits by place value.
## Units place.
Among the numbers from 1 to 321, there are 32 complete tens:
- from 1 to 10;
- from 11 to 20;
- from 21 to 30;
- ...
- from 311 to 320.
In each ten, the sum of the odd digits in the units place is $1+3+5+7+9=25... | 1727 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,239 |
Problem 1. In the morning, 5 foreign cars were parked along the road. By noon, 2 domestic cars were parked between each pair of foreign cars. And by evening, a motorcycle was parked between each pair of adjacent cars. How many motorcycles were parked in total $?$
Answer: 12 . | Solution. Between 5 foreign cars there are 4 gaps, so there were $4 \cdot 2=8$ domestic cars parked there; that is, a total of $5+8=13$ cars were parked. Between them, 12 motorcycles were parked. | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,240 |
Problem 2. At an open evening at the conservatory, four quartets, five duets, and six trios were supposed to perform (a quartet consists of four musicians, a trio of three, and a duet of two; each musician is a member of only one musical group). However, one quartet and two duets unexpectedly went on tour, and a solois... | Answer: 35.
Solution. If no one was absent, then at the evening there would have been
$4 \cdot 4(4$ quartets $)+5 \cdot 2(5$ duets $)+6 \cdot 3$ ( 6 trios $)=44$ people.
But $4(1$ quartet) $+2 \cdot 2$ (2 duets) +1 (a soloist from one of the trios) $=9$ people were absent. Thus, at the evening, $44-9=35$ people perf... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,241 |
Problem 3. Athletes $A, B, C, D$ and $E$ participated in a race. Athlete $D$ arrived later than $A$, and athlete $B$ arrived earlier than $D$ and immediately after $C$. Athlete $C$ was not first, but arrived before $A$. In what order did the participants finish? Enter the letters $A, B, C, D, E$ without spaces or comma... | Solution. Note that athlete $B$ arrived right after $C$. In addition, it is known that $A$ arrived after $C$ (and thus after $B$ as well), and $D$ arrived after $A$. Therefore, we get that $C$ overtook at least three others, yet did not arrive first. This means that the first to arrive was $E$, followed by $C$, then $B... | ECBAD | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,242 |
Problem 4. Dima stood on one of the steps of the staircase and suddenly noticed that there were an equal number of steps above and below him. Then he climbed up 7 steps, and after that, he went down 15 steps. In the end, he found himself on the 8th step of the staircase (counting from the bottom). How many steps does t... | Answer: 31.
Solution. Since Dima ended up on the 8th step at the end, before that he was on the $8+15=23$ step. He got there by climbing up 7 steps, so he started from the $23-7=16$ step. Thus, we have that the 16th step is the middle of the ladder. Therefore, the ladder consists of 31 steps. | 31 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,243 |
Problem 5. On the table, there were cards with digits from 1 to 9 (a total of 9 cards). Katya chose four cards such that the product of the digits on two of them equals the product of the digits on the other two. Then Anton took one more card from the table. In the end, the cards with the digits $1,4,5,8$ remained on t... | # Answer: 7.
Solution. One of the cards that is not currently on the table has the number 7. Note that Katya could not have taken the 7, because then one of her products would be divisible by 7, while the other would not. Therefore, Anton took the 7. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,244 |
Problem 6. Yulia thought of a number. Dasha added 1 to Yulia's number, and Anya added 13 to Yulia's number. It turned out that the number obtained by Anya is 4 times the number obtained by Dasha. What number did Yulia think of? | Answer: 3.
Solution. Note that since Anya's number is 4 times greater than Dasha's number, the difference between these numbers is 3 times greater than Dasha's number. Thus, Dasha's number is $(13-1): 3=4$. Therefore, Yulia's number is 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,245 |
Problem 7. Aся, Borya, Vasilina, and Grisha bought tickets to the cinema for one row. It is known that:
- There are a total of 9 seats in the row, numbered from 1 to 9.
- Borya did not sit in seat 4 or 6.
- Aся sat next to Vasilina and Grisha, and no one sat next to Borya.
- There were no more than two seats between A... | Solution. Note that Asey, Vasilina, and Grisha occupy three seats in a row, and Borya sits one seat away from them. Let's seat another child, Dima, in the free seat. Then, 5 children sit in a row. Thus, someone is sitting in the central seat of the row (that is, seat number 5). For any other seat, we can come up with a... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,246 |
Problem 8. Masha braided her dolls: half of the dolls got one braid each, a quarter of the dolls got two braids each, and the remaining quarter of the dolls got four braids each. She tied a ribbon in each braid. How many dolls does Masha have if she needed 24 ribbons in total? | Answer: 12.
Solution. Note that since a quarter of the dolls have four braids, the total number of ribbons used on them is the same as the total number of dolls. Half of the dolls have one braid, so the number of ribbons used on them is half the total number of dolls. And a quarter of the dolls have two braids, so the... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,247 |
11.1 Prove that if $\sin x>0.9$, then $|\sin 2x|<0.9$. | Solution.
$$
\begin{aligned}
& |\sin 2 x|=|2 \sin x \cos x|=2 \sin x|\cos x| \leq 2|\cos x|= \\
& =2 \sqrt{1-\sin ^{2} x}<2 \sqrt{1-0.9^{2}}=2 \sqrt{0.19}<0.9
\end{aligned}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 13,248 |
11.2 The lengths of the lateral edges of a triangular pyramid are 1, 2, and 4. Prove that the triangle lying at the base of the pyramid is not equilateral. | Solution. We will prove that if the base of the pyramid is an equilateral triangle, the lengths of the lateral edges cannot form the triplet $1,2,4$. Assume the opposite. See the diagram.
We use the triangle inequality.
In triangle ASB: $S A + S B > A B$, i.e., $3 > a$.
In triangle ASC: $S A + A C > S C$, i.e., $a >... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,249 |
11.3 In a circle of radius 3, several circles are placed arbitrarily, the sum of whose radii is 25. Prove that there exists a line that intersects at least nine of these circles. | Solution. Let's project all circles onto an arbitrary diameter of the large circle. The sum of the lengths of the projections is equal to the sum of the diameters of the circles, i.e., 50. If each point on the large diameter is covered by projections no more than eight times, the sum of their lengths does not exceed $6... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,250 |
11.4 The sequence of natural numbers $\mathrm{q}_{1}<\mathrm{q}_{2}<\mathrm{q}_{3}<\ldots$ is such that $\mathrm{q}_{\mathrm{n}}<2 \mathrm{n}$ for any index $\mathrm{n}$. Prove that any natural number can be represented as the difference of two numbers from this sequence or as a number from the sequence itself. | Solution. Let $\mathrm{m}$ be an arbitrary natural number. Consider the first m terms of the sequence: $\mathrm{q}_{1}<\mathrm{q}_{2}<\ldots<\mathrm{q}_{\mathrm{m}}<2 \mathrm{~m}$. Then, if $\mathrm{q}_{\mathrm{k}}$ is divisible by $\mathrm{m}$, due to the inequality $\mathrm{q}_{\mathrm{k}}<2 \mathrm{~m}$, we simply h... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,251 |
11.5 The sequence of numbers $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}, \ldots$ is defined by the formulas: $\mathrm{a}_{1}=1$, $\mathrm{a}_{\mathrm{n}+1}=\mathrm{a}_{\mathrm{n}}+\frac{1}{\sqrt{\mathrm{n}} \mathrm{a}_{\mathrm{n}}}$ for $\mathrm{n}=1,2,3, \ldots$ Prove that this se... | Solution. We have $\mathrm{a}_{\mathrm{k}+1}^{2}=\left(\mathrm{a}_{\mathrm{k}}+\frac{1}{\sqrt{\mathrm{k}} \mathrm{a}_{\mathrm{k}}}\right)^{2}=\mathrm{a}_{\mathrm{k}}^{2}+\frac{2}{\sqrt{\mathrm{k}}}+\frac{1}{\mathrm{ka}_{\mathrm{k}}^{2}}>\mathrm{a}_{\mathrm{k}}^{2}+\frac{2}{\sqrt{\mathrm{k}}}$. Summing these inequalitie... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,252 |
9.1. In a quiz, 10 students participated and gave a total of 42 correct answers. Prove that at least 2 students gave the same number of correct answers (possibly zero). | Solution. Suppose that all students gave a different number of correct answers. Then 10 students gave no less than $0+1+2+3+4+5+6+7+8+9=45$ (correct answers), which contradicts the condition. Therefore, there will be students who gave the same number of correct answers. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,253 |
9.2. Electronic clocks display time from 00.00.00 to 23.59.59. How much time during the day does the number on the display that reads the same from left to right and from right to left light up? | Answer: 96 seconds.
Solution. If the digits on the display are $a b . c d . m n$, then $a=0,1,2,0 \leq b \leq 9,0 \leq c \leq 5$, $0 \leq d \leq 9,0 \leq m \leq 5,0 \leq n \leq 9$. Therefore, if $a=n, b=m, c=d$, the symmetrical number on the display is uniquely determined by the digits $a, b$ and $c$, where $a=0,1,2,0... | 96 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,254 |
9.3. Factor the expression $a^{3}(b-c)+c^{3}(a-b)-b^{3}(a-c)$. | Answer. $(a-b)(b-c)(a-c)(a+b+c)$. | (-b)(b-)(-)(+b+) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,255 |
9.4. Does there exist a composite 2021-digit natural number that remains composite when any three consecutive digits are replaced by any three arbitrary digits? | Answer. Yes, it exists.
Solution. Let the desired composite number be even. Then, by replacing any triplet of digits, except the last one, we will again get an even number (i.e., composite). Construct a number \( N \) that ends in 000, so it will be even. Replacing the last triplet of digits of \( N \) is the same as ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,256 |
9.5. $A L$ and $B M$ are the angle bisectors of triangle $A B C$. The circumcircles of triangles $A L C$ and $B M C$ intersect again at point $K$, which lies on side $A B$. Find the measure of angle $A C B$. | Answer. $\angle A C B=60^{\circ}$.
Solution. Draw the segment $C K . \angle L C K=\angle L A K$ (these angles are inscribed in the circle and subtend the same arc). Similarly, $\angle M C K$ $=\angle M B K$. Since $\angle A C B=\angle L C K+\angle M C K$, the desired angle $A C B$ is one-third of the sum of the angles... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,257 |
# 1. Option 1
On the sheet, three rectangles A, B, and C are drawn.
A
5
B
Rectangles A and B have the same width, and rectangles B and C have the same length (width - top to bottom, lengt... | Answer: 88.
Solution: Let rectangle A have a length of $a$ cm and a width of $b$ cm. If the length is increased by 2 cm, the area will increase by $2 b$ cm $^{2}$. Therefore, $2 b=22, b=11$. The area of rectangle B is 40 cm $^{2}$ less than that of rectangle B, so the length of rectangle B is $40: 4=10$ cm. Therefore,... | 88 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,258 |
# 2. Option 1
There are candies in three boxes. It is known that there are 2 times fewer candies in the first box than in the second. It is also known that there are a total of 24 candies in the first and third boxes, and a total of 34 candies in the second and third boxes. How many candies are there in total in the b... | Answer: 44.
Solution: From the condition, it follows that in the second box there are $34-24=10$ more candies than in the first. And this difference is equal to the number of candies in the first box. Therefore, there are 10 candies in the first box, 20 in the second, and 14 in the third.
## Variant 2
In three boxes... | 44 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,259 |
# 3. Option 1.
Vasya calculated two sums: the sum of consecutive odd numbers from 1 to 2021: $(1+3+5+\ldots+2021)$ and the sum of consecutive even numbers from 2 to 2020: $(2+4+6+\ldots+2020)$. After that, he subtracted the smaller sum from the larger one. What result did he get? | Answer: 1011.
Solution. Let's find the difference between the sums of odd and even numbers: $(1+3+5+\ldots+2021)-(2+4+6+\ldots+2020)=(2021-2020)+(2019-2018)+\ldots+(3-2)+1=1011$. | 1011 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,260 |
# 4. Variant 1.
A square piece of paper is folded as follows: the four corners are folded to the center so that they meet at one point (see figure),
resulting in a square again. After perfo... | Answer: 12.
Solution: After each operation, the thickness of the square doubles, and the area is halved. Since the thickness has become 16 sheets, the operation was applied 4 times. In this process, the area decreased by a factor of 16 and became equal to 9 square centimeters. Therefore, the area of the original squar... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,261 |
# 5. Variant 1
Given nine cards with the numbers $5,5,6,6,6,7,8,8,9$ written on them. From these cards, three three-digit numbers $A$, B, V were formed, each of which has all three digits different. What is the smallest value that the expression A + B - V can have? | Answer: 149.
Solution. By forming the smallest sum of numbers A and B, as well as the largest number C, we get the smallest value of the expression A + B - C: $566 + 567 - 988 = 145$. However, this combination is not suitable: two numbers have the same digits. By swapping the digits 6 and 8 in the units place, we get ... | 149 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,262 |
# 6. Variant 1
Given the road map of the kingdom. The cities are marked with numbers, and the segments represent roads. One day, a traveling knight started his journey from one of the cities in the kingdom and managed to construct his route in such a way that he traveled each road exactly once. In which city, marked w... | Answer: 2, 5.
Solution: If a city is not the beginning or the end of the knight's journey, then every time he enters through one road, he must exit through another road. This means that the roads from such a city come in "pairs," and there is an even number of them in total. Therefore, cities 2 and 5, from which an od... | 2,5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,263 |
# 8. Variant 1.
Each of the 10 students came up with 5 natural numbers. It turned out that each number was thought of by at least three students. What is the maximum number of different numbers that could have been thought of? | Answer: 16.
Solution: In total, the students came up with 50 numbers, with each number being counted at least 3 times. We will prove that there could not have been more than 16 different numbers. If at least 17 different numbers were thought of and each by at least three students, then a total of no less than $17 \cdo... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,265 |
4. What is the maximum area of the orthogonal projection of a unit cube onto a plane
Answer. $S_{\max }=2 \sqrt{3}$. | The center of a cube is its center of symmetry. The edges of the cube can be divided into three classes, with each class containing edges that are parallel to each other. When orthogonally projecting the cube onto a plane in any direction, all edges of the same class are depicted as parallel segments. The result of the... | 2\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,266 |
5. Twelve pencils are sharpened so that they all have different lengths. Masha wants to put the pencils in a box in two rows of 6 each, so that in each row the lengths of the pencils decrease from left to right, and each pencil in the second row lies on a longer pencil. In how many ways can she do this?
Answer: 132. | A layout of pencils that satisfies the conditions in the problem will be called correct. We will stack the pencils in a box in descending order of length. The order of placement will be recorded in the form of a broken path from the point (0,0) in a rectangular Cartesian coordinate system on a plane to the point $(6,6)... | 132 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,267 |
1. It is known that the sequence of numbers $a_{1}, a_{2}, \ldots$, is an arithmetic progression, and the sequence of numbers $a_{1} a_{2}, a_{2} a_{3}, a_{3} a_{4} \ldots$, is a geometric progression. It is known that $a_{1}=1$. Find $a_{2017}$. | Answer: $a_{2017}=1$;
Trunov K.V.
## Solution:
Since the sequence of numbers $a_{1} a_{2}, a_{2} a_{3}, a_{3} a_{4} \ldots$, is a geometric progression, then $\left(a_{n} a_{n+1}\right)^{2}=\left(a_{n-1} a_{n}\right)\left(a_{n+1} a_{n+2}\right)$ for $n \geq 2$. From this, we obtain that $a_{n} a_{n+1}=a_{n-1} a_{n+2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,270 |
# 2. Prove that for any distinct prime numbers p, q, t
the number $2016^{p}+2017^{q}+2018^{t}$ is composite. Trunov K.V.
# | # Solution:
Since $p, q, t$ are distinct prime numbers, then exactly one of them can be equal to 2, while the others are odd prime numbers.
1) Let $t$ be an odd prime number, then we get that $2016=0(\bmod 3), 2017 \equiv 1(\bmod 3), 2018 \equiv-1(\bmod 3) \Rightarrow 2016^{p} \equiv 0(\bmod 3)$, $2017^{q}=1(\bmod 3)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,271 |
3. Find the number of triples of natural numbers $a, b$, and $c$ not exceeding 2017 such that the polynomial $x^{11} + a x^{7} + b x^{3} + c$ has a rational root.
Trunov K.V. | Answer: 2031120
Solution: Since all coefficients of the polynomial are natural and the leading coefficient is 1, any rational root must be an integer.
Notice that for $x \geq 0$, $x^{11} + a x^{7} + b x^{3} + c \geq c \geq 1$. Therefore, there are no integer roots for $x \geq 0$. If $x \leq -2$, then $x^{11} + a x^{7... | 2031120 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,272 |
# 4. A circle with center at point $O$ is inscribed in quadrilateral $A B C D$. Segments $O A, O B, O C$ and $O D$ intersect the circle at points $K, L, M$ and $N$ respectively. $P$ is the intersection point of the diagonals of quadrilateral $K L M N$, and $Q$ is the midpoint of segment $K L$. Prove that lines $P Q$ an... | # Solution.
$K O$ is the bisector of $\angle L K N, L O$ is the bisector of $\angle K L M$. Therefore, $\angle K O L=180^{\circ}-\angle L K O-\angle K L O=180^{\circ}-\frac{1}{2}(\angle L K N+\angle K L M)$.
Similarly, $\angle M O N=180^{\circ}-\frac{1}{2}(\angle L M N+\angle K N M)$.
Then $\angle K O L+\angle M O N... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,273 |
5. Fyodor starts writing pairs of natural numbers $(a, b)$ on the board, where $a<b$ and each does not exceed 2018. Moreover, if a pair of numbers $(a, b)$ is already written on the board, he cannot write any pair of the form $(c, a)$ or $(b, d)$. What is the maximum number of pairs of numbers he can write on the board... | Answer: $1009^{2}=1018081$.
Solution: Let A be the set of all the first numbers in the pairs, and B be the set of all the second numbers in the pairs. Then, from the condition of the problem, it follows that
$\mathrm{A} \cap \mathrm{B}=\varnothing$. Suppose there are $n$ elements in set A and $k$ elements in set B. Th... | 1009^2=1018081 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,274 |
5.1. Provide any solution to the riddle: 2014 + YEAR = SOCHI. (Different letters represent different digits.) | Answer: $2014+891=2905$ or $2014+893=2907$ or $2014+896=2910$.
Notice that from the addition of the hundreds place, it follows that the addition of the tens place is carried over, and therefore $\mathrm{O}=8$ (and there was a carry in the units place) or $\mathrm{O}=9$.
# Evaluation criteria:
+ one or more correct a... | 2014+891=2905or2014+893=2907or2014+896=2910 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,275 |
5.3. Nadya baked pies with raspberries, blueberries, and strawberries. The number of pies with raspberries was half of the total number of pies; the number of pies with blueberries was 14 less than the number of pies with raspberries. And the number of pies with strawberries was half the number of pies with raspberries... | Answer: 21 raspberry pies, 7 blueberry pies, and 14 strawberry pies.
Solution. First method. Since the number of raspberry pies is half of the total, the number of blueberry and strawberry pies together is the same as the number of raspberry pies (see Fig. 5.3). Moreover, there are 14 fewer blueberry pies than raspber... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,277 |
5.4. Postman Pechkin delivers mail by bicycle to four villages: Prostokvashino, Smetanino, Tvorozhnoye, and Molochnoye (see figure). He knows that the distance from Prostokvashino to Tvorozhnoye is 9 km, from Prostokvashino to Smetanino is 13 km, from Tvorozhnoye to Smetanino is 8 km, and from Tvorozhnoye to Molochnoye... | Answer: 19 km.
Solution. The distance from Prostokvashino to Smetanino with a detour to Tvorozhnoye is $9+8=17$ (km), and without the detour - 13 km. Therefore, the detour to Tvorozhnoye (from the highway and back) is 4 km. The distance from Prostokvashino to Molochnoye with a detour to Tvorozhnoye is $9+14=23$ (km), ... | 19 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,278 |
5.5. Out of five coins - two are counterfeit. One of the counterfeit coins is lighter than a genuine one, and the other is equally heavier than a genuine one. Explain how to find both counterfeit coins in three weighings on a balance scale without weights. | Solution. First method. Set aside one coin, and place two coins on each pan of the scales. There are two possible cases: 1) the scales are in balance; 2) one of the pans is heavier.
1) Since there are no four genuine coins, balance is only possible in one case: both counterfeit coins are on one pan. Therefore, the nex... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,279 |
10.1. In the product of five natural numbers, each factor was decreased by 3. Could the product have increased exactly 15 times as a result?
(N. Agakhanov, I. Bogdanov) | Answer. Yes, it could.
Solution. The product $1 \cdot 1 \cdot 1 \cdot 1 \cdot 48$ serves as an example. After the specified operation, we get
$$
(-2) \cdot(-2) \cdot(-2) \cdot(-2) \cdot 45=720=15 \cdot 48
$$
Remark. The given example is the only one. We will explain how to come up with it. Suppose that four of the f... | 1\cdot1\cdot1\cdot1\cdot48 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,280 |
10.2. A circle with center at point $I$ is inscribed in quadrilateral $A B C D$. Rays $B A$ and $C D$ intersect at point $P$, and rays $A D$ and $B C$ intersect at point $Q$. It is known that point $P$ lies on the circumcircle $\omega$ of triangle $A I C$. Prove that point $Q$ also lies on the circumcircle $\omega$.
(... | Solution. Since the quadrilateral $A I C P$ is inscribed, then $\angle P C I=180^{\circ}-\angle P A I=\angle B A I ;$ in other words, $\angle D C I=$ $=\angle B A I$ (see Fig. 4). The center $I$ of the inscribed circle of the quadrilateral lies on the bisectors of its angles, so $\angle D C I=$ $=\angle B C I$ and $\an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,281 |
10.3. Pasha chose 2017 (not necessarily distinct) natural numbers $a_{1}, a_{2}, \ldots, a_{2017}$ and plays the following game with himself. Initially, he has an unlimited supply of stones and 2017 large empty boxes. In one move, Pasha adds $a_{1}$ stones to any box (of his choice), $a_{2}$ stones to any of the remain... | Answer. Yes, he could.
Solution. Note that $2017=43 \cdot 46+39$. Let's provide an example of Pasha's numbers for which the required condition is met. Suppose among his numbers there are 39 twos, 46 numbers equal to 44, and the rest are ones.
To achieve the required in 43 moves, Pasha selects 39 boxes, into which he ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,282 |
10.4. The teacher is going to give the children a problem of the following type. He will inform them that he has thought of a polynomial $P(x)$ of degree 2017 with integer coefficients, the leading coefficient of which is 1. Then he will tell them $k$ integers $n_{1}, n_{2}, \ldots, n_{k}$, and separately inform them o... | Answer. For $k=2017$.
Solution. First, we prove that $k>2016$. Suppose the teacher used some $k \leqslant 2016$, thinking of the polynomial $P(x)$. Consider the polynomial $Q(x)=P(x)+\left(x-n_{1}\right)\left(x-n_{2}\right) \ldots(x-n_{k})$. Note that the degree of the polynomial $Q(x)$ is also 2017, and its leading c... | 2017 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,283 |
11.1. Petya wrote ten natural numbers on the board, none of which are equal. It is known that among these ten numbers, three can be chosen that are divisible by 5. It is also known that among the ten numbers written, four can be chosen that are divisible by 4. Can the sum of all the numbers written on the board be less... | Answer. It can.
Solution. Example: $1,2,3,4,5,6,8,10,12,20$. In this set, three numbers $(5,10,20)$ are divisible by 5, four numbers $(4,8,12,20)$ are divisible by 4, and the total sum is 71.
Remark. It can be proven (but, of course, this is not required in the problem), that in any example satisfying the problem's c... | 71 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,284 |
11.2. Given a quadratic trinomial $P(x)$. Prove that there exist pairwise distinct numbers $a, b$, and $c$ such that the following equalities hold:
$$
P(b+c)=P(a), P(c+a)=P(b), P(a+b)=P(c)
$$
(N. Agakhanov) | Solution. Let $d$ be the abscissa of the vertex of the parabola $y=P(x)$, so that the line $x=d$ is the axis of symmetry of the parabola. Then for any numbers $t$ and $s$ with sum $2 d$ (i.e., such that the points $t$ and $s$ are symmetric with respect to $d$), it holds that $P(t)=P(s)$. Thus, any triple of pairwise di... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,285 |
11.3. In the triangular pyramid $A B C D$, points $A^{\prime}$ and $B^{\prime}$ were found on its faces $B C D$ and $A C D$ respectively, such that $\angle A B^{\prime} C=$ $=\angle A B^{\prime} D=\angle B A^{\prime} C=\angle B A^{\prime} D=120^{\circ}$. It is known that the lines $A A^{\prime}$ and $B B^{\prime}$ inte... | Solution. From the condition of the problem, it follows that points $A, A^{\prime}, B, B^{\prime}$ lie in the same plane, so lines $B A^{\prime}$ and $A B^{\prime}$ intersect edge $C D$ at one point $X$. From the condition, it follows that these lines are the bisectors of angles $C A^{\prime} D, C B^{\prime} D$ respect... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,286 |
11.4. In a company, some pairs of people are friends (if $A$ is friends with $B$, then $B$ is also friends with $A$). It turned out that for any selection of 101 people from this company, the number of pairs of friends among them is odd. Find the largest possible number of people in such a company.
(E. Bakayev, I. Bog... | Answer: 102.
Solution: In all solutions below, we consider the friendship graph, where vertices are people in the company, and two people are connected by an edge if they are friends.
Consider 102 vertices, and construct the following graph on them. Connect one vertex $x$ to three others $v_{1}, v_{2}, v_{3}$. Divide... | 102 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,287 |
11.5. Let $S$ be a 100-element set consisting of natural numbers not exceeding 10000. Mark all points in space where each of the coordinates belongs to the set $S$. Attach to each of the 1000000 marked points $(x, y, z)$ a ball with the number $\frac{x^{2}+y^{2}+z^{2}}{x y+y z+z x}$ written on it. What is the maximum n... | Answer. $3 \cdot C_{100}^{2}=14850$.
Solution. Let's call a triplet of natural numbers $(x, y, z)$, whose elements belong to $S$, good if
$$
x^{2}+y^{2}+z^{2}=2(x y+y z+z x)
$$
Thus, we need to find the maximum possible number of good triplets.
Let's determine when a triplet is good. Rewrite (*) as a quadratic equa... | 14850 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,288 |
2. Buratino accurately calculated the time and left Papa Carlo's house at 13:40 to arrive at the Field of Wonders and plant 4 soldi exactly at sunset. If he had walked 25% faster, he would have arrived at the Field of Wonders 1.5 hours earlier and waited. At what time did Buratino arrive at the Field of Wonders? | Solution. Since the distance is the same in both cases, and the speed is $\frac{5}{4}$ times greater (25%), the time spent will be $\frac{5}{4}$ times less. We get the equation $\frac{4}{5} t=t-\frac{3}{2} \Rightarrow t=7.5$ hours. Therefore, Buratino arrived at the Field of Wonders at $21:10$.
Answer 21:10.
## Gradi... | 21:10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,290 |
3. Let's call a four-digit number accompanying the year $\overline{20 a b}$ if it also ends in $\overline{a b}$ and, in addition, is divisible by $\overline{a b}$ (a two-digit number), for example, the number 4623 accompanies the year 2023. How many numbers accompany the year $2022?$ | Solution. $\overline{m n 22}=\overline{m n} \cdot 100+22 \Rightarrow \overline{m n} \cdot 100: 22 \Rightarrow \overline{m n}: 11: 11,22,33,44,55,66,77$, 88,99 - 9 numbers.
Answer 9.
## Grading Criteria.
Correct answer with valid reasoning - 7 points.
Valid reasoning for divisibility by 11 and description of the set... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,291 |
5. In a round-robin football tournament (where each team plays every other team once), 12 teams participated. The tournament was not yet completed by November. Could it be that one team played exactly 11 games, three teams played exactly 9 games each, two teams played exactly 6 games each, four teams played exactly 4 g... | # Solution.
The team that played 11 games played with all the teams, which means the two teams that played only 1 game each played only with this team and no one else.
The teams that played 9 games (3 teams) played with all the teams except the two mentioned above (who played 1 game each). Therefore, the teams that p... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,293 |
6. How many solutions does the rebus $\frac{B+O+C+b+M+O+\breve{U}}{K+J+A+C+C}=\frac{22}{29}$ have, where different letters represent different digits, and the same letters represent the same digits? It is known that the digit 0 is not used. | Solution. In the rebus, 9 different letters are used, i.e., all non-zero digits are used. $1+2+3+0+0 \leq K+Л+A+C+C \leq 6+7+8+9+9$, i.e.
$$
6 \leq K+Л+A+C+C \leq 39 . \quad \text { Since the fraction } \frac{22}{29} \text { is irreducible, the number }
$$
$K+Л+A+C+C$ must be divisible by 29, the only case that satis... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,294 |
10.1. Different positive numbers $x, y, z$ satisfy the equations
$$
x y z=1 \quad \text{and} \quad x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}
$$
Find the median of them. Justify your answer. | # Solution:
Method 1. Consider the polynomial $P(t)=(t-x)(t-y)(t-z)$. The numbers $x, y, z$ are its roots. Expanding the brackets or using Vieta's theorem for cubic polynomials, we get $P(t)=t^{3}+a t^{2}+b t+c$, where $a=-(x+y+z)$, $b=xy+xz+yz$, and $c=-xyz$. The condition of the problem shows that $a=-b$ and $c=-1$.... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,295 |
10.2. Circles with centers $O_{1}$ and $O_{2}$ intersect at points $A$ and $B$, with $O_{1} O_{2} > O_{1} A > O_{2} B$. Let $P$ and $Q$ be the points of intersection of the circumcircle of triangle $O_{1} A O_{2}$ with the first and second circles, respectively. Prove that the segments $O_{1} Q$ and $O_{2} P$ intersect... | Solution: The condition $O_{1} O_{2}>O_{1} A>O_{2} B$ means that points $O_{1}$ and $O_{2}$ cannot lie inside the second and first circles, respectively. Therefore, their positions are as indicated in the diagram. We will show that the line $O_{2} P$ passes through point $B$ (the fact that the line $O_{1} Q$ passes thr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,296 |
10.3. A group of 25 children from kindergarten returned from a walk. To dry, the children laid out all their 50 mittens in a row on the radiator. The group's teacher noticed that the mittens of each child were not next to each other. Moreover, it turned out that between the mittens of any child, there lay a prime numbe... | # Solution:
Method 1. Assume the opposite, that is, let there be a prime number of other mittens between each child's mittens, greater than two. Number the mittens from 1 to 50 in the order they are placed on the radiator (from left to right). Then the total sum of the mitten numbers is $1+2+\ldots+50$ - an odd number... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,297 |
10.4. The graph of the quadratic trinomial
$$
y=\frac{2}{\sqrt{3}} x^{2}+b x+c
$$
intersects the positive half-axis of the y-axis at point $M$, and the positive half-axis of the x-axis at points $K$ and $L$ (point $L$ is to the right of point $K$). It turns out that $\angle L K M=120^{\circ}$, and $K L=K M$. Find the... | Solution: Consider the triangle $K O M$ ($O$ - the origin). It is a right triangle, and its angle $K$, being adjacent to angle $L K M$, is $60^{\circ}$. Then $O M = O K \cdot \sqrt{3}$ and $K L = K M = 2 O K$. Let the abscissa of point $K$ be $t > 0$. Then the abscissa of point $L$ is $3 t$. The numbers $t$ and $3 t$ a... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,298 |
10.5. Professor Potapov exchanges a chisel for soap, Academician Agatov exchanges 4 soaps for 1 chisel, and Associate Professor Dolmatov exchanges 1 soap for 5 chisels (but not vice versa). After several exchanges, Student Sidorov ended up with the same number of chisels and soaps as he had at the beginning. Prove that... | Solution: Let Sidorov have made $p$ exchanges with the professor, $a-\mathrm{c}$ with the academician, and $d$ with the associate professor. Then he gave them $p$ soaps and $a+5 d$ shillings, and received from them $p$ shillings and $4 a+d$ soaps. According to the condition, the amount of soap given and received is the... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,299 |
10.6. In an equilateral triangle $ABC$, a point $P$ is chosen such that $AP=10, BP=8, CP=6$. Find the area of this triangle.
# | # Solution:
Method 1. Rotate triangle $A P B$ by $60^{\circ}$ around point $A$, triangle $B P C$ by $60^{\circ}$ around point $B$, and triangle $C P A$ by $60^{\circ}$ around point $C$ (all rotations counterclockwise) - see the left figure.
$ that simultaneously satisfy the following conditions: 1) in each pair, $x$ and $y$ do not coincide; 2) in each subsequent pair, the number $x$ is 1 greater than the number $x$ of the previous pair; 3) in each subsequent pair, the number $y$ is 1 greater than the... | Solution. Such pairs do exist. It is sufficient to provide an example.
$$
(2018!+1 ; 1),(2018!+2 ; 2),(2018!+3 ; 3), \ldots,(2018!+2018 ; 2018)
$$
Answer. They do exist. | Theydoexist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,302 |
3. In triangle $ABC$, angle $A$ is equal to $\alpha$, and $BC$ is the shortest side. On side $AB$, point $P$ is marked, and on side $AC$, point $Q$ is marked such that $PB = BC = CQ$. Segments $BQ$ and $CP$ intersect at point $M$. Find the measure of angle $BMC$. | Solution.
Triangles $P B C$ and $Q B C$ are isosceles, with corresponding sides marked on the diagram. Let the angles be: $\angle Q B C=\angle B Q C=x, \angle P B C=\angle B P C=$ $y, \angle A B Q=z, \angle A C P=v$. By the property of the exterior angle, applied to triangles $A B Q$ and $A P C$,
$$
\left\{\begin{arr... | 90-\frac{1}{2}\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,303 |
4. A company is called public if it has at least 15 shareholders. A shareholder of a company is called a minority shareholder if they own no more than $25 \%$ of the shares of that company. On the stock exchange where the shares are traded, one sixth of the companies are public. Prove that among all shareholders partic... | Solution. Let the number of firms on the exchange be N. We will call a shareholder who owns more than $25 \%$ of a firm's shares a real shareholder. The number of real shareholders in one firm cannot exceed three. (If there are four, then one of them cannot own more than $25 \%$ of the shares). Therefore, the total num... | 20 | Combinatorics | proof | Yes | Yes | olympiads | false | 13,304 |
5. Can the square of some natural number be the least common multiple of two consecutive natural numbers? Justify your answer. | Solution. No, the least common multiple (LCM) of consecutive numbers cannot be a square of a natural number. Note that two consecutive numbers, $n$ and $n+1$, are coprime. (This fact is considered known, and if necessary, it can be proven as follows: if $n$ has a divisor $d, d>1$, then $n+1$ cannot have the divisor $d$... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,305 |
1. Kolya, Seryozha, and Vanya regularly go to the cinema: Kolya goes there every 4th day, Seryozha - every 5th day, and Vanya - every 6th day. Today all the boys were at the cinema. When will all three meet at the cinema again? | 1. Answer: 60.
Let's start numbering the days beginning with tomorrow (today's day is "zero"). Kolya will go to the cinema on the days whose numbers are divisible by 4, Seryozha - on the days whose numbers are divisible by 5, Vanya - on the days whose numbers are divisible by 6. For them to all be at the cinema togeth... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,306 |
2. There are 25 people in the class. Is it possible that 6 of them have exactly 3 friends, 10 have exactly 4 friends, and 9 have exactly 5 friends in this class? | 2. Answer. No.
Let's count the number of "friendships". It should be equal to ( $6 \cdot 3+10 \cdot 4+9 \cdot 5$ ) / 2 (if we simply add up the number of friends each person has, each "friendship" will be counted twice), but this is not an integer. Therefore, the situation described in the problem is impossible. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,307 |
3. Natural numbers from 1 to 6 are arranged in a circle in order. It is allowed to add 1 to any three consecutive numbers or subtract 1 from any three numbers that are spaced one apart. Is it possible to make all the numbers equal using several such operations? | 3. Answer: No.
Consider three sums - the first and fourth, the second and fifth, and the third and sixth numbers. Initially, these sums were $1+4=5$, $2+5=7$, and $3+6=9$. When the first operation is performed, each of these three sums increases by 1, and when the second operation is performed, each of these sums decr... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,308 |
4. Four friends came back from fishing. Every two of them counted the sum of their catches. Six numbers were obtained: $7,9,14,14,19,21$. How many fish were caught in total? | 4. 28 fish.
Note that the catch of each person is counted in exactly three sums. Therefore, if we add up all six numbers, we get the tripled total catch. Thus, the total number of fish caught is $(7+9+14+14+19+21): 3=28$ fish. | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,309 |
5. A pile contains 50 stones. Two players take turns adding from 1 to 9 stones to it. The player who brings the total number of stones to 200 wins. Who will it be - the first or the second? | 5. Answer: the second wins.
The second has a winning strategy. The second can always complement the first's move to make the sum 10. Therefore, after the first's move, the last digit will be non-zero, and after the second's move, it will be zero. With this strategy, the second always has a move as long as there are fe... | thewins | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,310 |
8.1. From a $7 \times 7$ square grid, an equal number of $2 \times 2$ squares and $1 \times 4$ rectangles were cut out along the grid lines. What is the maximum number of these figures that could have been cut out? | Answer: 12.
Solution: Both the square and the rectangle consist of 4 cells. Therefore, the number of cut-out figures is no more than 49/4, that is, no more than 12. There are an equal number of figures of both types, so there are no more than 6 squares $2 \times 2$ and rectangles $1 \times 4$. The diagram shows how to... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,311 |
8.2. Find the value of the expression $a^{3}+12 a b+b^{3}$, given that $a+b=4$. | Answer: 64.
Solution. $a^{3}+12 a b+b^{3}=a^{3}+b^{3}+12 a b=(a+b)\left(a^{2}-a b+b^{2}\right)+12 a b=4 a^{2}-4 a b+4 b^{2}+12 a b=4(a+b)^{2}=64$. | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,312 |
8.3. On an island, there live knights and liars. Knights always tell the truth, and liars always lie. Some of the island's inhabitants claimed that there is an even number of knights on the island, while all the other inhabitants claimed that there is an odd number of liars on the island. Could there be exactly 2021 in... | Answer: cannot.
Solution. If the number of island inhabitants is odd, then either there is an even number of knights and an odd number of liars, or an odd number of knights and an even number of liars. In the first case, it turns out that all the islanders told the truth. But then all of them are knights, and the numb... | cannot | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,313 |
8.4. Point $D$ lies on the hypotenuse $B C$ of a right triangle $A B C$, but does not coincide with its midpoint. Prove that among the segments $A D, B D$, and $C D$ there are no equal ones. | First solution. We complete the right triangle $ABC$ to a rectangle $ABEC$. The point $O$ of intersection of the diagonals of this rectangle is the midpoint of the hypotenuse of the triangle. Since the diagonals of a rectangle are equal, it follows that $AO = BO = CO$. Considering now the isosceles triangles $AOB$ and ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,314 |
8.5. A natural number is called interesting if all its digits are different, and the sum of any two adjacent digits is a square of a natural number. Find the largest interesting number. | Answer: 6310972.
Solution. Let's mark 10 points on the plane, representing the digits from 0 to 9, and connect those points whose sum is a square of a natural number.
On this diagram, we ne... | 6310972 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,315 |
9.8. In a state, there are $n$ cities, and an express train runs between each pair of them (in both directions). For any express train, the ticket prices for the "outbound" and "return" trips are the same, and for any two different express trains, these prices are different. Prove that a traveler can choose an initial ... | First solution. Remove all express trains, and then start launching them back one by one in ascending order of price (i.e., the first one launched is the cheapest, the second one is the cheapest of the remaining, and so on). At each moment, in each city, we will write the maximum number of express trains that can be se... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,316 |
10.1. At each point $A$ in the plane, there stands a real number $f(A)$. It is known that if $M$ is the point of intersection of the medians of triangle $ABC$, then $f(M)=f(A)+f(B)+f(C)$. Prove that $f(A)=0$ for all points $A$. | Solution. Let's take an arbitrary point $M$ on the plane and prove that $f(M)=0$. For this, consider an arbitrary triangle $ABC$ for which point $M$ is the centroid. Denote by $D$, $E$, and $F$ the centroids of triangles $BCM$, $CAM$, and $ABM$, respectively (see Fig. 2).
(A. S. Golvachev)
Note that point $M$ is also... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,317 |
10.2. Pasha and Vova are playing the following game, taking turns. Pasha starts. Initially, the boys have a large piece of plasticine in front of them. On his turn, Pasha can cut any of the existing pieces of plasticine into three parts (not necessarily equal). Vova, on his turn, chooses two pieces and combines them. P... | Answer. No, he cannot.
First solution. We will provide an algorithm that allows Pasha to win. Let the mass of the initial piece be 1 kg. Pasha will cut off two pieces, each weighing 0.01 g, from the largest of the available pieces with each move. We will prove that Pasha will win no later than after 10,000 moves.
Sup... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,318 |
10.3. In an intergalactic hotel, there are 100 rooms with capacities of
$101, 102, \ldots, 200$ people. In these rooms, a total of $n$ people live. A VIP guest has arrived at the hotel, and a whole room needs to be vacated for them. To do this, the hotel manager selects one room and relocates all its occupants to anot... | Answer: 8824.
Solution: Suppose that with 8824 guests, the director cannot carry out the relocation. Let's divide the rooms into pairs by capacity: $101-200, 102-199, \ldots, 150-151$. Note that for each pair of rooms, the total number of people living in the two rooms is greater than the capacity of the larger room i... | 8824 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,319 |
10.4. Given an acute-angled triangle $A B C$, where $A C < B C$. A circle passes through points $A$ and $B$ and intersects segments $C A$ and $C B$ again at points $A_{1}$ and $B_{1}$, respectively. The circumcircles of triangles $A B C$ and $A_{1} B_{1} C$ intersect again at point $P$. Segments $A B_{1}$ and $B A_{1}$... | Solution. Using circles $(P A B C)$ and $\left(P A_{1} B_{1} C\right)$, we get $\angle P A B=180^{\circ}-\angle P C B=\angle P A_{1} B_{1}$ and similarly $\angle P B A=\angle P B_{1} A_{1}$. Thus, $\triangle P A B \sim \triangle P A_{1} B_{1}$. From this similarity, it follows that $\angle A P A_{1}=\angle A P B \pm \a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,320 |
10.6. In an acute-angled triangle $A B C$, the bisector $B L$ is drawn. Points $D$ and $E$ are the midpoints of the smaller arcs $A B$ and $B C$ of the circumcircle $\omega$ of triangle $A B C$. On the extension of segment $B D$ beyond point $D$, point $P$ is marked, and on the extension of segment $B E$ beyond point $... | Solution. Let $\angle B A C=2 \alpha, \angle A C B=2 \gamma$. Without loss of generality, assume that $\alpha \geqslant \gamma$. Since points $D$ and $E$ are the midpoints of arcs $A B$ and $A C$ of the circle $\omega$, we have $\angle A B D=\frac{\angle A C B}{2}=\gamma$ and $\angle C B E=\alpha$.
Let $M$ and $N$ be ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,322 |
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