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1. Variant 1. Given two natural numbers. One number was increased by 3, and the other was decreased by 3. As a result, their product increased by 600. By how much will the product decrease if the opposite is done: the first number is decreased by 3, and the second is increased by 3? | Answer: 618.
Solution: Let these numbers be $a$ and $b$. Then, according to the condition, $(a-3)(b+3)-ab=600$. Expanding the brackets: $ab+3a-3b-9-ab=600$, so $a-b=203$. We need to find the difference $ab-(a+3)(b-3)=$ $ab-ab+3a-3b+9=3(a-b)+9=3 \cdot 203+9=618$. | 618 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,972 |
2. Variant 1. At the intersection of perpendicular roads, a highway from Moscow to Kazan and a road from Vladimir to Ryazan intersect. Dima and Tolya set out with constant speeds from Moscow to Kazan and from Vladimir to Ryazan, respectively. When Dima passed the intersection, Tolya had 900 meters left to reach it. Whe... | Answer: 1500.
Solution: When Tolya has traveled 900 meters, Dima will have traveled 600 meters, so at the moment when Tolya is 900 meters from the intersection, Dima will be 1200 meters from the intersection. According to the Pythagorean theorem, the distance between the boys is 1500 meters. | 1500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,973 |
3. Variant 1. Petya and Masha take candies from a box in turns. Masha took one candy, then Petya took 2 candies, Masha - 3 candies, Petya - 4 candies, and so on. When the number of candies in the box became less than needed for the next turn, all the remaining candies went to the one whose turn it was to take candies. ... | Answer: 110.
Solution. Since $1+3+5+7+9+11+13+15+17+19=100101$, then Masha got the last candy. Then Petya took for himself $2+4+6+8+10+12+14+16+18+20=2(1+2+3+4+5+6+7+8+9+10)=110$ candies. | 110 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,974 |
4. Variant 1. An ant, starting from point A, goes $1+\frac{1}{10}$ cm north, then $2+\frac{2}{10}$ cm west, then $3+\frac{3}{10}$ cm south, then $4+\frac{4}{10}$ cm east, then $5+\frac{5}{10}$ cm north, then $6+\frac{6}{10}$ cm west, and so on. After 1000 steps, the ant is at point B. Find the distance between points A... | Answer: 605000.
Solution. Let's divide 1000 steps into quartets. After each quartet, the ant will move southeast relative to its current position, by a distance equal to the diagonal of a square with side 2.2, i.e., $\sqrt{2.2^{2}+2.2^{2}}=\sqrt{9.68}$. After 250 such quartets, the ant will be at a distance of $250 \c... | 605000 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,975 |
# 5. Option 1.
It is known that the equations $x^{2}+(2 a-5) x+a^{2}+1=0$ and $x^{3}+(2 a-5) x^{2}+\left(a^{2}+1\right) x+a^{2}-4=0$ have common roots. Find the sum of these roots. | Answer: 9.
Solution. The left side of the second equation is obtained from the left side of the first equation by multiplying by $x$ and adding the expression $a^{2}-4$. Therefore, $a^{2}-4=0$, from which $a=2$ or $a=-2$. If $a=2$, then the first equation has no roots. Therefore, $a=-2$, and then we get that the first... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,976 |
6. Variant 1. Grisha thought of such a set of 10 different natural numbers that their arithmetic mean is 16. What is the maximum possible value of the largest of the numbers he thought of? | Answer: 115.
Solution: The sum of the given numbers is $10 \cdot 16=160$. Since all the numbers are distinct, the sum of the 9 smallest of them is no less than $1+2+\cdots+9=45$. Therefore, the largest number cannot be greater than $160-45=115$. This is possible: $(1+2+\cdots+9+115): 10=16$. | 115 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,977 |
7. Variant 1. Roma decided to create his own multiplication table. The rows correspond to the numbers 12, $13,14, \ldots, 60$, and the columns - to the numbers $15,16,17, \ldots, 40$. In the cells of the table, he wrote the products of the pairs of numbers from the row and column. How many of these products will be eve... | Answer: 962.
Solution. Note that the product of two numbers is odd if and only if both factors are odd, and even in all other cases. In total, the table contains 49$\cdot$26 products. Note that among the numbers from 12 to 60, there are 24 odd numbers, and among the numbers from 15 to 40, there are 13 odd numbers. The... | 962 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,978 |
8. Variant 1. In triangle $A B C$, the bisector $A L$ and the median $B M$ are drawn. It turns out that $A B=2 B L$. What is the measure of angle $B C A$, if $\angle L M A=127^{\circ}$? | Answer: $74^{\circ}$.
Solution.
By the property of the angle bisector, $\frac{A C}{C L}=\frac{A B}{B L}=\frac{2}{1}$, so $A C=2 C L$. Since $M$ is the midpoint of $A C$, then $A M=$ $M C=C ... | 74 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,979 |
1. Arrange the numbers $\sin \frac{20}{19}, \cos \frac{20}{19}, \tan \frac{20}{19}, \cot \frac{20}{19}$ in ascending order. Justify your answer. | 1. Since $0\cos \frac{20}{19}$ and $\operatorname{tg} \frac{20}{19}=\sin \frac{20}{19}: \cos \frac{20}{19}>\sin \frac{20}{19}$. Since $\frac{20}{19}>\frac{\pi}{3}$, then
$$
\operatorname{ctg} \frac{20}{19}<\operatorname{ctg} \frac{\pi}{3}=\frac{1}{\sqrt{3}}<\frac{\sqrt{3}}{2}=\sin \frac{\pi}{3}<\sin \frac{20}{19}
$$
... | \cos\frac{20}{19},\operatorname{ctg}\frac{20}{19},\sin\frac{20}{19},\operatorname{tg}\frac{20}{19} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,980 |
2. Solve the equation in natural numbers: $1+x+x^{2}+x^{3}=2^{y}$. | 2. Transform the equation to the form: $\left(1+x^{2}\right)(1+x)=2^{y}$, from which $1+x^{2}=2^{m}$, where $m$ is a non-negative integer, and $1+x=2^{n}$, where $n$ is a non-negative integer. Since $x=2^{n}-1$, then $x^{2}=\left(2^{n}-1\right)^{2}=2^{2 n}-2 \cdot 2^{n}+1$. Considering that $1+x^{2}=2^{m}$, we get $2^{... | 1,2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,981 |
3. The lengths of the sides of a triangle are consecutive natural numbers, $R$ and $r$ - the radii of the circumscribed and inscribed circles, respectively. Prove that $R=2 r+\frac{1}{2 r}$. | 3. Let the sides of the triangle be $n-1, n, n+1$. Then its semi-perimeter $p=\frac{n-1+n+n+1}{2}=\frac{3 n}{2}$. Let $S$ be the area of the triangle. By Heron's formula, we get: $S^{2}=\frac{3 n}{2} \cdot\left(\frac{3 n}{2}-n+1\right) \cdot\left(\frac{3 n}{2}-n\right) \cdot\left(\frac{3 n}{2}-n-1\right)=\frac{3 n}{2} ... | 2r+\frac{1}{2r} | Geometry | proof | Yes | Yes | olympiads | false | 12,982 |
4. There are 20, 1, and 9 stones in three piles. In one move, it is allowed to take one stone from any two piles and place them in the third pile. Is it possible to gather all the stones in one pile after several moves? | 4. The number 1 is divided by 3 with a remainder of 1, the number 9 is divided by 3 with no remainder, and the number 20 is divided by 3 with a remainder of 2. That is, the initial numbers have different remainders when divided by 3.
What will happen to these remainders in one move? They will all change, but they will... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,983 |
5. For the Day of the Russian Flag, the seller decided to decorate the store window with 10 horizontal strips of fabric in three colors. At the same time, he follows two conditions:
1) strips of the same color should not hang next to each other;
2) each blue strip must hang between a white and a red one.
In how many w... | 5. Let's call the white and red strips of fabric green. Denote by $T_{n}$ the number of ways to decorate the shop window with $n$ strips of fabric. The first strip can only be a green strip. If the second strip is blue, then the following strips must be such that the first one is green. The total number of ways to hang... | 110 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,984 |
9.6. For a natural number $n$, denote by $S_{n}$ the least common multiple of all numbers $1,2, \ldots, n$. Does there exist a natural number $m$ such that $S_{m+1}=4 S_{m}$?
(A. Kuznetsov) | Answer. No.
Solution. Suppose the opposite. Let $S_{m+1}$ be divisible by $2^{s}$ but not by $2^{s+1}$; then $s \geqslant 2$. This means that among the numbers $1,2, \ldots, m+1$ there is a number $a$ that is divisible by $2^{s}$. But then the number $a / 2$ does not exceed $m$ and is divisible by $2^{s-1}$; hence, $S... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,985 |
9.7. On the board, 99 numbers were written, none of which are equal. In a notebook, $\frac{99 \cdot 98}{2}$ numbers were written—all differences between two numbers from the board (each time subtracting the smaller number from the larger one). It turned out that the number 1 was written exactly 85 times in the notebook... | Answer. $d=7$.
Solution. We will prove that $d \geqslant 7$. All numbers on the board can be divided into chains of numbers of the form $a, a+1, a+2, \ldots, a+t$ such that numbers from different chains do not differ by exactly 1. Such a partition is not difficult to construct by connecting any two numbers that differ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,986 |
9.8. Given an acute-angled triangle $A B C$, in which $A B < B C$. Let $M$ and $N$ be the midpoints of sides $A B$ and $A C$ respectively, and $H$ be the foot of the altitude from vertex $B$. The incircle touches side $A C$ at point $K$. The line passing through $K$ and parallel to $M H$ intersects segment $M N$ at poi... | First solution. Perform a homothety with center $A$ and coefficient 2. Under this homothety, points $M$ and $N$ map to $B$ and $C$ respectively; let points $K$ and $P$ map to $K^{\prime}$ and $P^{\prime}$ respectively (see Fig. 1). It is sufficient to prove that the quadrilateral $A B P^{\prime} K^{\prime}$ is circumsc... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,987 |
9.9. Find the largest number $m$ such that for any positive numbers $a, b$, and $c$, the sum of which is 1, the inequality
$$
\sqrt{\frac{a b}{c+a b}}+\sqrt{\frac{b c}{a+b c}}+\sqrt{\frac{c a}{b+c a}} \geqslant m
$$
holds.
(l. Emelyanov) | Answer. $m=1$.
First solution. First, we prove that $m=1$ satisfies the requirements of the problem. Notice that $ab + c = ab + c(a + b + c) = (c + a)(c + b)$. Therefore,
\[
\begin{aligned}
& \sqrt{\frac{ab}{c + ab}} + \sqrt{\frac{bc}{a + bc}} + \sqrt{\frac{ca}{b + ca}} = \\
& = \sqrt{\frac{ab}{(c + a)(c + b)}} + \sq... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 12,988 |
9.10. A cube $100 \times 100 \times 100$ is divided into a million unit cubes; each cube contains a light bulb. Three faces of the large cube, sharing a common vertex, are painted: one red, one blue, and one green. We will call a column a set of 100 cubes forming a $1 \times 1 \times 100$ block. Each of the 30000 colum... | Solution. It is clear that the result of pressing several switches does not depend on the order in which these presses were made - the number of switches for each bulb does not depend on this order. In particular, we can assume that Petya used each switch no more than once.
The entire cube is divided into 100 layers p... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,989 |
8.2. Do there exist five two-digit composite numbers such that any two of them are coprime
---
Translation:
8.2. Do there exist five two-digit composite numbers such that any two of them are coprime | Answer: They do not exist.
Solution: Each composite number is a product of at least two prime numbers. In each such product, there cannot be more than one two-digit factor, otherwise the product would be at least three digits. Therefore, in the prime factorization of each of the sought two-digit numbers, there must be... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,991 |
8.3. In an isosceles triangle \(ABC\) with base \(AB\), a point \(D\) is chosen on side \(CB\) such that \(CD = AC - AB\). Point \(M\) is the midpoint of \(AD\). Prove that angle \(BMC\) is obtuse. | Solution. Since $C D=A C-A B=B C-A B$, we obtain that $D B=A B$, which means that triangle $A B D$ is isosceles. Then its median $B M$ is also an altitude, i.e., angle $B M D$ is right. Therefore, $\angle B M C=\angle B M D+\angle D M C=90^{\circ}+\angle D M C>90^{\circ}$ is obtuse.
## Criteria.
$ times.
Thus, we have the equation: $K=10 \frac{K}{n-1}$.
From this - the answer | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,995 |
3. Is it possible to color the cells of an 8 x 8 square in 16 colors so that for any two colors, there are cells of these colors that share a side? | Answer: No.
## Solution:
The total number of different color pairs: $16 \cdot 15 / 2=120$. And there are 112 borders between cells. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,996 |
4. Let $p(x)=2 x^{3}-3 x^{2}+1$. How many squares of integers are among the numbers $p(1), p(2), \ldots, p(2016)$? | # Answer: 32.
## Solution:
Notice that $p(x)=(x-1)^{2}(2 x+1)$.
For an integer $x(1 \leq x \leq 2016)$, the number $p(x)=(x-1)^{2}(2 x+1)$ is a perfect square of an integer either when $x=1(p(1)=0)$, or (for $x \geq 2$) when the number $2 x+1$ is a perfect square.
Note that for $x \geq 2$, the inequality holds: $5 ... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,997 |
5. The letters А, Б, К, М, П, У, Ш were encoded with sequences of zeros and ones (each with its own). Then, in the word ПАПАМАМАБАБУШКА, the letters were replaced with their codes. Could the length of the resulting sequence be shorter than 40 characters, if the sequence can be uniquely decoded?
保留源文本的换行和格式，所以翻译结果如下：
... | Answer: She could.
## Solution:
Here is an example of a code table:
| A | Б | К | М | П | У | Ш |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | 110 | 1111 | 100 | 101 | 11100 | 11101 |
The word will look like this:
10101010100010001100110111001110111110 - a total of 38 characters. Decoding is una... | 38 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,998 |
7.1. Write the number 1.11 ten times and the number 1.01 eleven times. Cross out one or several numbers so that the sum of the remaining numbers is equal to 20.19. | Answer: 1,$11 ; 1,11 ; 1,11 ; 1,11 ; 1,11 ; 1,11 ; 1,11 ; 1,11 ; 1,11 ; 1,11 ; 1,01 ; 1,01 ; 1,01 ; 1,01 ;$ 1,$01 ; 1,01 ; 1,01 ; 1,01 ; 1,01$. (Struck out numbers 1,01 and 1,01.)
Solution. The sum of all the written numbers is 10$\cdot$1,11 + 11$\cdot$1,01 = 22,21. This is 2,02 more than the required sum. That is, it... | 20.19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,999 |
7.3. In the train, there are 18 identical cars. In some cars, exactly half of the seats are free, in some others - exactly one third of the seats are free, and in the rest, all seats are occupied. At the same time, in the entire train, exactly one ninth of all seats are free. In how many cars are all seats occupied? | Answer: in 13 carriages.
Solution. Let the number of passengers in each carriage be taken as a unit. We can reason in different ways.
First method. Since exactly one ninth of all seats in the train are free, this is equivalent to two carriages being completely free. The number 2 can be uniquely decomposed into the su... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,001 |
7.4. In grandmother's garden, apples have ripened: Antonovka, Grushovka, and White Naliv. If there were three times as many Antonovka apples, the total number of apples would increase by $70 \%$. If there were three times as many Grushovka apples, it would increase by $50 \%$. By what percentage would the total number ... | Answer: increased by $80 \%$.
Solution. First method. If the amount of each type of apple were three times as much, the total number of apples would increase by $200 \%$. Of this, $70 \%$ is due to the increase in Antonovka, and $50 \%$ is due to the increase in Grushovka. Therefore, the increase due to White Naliv wo... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,002 |
7.5. Vera has 27 cubes with an edge of 1 cm: 9 red and 18 blue. She assembled them into a cube with an edge of 3 cm. Can the number of red squares with a side of 1 cm on the surface of the cube be equal to the number of such blue squares? | Answer: No.
Solution. In total, there are $9 \cdot 6 = 54$ squares on the surface of the resulting cube. Three faces of a small cube will be on the surface if the cube is in a corner, two faces if the cube touches an edge of the large cube, and one face if the cube is in the center of a face. The maximum number of red... | No | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,003 |
7.6. In each cell of a $5 \times 5$ square, exactly one diagonal has been drawn. A vertex of a cell is free if it is not the endpoint of any of the drawn diagonals. Find the maximum possible number of free vertices. | Answer: 18 vertices.
Solution. Example. See Fig. 7.6a. On each of the six horizontal lines, three vertices are free.
Estimate. The total number of cell vertices: $6 \cdot 6=36$. Let's select nine cells that do not share any vertices (see Fig. 7.6b). They contain all 36 vertices. In each cell, a diagonal is drawn, so ... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,004 |
4. $\frac{1}{a}=\frac{a+b+c+d+m}{a}=1+\frac{b}{a}+\frac{c}{a}+\frac{d}{a}+\frac{m}{a}$
$$
\begin{aligned}
& \frac{1}{b}=\frac{a+b+c+d+m}{b}=\frac{a}{b}+1+\frac{c}{b}+\frac{d}{b}+\frac{m}{b} \\
& \frac{1}{c}=\frac{a+b+c+d+m}{c}=\frac{a}{c}+1+\frac{b}{c}+\frac{d}{c}+\frac{m}{c}
\end{aligned}
$$
$\frac{1}{d}=\frac{a+b+c... | Answer: the inequality is valid. | 25 | Inequalities | proof | Yes | Yes | olympiads | false | 13,009 |
5. Let the mass of the new alloy be 1, and the relative masses of the 1st, 2nd, and 3rd alloys in it be \( x, y, z \) respectively; the content of chromium in the new alloy is \( k \). We obtain the system of equations:
$$
\left\{\begin{array}{c}
x+y+z=1 \\
0.9 x+0.3 z=0.45 \\
0.4 x+0.1 y+0.5 z=k
\end{array}\right.
$$... | Answer: the minimum percentage content of chromium in the new alloy is $25 \%$, and the maximum is $40 \%$. | 2540 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,010 |
3. On Eeyore's Birthday, Winnie-the-Pooh, Piglet, and Owl came to visit. When Owl left, the average age in this company decreased by 2 years, and when Piglet left, the average age decreased by another 1 year. How many years older is Owl than Piglet?
Answer: Owl is 6 years older than Piglet. | Solution. Let the average age of those who remained after Piglet be x, and Piglet's age be y. Then $2x + y = 3(x + 1)$, which means $y = x + 3$. Let Owl's age be K. Then $3(x + 1) + K = 4(x + 3)$, which means $K = x + 9$. Therefore, Owl is older than Piglet by $(x + 9) - (x + 3) = 6$ years.
Criteria. If the solution i... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,014 |
4. Vova listed the first million natural numbers that are not divisible by 4. Dima calculated the sum of 1000 consecutive numbers in Vova's list. Could the result have been 20172018?
Answer: No. | Solution. From any three consecutive numbers in Vova's record, one has a remainder of 1 when divided by 4, another has a remainder of 2, and the remaining one has a remainder of 3. Therefore, their sum gives a remainder of 2 when divided by 4. Among the first 999 of Dima's numbers, there are exactly 333 such triples, t... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,015 |
6.1. In a two-digit number $A$, the digits were swapped to obtain the number $B$. Find such an $A$ so that the sum $A+B$ is divisible by 17. | Answer: $A=89$ or $A=98$.
Solution: $A=89$ or $A=98$ will work. In both cases, $A+B=187=17 \cdot 11$.
Comment: Either of the two correct answers earns 7 points. | A=89orA=98 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,017 |
6.2. Solve the rebus DON + OKA + LENA + VOLGA $=$ ANGARA or explain why the rebus has no solution. (Identical letters represent identical digits, different letters represent different digits). | Answer. The rebus has no solution.
Solution. Considering the last digits of the rebus, we get that the sum $\mathrm{H}+\mathrm{A}+\mathrm{A}+\mathrm{A}$ ends in $\mathrm{A}$, which means that $\mathrm{H}+\mathrm{A}+\mathrm{A}$ ends in 0. The sum of the numbers DON + OKA + LENA + VOLGA is less than $999+999+9999+99999<... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,018 |
6.3. In the park, all bicycle paths run from north to south or from west to east. Petya and Kolya simultaneously started from point $A$ and rode their bicycles at constant speeds: Petya - along the route $A-B-C$, Kolya - along the route $A-D-E-F-C$ (see fig.), and both spent 12 minutes on the road. It is known that Kol... | Answer: 1 minute.
Solution. Draw the segment $D H$, as shown in Fig. 2. Kolya travels 1.2 times faster than Petya, so it would take him $12 / 1.2=10$ minutes to travel the route $A-B-C$. The difference in time
. Then each of them said: "Among my neighbors, there is a liar." What is the maximum number of people sitting at the table who can say: "Among my neighbors, there is a k... | # Answer: 8.
Solution. Note that two liars cannot sit next to each other (otherwise, each of them would be telling the truth). Therefore, no liar can say the second phrase.
On the other hand, 3 knights also cannot sit next to each other (otherwise, the middle one would have lied by saying that he has a neighbor who i... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,021 |
1. Let the desired number be $\overline{a b c d}=1000 a+100 b+10 c+d$. Then the other number will be $\overline{d c b a}=1000 d+100 c+10 b+a$. The sum of these numbers will be $1001 a+110 b+110 c+1001 d$. Since the division of this sum by 143 resulted in 136, we have the equation: $1001(a+d)+110(b+c)=19448$. Since $(b+... | Answer: 9949, 9859, 9769, 9679, 9589, 9499. | 9949,9859,9769,9679,9589,9499 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,022 |
2. Let's represent the left side of the equation as follows:
$$
\begin{gathered}
x^{4}+4 x-1=\left(x^{2}\right)^{2}+2 x^{2}-2 x^{2}+1-1+4 x-1= \\
=\left(x^{2}+1\right)^{2}-\left(2 x^{2}-4 x+2\right)=\left(x^{2}+1\right)^{2}-\left((\sqrt{2}(x-1))^{2}=\right. \\
=\left(x^{2}+1-\sqrt{2} x+\sqrt{2}\right)\left(x^{2}+1+\sq... | Answer: $x_{1,2}=\frac{-1 \pm \sqrt{2 \sqrt{2}-1}}{\sqrt{2}}$.
Let the legs of the right triangles be $x$ and $y$, with $x > y$. Then the side of the smaller square will be $x - y$. According... | \sqrt{3}-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,023 |
4. Let $a=\sin 1^{\circ} \cdot \sin 2^{\circ} \cdot \sin 3^{\circ} \cdot \ldots \cdot \sin 89^{\circ}$ and $b=\sin 1^{\circ} \cdot \sin 3^{\circ} \cdot \ldots \cdot \sin 87^{\circ} \cdot \sin 89^{\circ}$. Then $\frac{a}{b}=\sin 2^{\circ} \cdot \sin 4^{\circ} \cdot \sin 6^{\circ} \cdot \ldots \cdot \sin 88^{\circ} \cdot... | Answer: $\frac{\sqrt{2}}{2^{45}}$. | \frac{\sqrt{2}}{2^{45}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,024 |
1. Five consecutive natural numbers were written on the board, and then one number was erased. It turned out that the sum of the remaining four numbers is 2015. Find the smallest of these four numbers. | Answer: 502
Solution. If the smallest number is K, then the sum of all numbers is not less than $\mathrm{K}+(\mathrm{K}+1)+(\mathrm{K}+2)+ (\mathrm{K}+3)=4 \mathrm{~K}+6$ and not more than $\mathrm{K}+(\mathrm{K}+2)+(\mathrm{K}+3)+(\mathrm{K}+4)=4 \mathrm{~K}+9$. Therefore, $4 \mathrm{~K}+6 \leq 2015 \leq 4 \mathrm{~K... | 502 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,025 |
2. Through point P, lying on the common chord of two intersecting circles, chords KM of the first circle and LN of the second circle are drawn. Prove that the quadrilateral with vertices at points K, L, N, and M is inscribed. | Solution: Let the ends of the common chord be points A and B. Then, according to the theorem of chord segments from the first circle, $K P \cdot P M = A P \cdot P B$, and from the second circle, $LP \cdot PN = AP \cdot PB$. But then $KP \cdot PM = LP \cdot PN$, which means points $\mathrm{K}, \mathrm{L}, \mathrm{N}$, a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,026 |
3. What can the value of the expression $p^{4}-3 p^{3}-5 p^{2}+16 p+2015$ be if $p$ is a root of the equation $x^{3}-5 x+1=0$?
Answer: 2018 | Solution: $\mathrm{p}^{4}-3 \mathrm{p}^{3}-5 \mathrm{p}^{2}+16 \mathrm{p}+2015=\left(\mathrm{p}^{3}-5 \mathrm{p}+1\right)(\mathrm{p}-3)+2018$. Since $\mathrm{p}$ is a root of the polynomial in the first parenthesis, the entire expression equals 2018. | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,027 |
4. There are 400 students in a school. For New Year, each student sent 200 greetings to other students. What is the minimum number of pairs of students who could have greeted each other? Answer: 200. | Solution: The total number of greetings was $200 * 400=80000$. And the number of different pairs of students is $-400 * 399 / 2=79800$, which is 200 less than the number of greetings. Therefore, at least 200 pairs had 2 greetings. We can show that exactly 200 pairs can be: Let the 1st student greet everyone from the 2n... | 200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,028 |
1. All house spirits love to play. All house spirits love cleanliness and order. Everyone who loves cleanliness and order does not like to play. Then, does it necessarily mean that house spirits do not exist?
# | # Solution
1st method. 1) From the conditions "All domovoi love cleanliness and order" and "Everyone who loves cleanliness and order does not like to play pranks," it follows that "All domovoi do not like to play pranks" (From "If A, then B" and "If B, then C" it follows that "If A, then C").
2) From the conditions "A... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,029 |
2. Three candles were lit simultaneously. When the first candle burned out, $\frac{2}{5}$ of the second candle remained, and $\frac{3}{7}$ of the third candle remained. What part of the third candle will remain when the second candle burns out? | # Solution
In time $t$, $\frac{3}{5}$ of the second candle and $\frac{4}{7}$ of the third candle have burned.
The burning rate of the second candle is $\frac{3}{5 t}$, and the burning rate of the third candle is $\frac{4}{7 t}$.
$\frac{2}{5}$ of the second candle will burn in $\frac{2}{5} : \frac{3}{5 t} = \frac{2 t... | \frac{1}{21} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,030 |
3. Solve the equation $5 x+2|x|=3 x$
$(|x|=x$ when $x \geq 0$ and $|x|=-x$ when $x<0)$.
# | # Solution
Transform the equation to the form
$$
2 x+2|x|=0
$$
1) Solve the equation for $x \geq 0$.
For $x \geq 0$, we get
$$
\begin{gathered}
2 x+2 x=0 \\
4 x=0 \\
x=0
\end{gathered}
$$
The obtained value of $x$ satisfies the condition $x \geq 0$. Therefore, $x=0$ is not a solution to the equation.
2) Solve th... | x\leq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,031 |
4. Is it possible to divide a convex heptadecagon (17-sided polygon) into black and white triangles such that any two triangles either share a side when colored differently, share a vertex, or have no common points, and each side of the heptadecagon is a side of one of the black triangles?
# | # Solution
Let $n$ be the number of white triangles. Then, there are $3 n$ sides of white triangles, since it follows from the condition that monochromatic triangles do not share sides.
Then, there are $3 n + 17$ sides of black triangles, since any side of a white triangle is also a side of a black one. However, $3 n... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,032 |
5. There is a pile of 100 stones. Two people are playing. In one move, you can take 1, 2, 3, or 5 stones from the pile. The one who takes the last stones wins. How to play to win
# | # Solution
The second player wins, by leaving a number of stones in the pile that is a multiple of four with each of their moves:
1) if the first player takes 1 stone, the second player takes 3 stones;
2) if the first player takes 2 stones, the second player takes 2 stones;
3) if the first player takes 3 stones, the ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,033 |
9.1. On some cells of a $4 \times 4$ square board, golden coins are placed in a stack, and on the remaining cells, silver coins are placed. Is it possible to arrange the coins so that in each $3 \times 3$ square, there are more silver coins than golden ones, and on the entire board, there are more golden coins than sil... | Answer: Yes, it can be done.
For example, place a stack of nine silver coins on one of the cells of the central $2 \times 2$ square, and place one gold coin on each of the other cells of the board. Then, in each $3 \times 3$ square, there will be 9 silver coins and 8 gold coins, and on the entire board, there will be ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,034 |
9.2. A merchant bought several bags of salt in Tver and sold them in Moscow with a profit of 100 rubles. With all the money earned, he again bought salt in Tver (at the Tver price) and sold it in Moscow (at the Moscow price). This time the profit was 120 rubles. How much money did he spend on the first purchase? | Answer: 500 rubles.
First method. Let the cost of a kilogram of salt in Tver be $x$ rubles, and in Moscow $-y$ rubles, and let the merchant buy $a$ kg of salt the first time. Then, according to the condition, $a(y-x)=100$.
The revenue amounted to $ay$ rubles, so the second time the merchant was able to buy $\frac{ay}... | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,035 |
9.3. In parallelogram $A B C D$, diagonal $A C$ is twice the length of side $A B$. A point $K$ is chosen on side $B C$ such that $\angle K D B = \angle B D A$. Find the ratio $B K: K C$. | Answer: $2: 1$.
Let $O-$ be the point of intersection of the diagonals of the parallelogram. From the problem statement, it follows that $A B=A O=O C=C D$ (see Fig. 9.3). Since $\angle K D B=\angle B D A=\angle D B K$, then $B K=K D$, therefore
 x+q=0$ be integers if: a) $q>0$; b) $q<0$? | Answer: a) yes; b) no.
a) For example, $x^{2}-7 x+12=0$ and $x^{2}-8 x+12=0$. The roots of the first equation are 3 and 4, and the roots of the second are 2 and 6.
There are other examples as well.
b) Let $q0$ and $x_{2}0$ and $x_{4}0$. Moreover, $x_{1} \neq x_{3}$, otherwise $x_{2}=x_{4}$, and the equations cannot ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,038 |
2. The number 11 can be represented as the sum of four squares of numbers in only one way, disregarding the order of the addends:
$$
11=9+1+1+0=3^{2}+1^{2}+1^{2}+0^{2}
$$
Can the number 99 be represented as the sum of four squares in two different ways? | Answer. Yes, it can be done. There are even more than two ways:
$$
\begin{gathered}
99=7^{2}+7^{2}+1^{2}+0^{2}=7^{2}+5^{2}+5^{2}+0^{2}=8^{2}+5^{2}+3^{2}+1^{2}=9^{2}+4^{2}+1^{2}+1^{2}= \\
=9^{2}+3^{2}+3^{2}+0^{2} .
\end{gathered}
$$
## Grading Criteria.
- Correct answer (two or more ways are provided), explanations a... | 99=7^{2}+7^{2}+1^{2}+0^{2}=7^{2}+5^{2}+5^{2}+0^{2}=8^{2}+5^{2}+3^{2}+1^{2}=9^{2}+4^{2}+1^{2}+1^{2}=9^{2}+3^{2}+3^{2}+0^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,043 |
3. Sixteen boys gathered for fishing. It is known that every boy who put on boots also put on a cap. Without boots, there were 10 boys, and without a cap - two. Which boys are more and by how many: those who wore a cap but no boots, or those who put on boots? Be sure to explain your answer. | Answer. Those who were in caps but without boots were 2 more than those who were in boots.
Solution. Out of 16 boys, 10 were without boots, which means 6 were in boots. Two were without caps, so 14 were in caps. Since everyone who wore boots also wore a cap, out of the 14 who wore caps, 6 also wore boots, and the rema... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,044 |
4. Sasha went to visit his grandmother. On Saturday, he got on the train, and after 50 hours on Monday, he arrived at his grandmother's city. Sasha noticed that on this Monday, the date matched the number of the carriage he was in, that the number of his seat in the carriage was less than the number of the carriage, an... | Answer. Carriage No. 2, seat No. 1.
Solution. Since the carriage number on Saturday was less than the date, and on Monday it was equal to the date, it is clear that Saturday and Monday belong to different months. Since Sasha traveled for a little more than two days, Monday is the first or second day of the month. Ther... | CarriageNo.2,seatNo.1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,045 |
# 5. Rectangle $A B C D$ was divided into four
smaller rectangles with equal perimeters (see figure). It is known that $A B=18$ cm, and $B C=16$ cm. Find the lengths of the sides of the other rectangles. Be sure to explain your answer. | Answer. 2 cm and 18 cm are the lengths of the sides of rectangle $A B L E$, 6 cm and 14 cm are the lengths of the sides of the other rectangles.
Solution. Since the perimeters of the three vertical rectangles are equal and the segments $E D, F G, K H$ and $L C$ are also equal, the segments $E F$, $F K$ and $K L$ are a... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,046 |
1. For which eleven consecutive integers is the sum of the first six numbers equal to the sum of the last five? | Answer: $25,26,27,28,29,30,31,32,33,34,35$.
## Solution:
There are no other solutions, because shifting these values along the number line by $k$ units changes the sum of the first six numbers by $6k$, and the sum of the last five by $5k$.
How can we find this solution? First, sum the first 6 numbers (from 1 to 6) a... | 25,26,27,28,29,30,31,32,33,34,35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,047 |
2. For what values of $a$ does the equation $x \cdot|x-a|=1$ have three distinct solutions? | Answer: $a>2$.
## Solution:
We will construct the graph of the function $y=x|x-a|$ and find the number of intersection points with the horizontal line $y=1$ for different values of the parameter (more precisely, we will determine for which values of the parameter there will be three intersection points).
When $a=0$,... | >2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,048 |
3. A point in a triangle is connected to the vertices by three segments. What is the maximum number of these segments that can equal the opposite side?
# | # Answer: One.
## Solution:
Let $B M = A C$ and $A B = M C$ (see fig.). Triangles $A B M$ and $M C A$ are equal by three sides. Therefore, angle $B A M$ is equal to angle $A M C$, which means $A B \parallel M C$.
Similarly, $A C \parallel M B$.
Thus, $A B M C$ is a parallelogram, but this is not the case, because a... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,049 |
4. There is a pile of 2016 stones. Two players are playing a game. The first player can take either 1, 3, or 4 stones on each of their turns. The second player can take either 1, 2, or 3 stones on each of their turns. The player who cannot make a move loses. Who can win, no matter how the opponent plays? | Answer: The second.
## Solution:
The second player's strategy: each time leave a number of stones in the pile that gives a remainder of 2 when divided by 3. Note that he can do this after any move by the opponent, as his set of moves allows him to change the remainder of the remaining stones to any value. The first p... | Theplayerwins | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,050 |
5. Let $p(x)=2 x^{3}-3 x^{2}+1$. How many squares of integers are among the numbers $p(1), p(2), \ldots$, $p(2016) ?$
# | # Answer: 32.
## Solution:
Notice that $p(x)=(x-1)^{2}(2 x+1)$.
For an integer $x(1 \leq x \leq 2016)$, the number $p(x)=(x-1)^{2}(2 x+1)$ is a perfect square of an integer either when $x=1(p(1)=0)$, or (for $x \geq 2$) when the number $2 x+1$ is a perfect square.
Note that for $x \geq 2$, the inequality holds: $5 ... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,051 |
Problem 9.1. Find the largest five-digit number, the product of whose digits is 120. | Answer: 85311.
Solution. The largest single-digit divisor of the number $120-8$, so the desired number definitely starts with this digit. The product of all the remaining digits is 15.
The largest single-digit divisor of the number $15-5$, so the digit in the thousands place will be this digit. The product of the las... | 85311 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,053 |
Problem 9.2. During the first half of the year, lazy Pasha forced himself to solve math problems. Every day he solved no more than 10 problems, and if on any day he solved more than 7 problems, then for the next two days he solved no more than 5 problems per day. What is the maximum number of problems Pasha could solve... | Answer: 52.
Solution. Suppose Pasha solved at least 8 tasks (but no more than 10) in one of the first five days, then in the next two days he solved no more than 5 tasks per day. Thus, in these three days, he solved no more than $20(10+5+5)$ tasks. If he solved 7 tasks each day, it would be more.
It turns out that in... | 52 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,054 |
Problem 9.3. Given a convex quadrilateral $ABCD$, $X$ is the midpoint of diagonal $AC$. It turns out that $CD \parallel BX$. Find $AD$, if it is known that $BX=3, BC=7, CD=6$.
 | Answer: 14.
Solution. Double the median $B X$ of triangle $A B C$, to get point $M$. Quadrilateral $A B C M$ is a parallelogram (Fig. 1).
Notice that $B C D M$ is also a parallelogram, since segments $B M$ and $C D$ are equal in length (both 6) and parallel. This means that point $M$ lies on segment $A D$, since $A M... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,055 |
Problem 9.4. For positive numbers $a, b, c$, it is known that
$$
\frac{a+b+c}{a+b-c}=7, \quad \frac{a+b+c}{a+c-b}=1.75
$$
What is the value of $\frac{a+b+c}{b+c-a}$? | Answer: 3.5.
Solution. Let $\frac{a+b+c}{b+c-a}=x$. Notice that
$$
\begin{gathered}
\frac{1}{7}+\frac{1}{1.75}+\frac{1}{x}=\frac{a+b-c}{a+b+c}+\frac{a+c-b}{a+b+c}+\frac{b+c-a}{a+b+c}=1 \\
\frac{1}{7}+\frac{4}{7}+\frac{1}{x}=1 \\
x=3.5
\end{gathered}
$$ | 3.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,056 |
Problem 9.5. Point $M$ is the midpoint of side $B C$ of triangle $A B C$, where $A B=17$, $A C=30, B C=19$. A circle is constructed with side $A B$ as its diameter. An arbitrary point $X$ is chosen on this circle. What is the minimum value that the length of segment $M X$ can take?
, as shown in Fig. 2.
 \ldots))+\frac{1}{2}=0
$$
have, where $f(x)=|x|-1$? | Answer: 20.
Solution. Let
$$
f_{n}(x)=\overbrace{f(f(\ldots(f}^{n \text { times } f}(x) \ldots))
$$
We will solve the equation $f_{10}(x)=-\frac{1}{2}$ graphically.
We will use the fact that the graph of $y=f_{k}(x)$ can be obtained from the graph of $y=f_{k-1}(x)$, based on the relation $f_{k}(x)=\left|f_{k-1}(x)\... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,062 |
Problem 10.3. Anton wrote three natural numbers $a, b$, and $c$ on the board. Ira drew three rectangles $a \times b, a \times c$, and $b \times c$ on the board. It turned out that the difference in the areas of one pair of rectangles is 1, and the difference in the areas of another pair of rectangles is 49. What can $a... | Answer: 16.
Solution. Without loss of generality, we assume that $1=ac-ab=a(c-b)$, then $a=1$, $c=b+1$. Thus, the numbers written on the board were $1, b, b+1$.
Notice that either $b(b+1)-b \cdot 1=49$, or $b(b+1)-(b+1) \cdot 1=49$.
In the first case, we get $b^{2}=49, b=7$, and $a+b+c=1+b+(b+1)=16$.
In the second ... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,063 |
Problem 10.4. An isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ is such that $\angle ADC = 2 \angle CAD = 82^{\circ}$. Inside the trapezoid, a point $T$ is chosen such that $CT = CD, AT = TD$. Find $\angle TCD$. Give your answer in degrees.
.
![](https://cdn.mathpix.com/cropped/2024_05_06_2befe97065... | 38 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,064 |
Problem 10.5. Integers $a$ and $b$ are such that the quadratic trinomials $x^{2}+a x+b$ and $x^{2}+$ $b x+1100$ have a common root, which is a prime number. Find $a$. List all possible options. | Answer: 40 and 274.
Solution. Let $p$ be the common root of the quadratic trinomials, which is a prime number. Then $p^{2}+a p+b=0$, from which we can conclude that $b \vdots p$.
Now let's look at the second quadratic trinomial: $p^{2}+b p+1100=0$. From the fact that $b \vdots p$, we conclude that $1100=2^{2} \cdot 5... | 40274 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,065 |
Problem 10.6. On an island, there are two tribes: knights and liars. Knights always tell the truth, while liars always lie. One day, 80 people sat around a round table, and each of them stated: "Among the 11 people sitting after me in a clockwise direction, there are at least 9 liars." How many knights are sitting at t... | Answer: 20.
Solution. First, we prove that among 12 consecutive people, there are no more than 3 knights. Suppose this is not the case. Consider the first knight in this group. Among the 11 people sitting after him in a clockwise direction, there are at least 3 knights, which contradicts the problem's condition.
Next... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,066 |
Problem 10.7. Given a right triangle $ABC$ with legs $AB=42$ and $BC=56$. A circle passing through point $B$ intersects side $AB$ at point $P$, side $BC$ at point $Q$, and side $AC$ at points $K$ and $L$. It is known that $PK=KQ$ and $QL: PL=3: 4$. Find $PQ^2$.
. From this similarity and the cyclic nature of the pentagon
 into a bag, and the second bandit chooses who gets this bag; then this action is repeated several times. The division ends when
- either all the money is gone,
- or someone gets 11 ba... | Answer: 146.
Solution. First, we will show that the first bandit can get at least 146 coins. His strategy will be to put 14 coins in each bag. First, note that he will always be able to do this: since $21 \cdot 14=294$, he will do this at least 21 times, and when the coins start to run out, the process will certainly ... | 146 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,068 |
Problem 11.1. The teacher wrote a two-digit number on the board. Each of the three boys made two statements.
- Andrey: “the number ends with the digit 6” and “the number is divisible by 7”.
- Borya: “the number is greater than 26” and “the number ends with the digit 8”.
- Sasha: “the number is divisible by 13” and “th... | Answer: 91.
Solution. Let's consider Sasha's statements. If his second statement that the number is less than 27 is true, then Borya's first statement is definitely false, so the number must end in 8. The only two-digit number that meets these conditions is 18, but then none of Andrei's statements are true. Contradict... | 91 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,069 |
Problem 11.2. Vera has a set of weights of different masses, each weighing an integer number of grams. It is known that the lightest weight in the set weighs 71 times less than all the other weights combined. It is also known that the two lightest weights in the set together weigh 34 times less than all the other weigh... | Answer: 35.
Solution. All weights in the solution are expressed in grams.
Let the lightest weight be $m$, then the other weights are $71 m$, and the total weight is $72 \mathrm{~m}$. Let also the two lightest weights together weigh $n$, then the other weights weigh $34 n$, and the total weight is $35 n$, which is div... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,070 |
Problem 11.3. On the coordinate plane, all points $(x, y)$ such that $x$ and $y$ are integers satisfying the inequalities $0 \leqslant x \leqslant 2$ and $0 \leqslant y \leqslant 26$ are marked. How many lines exist that pass through exactly 3 of the marked points? | Answer: 365.
Solution. All marked points are located on three vertical lines $x=0, x=1, x=2$. Let's call these lines the left, middle, and right lines, respectively.
Consider any three marked points lying on one line. If at least two of them lie on a vertical line, then there are more than 3 marked points on such a l... | 365 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,071 |
Problem 11.4. On the side $AC$ of triangle $ABC$, points $M$ and $N$ are marked ($M$ lies on the segment $AN$). It is known that $AB = AN$, $BC = MC$. The circumcircles of triangles $ABM$ and $CBN$ intersect at points $B$ and $K$. How many degrees does the angle $AKC$ measure if $\angle ABC = 68^\circ$?
 \cos \left(\frac{\angle A-\angle B}{2}\right) \\
2-\sqrt{2} & =\cos \angle A+\cos \angle B=2 \cos \left(\frac{\angle A+\angle B}{2}\... | 168 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,074 |
Problem 11.7. Natural numbers $a$ and $b$ are such that $a^{a}$ is divisible by $b^{b}$, but $a$ is not divisible by $b$. Find the smallest possible value of the number $a+b$, given that the number $b$ is coprime with 210. | Answer: 374.
Solution. Obviously, $b \neq 1$. Let $p$ be a prime divisor of the number $b$; then $p \geqslant 11$, since $b$ is coprime with $210=2 \cdot 3 \cdot 5 \cdot 7$. Since $a^{a}$ is divisible by $b^{b}$, which is divisible by $p$, then $a$ is also divisible by $p$. From this, it immediately follows that the n... | 374 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,075 |
Problem 11.8. Inside the tetrahedron $ABCD$, points $X$ and $Y$ are given. The distances from point $X$ to the faces $ABC, ABD, ACD, BCD$ are $14, 11, 29, 8$ respectively. And the distances from point $Y$ to the faces $ABC, ABD, ACD, BCD$ are $15, 13, 25, 11$ respectively. Find the radius of the inscribed sphere of the... | Answer: 17.
Solution. Consider a point $Z$ lying on the ray $XY$ such that $XY: YZ = 1: 2$. We will prove that this point is the center of the inscribed sphere of the tetrahedron.
Drop perpendiculars $X_{\alpha}, Y_{\alpha}, Z_{\alpha}$ from points $X, Y, Z$ to the plane $\alpha$ - obviously, they will lie in the sam... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,076 |
1. Find 10 consecutive natural numbers, the sum of all digits of which equals 145. | Solution. Note that among ten consecutive natural numbers, each digit from 0 to 9 appears exactly once in the units place. Therefore, the sum of their last digits is always 45, and we need to make up the sum of 100 using the other digits. This can be easily done by taking all numbers from the same decade such that the ... | 190to199 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,077 |
3. The seller has a balance scale. Help the seller come up with a set of 4 weights that can be used to weigh any whole number of kilograms from 1 to 12 on these scales. No more than two weights can be used for each weighing; weights can be placed on different pans of the scale. | Solution. For example, the following set of weights will do: 1 kg, 2 kg, 5 kg, and 10 kg. The table in the figure shows how to place the weights on the scales. (It is assumed that the item being weighed is on the left pan.)
Remark. The example provided is not the only one.
| Weight of the item | Left pan | Right pan ... | 1,2,5,10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,078 |
4. Cut an equilateral triangle into 4 convex polygons: a triangle, a quadrilateral, a pentagon, and a hexagon. (A polygon is convex if and only if any of its diagonals lies entirely inside it.)
# | # Solution.
Answer: see the figure
Note. Since all polygons must be convex, no two of them can intersect except along part of a side. Therefore, among the sides of the hexagon, no more than... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,079 |
5. Misha and Masha had the same multi-digit integer in their notebooks, ending in 9876. Masha placed a plus sign between the third and fourth digits from the right, while Misha placed a plus sign between the fourth and fifth digits from the right. To the surprise of the schoolchildren, both resulting sums turned out to... | Solution. Let the written number have the form $\overline{x 9876}$, where $x$ is also some natural number. Then Misha got the sum $x+9876$, and Masha got the sum $10 x+9+876$. From the equality $x+9876=10 x+9+876$ we find that $x$ $=999$.
Answer: 9999876 and there is no other number.
## CONDITION | 9999876 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,080 |
6. Student Vasya, living 50 km from his institute, rides a bicycle to classes and always arrives exactly on time. However, today he rode at the planned speed for only the first 10 km, after which his bicycle broke down, and Vasya had to walk. After some time, Vasya got lucky, and the last 24 km he traveled by a passing... | Solution. Vasya walked $50-10-24=16$ km. In the time spent walking, Vasya would have traveled 2.5 times more on a bicycle, i.e., 16 * $2.5=40$ km. This means that his classes at the institute started just when Vasya managed to get a ride. Conclusion: Vasya was late for his classes.
Answer: no, he did not make it. | no,hedidnotmakeit | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,081 |
1. The sum of 2020 natural numbers is an odd number. Is the product of these numbers even or odd? Justify your answer in detail. | Solution. Since the sum of an even number of addends is odd, there is at least one even number among them. The product of any number of numbers is even if there is at least one even factor. Answer: an even number. | aneven | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,082 |
4. Buratino buried two ingots on the Field of Wonders: a gold one and a silver one. On the days when the weather was good, the gold ingot increased by $30 \%$, and the silver one by $20 \%$. On the days when the weather was bad, the gold ingot decreased by $30 \%$, and the silver one by $20 \%$. After a week, it turned... | Solution. Increasing a number by $20 \%$ is equivalent to multiplying it by 1.2, and decreasing a number by $20 \%$ is equivalent to multiplying it by 0.8 (for $30 \%$ - by 1.3 and 0.7, respectively). Therefore, the result does not depend on the sequence of good and bad weather days, but only on the number of good and ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,085 |
5. Five friends Sonya, Tanya, Lena, Galia, and Vika were born in five cities: Riga, Penza, Kazan, Belgorod, and Moscow. Each of them loves candies produced in one of these cities. It is known that no one likes candies produced in their hometown. Sonya loves candies from Riga. Tanya is from Riga, and her favorite candie... | Solution. From the condition, we can determine who likes which candies:
Sonia - from Riga
Tanya - from Penza;
Vika - from Moscow;
Galina - from Belgorod; therefore, Lena - from Kazan.
It is known that no one likes candies produced in their hometown. Let's mark this condition in the table, as well as the conditions... | Sonia | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,086 |
10.1. Place five natural numbers in the circles located at the vertices of a square and at its center, so that any two numbers connected by a segment have a common divisor greater than 1, while any two numbers not connected by a segment are coprime. | Answer: one possible example is shown in Figure 10.1a.
Fig. $10.1 \mathrm{a}$
. It is sufficient to show that the circumcircle of triangle $O C I$ also passes through point $M$, since in this ca... | The second solution. Since $\angle O A I=\angle O B I=90^{\circ}$, the quadrilateral $O A I B$ is inscribed in some circle $\Omega$. Denote
Fig. 2
-x\left(x^{x}-1\right)=x\left(x^{x}-1\right)\left(x^{x-1}-1\right)$.
Thus, $x=1$.
## Answer $x=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,090 |
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