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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 |
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 |
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 |
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 |
# 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 |
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 |
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 |
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 |
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 |
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 |
Problem 9.6. A white checkered $8 \times 8$ table is given. In it, 20 cells are painted black. What is the minimum number of pairs of adjacent white cells that could remain? | Answer: 34.
Solution. We will call a pair of cells that are adjacent by side simply a pair. Let's first count the total number of pairs. There are 7 pairs in each row and column, so there are a total of $7 \cdot 8 \cdot 2 = 112$ pairs.
We will call cells that touch the border of the table boundary cells, and those th... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. The sides of the square $A B C D$ are parallel to the coordinate axes, with $A B$ lying on the y-axis, and the square is positioned as shown in the figure. The parabola defined by the equation
$$
y=\frac{1}{5} x^{2}+a x+b
$$
passes through points $B$ and $C$. Additionally, the vertex of this parabola (po... | Answer: 20.
Solution. Note that the parabola is symmetric with respect to the vertical axis passing through its vertex, point $E$. Since points $B$ and $C$ are on the same horizontal line, they are symmetric with respect to this axis. This means that this axis passes through the midpoint of $B C$, and therefore, throu... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. 73 children are standing in a circle. A mean Father Frost walks around the circle clockwise and distributes candies. At first, he gave one candy to the first child, then skipped 1 child, gave one candy to the next child, then skipped 2 children, gave one candy to the next child, then skipped 3 children, an... | Answer: 36.
Solution. Let's number the children clockwise from 0 to 72. Initially, Santa Claus gives a candy to the child with number 1.
Consider the sequence of numbers $a_{n}=1+2+3+\ldots+n$, where $n=1,2,3, \ldots, 2020$. Notice that the $n$-th candy is given to the child whose number is the remainder of the divis... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 10.2. How many roots does the equation
$$
\overbrace{f(f(\ldots f}^{10 \text { times } f}(x) \ldots))+\frac{1}{2}=0
$$
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
1. Find all positive roots of the equation $x^{x}+x^{1-x}=x+1$.
# | # Solution
Since $x>0$, then
$0=x^{2 x}+x-x^{x+1}-x^{x}=x^{x}\left(x^{x}-1\right)-x\left(x^{x}-1\right)=x\left(x^{x}-1\right)\left(x^{x-1}-1\right)$.
Thus, $x=1$.
## Answer $x=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. On the first day, Masha collected $25 \%$ fewer berries than Vanya, and on the second day, $20 \%$ more than Vanya. Over the two days, Masha collected $10 \%$ more berries than Vanya. What is the smallest number of berries they could have collected together? | # Solution
Masha collected $3 / 4$ on the first day and $-6 / 5$ of the number of berries collected by Vanya over these days. Let Vanya collect $4 x$ berries on the first day and $5 y$ on the second day, then Masha collected $3 x$ and $6 y$ berries respectively. According to the condition, $3 x + 6 y = 11 / 10 (4 x + ... | 189 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task No. 1.1
# Condition:
The figure shows 4 circles.
Find the sum of the numbers that are in exactly two circles.
# | # Answer: 22
## Exact match of the answer -1 point
## Solution.
The gray areas on the diagram represent the regions that are included in exactly two circles. The number 10 is in the orange and blue circles, the number 2 is in the blue and brown circles, the number 1 is in the green and brown circles, and the number ... | 59 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. Several numbers are written on the board. It is known that the square of any written number is greater than the product of any two other written numbers. What is the maximum number of numbers that can be on the board
# | # Answer. 3 numbers.
Solution. Suppose there are at least four numbers, and $a-$ is the number with the smallest absolute value. Among the remaining numbers, at least two have the same sign (both non-negative or both non-positive). Let these numbers be $b$ and $c$; then $bc = |bc| \geqslant |a|^2 = a^2$, which contrad... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. (a) (2 points) A natural number $n$ is less than 120. What is the largest remainder that the number 209 can give when divided by $n$?
(b) (2 points) A natural number $n$ is less than 90. What is the largest remainder that the number 209 can give when divided by $n$? | Answer: (a) 104. (b) 69.
Solution. Let $209=n k+r$, where $k-$ is the quotient, and $r$ is the remainder of the division. Since $rn k+r=209=n k+r>r k+r=r(k+1)$, hence
$$
k+1>\frac{209}{n} \quad \text { and } \quad r\frac{209}{119}$, i.e., $k \geqslant 1$. Then $r\frac{209}{89}$, i.e., $k \geqslant 2$. Then $r<\frac{2... | 104 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.3. On the board, natural numbers $a, b, c, d$ are written. It is known that among the six sums
$$
a+b, \quad b+c, c+d, d+a, a+c, b+d
$$
three are equal to 23, and the other three are equal to 34.
(a) (1 point) What is the value of $a+b+c+d$?
(b) (3 points) What is the smallest of the numbers $a, b, c, d$... | Answer: (a) 57. (b) 6.
Solution. (a) Let's add all 6 sums $a+b, b+c, c+d, d+a, a+c, b+d$. Since three of them are equal to 23 and the other three are equal to 34, we get $23 \cdot 3 + 34 \cdot 3$. On the other hand, we get $3(a+b+c+d)$. Therefore,
$$
a+b+c+d=\frac{23 \cdot 3 + 34 \cdot 3}{3}=57
$$
(b) Suppose that a... | 57 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.4. Given a parallelogram $A B C D$, point $M$ is the midpoint of side $B C$. On side $A D$, there is a point $K$ such that $B K=B M$ and the quadrilateral $K B M D$ is cyclic.
(a) (2 points) What is the length of segment $M D$, if $A D=17$?
(b) (2 points) How many degrees does the angle $K M D$ measure, if... | Answer: (a) 8.5. (b) 48.
Solution. (a) Note that KBMD is a cyclic trapezoid, so it is isosceles, i.e., $M D=K B$. Therefore,
$$
M D=K B=\frac{B C}{2}=\frac{A D}{2}=8.5
$$
Fig. 7: to the so... | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.5. One winter day, 43 children were throwing snowballs at each other. Each of them threw exactly one snowball at someone else. It is known that:
- the first threw a snowball at the one who threw a snowball at the second,
- the second threw a snowball at the one who threw a snowball at the third,
- the forty... | # Answer: 24.
Solution. First, note that not only did each throw exactly one snowball, but each was hit by exactly one snowball. Indeed, from the phrase "the first threw a snowball at the one who threw a snowball at the second," it follows that someone threw a snowball at the second; similarly, it is established that ... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. A pair of natural numbers ( $a, p$ ) is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 13)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 20. | Answer: (a) any of the numbers $14, 26, 182$. (b) 24.
Solution. Since $a^{3}+p^{3}=(a+p)\left(a^{2}-a p+p^{2}\right)$, and $a^{2}-p^{2}=(a+p)(a-p)$, the condition of divisibility is equivalent to $a^{2}-a p+p^{2}=a(a-p)+p^{2}$ being divisible by $a-p$. Note that $a(a-p)$ is divisible by $a-p$, so $p^{2}$ must be divis... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. At one meal, Karlson can eat no more than 5 kg of jam. If he opens a new jar of jam, he must eat it completely during this meal. (Karlson will not open a new jar if he has to eat more than 5 kg of jam together with what he has just eaten.)
Little Boy has several jars of raspberry jam weighing a total of ... | Answer: 12.
Solution. We will prove that in 12 meals, Karlson will always be able to eat all the jam.
We will distribute the jars into piles according to the following algorithm. In each pile (starting with the first, then the second, and so on), we will place jars one by one until the pile contains more than 5 kg of... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Variant 10.4.2. Given a parallelogram $A B C D$, point $M$ is the midpoint of side $B C$. On side $A D$, there is a point $K$ such that $B K = B M$ and quadrilateral $K B M D$ is cyclic.
(a) (2 points) What is the length of segment $M D$, if $A D = 19$?
(b) (2 points) How many degrees does angle $K M D$ measure, if $... | Answer: (a) 9.5. (b) 42. | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 10.6.1. A pair of natural numbers $(a, p)$ is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 13)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 20. | Answer: (a) any of the numbers $14,26,182$. (b) 24. | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Variant 10.6.2. A pair of natural numbers $(a, p)$ is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 17)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 18. | Answer: (a) any of the numbers $18,34,306$. (b) 21. | 21 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Variant 10.6.3. A pair of natural numbers ( $a, p$ ) is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 19)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 24. | Answer: (a) any of the numbers $20,38,380$. (b) 27. | 27 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Variant 10.6.4. A pair of natural numbers $(a, p)$ is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.
(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 11)$ is good.
(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 16. | Answer: (a) any of the numbers $12, 22, 132$. (b) 18. | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. Several consecutive natural numbers are written on the board. Exactly $52 \%$ of them are even. How many even numbers are written on the board | # Answer: 13.
Solution. Since the natural numbers written down are consecutive, even and odd numbers alternate. According to the condition, there are more even numbers, which means the sequence starts and ends with even numbers.
First method. Let $n$ be the number of even numbers, then the number of odd numbers is $(... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.2. The graph of the function $y=x^{2}+a x+b$ is shown in the figure. It is known that the line $A B$ is perpendicular to the line $y=x$. Find the length of the segment $O C$.
---
The text has been translated while preserving the original formatting and line breaks. | Answer: 1.
Solution. Since $y(0)=b$, then $B(0 ; b)$. Now let's find the length of the segment $O A$.
First method. Since the line $A B$ is perpendicular to the line $y=x$, it is parallel to the line $y=-x$. Moreover, this line passes through the point $B(0 ; b)$. Therefore,
. Then $\angle F A H=\angle H B F=\alpha$ (inscribed angles subtending the same arc). From the right triangle $A D C: \angle C=90^{\circ}-\alpha$, and from the right triangle $E C B: \angle E B C=90^{\circ}-\angle C=\alpha$.
Thus, $B E$ is the height an... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.6. Twenty-five coins are arranged into piles as follows. First, they are arbitrarily divided into two groups. Then any of the existing groups is again divided into two groups, and so on until each group consists of one coin. Each time a group is divided into two, the product of the number of coins in the two resultin... | Answer: 300.
Solution. First method. Let's represent the coins as points and connect each pair of points with a segment. We will get $\frac{25(25-1)}{2}$ $=300$ segments. With each division of one group of coins into two, we will erase all segments connecting points corresponding to coins that end up in different grou... | 300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.5. To a natural number $N$, the largest divisor of $N$ less than $N$ was added, and the result was a power of ten. Find all such $N$. (N. Agakhanov) | Answer: 75.
Solution. Let $m$ be the greatest divisor of the number $N$, less than $N$. Then $n=mp$, where $p$ is the smallest prime divisor of the number $N$. We have $N+m=10^{k}$, that is, $m(p+1)=10^{k}$. The number on the right side is not divisible by 3, so $p>2$. From this, it follows that $N$ is an odd number, ... | 75 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. To a natural number $N$, the largest divisor of $N$ less than $N$ was added, and the result was a power of ten. Find all such $N$.
(N. Agakhanov) | Answer: 75.
Solution. Let $m$ be the greatest divisor of the number $N$, less than $N$. Then $n=m p$, where $p$ is the smallest prime divisor of the number $N$. We have $N+m=10^{k}$, that is, $m(p+1)=10^{k}$. The number on the right side is not divisible by 3, so $p>2$. From this, it follows that $N$ is an odd number,... | 75 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.5. A natural number $n$ is called good if each of its natural divisors, increased by 1, is a divisor of the number $n+1$. Find all good natural numbers.
(S. Berlov) | Answer. One and all odd prime numbers.
Solution. It is clear that $n=1$ satisfies the condition. Also, all odd primes satisfy it: if $n=p$, then its divisors increased by 1 are 2 and $p+1$; both of them divide $p+1$. On the other hand, any number $n$ that satisfies the condition has a divisor 1; hence, $n+1$ is divisi... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.1. Find the sum $\sin x + \sin y + \sin z$, given that $\sin x = \tan y$, $\sin y = \tan z$, $\sin z = \tan x$ | Answer: 0.
First solution. From $\sin x = \operatorname{tg} y$, we get $\sin x \cos y = \sin y$. Therefore, $|\sin x| \cdot |\cos y| = |\sin y|$. This means $|\sin x| \geq |\sin y|$, and the inequality becomes an equality only if either $\sin y = \sin x = 0$ or $|\cos y| = 1$ (which again implies $\sin y = \sin x = 0$... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.4. In the castle, there are 16 identical square rooms forming a $4 \times 4$ square. Sixteen people, who are either liars or knights (liars always lie, knights always tell the truth), have settled in these rooms, one person per room. Each of these 16 people said: "At least one of the rooms adjacent to mine is occupi... | Answer: 12 knights.
Solution: Note that for each knight, at least one of their neighbors must be a liar. We will show that there must be no fewer than 4 liars (thus showing that there are no more than 12 knights). Suppose there are no more than 3 liars, then there will be a "vertical row" of rooms where only knights l... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. First, we will calculate the number of ways to choose three points out of the 60 available. The first point can be chosen in 60 ways, the second in 59 ways, and the third in 58 ways; as a result, we get $205320=60 \cdot 59 \cdot 58$ options. Since the selections of vertices ABC, ACB, CBA, CAB, BAC, and BCA all give ... | Answer: 34190
Recommendations for evaluating solutions: to determine the number of triplets of points, one can use the combination formula $C_{60}^{3}=\frac{60!}{3!57!}=34220$. | 34190 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that the quadratic function $f(x)=x^{2}+a x+b$ has zeros $x_{1}$, $x_{2}$ and $f(2014)=f(2016)$. Find $\frac{x_{1}+x_{2}}{2}$. | # Answer: 2015
## First Solution.
Since the graph of the quadratic function $f(x)$ is symmetric with respect to the line $x=-a / 2$ and $f(2014)=f(2016), f\left(x_{1}\right)=f\left(x_{2}\right)=0$, then
$-\frac{a}{2}=\frac{x_{1}+x_{2}}{2}=\frac{2014+2016}{2}=2015$.
## Second Solution.
Since $f(2014)=f(2016)$, then... | 2015 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In a square grid of cells, some cells are painted black. It turned out that no black cell shares a side with more than one other black cell? What is the maximum number of cells that could be painted black?
In a square of cells, some cells are painted black. It turned out that no black cell shares a side with more t... | Solution. An example of properly coloring 8 squares is shown in Fig. 1a. Suppose more than 8 were colored, then at least one of the four $2 \times 2$ squares would have at least three cells colored (Fig. 1b). But then at least one of them would share adjacent sides with two other black cells. This contradiction complet... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Three bees collect nectar from 88 garden flowers. Each flower was visited by at least one bee. Each bee visited exactly 54 flowers. We will call a flower sweet if it was visited by all three bees, and bitter if it was visited by exactly one. Which of these 88 flowers are more: sweet or bitter, and by how many? | Solution. Let among the 88 garden flowers, $s$ be sweet and $g$ be bitter. Then the number of flowers visited by two bees is exactly $88-s-g$. On one hand, the total number of bee landings on the flowers is $3 \cdot 54=162$ (each bee visited 54 flowers), and on the other hand, it is $3s + 2(88-s-g) + g = s - g + 176$. ... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the tournament, 15 volleyball teams are playing, and each team plays against all other teams only once. Since there are no draws in volleyball, there is a winner in each match. A team is considered to have performed well if it loses no more than two matches. Find the maximum possible number of teams that performe... | Answer: 5.
Solution.
Evaluation. If the number of teams that played well is not less than 6, then consider six of them. They could have lost no more than $6 \times 2=12$ matches. But the games between them amounted to $6 \times 5 / 2=15$. Thus, they lost no fewer than 15 matches in total. Contradiction.
Example. Pla... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Let there be $x$ seventh-graders participating in the tournament, who together scored $n$ points. Then the number of eighth-graders participating in the tournament is $10 * x$ people, and the total points they scored is $4.5 * n$ points. Therefore, a total of $11 * x$ students participated in the tournament, and the... | Answer: 1 student from 7th grade participated in the tournament and scored 10 points. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The bisectors $\mathrm{AD}$ and $\mathrm{BE}$ of triangle $\mathrm{ABC}$ intersect at point I. It turns out that the area of triangle ABI is equal to the area of quadrilateral CDIE. Find the greatest possible value of angle ACB. | Answer: $60^{\circ}$.
Solution. Let $\mathrm{S}(\mathrm{CDIE})=\mathrm{S}_{1}, \mathrm{~S}(\mathrm{ABI})=\mathrm{S}_{2}$, $S(B D I)=S_{3}, S(A I E)=S_{4}$ (see figure). Since the ratio of the areas of triangles with a common height is equal to the ratio of the bases, and the angle bisector divides the opposite side in... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) In Sun City, they exchange a honey cake for 6 pretzels, and for 9 pretzels, they give 4 doughnuts. How many doughnuts will they give for 3 honey cakes? Explain your answer. | Answer: 8.
Solution.
If for one cookie you get 6 pretzels, then for 3 cookies you will get $3 \times 6=18$ pretzels. 18 pretzels is 2 times 9 pretzels. Therefore, for them, you will get 2 times 4 gingerbread cookies, i.e., 8 gingerbread cookies.
## Grading Criteria.
- Any correct and justified solution - 7 points.
... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. In the cells of an $8 \times 8$ board, the numbers 1 and -1 are placed (one number per cell). Consider all possible placements of a four-cell figure
$\square$ on the board (the figure can be rotated, but its cells must not go beyond the board's boundaries). A placement is called unsuccessful if the sum of the numbe... | Answer: 36. Solution: We will show that in each "cross" of five cells on the board, there will be at least one unsuccessful placement. Suppose the opposite; let the numbers in the outer cells of the cross be \(a, b, c, d\), and in the central cell be \(e\); denote the sum of all these five numbers by \(S\). Then, accor... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. On the day when Dima's brother and sister were congratulating him on his birthday, Dima said: "Look how interesting, I am now twice as old as my brother and three times as old as my sister!" - "And your average age is 11 years," - added Dad. How old did Dima turn? | Answer: 18 years.
Solution. The first method. According to the problem, we can form an equation. Let Dima's age be $x$ years, then his sister's age is $x / 3$, and his brother's age is $-x / 2 ;(x+x / 3+x / 2): 3=11$. After solving this equation, we get that $x=18$. Dima is 18 years old.
It will be useful to provide ... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
# 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 |
# 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 |
# 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 |
# 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 |
# 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... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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