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8.2. Through the vertices $A$ and $C$ of triangle $ABC$, lines are drawn perpendicular to the bisector of angle $ABC$ and intersect the lines $CB$ and $BA$ at points $K$ and $M$ respectively. Find $AB$, if $BM=10, KC=2$. | Solution. Let point $K$ be located on side $BC$ of triangle $ABC$. Triangles $ABK$ and $MBC$ are isosceles (the bisector drawn from vertex $B$ is also an altitude), so $AB = BK = BC - CK = 10 - 2 = 8$.
If point $K$ is located on the extension of segment $BC$ beyond point $C$, then similarly we find that $AB = 12$.
An... | 8or12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,945 |
8.4. Petya wrote down a line of three positive numbers, below it - a line of their pairwise sums, and below that - a line of the pairwise products of the numbers in the second line. The numbers in the third line coincided (in some order) with the numbers in the first line. Find these numbers. | Solution. Let Pete write down the numbers $a \leq b \leq c$, then the numbers in the third row are $(a+b)(a+c) \leq(a+b)(b+c) \leq(a+c)(b+c)$. Therefore, $a=(a+b)(a+c), b=(a+b)(b+c)$. Adding these equations, we get $a+b=(a+b)(a+b+2 c)$, from which $a+b+2 c=1$.
Similarly, $a+2 b+c=1, 2 a+b+c=1$, from which $a=b=c=\frac... | \frac{1}{4};\frac{1}{4};\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,946 |
Task 1. Fill in the squares with numbers from 1 to 5 to make the equation true (each number is used exactly once):
$$
\square+\square=\square \cdot(\square-\square)
$$
It is sufficient to provide one example. | Answer: $1+2=3 \cdot(5-4)$.
Note. Other examples are possible.
## Criteria
## 4 p. A correct example is provided. | 1+2=3\cdot(5-4) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,947 |
Problem 2. Two painters are painting a 15-meter corridor. Each of them moves from the beginning of the corridor to its end and starts painting at some point until the paint runs out. The first painter has red paint, which is enough to paint 9 meters of the corridor; the second has yellow paint, which is enough for 10 m... | Answer: 5 meters.
Solution. The first painter starts painting 2 meters from the beginning of the corridor and finishes at $2+9=11$ meters from the beginning of the corridor.
The second painter finishes painting 1 meter from the end of the corridor, which is 14 meters from the beginning of the corridor, and starts at ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,948 |
Problem 3. Lёsha bought a chocolate bar in the shape of a heart (see the picture on the right). Each whole small square of the bar weighs 6 g. How much does the entire bar weigh?
Answer: 240... | Solution. The tile consists of 32 whole squares and 16 triangles. Each triangle is half a square, meaning it weighs $6: 2=3$ g. Therefore, the weight of the chocolate tile is calculated as follows:
$$
32 \cdot 6+16 \cdot 3=240_{\Gamma} \text {. }
$$
## Criteria
4 p. The correct answer is obtained. | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,949 |
Problem 4. As is known, balance scales come to equilibrium when the weight on both pans is the same. On one pan, there are 9 identical diamonds, and on the other, 4 identical emeralds. If one more such emerald is added to the diamonds, the scales will be balanced. How many diamonds will balance one emerald? The answer ... | Answer: 3 diamonds.
Solution. From the condition of the problem, it follows that 9 diamonds and 1 emerald weigh as much as 4 emeralds. Thus, if we remove one emerald from each side of the scales, the equality will be preserved, that is, 9 diamonds weigh as much as 3 emeralds. This means that 3 diamonds weigh as much a... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,950 |
Problem 5. Six gnomes are sitting around a round table. It is known that exactly two gnomes always tell the truth, and they sit next to each other. Additionally, exactly two gnomes always lie, and they also sit next to each other. The remaining two gnomes can either lie or tell the truth, and they do not sit next to ea... | Answer: in the cave.
Solution. We have two gnomes who always tell the truth. Let's call them truthful gnomes (denoted by the letter T in the diagram).
There are two gnomes who always lie. Let's call them liars (denoted by the letter L in the diagram).
And there are two gnomes who can both lie and tell the truth. Let... | inthecave | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,951 |
Task 1. For what values of p is one of the roots of the equation $x^{2}+p x+18=0$ twice the other? | Answer: 9 or -9.
Solution. Let the roots of the equation be $a$ and $2a$. By Vieta's theorem, $a + 2a = -p$, $a \cdot 2a = 18$. Therefore, $a = \pm 3$, $p = -3a$.
Criteria. For losing the second solution - 2 points are deducted. For an answer without justification - 0 points. A solution without using Vieta's theorem ... | 9or-9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,952 |
2. It is known that the number $a=\frac{x}{x^{2}-x+1}$ is rational. Prove that the number $b=\frac{x^{2}}{x^{4}-x^{2}+1}$ is also rational. | Solution. When $x=0$, the number $b=0$ is rational. If $x \neq 0$, then $a \neq 0, b \neq 0$. Then we can write $\frac{1}{a}=x-1+\frac{1}{x}, \frac{1}{b}=x^{2}-1+\frac{1}{x^{2}}$. Therefore, $x+\frac{1}{x}=\frac{1}{a}+1$. Squaring this equality, we get $x^{2}+2+\frac{1}{x^{2}}=\frac{1}{a^{2}}+\frac{2}{a}+1$, from which... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,953 |
3. A natural number $n$ is such that the numbers $2 n+1$ and $3 n+1$ are squares. Can the number $n$ be prime in this case? | Answer. No, it cannot.
Solution. Let $2 n+1=a^{2}$ and $3 n+1=b^{2}$, then
$$
n=(3 n+1)-(2 n+1)=b^{2}-a^{2}=(b-a)(b+a)
$$
If the number $n$ is prime, then $b-a=1$ and $b+a=n$. From these equations, it is easy to express the numbers $a$ and $b$ in terms of $n: a=\frac{n-1}{2}$ and $b=\frac{n+1}{2}$. Substituting the ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,954 |
4. The angle at the vertex B of the isosceles triangle $A B C$ is $108^{\circ}$. Prove that the bisector of angle A is twice the bisector of angle $B$.
 | Solution. Let $A D$ and $B E$ be the angle bisectors of isosceles triangle $A B C$. Draw a segment $E F$ parallel to $A D$ through point $E$, then $E F$ is the midline of triangle $A C D$, and $E F=\frac{1}{2} A D$. By a simple calculation of angles, we get $\angle F B E=\angle B F E=54 \circ$, and thus, triangle $B E ... | AD=2\cdotBE | Geometry | proof | Yes | Yes | olympiads | false | 14,955 |
1. Do there exist four distinct natural numbers $a, b, c, d$ such that the numbers $a^{2}+2 c d+b^{2}$ and $c^{2}+2 a b+d^{2}$ are squares of natural numbers? | Solution. We want to obtain two squares of sums. To use the formula for abbreviated multiplication, it is enough to make $a b=c d$. We can choose any quartet of numbers, for example, $a b=2 \cdot 12=24=3 \cdot 8=c d$.
The equality $2^{2}+2 \cdot 3 \cdot 8+12^{2}=3^{2}+2 \cdot 2 \cdot 12+8^{2}$, which is the same as $2... | Yes,suchexist,anexamplecanbeprovided | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,958 |
2. There is a ten-digit number, for which all ten digits are used, and zero does not stand in the first place. The digits of this number were rewritten in reverse order and the resulting sequence was appended to the right of the original number. Prove that the resulting twenty-digit number is divisible by 99. | Solution. The sum of all digits of the first ten-digit number is 45, and the sum of all digits of the obtained number is 90. Therefore, it is divisible by 9. Each digit of the first ten-digit number, when the number is written in reverse order, will change its position to a new position of a different parity - the firs... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,959 |
3. Given a circle with center at point $O$ and radius $R$. On the circle, two points $A$ and $B$ are marked such that $\angle A O B=20^{\circ}$. Prove that $A B>\frac{1}{3} R$. | Solution.
Construct two more angles equal to $20^{\circ}-\angle B O C$ and $\angle C O D$. We get $\angle A O D=60^{\circ}$, triangle $A O D$ is equilateral, $A D=R$. Also, $A B=B C=C D$. $A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,960 |
4. Once, a team of Knights and a team of Liars met in the park and decided to ride the circular carousel, which can accommodate 40 people (the "Chain" carousel, where everyone sits one behind the other). When they sat down, each person saw two others, one in front of them and one behind them, and said: "At least one of... | Solution. Let's consider the initial seating arrangement. From the statement of each Liar, it follows that none of those sitting in front of or behind him is a Liar, i.e., each Liar is surrounded by two Knights. Knights and Liars cannot alternate, as in front of or behind each Knight, there must be at least one Knight.... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,961 |
5. There are 12 positive real numbers. It is known that the ratio of any two numbers from this set does not exceed the number 2. Prove that they can be divided into six pairs such that if the sums of the numbers in each pair are calculated, the ratio of any two of the resulting six sums will not exceed $\frac{3}{2}$. | Solution. Arrange the numbers in non-decreasing order: $x_{1} \leq x_{2} \leq \ldots \leq x_{6} \leq x_{7} \leq \ldots \leq x_{11} \leq x_{12}$. For convenience in the proof, denote the numbers $x_{1}=a, x_{6}=b, x_{7}=c, x_{12}=d$.
By the condition, $\frac{d}{b} \leq 2$, i.e., $d \leq 2 b$. Similarly, $\frac{b}{a} \l... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,962 |
1. Variant 1.
Find the smallest natural number whose sum of digits is 47. | Answer: 299999.
Solution. To find the smallest number, you need to get by with as few digits as possible. The largest digit is 9, so you can't do with fewer than 6 digits ( $5 \cdot 9<47$ ). We cannot put less than 2 in the first place, and by taking 2, the other five digits are uniquely determined: they are nines. | 299999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,964 |
2. Variant 1.
Athletes started in groups of 3 people with a delay between groups of several seconds. Petya, Vasya, and Kolya started simultaneously, and they were in the seventh trio from the beginning and the fifth trio from the end. How many athletes participated in the race? | Answer: 33.
Solution. After Petya, Kolya, and Vasya started, 4 more triples of athletes started. Then the total number of triples was $7+4=11$, i.e., a total of 33 athletes participated in the race. | 33 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,965 |
3. Variant 1.
In the apartment, there are four square rooms, which are marked as room №1, №2, №3, №4, and a corridor (№5). The perimeter of room №1 is 16 m, and the perimeter of room №2 is 24 m. What is the perimeter of the corridor (№5)? Give your answer in meters.
, and the side of room No.2 is $24: 4=6$ meters. Then, the side of room No.3 is $6+4=10$ meters, so the side of room No.4 is $10+4=14$ meters. Therefore, the longer side of the corridor is $14+4=18$ meter... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,966 |
4. Variant 1.
The numbers $96, 28, 6, 20$ were written on the board. One of them was multiplied, another was divided, a third was increased, and a fourth was decreased by the same number. As a result, all the numbers became equal to one number. Which one? | Answer: 24.
Solution: Addition and multiplication increase numbers, while subtraction and division decrease them, so the two larger numbers have decreased, and the smaller ones have increased. Now let's consider the two middle numbers 20 and 28. It is clear that a number was subtracted from 28, and the same number was... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,967 |
6. Variant 1.
Petya wrote the numbers from 1 to 10 on cards and laid them out along the edge of a $3 \times 4$ rectangle. At first, one of the cards - with the number 7 - was opened (see the figure). When the other cards were opened, it turned out that the sum of the numbers in the top and bottom horizontal rows is th... | Answer: 23.
Solution. Let $B$ be the last number in the second horizontal row. Since the sum of all numbers from 1 to 10 is 55, we have $2A + B + 7 = 55$ or $B = 48 - 2A$. It follows that $B$ is an even number. To make $A$ the largest, $B$ needs to be the smallest. The smallest even number among the given numbers is 2... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,968 |
# 7. Variant 1.
100 natural numbers are written in a circle. It is known that among any three consecutive numbers, there is an even number. What is the smallest number of even numbers that can be among the written numbers? | Answer: 34.
Solution. Consider any 3 consecutive numbers. Among them, there is an even number. Fix it, and divide the remaining 99 into 33 groups of 3 consecutive numbers. In each such group, there is at least one even number. Thus, the total number of even numbers is at least $1+33=34$. Such a situation is possible. ... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,969 |
Variant 2.
Given a square. Inside it, a point is taken that is removed from three sides at distances of $4, 7, 13$ centimeters. What can the distance to the fourth side be? List all possible options. Express your answer in centimeters. | Answer: 10 and 16.
Option 3
A square is given. Inside it, a point is taken that is 4, 8, and 13 centimeters away from three sides. What can the distance to the fourth side be? List all possible options. Express your answer in centimeters.
Answer: 9 and 17. | 917 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,971 |
6.1. How to transport a goat, a cabbage, two wolves, and a dog from one bank to the other in a boat, given that a wolf cannot be left unattended with the goat or the dog, the dog is in a "dispute" with the goat, and the goat is "partial" to the cabbage? The boat has only three places (one of which is for the boatman), ... | Solution: The algorithm is as follows.
1) Transport the goat and the dog.
2) Return with the dog.
3) Pick up both wolves.
4) Return with the goat.
5) Transport the dog and the cabbage.
6) Return with the dog.
7) Transport the dog and the goat.
Note: The first and second transportations are strictly unique, as are t... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,972 |
6.2. Can six lines be drawn on a plane so that there are exactly six intersection points? (A point of intersection can be passed through by two or more lines). Justify your answer. | Solution: There are several ways to complete the task. Three of them are illustrated in the figure.
1) Through one point, draw 5 lines, and the 6th one - not parallel to any of them and not passing through the specified point (left figure).
2) Three lines form a triangle, and the other three are parallel to each other... | Yes | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,973 |
6.3. Grandma has two balls of wool: a large one and a small one. From the large one, she can knit either a sweater and three socks, or five identical caps. $A$ from the small one - either half a sweater, or two caps. (In both cases, all the wool will be used.) What is the maximum number of socks that Grandma can knit u... | # Solution:
Method 1. Half a sweater requires as much wool as 2 hats, so a sweater requires as much wool as 4 hats. Then 4 hats and 3 socks require as much wool as 5 hats. Therefore, one hat is equivalent to three socks. In total, 5 + 2 = 7 hats or 21 socks can be knitted.
Method 2. Let the amount of wool required fo... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,974 |
6.4. A mushroom is called bad if it contains no fewer than 10 worms, and good otherwise. In the basket, there are 100 bad and 11 good mushrooms. Can all the mushrooms become good after some worms crawl from the bad mushrooms to the good ones? Justify your answer. | # Solution:
Method 1. Assume the opposite. Then from each bad mushroom, at least one worm must crawl to a good mushroom. In total, at least 100 worms will crawl to the good mushrooms. On the other hand, no more than 9 worms can crawl into each good mushroom, so in total, no more than \(11 \cdot 9 = 99\) worms will cra... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,975 |
6.5. Ten different natural numbers are such that the product of any five of them is even, and the sum of all ten is odd. What is the smallest possible sum of all these numbers? Justify your answer. | Solution: There cannot be more than four odd numbers (five odd numbers in a product give an odd number), and, moreover, their quantity must be odd. Therefore, there are either 3 or 1 odd numbers, and 7 or 9 even numbers, respectively. The sets with the smallest sum are 1, 3, 5, 2, 4, 6, 8, 10, 12, 14 and 1, 2, 4, 6, 8,... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,976 |
1. Find all representations of the number 2022 as the sum of several consecutive natural numbers. | Solution. Let $a$ be the first number of the desired sequence, then
$a+(a+1)+\cdots+(a+p-1)=2022 \quad$ and $\quad p a+\frac{p(p-1)}{2}=\frac{p(2 a+p-1)}{2}=2022$.
Notice that $2022=2 \cdot 3 \cdot 337$, therefore $p(2 a-1+p)=2^{2} \cdot 3 \cdot 337$. Since $p+2 a-1>p$ for natural $a$, then $p<337$ and $p=2,3,4,6$ or... | 673+674+675;\quad504+505+506+507;\quad163+164+\ldots+174 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,977 |
2. Solve the equation $8 \sin (x)+12 \sin ^{3}(x)+2022 \sin ^{5}(x)=8 \cos (2 x)+$ $12 \cos ^{3}(2 x)+2022 \cos ^{5}(2 x)$ | Solution. Let's introduce the function $f(t)=8 t+12 t^{3}+2022 t^{5}$, and then the original equation can be represented as $f(\sin (x))=f(\cos (2 x))$.
Note that the introduced function $f(t)$ is monotonically increasing (as the sum of three monotonically increasing functions). The monotonicity of the function $f(t)$... | -\frac{\pi}{2}+2\pik,(-1)^{k}\frac{\pi}{6}+\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,978 |
3. Prove that $3(a b+b c+c a) \leq(a+b+c)^{2}<4(a b+b c+c a)$, where $a, b$ and $c$ are the sides of a triangle. In which cases is equality achieved? | Solution. Let's prove the left part of the double inequality.
Write down the known inequalities for arbitrary $a, b$, and $c$:
$2 a b \leq a^{2}+b^{2}, 2 a c \leq a^{2}+c^{2}$, and $2 b c \leq c^{2}+b^{2}$.
By adding the right and left parts of these inequalities, we get
$$
2(a b+b c+c a) \leq 2\left(a^{2}+b^{2}+c^... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 14,979 |
4. In 1011 open chests lie 2022 coins. Vasya and Petya take one coin each in turn, with Vasya choosing first. Prove that Petya can choose the coins in such a way that the last two coins end up from the same chest. | Solution. After Vasya's choice, Petya can take his first coin in such a way that after this, at least one chest will be empty. Indeed, after Vasya's first move, there are 2022 - 1 coins left in 1011 chests, and therefore, in some chest, there is no more than one coin. If there is one coin in this chest, Petya will take... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 14,980 |
5. In triangle $\mathrm{ABC}$, the sine of angle $\mathrm{A}$ is equal to $\frac{\sqrt{3}}{2}$. On side $\mathrm{AC}$, point $\mathrm{M}$ is taken such that $\mathrm{CM}=b$, on side $\mathrm{AB}$, point $\mathrm{N}$ is taken such that $\mathrm{BN}=a, \mathrm{~T}$ is the midpoint of $\mathrm{NC}$, and $\mathrm{P}$ is th... | Solution. Method 1. Since we know the lengths of vectors $\overrightarrow{M C}, \overrightarrow{N C}$ and the angle between them, it is obvious that we need to express the vector $\overrightarrow{P T}$ in terms of vectors $\overrightarrow{M C}$ and $\overrightarrow{N C}$. This can be done in different ways. We will sho... | PT=\frac{1}{2}\sqrt{b^{2}+^{2}\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,981 |
6. On the faces $BCD, ACD, ABD$, and $ABC$ of the tetrahedron $ABCD$, points $A_{1}, B_{1}, C_{1}$, and $D_{1}$ are marked, respectively. It is known that the lines $AA_{1}, BB_{1}, CC_{1}$, and $DD_{1}$ intersect at point $P$, and $\frac{AP}{A_{1}P}=\frac{BP}{B_{1}P}=\frac{CP}{C_{1}P}=\frac{DP}{D_{1}P}=r$. Find all po... | Solution. Let V be the volume* of tetrahedron ABCD. We introduce the consideration of the partition of the original tetrahedron ABCD into tetrahedra PBCD, PACD, PABD, and PABC. Then for the volumes of the specified tetrahedra, the following is true:
$$
V=V_{\mathrm{PBCD}}+V_{\mathrm{PACD}}+V_{\mathrm{PABD}}+V_{\mathrm... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,982 |
3. Plot on the plane the set of points whose coordinates $(x, y)$ satisfy the condition $y=\frac{\sin x}{|\sin x|}$. | 3. Solution. Rewrite the function $y$, expanding the modulus:
$$
y=\left\{\begin{array}{c}
1, \text { if } \sin x>0 \\
-1 \text { if } \sin x<0
\end{array}\right.
$$
$\sin x>0$ on intervals of the form $(\pi(2 n-1) ; 2 \pi n)$, where $n \in \mathbb{Z}$,
$\sin x<0$ on intervals of the form $(2 \pi n ; \pi(2 n+1))$, w... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,985 |
4. In space, two skew lines $a$ and $b$ and some line $c$ are given (see the figure below). Construct a line that is parallel to line $c$ and intersects each of the lines $a$ and $b$. Justify the construction.
.
Fig. 1.
Through two intersecting lines, a plane can be drawn (Fig. 2). This plane is parall... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,986 |
# 7.1. (7 points)
Find the value of the expression
$$
\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\left(1-\frac{1}{5}\right) \ldots\left(1+\frac{1}{2 m}\right)\left(1-\frac{1}{2 m+1}\right)
$$ | Answer: 1.
Solution: Notice that
$$
\begin{array}{r}
\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=\frac{3}{2} \cdot \frac{2}{3}=1,\left(1+\frac{1}{4}\right)\left(1-\frac{1}{5}\right)=\frac{5}{4} \cdot \frac{4}{5}=1, \ldots \\
\left(1+\frac{1}{2 m}\right)\left(1-\frac{1}{2 m+1}\right)=\frac{2 m+1}{2 m} \cdot \... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,988 |
# 7.2. (7 points)
Two pedestrians set out at dawn. Each walked at a constant speed. One walked from $A$ to $B$, the other from $B$ to $A$. They met at noon and, without stopping, arrived: one - in $B$ at 4 PM, and the other - in $A$ at 9 PM. At what hour was dawn that day? | Answer: at 6 o'clock.
Solution: Let $x$ be the number of hours from dawn to noon. The first pedestrian walked $x$ hours before noon and 4 after, the second - $x$ before noon and 9 after. Note that the ratio of times is equal to the ratio of the lengths of the paths before and after the meeting point, so $\frac{x}{4}=\... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,989 |
7.3. (7 points)
Prove that if in a six-digit number the first and fourth digits are equal, the second and fifth are equal, and the third and sixth are equal, then this number is divisible by 7, 11, and 13.
# | # Solution:
$$
\overline{x y z x y z}=100000 x+10000 y+1000 z+100 x+10 y+z=
$$
$=100100 x+10010 y+1001 z=1001(100 x+10 y+z)$.
$1001=7 \cdot 11 \cdot 13$
# | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,990 |
# 7.4. (7 points)
Can the cells of a $6 \times 6$ square be colored in two colors such that there are more cells of one color than the other, and in each rectangle of size $1 \times 4$ there are an equal number of cells of each color? | Answer: Yes.
Solution: For example, we can divide a $6 \times 6$ square into nine $2 \times 2$ squares, which we will color in a checkerboard pattern (see the figure), then the condition of the problem is satisfied.
In a family, there are six children. Five of them are older than the youngest by 2, 6, 8, 12, and 14 years, respectively. How old is the youngest if the ages of all the children are prime numbers? | Answer: 5 years.
Solution. First, let's check the prime numbers less than six. It is obvious that the number we are looking for is odd. The number 3 does not satisfy the condition because $3+6=9-$ is not a prime number. The number 5 satisfies the condition because $5+2=7, 5+6=11, 5+8=13, 5+12=17, 5+14=19$, which means... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,992 |
1. The second term of an infinite decreasing geometric progression is 3. Find the smallest possible value of the sum $A$ of this progression, given that $A>0$. | Answer: 12.
Solution. Let the first term of the progression be $a$, and the common ratio be $q$. The sum of the progression $A$ is $\frac{a}{1-q}$. From the condition, we have $3=a q$, from which $a=3 / q$. Therefore, we need to find the minimum value of $A=\frac{3}{q(1-q)}$. Note that from the condition it follows: $... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,993 |
2. Find the geometric locus of points $\left(x_{z} ; y_{z}\right)$, which are the vertices of parabolas $y=a x^{2}+z x+$ $c$, where $a$ and $c$ are fixed positive numbers, and $z$ takes all real values. | Answer: parabola $y_{z}=-a x_{z}^{2}+c$.
Solution. The vertex of the parabola $y=a x^{2}+z x+c$ has coordinates $x_{z}=-\frac{z}{2 a}, y_{z}=-a\left(\frac{z}{2 a}\right)^{2}+$ $c=-a x_{z}^{2}+c$. Thus, any vertex lies on the parabola $y_{z}=-a x_{z}^{2}+c$. Now we need to show that any point on this line is the vertex... | y_{z}=-x_{z}^{2}+ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,994 |
3. In the chess tournament for the cup of the city of Krasnoyarsk, boys and girls from schools in the city participated. Each player played one game with each other, 1 point was awarded for a win, 0.5 for a draw, and 0 for a loss. At the end of the tournament, it turned out that each participant scored exactly half of ... | Solution. Let $n$ be the number of boys and $k$ be the number of girls participating in the tournament. Since each boy scored half of his points in matches with boys, the number of points scored by boys in matches with girls is equal to the number of points scored by boys in matches among themselves, which is $\frac{1}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,995 |
4. The base of a quadrilateral pyramid is a square $ABCD$, with a side length of 2, and the lateral edge $SA$ is perpendicular to the base plane and also equals 2. A plane is drawn through the lateral edge $SC$ and a point on the side $AB$, such that the resulting cross-section of the pyramid has the smallest perimeter... | Answer: $\sqrt{6}$.
Solution. The length of the edge $S C=\sqrt{2^{2}+(2 \sqrt{2})^{2}}=2 \sqrt{3}$. Let the section be made through point $M$ on side $A B$. To minimize the perimeter of the section, it is necessary that the sum $S M+M C$ be minimized. Consider the unfolding of faces $S A B$ and $A B C D$ in the form ... | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,996 |
5. A positive integer $N$ and $N^{2}$ end with the same sequence of digits $\overline{a b c d}$, where $a-$ is a non-zero digit. Find $\overline{a b c}$. | Answer: 937.
Solution. Let's represent the number $N$ as $10000 M+k$, where $k$ and $M$ are natural numbers, and $1000 \leq k \leq 9999$. Since $N$ and $N^{2}$ end with the same sequence of digits, the difference
$$
N^{2}-N=\left(10^{4} M+k\right)^{2}-\left(10^{4} M+k\right)=10^{4}\left(10^{4} M^{2}+2 M k-M\right)+k^... | 937 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,997 |
11.1. A number is inserted into each cell of a $2 \times 2$ square. All numbers are pairwise distinct, the sum of the numbers in the first row is equal to the sum of the numbers in the second row, and the product of the numbers in the first column is equal to the product of the numbers in the second column. Prove that ... | # Solution:
Method 1. Let the numbers in the top row of the table be \(a\) and \(b\), and in the second row \(-c\) and \(d\); the numbers \(a\) and \(c\) are in the same column. Then the condition of the problem is equivalent to the system
\[
\left\{\begin{aligned}
a+b & =c+d \\
a c & =b d
\end{aligned}\right.
\]
Ex... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,998 |
11.2. In the city of Perpendicularinsk, it was decided to build new multi-story houses (some of them may be single-story), but in such a way that the total number of floors would be 30. The city architect, Parallelnikov, proposed a project according to which, if after construction one climbs to the roof of each new hou... | Solution: 1) We will show that the project does not involve building houses with more than two floors. Assume the opposite, that such houses are planned. Take the lowest of them and reduce it by one floor, building an additional one-story house as a result. The sum of the numbers in question will decrease by the number... | 112 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,999 |
11.3. This time, the pizza was baked in the shape of an irregular triangle and cut into 7 pieces with three straight cuts, as shown in the figure (each cut was made from a vertex to a point dividing the opposite side in the ratio 1:2).
Billy, Willie, and Dilly each took a triangular piece from the corners (they are ma... | Solution: First, we need to establish the ratio in which the segments connecting the vertices of a triangle to the points dividing the sides into three parts (such segments are called trisectors) divide each other.
Let the trisectors $A M$ and $C N$ of triangle $A B C$ intersect at point $K$ - see the diagram. $2 B M ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,000 |
11.4. Solve the system of equations
$$
\left\{\begin{array}{l}
x \sqrt{1-y^{2}}=\frac{1}{4}(\sqrt{3}+1) \\
y \sqrt{1-x^{2}}=\frac{1}{4}(\sqrt{3}-1)
\end{array}\right.
$$ | # Solution:
Method 1. From the domain of the system, it follows that $-1 \leqslant x, y \leqslant 1$. Moreover, both these numbers are positive, since the right-hand sides of the equations are positive. Therefore, we can assume that $x=\sin \alpha, y=\sin \beta$ for some $\alpha$ and $\beta$ from the interval $\left(0... | {(\frac{\sqrt{2+\sqrt{3}}}{2};\frac{\sqrt{2}}{2}),(\frac{\sqrt{2}}{2};\frac{\sqrt{2-\sqrt{3}}}{2})} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,001 |
11.5. In space, a line $l$ is drawn and a point $P$ is marked, not lying on this line. Find the geometric locus of points - projections of point $P$ onto all possible planes passing through the line $l$. | Solution: Let the given line and point be the line $l$ and point $P (P \notin l)$. Draw a plane $\alpha$ through $P$ perpendicular to the line $l$, and let it intersect the line $l$ at point $Q$. Any plane $\beta$ passing through the line $l$ intersects the plane $\alpha$ along some line $m(\beta)$, and $l \perp m(\bet... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,002 |
11.6. The sheriff believes that if the number of bandits he catches on a certain day is a prime number, then he is lucky. On Monday and Tuesday, the sheriff was lucky, and starting from Wednesday, the number of bandits he caught was equal to the sum of the number from the day before yesterday and twice the number from ... | Solution: Let's say the sheriff caught 7 bandits on Monday and 3 on Tuesday. Then on Wednesday, Thursday, and Friday, he caught 13, 29, and 71 bandits, respectively. All these numbers are prime, and the sheriff has been lucky for five days in a row.
We will show that the sheriff cannot be lucky for six days in a row. ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,003 |
11.6. Petya chose a natural number $n$ and wrote the following $n$ fractions on the board:
$$
\frac{0}{n}, \frac{1}{n-1}, \frac{2}{n-2}, \frac{3}{n-3}, \ldots, \frac{n-1}{n-(n-1)}
$$
Let the number $n$ be divisible by a natural number $d$. Prove that among the written fractions, there will be a fraction equal to the ... | Solution. Let $n=k d$. Then the fraction on the board is
$$
\frac{n-k}{k}=\frac{k d-k}{k}=d-1
$$
which is what we needed to prove. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,004 |
11.7. A function $f(x)$ defined on the entire number line satisfies the condition for all real $x$ and $y$
$$
f(x)+f(y)=2 f\left(\frac{x+y}{2}\right) f\left(\frac{x-y}{2}\right)
$$
Is it true that the function $f(x)$ is necessarily even? (O. Podlipsky) | Answer. Correct.
Solution. Substitute $-y$ for $y$ in the given equation. We get
$$
f(x)+f(-y)=2 f\left(\frac{x-y}{2}\right) f\left(\frac{x+y}{2}\right) \text {. }
$$
Thus, $f(x)+f(y)=f(x)+f(-y)$, from which we obtain $f(y)=f(-y)$ for all real $y$. This means that the function $f(x)$ is even.
Remark. There exist no... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,005 |
11.9. In the company, there are 100 children, some of whom are friends (friendship is always mutual). It is known that by selecting any child, the remaining 99 children can be divided into 33 groups of three such that in each group, all three are pairwise friends. Find the minimum possible number of pairs of friends.
... | Answer: 198.
Solution: Let's translate the problem into the language of graphs, associating each child with a vertex and each friendship with an edge. Then we know that in this graph with 100 vertices, after removing any vertex, the remaining vertices can be divided into 33 triples such that the vertices in each tripl... | 198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,007 |
11.10. On the sphere $\omega_{1}$, a fixed point $A$ is marked, and on the sphere $\omega_{2}$, a fixed point $B$ is marked. On the sphere $\omega_{1}$, a variable point $X$ is chosen, and on the sphere $\omega_{2}$, a variable point $Y$ is chosen such that $A X \| B Y$. Prove that the midpoints of all such constructed... | Solution. Let $O_{1}$ and $O_{2}$ be the centers of spheres $\omega_{1}$ and $\omega_{2}$, respectively. Mark the (fixed) midpoints $S$ and $K$ of segments $O_{1} O_{2}$ and $A B$, respectively, as well as the (variable) midpoints $L, M$, and $N$ of segments $B Y, X Y$, and $A X$, respectively (see Fig. 5). Since segme... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,008 |
1. On eight cards, the digits $1,2,5,7,9,0$ and the signs “+" and "=" are written. Is it possible to form a correct addition example using all the specified cards? | Answer: possible.
Example: $95+7=102$.
Criteria. Any explanation that it is not possible: 0 points.
Answer "possible" without an example: 0 points. | 95+7=102 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,009 |
3. 9 knights and liars stood in a row. Each said that there is exactly one liar next to him. How many liars are there among them, if knights always tell the truth, and liars always lie? | Answer: 3 liars.
Sketch of the solution. Consider the partition of all into groups of people of the same type standing in a row, with people of different types in adjacent groups. In such a group, there can only be one liar. If there are at least two, the extreme liars tell the truth. There are no more than two knight... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,011 |
4. Will Katya be able to write a ten-digit number on the board where all digits are different and all differences between two adjacent digits are different (when finding the difference, the larger is subtracted from the smaller)? | Answer: will be able to.
Example: 9081726354.
Criteria. 7 points for any correct example. | 9081726354 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,012 |
6. There is a paper rectangle $3 \times 100$, divided into 300 cells $1 \times 1$. What is the maximum number of pairs consisting of one corner and one $2 \times 2$ square that can be cut out along the grid lines? (A corner is obtained from a $2 \times 2$ square by removing one of its corner cells). | Answer: 33.
Sketch of the solution. The square in the middle row occupies two cells, and the corner - at least one, so the pair occupies at least three cells in the middle row. If there are no fewer than 34 pairs, then they occupy at least $34 \times 3=102$ cells in the middle row, while there are only 100 there.
Exa... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,014 |
11.1. Parallelogram $A B C D$ is such that $\angle B<90^{\circ}$ and $A B<B C$. Points $E$ and $F$ are chosen on the circle $\omega$ circumscribed around triangle $A B C$ such that the tangents to $\omega$ at these points pass through $D$. It turns out that $\angle E D A=\angle F D C$. Find the angle $A B C$.
(A. Yaku... | Answer: $60^{\circ}$.
Solution. Let $\ell$ be the bisector of angle $E D F$. Since $D E$ and $D F$ are tangents to $\omega$, the line $\ell$ passes through the center $O$ of the circle $\omega$.
$. For which natural numbers $n>1$ do the numbers $f(n)$ and $f(2015 n)$ have different parit... | Answer. For all natural $n>1$.
First solution. Let $n=2^{t} \cdot m$, where $t \geqslant 0$, and the number $m$ is odd.
Consider an arbitrary fraction $\frac{k}{n}$. If $k$ is divisible by $2^{t+1}$, then the numerator of this fraction after reduction will be even; otherwise, it will be odd. Among the numbers $1,2, \... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,016 |
11.3. In a volleyball tournament, 110 teams participated, each playing exactly one game against each of the others (there are no ties in volleyball). It turned out that in any group of 55 teams, there is one team that lost to no more than four of the other 54 teams in this group. Prove that in the entire tournament, th... | Lemma. Let $k \geqslant 55$, and suppose that among any $k$ teams, there is one that has lost to no more than four of the remaining $k-1$ teams. Then among any $k+1$ teams, there is one that has lost to no more than four of the remaining $k$ teams.
Proof. Assuming the contrary, consider a set of $k+1$ teams $M=\left\{... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,017 |
11.4. Given a natural number $N \geqslant 3$. We call a set of $N$ points on the coordinate plane admissible if their abscissas are distinct, and each of these points is colored either red or blue. We will say that a polynomial $P(x)$ separates an admissible set of points if either there are no red points above the gra... | Answer. $k=N-2$.
Solution. We will prove that $k=N-2$ works. Take any $N-1$ of the given $N$ points; there exists a polynomial of degree not greater than $N-2$ whose graph passes through them. This polynomial, obviously, separates our points.
It remains to construct an example of an admissible set that cannot be sepa... | N-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,018 |
11.5. Let $n$ be a natural number. On $2 n+1$ cards, non-zero integers are written; the sum of all numbers is also non-zero. It is required to replace the asterisks in the expression $* x^{2 n}+* x^{2 n-1}+\ldots+* x+*$ with these cards so that the resulting polynomial has no integer roots. Is it always possible to do ... | Answer. Yes, necessarily.
Solution. Let $p_{0}, p_{1}, \ldots, p_{2 n}$ be the numbers on the cards, with $p_{2 n}$ being the largest in absolute value among them. We will set $p_{i}$ as the coefficient of $x^{i}$. Then, if $a$ is an integer, modulo not less than two, we have
$$
\begin{aligned}
& \left|p_{2 n} a^{2 n... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,019 |
11.6. In the country, there are $n>1$ cities, some pairs of which are connected by two-way non-stop flights. Moreover, between any two cities, there exists a unique air route (possibly with layovers). The mayor of each city $X$ calculated the number of ways to number all cities with numbers from 1 to $n$ such that on a... | Solution. Let's call some city $A$ the capital. We will call a city even if the route from $A$ to it contains an even number of flights, and odd otherwise (thus, city $A$ is even). Then the parity of any two cities connected by a flight is different. We will prove that the sum of the numbers obtained by the mayors of e... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,020 |
11.7. The sum of positive numbers $a, b, c$ and $d$ is 3. Prove the inequality
$$
\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}+\frac{1}{d^{3}} \leqslant \frac{1}{a^{3} b^{3} c^{3} d^{3}}
$$
(A. Khryabrov) | Solution. Multiply the inequality to be proved by $a^{3} b^{3} c^{3} d^{3}$, we get
$$
a^{3} b^{3} c^{3}+a^{3} b^{3} d^{3}+a^{3} c^{3} d^{3}+b^{3} c^{3} d^{3} \leqslant 1
$$
Since the inequality is symmetric, we can assume that $a \geqslant b \geqslant c \geqslant d$. By the inequality of means for the numbers $a, b$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,021 |
11.8. In triangle $A B C$, the medians $A M_{A}, B M_{B}$, and $C M_{C}$ intersect at point $M$. Construct the circle $\Omega_{A}$ passing through the midpoint of segment $A M$ and tangent to segment $B C$ at point $M_{A}$. Similarly, construct circles $\Omega_{B}$ and $\Omega_{C}$. Prove that the circles $\Omega_{A}, ... | Lemma. Let triangles $A_{1} B C, A B_{1} C$, and $A B C_{1}$ be constructed outwardly on the sides of triangle $A B C$ such that the sum of their angles at vertices $A_{1}, B_{1}$, and $C_{1}$ is a multiple of $180^{\circ}$. Then the circumcircles of triangles $A_{1} B C, A B_{1} C$, and $A B C_{1}$ intersect at one po... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,022 |
11.5. We will call a non-empty (finite or infinite) set $A$, consisting of real numbers, complete if for any real numbers $a$ and $b$ (not necessarily distinct and not necessarily in $A$) such that $a+b$ is in $A$, the number $ab$ is also in $A$. Find all complete sets of real numbers.
(N. Agakhanov) | Answer. There is only one such set: the set $\mathbb{R}$ of all real numbers.
First solution. Let $A$ be a complete set. Since it is non-empty, we can choose an element $a \in A$. Then $a+0=$ $=a \in A$, so $a \cdot 0=0 \in A$. Since $(-x)+x=0 \in A$, we now get that $(-x) \cdot x=-x^{2} \in A$ for all real $x$. By th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,023 |
11.6. In space, there are 2016 spheres, no two of which coincide. Some of the spheres are red, and the rest are green. Each point of contact between a red and a green sphere was painted blue. Find the maximum possible number of blue points.
( $A$. Kuznetsov) | Answer. $1008^{2}=1016064$ points.
Solution. Let there be $r$ red spheres and $2016-r$ green spheres. Since any two spheres can touch at most at one point, the number of blue points does not exceed $r(2016-r)=1008^{2}-$ $-(1008-r)^{2} \leqslant 1008^{2}$.
We will provide an example with this number of blue points. Le... | 1008^2=1016064 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,024 |
11.7. There are $n$ boys and $n$ girls standing in a circle. We will call a pair of a boy and a girl good if there are an equal number of boys and girls on one of the arcs between them (in particular, a boy and a girl standing next to each other form a good pair). It turns out that there is a girl who participates in e... | Solution. Note immediately that on any arc between members of a good pair, there are an equal number of girls and boys.
Let $D$ be a girl participating in 10 good pairs. Denote all the children clockwise as $K_{1}, K_{2}, \ldots, K_{2 n}$ such that $K_{1}$ is $D$, and continue the numbering cyclically (for example, $K... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,025 |
11.8. A natural number $N$ is represented in the form $N=a_{1}-a_{2}=b_{1}-$ $-b_{2}=c_{1}-c_{2}=d_{1}-d_{2}$, where $a_{1}$ and $a_{2}$ are squares, $b_{1}$ and $b_{2}$ are cubes, $c_{1}$ and $c_{2}$ are fifth powers, and $d_{1}$ and $d_{2}$ are seventh powers of natural numbers. Is it necessarily true that among the ... | Answer. No, it is not necessarily so.
Solution. We will provide an example of a number $N$ for which all the specified numbers will be different. Let
$$
N=\left(3^{2}-2^{2}\right)^{105}\left(3^{3}-2^{3}\right)^{70}\left(3^{5}-2^{5}\right)^{126}\left(3^{7}-2^{7}\right)^{120}
$$
Then
$$
N=M_{2}^{2}\left(3^{2}-2^{2}\r... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,026 |
9.1. Find the smallest integer $x$ that satisfies the inequality $\frac{100}{|x|}>x^{2}+1$. | Answer: -4.
If $|x| \geq 5$, then $\frac{100}{|x|} \leq \frac{100}{5}=20 \leq 17 = (-4)^{2} + 1$ we verify the correctness of the answer.
Comment. The correct answer without a rigorous justification up to 2 points. The inequality is solved, but the wrong answer is chosen 2 points. | -4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 15,028 |
9.2. Two cars simultaneously departed from the same point and are driving in the same direction. One car was traveling at a speed of 50 km/h, the other at 40 km/h. Half an hour later, a third car departed from the same point and in the same direction, which overtook the first car one and a half hours later than the sec... | Answer: 60 km/h.
In half an hour, the first car will travel 25 km, and the second car will travel 20 km. Let $x$ be the speed of the third car. The time it takes for the third car to catch up with the first car is $\frac{25}{x-50}$, and the time to catch up with the second car is $\frac{20}{x-40}$. We get the equation... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,029 |
9.5. On the table, there is a pile of 2013 coins. One coin is removed from it and the pile is divided into two (not necessarily equal). Then, from any pile containing more than one coin, one coin is removed again and the pile is divided into two. And so on. Is it possible to leave only piles consisting of three coins o... | Answer: No.
After each procedure, the number of coins decreases by 1, and the number of piles increases by 1. Since there were originally 2013 coins and one pile, after $n$ procedures, the number of coins will be $2013-n$, and the number of piles will be $n+1$. The problem requires that the equation $2013-n=3(n+1)$ or... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,030 |
9.3. At a drama club rehearsal, 8 people gathered. Some of them (honest people) always tell the truth, while the others always lie. One of those present said: "There is not a single honest person here." The second said: "There is no more than one honest person here." The third said: "There is no more than two honest pe... | 3. The standard evaluation methodology for solutions is provided below.
| Points | Correctness (Incorrectness) of the Solution |
| :---: | :--- |
| 7 | Fully correct solution. |
| $6-7$ | Correct solution, but with minor flaws that do not significantly affect the solution. |
| $5-6$ | The solution is generally correct... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,033 |
1. Two-headed and seven-headed dragons came to a meeting. At the very beginning of the meeting, one of the heads of one of the seven-headed dragons counted all the other heads. There were 25 of them. How many dragons in total came to the meeting? | Answer: 8 dragons. Solution. Subtract the 6 heads belonging to the seven-headed dragon from the 25 heads counted by the seven-headed dragon. 19 heads remain. The remaining dragons cannot all be two-headed (19 is an odd number). There can only be one more seven-headed dragon (if there were two, an odd number of heads wo... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,034 |
2. A three-digit number $X$ was written with three different digits $A B C$. Four schoolchildren made the following statements. Petya: “The largest digit in the number $X$ is $B$”. Vasya: “$C=8$”. Tolya: “The largest digit is $C$”. Dima: “$C$ is the arithmetic mean of $A$ and $B$”. Find the number $X$, given that exact... | Answer: 798. Solution. If Tolya told the truth and the largest number is $-C$, then immediately two people (Petya and Dima) would be wrong, which cannot be the case, so Tolya must be wrong, and the others told the truth. Thus, the number $C=8 ; B$ is the largest number, i.e., $B=9 ; C$ is the average of numbers $A$ and... | 798 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,035 |
3. Vovodya is running on a circular track at a constant speed. There are two photographers standing at two points on the track. After the start, Vovodya was closer to the first photographer for 2 minutes, then closer to the second photographer for 3 minutes, and then closer to the first photographer again. How long did... | Solution. Let's divide both arcs of the circle between the photographers in half. The halves of the arcs adjacent to the second photographer make up half the distance. Vasya ran this half in 3 minutes, so he will run the entire circle in 6 minutes.
Answer: in 6 minutes.
Grading criteria: Full solution - 7 points. Onl... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,036 |
5. There are 30 students in the class. They sit at 15 desks such that exactly half of all the girls in the class sit with boys. Prove that it is impossible to rearrange them (at the same 15 desks) so that exactly half of all the boys in the class sit with girls. | Solution. Since there are 15 desks and 30 students, all students sit in pairs, and no one sits alone. Exactly half of the girls sit with boys, which means the other half of the girls sit with each other. Since half of all the girls is an even number, we get that the number of girls is divisible by 4. If we could arrang... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,038 |
11.1. Without using a calculator, determine the sign of the number
$$
(\cos (\cos 1)-\cos 1)(\sin (\sin 1)-\sin 1) .
$$ | Answer: a negative number.
Solution. The function $y=\sin x$ is increasing on the interval $\left[0 ; \frac{\pi}{2}\right]$, so from the inequality $0<\cos 1<1<\frac{\pi}{2}$, it follows that $\sin (\sin 1) - \sin 1 < 0$.
Since $0<\cos 1<1<\frac{\pi}{2}$, and the function $y=\cos x$ is decreasing on the interval $\le... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,039 |
11.2. What is the minimum number of factors that need to be crossed out from the number $99!=1 \cdot 2 \cdot \ldots \cdot 99$ so that the product of the remaining factors ends in $2?$ | Answer: 20 factors.
Solution. From the number 99! it is necessary to remove all factors that are multiples of 5, otherwise the product will end in 0. There are a total of 19 such factors (ending in 0 or 5).
The product of the remaining factors ends in 6. Indeed, since the product $1 \times 2 \cdot 3 \cdot 4 \cdot 6 \... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,040 |
11.3. Does there exist a tetrahedron $A B C D$ in which $A B=A C=A D=B C$, and the sums of the plane angles at each of the vertices $B$ and $C$ are both $150^{\circ}$? | Answer: No, it does not exist.
Solution. Suppose such a tetrahedron exists. Then its faces $DAB$ and $DAC$ are isosceles triangles. Let $\angle ADB = \angle ABD = \alpha$, $\angle DBC = \beta$, $\angle DCB = \gamma$, $\angle ADC = \angle ACD = \delta$, then $\angle BDC = 180^{\circ} - \beta - \gamma$ (see Fig. 11.3).
... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,041 |
11.4. For what values of $x$ and $y$ is the equality
$$
x^{2}+(1-y)^{2}+(x-y)^{2}=\frac{1}{3} ?
$$ | Answer: when $x=\frac{1}{3}, y=\frac{2}{3}$.
Solution. First method. After expanding the brackets and combining like terms, we get:
$$
\begin{aligned}
& x^{2}+(1-y)^{2}+(x-y)^{2}=\frac{1}{3} \quad \Leftrightarrow \quad 2 x^{2}+2 y^{2}-2 x y-2 y+\frac{2}{3}=0 \quad \Leftrightarrow \\
& \Leftrightarrow \quad \frac{1}{2... | \frac{1}{3},\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,042 |
11.5. Given an acute-angled triangle $A B C$. Circles centered at $A$ and $C$ pass through point $B$, intersect again at point $F$, and intersect the circumcircle $w$ of triangle $A B C$ at points $D$ and $E$. Segment $B F$ intersects circle $w$ at point $O$. Prove that $O$ is the center of the circumcircle of triangle... | Solution. First method. First, we will prove that $O E = O F$. For this, the circle centered at point $A$ is not needed. Consider the drawing without it (see Fig. 11.5a). Let the central angle $B C E$ in the circle centered at $C$ be $2 \alpha$. Then the inscribed angle $E F B$ is $\alpha$.
In the circle $\omega$, ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,043 |
11.6. On the computer screen, a certain finite sequence of zeros and ones has been generated. The following operation can be performed with it: the digit sequence "01" can be replaced by the digit sequence "1000". Can such a process of replacements continue indefinitely or will it inevitably come to an end? | Answer: the process will inevitably stop.
Solution. First method. Let the original sequence contain $n$ ones, then it is convenient to write it in the form: $\underbrace{0 \ldots 0}_{a_{1}} 1 \underbrace{0 \ldots 0}_{a_{2}} 1 \ldots 1 \underbrace{0 \ldots 0}_{a_{n}} 1 \underbrace{0 \ldots 0}_{a_{n+1}}$, where the numb... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,044 |
7.1. Does there exist a ten-digit number, divisible by 11, in which all digits from 0 to 9 appear? | Answer. Yes, for example, 9576843210.
Solution. Let's consider a possible way to find the required number. Note that a number is divisible by 11 if the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions is divisible by 11. If we write all ten digits in descend... | 9576843210 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,045 |
7.2. There are pan scales without weights and 11 visually identical coins, among which one may be counterfeit, and it is unknown whether it is lighter or heavier than the genuine coins (genuine coins have the same weight). How can you find at least 8 genuine coins in two weighings? | Solution. Let's divide the coins into three piles of three coins each. Compare pile 1 and pile 2, and then compare pile 2 and pile 3. If all three piles weigh the same, then all the coins in them are genuine, and we have found 9 genuine coins. Otherwise, one of the piles differs in weight from the others, and the count... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,046 |
7.3. From the set of numbers $1,2,3,4, \ldots, 2021$, one number was removed, after which it turned out that the sum of the remaining numbers is divisible by 2022. Which number was removed? | Answer: 1011.
Solution. Let's write out the sum and perform grouping:
$1+2+3+\cdots+2019+2020+2021=(1+2021)+(2+2020)+$ $\cdots+(1012+1010)+1011$.
All the terms enclosed in parentheses are divisible by 2022. If we remove the last ungrouped term from the sum, the sum will be a multiple of 2022.
Comment. Answer only -... | 1011 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,047 |
7.4. Can several rectangles, the sum of the perimeters of which is 100, be combined to form a square with a side length of 16? | Answer: Yes.
Solution. Let's take, for example, two rectangles $8 \times 2$ and one rectangle $14 \times 16$ and arrange them as shown in the figure (the size of one cell is $4 \times 4$).
 | Yes | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,048 |
7.5. On each cell of a chessboard, there is a stack of several chips. In one move, one of the two operations is allowed:
- remove one chip from each stack in any vertical column;
- double the number of chips in each stack in any horizontal row.
Is it possible to clear the entire board of chips in some number of such ... | Answer: Yes.
Solution. Let's record the number of chips in each cell of the board in an $8 \times 8$ table. In this case, there are two types of operations possible: subtracting one from each cell in any column; doubling any row. Our goal is to obtain a table filled with zeros.
First, we will turn all the numbers in ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,049 |
1. There are 20 apples, and the weight difference between any two apples does not exceed 40 grams. Prove that these apples can be divided into two piles of 10 apples each, such that the weight difference between these piles does not exceed 40 grams. | Solution. Let's take four apples and arrange their weights in non-decreasing order: $a \leq$ $b \leq c \leq d$. We will form two piles, each with two apples, by grouping the heaviest with the lightest. According to the condition, the following two inequalities hold.
$$
\begin{aligned}
& 0 \leq d-a \leq 40 \\
& 0 \leq ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,050 |
2. Prove that for all real $a, b, c, d$ the inequality
$$
(a b+1)^{2}+(c d+1)^{2}+(a c)^{2}+(b d)^{2} \geq 1
$$
holds. | Solution. Let's expand the brackets.
$$
\begin{gathered}
(a b+1)^{2}+(c d+1)^{2}+(a c)^{2}+(b d)^{2} \geq 1 \Leftrightarrow \\
(a b)^{2}+2 a b+1+(c d)^{2}+2 c d+1+(a c)^{2}+(b d)^{2} \geq 1 \Leftrightarrow
\end{gathered}
$$
We aim to isolate a square of a sum. Let's add and subtract $2 a b c d$. Subtract 1 from both ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,051 |
3. Two equal circles $S_{1}$ and $S_{2}$ touch the circle $S$ from the inside at points $K$ and $M$. Point $P$ is an arbitrary point on the circle $S$. Points $Q$ and $R$ are the points of intersection of $K P$ and $M P$ with circles $S_{1}$ and $S_{2}$, respectively. Prove that the lines $K M$ and $Q R$ are parallel. | Solution.
Let $K M$ intersect circles $S_{1}$ and $S_{2}$ at points $X$ and $Y$. Circles $S$ and $S_{1}$ have a common tangent at point $K$ (This fact is considered known and does not need to... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,052 |
4. As is known, from five different numbers, you can get ten different groups of three numbers. Do there exist five such different natural numbers such that their sums of three numbers are ten consecutive numbers | Solution. Let the five numbers be $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$. For convenience, we list all ten sums.
$$
\begin{array}{lllll}
x_{1}+x_{2}+x_{3} & x_{1}+x_{2}+x_{4} & x_{1}+x_{2}+x_{5} & x_{1}+x_{3}+x_{4} & x_{1}+x_{3}+x_{5} \\
x_{1}+x_{4}+x_{5} & x_{2}+x_{3}+x_{4} & x_{2}+x_{3}+x_{5} & x_{2}+x_{4}+x_{5} & x_{3... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,053 |
5. Given two functions (all variables and coefficients are real)
$$
f(x)=x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}, \quad g(x)=x^{4}+b_{3} x^{3}+b_{2} x^{2}+b_{1} x+b_{0}
$$
It is known that the equality $f(x)=g(x)$ does not hold for any value of $x$. Prove that the equality $f(x+1)=g(x-1)$ holds for at least one v... | Solution. The equality $f(x)=g(x)$ can be rewritten as
$$
\begin{gathered}
a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}=b_{3} x^{3}+b_{2} x^{2}+b_{1} x+b_{0} \\
\left(a_{3}-b_{3}\right) x^{3}+\left(a_{2}-b_{2}\right) x^{2}+\left(a_{1}-b_{1}\right) x+\left(a_{0}-b_{0}\right)=0
\end{gathered}
$$
If $a_{3} \neq b_{3}$, then th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,054 |
10.1. In the National Basketball Association, there are 30 teams, each of which plays 82 games with other teams in the regular season. Can the Association's management divide the teams (not necessarily equally) into Eastern and Western Conferences and create a schedule such that games between teams from different confe... | Answer: No, it cannot.
Solution: Let $x$ and $y$ be the total number of matches played within the Eastern and Western conferences, respectively, and let $z$ be the number of matches between teams from different conferences. We need to prove that the equation $z=\frac{x+y+z}{2}$ is impossible.
Each of the $k$ teams in... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,055 |
10.2. Diagonals $A C$ and $B D$ of the inscribed quadrilateral $A B C D$ intersect at point $P$. Point $Q$ is chosen on segment $B C$ such that $P Q \perp A C$. Prove that the line passing through the centers of the circumcircles of triangles $A P D$ and $B Q D$ is parallel to the line $A D$.
( | Solution. Choose a point $T$ on the line $QP$ such that $DT \perp DA$ (see Fig. 3). Since $\angle APT = 90^{\circ} = \angle ADT$, the points $A, P, D$, and $T$ lie on the same circle. Therefore, the center of the circle $\omega_{1}$, circumscribed around triangle $APD$, lies on the perpendicular bisector $\ell$ of segm... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,056 |
10.3. Given a cubic polynomial $f(x)$. We will call a cycle a triplet of distinct numbers $(a, b, c)$ such that $f(a)=b, f(b)=c$ and $f(c)=a$. It is known that eight cycles $\left(a_{i}, b_{i}, c_{i}\right), i=1,2, \ldots, 8$, have been found, involving 24 different numbers. Prove that among the eight numbers of the fo... | Solution. Suppose the opposite; then for some four cycles $\left(a_{i}, b_{i}, c_{i}\right)$ the sums of the numbers are the same and equal to some $s$. For each of these cycles, we have
\[
\begin{aligned}
& s=a_{i}+b_{i}+c_{i}=a_{i}+f\left(a_{i}\right)+f\left(f\left(a_{i}\right)\right)= \\
& \quad=b_{i}+f\left(b_{i}\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,057 |
10.4. Inside a convex 100-gon, a point $X$ is chosen, not lying on any of its sides or diagonals. Initially, the vertices of the polygon are not marked. Petya and Vasya take turns marking unmarked vertices of the 100-gon, with Petya starting and marking two vertices on his first move, and then each marking one vertex p... | Solution. Let's color the sides of the 100-gon in black and white such that any two adjacent sides have different colors. Consider two same-colored sides $AB$ and $CD$ that form a convex quadrilateral $ABCD$; let its diagonals $AC$ and $BD$ intersect at point $K$. Suppose point $X$ lies in triangle $KBC$ (see Fig. 4). ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,058 |
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