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6.4. The child placed four identical cubes such that the letters on the sides of the cubes facing him spell out his name (see the figure). Draw how the other letters are arranged on the given net of the cube, and determine the child's name.
. Then, looking at the second cube, we draw the letters K and T on the net (see Fig. 6.4c). Fi... | Nika | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,516 |
6.5. Ladybugs gathered on a clearing. If a ladybug has 6 spots on its back, it always tells the truth, and if it has 4 spots, it always lies, and there were no other ladybugs on the clearing. The first ladybug said: "Each of us has the same number of spots on our backs." The second said: "Together, we have 30 spots on ... | Answer: 5 ladybugs.
If the first ladybug is telling the truth, then the second and third should also be telling the truth, as they should have the same number of spots as the first. However, the second and third ladybugs contradict each other, so at least one of them is lying, which means the first ladybug is also lyi... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,517 |
1. (7 points) In the equality $1-2-4-8-16=19$, place several modulus signs so that it becomes correct. | Answer. || $1-2|-| 4-8|-16|=19$.
There are other examples.
Comment. It is sufficient to provide one example. It is not necessary to explain how it was obtained.
## Evaluation criteria.
## - Any correct example -7 points. | ||1-2|-|4-8|-16|=19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,518 |
2. (7 points) Cheburashka and Gena ate a cake. Cheburashka ate twice as slowly as Gena, but started eating a minute earlier. In the end, they each got an equal amount of cake. How long would it take Cheburashka to eat the cake alone? | Answer. In 4 minutes.
Solution.
First method. If Cheburashka eats twice as slowly as Gena, then to eat the same amount of cake as Gena, he needs twice as much time. This means that the time Cheburashka ate alone (1 minute) is half of the total time it took him to eat half the cake. Thus, he ate half the cake in 2 min... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,519 |
4. (7 points) Three schoolchildren made two statements about natural numbers $a, b, c$:
Anton: 1) $a+b+c=34$
2) $a b c=56$
Boris: 1) $a b+b c+a c=311$
2) the smallest of the numbers is 5;
Nastya: 1) $a=b=c$
2) the numbers $a, b$ and $c$ are prime.
Each schoolchild has one true statement and one false statement. F... | Answer: 2, 13, 19 (in any order).
Solution. If the second statement of Anton is true, then both of Nastya's statements are false. Therefore, \(a + b + c = 34\). Thus, the second statement of Nastya is true. Since the sum of three prime numbers is 34, they cannot all be odd, and one of them must be 2. Therefore, the fi... | 2,13,19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,521 |
5. In an equilateral triangle $ABC$ with side length $a$, points $M, N, P, Q$ are positioned as shown in the figure. It is known that $MA + AN = PC + CQ = a$. Find the measure of angle $NOQ$.
 | Answer: $60^{\circ}$
Solution. According to the problem, $A N=a-A M$, hence $A N=M C$. Similarly, $A P=Q C$. From these equalities and the equality $\angle A=\angle C=60^{\circ}$, it follows that $\triangle A N P=\triangle C M Q$. Therefore, $\angle A N P=\angle Q M C, \angle A P N=\angle M Q C$. By the theorem on the... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,522 |
6. (7 points) On a chessboard, there were 21 kings. Each king was under attack by at least one of the others. After some kings were removed, no two of the remaining kings attack each other. What is the maximum number of kings that could remain?
a) Provide an example of the initial arrangement and mark the removed king... | Answer: b) 16.
Solution: Note that each king removed from the board could not have attacked more than 4 of the remaining ones (otherwise, some of the remaining ones would also attack each other). Therefore, the number of remaining kings cannot exceed the number of removed ones by more than 4 times, meaning it cannot b... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,523 |
3. In the sum $(1-2-3+4)+(5-6-7+8)+\ldots+(2009-2010-2011+2012)+$ (2013-2014-2015), all numbers that end in 0 were crossed out. Find the value of the resulting expression. | 3. Answer: -3026. It is easy to see that in all parentheses except the last one, the sum is zero. Then the entire sum before cancellation was equal to -2016. At the same time, numbers that end in 0 and are divisible by 4 were included in it with a plus, while numbers that end in 0 and are not divisible by 4 were includ... | -3026 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,524 |
4. The quadratic trinomial $a x^{2}+b x+c$ has two distinct roots, and its coefficients are non-zero integers. Moreover, the product of all its coefficients equals the product of all its roots. What values can this product take? Provide a complete and justified answer to this question. | 4. Answer: The product can be any negative integer. Solution From the condition of the problem and Vieta's theorem, it follows that $\mathrm{a}^{2} \mathrm{bc}=\mathrm{c}$. Since all coefficients are integers and non-zero, this means that $\mathrm{b}=1, \mathrm{a}= \pm 1$ and the quadratic trinomial has the form either... | any\negative\integer | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,525 |
5. Four circles with radius 1 and centers at $A, B, C, D$ are arranged on a plane such that each circle touches exactly two others. A fifth circle touches two of these circles externally and passes through the centers of the other two. Find all possible values of the area of quadrilateral ABCD. | 5. Answer: 4 or 3. Solution: Let's assume that the circle centered at $B$ touches the circles centered at $\mathrm{A}$ and $\mathrm{C}$, and the circle centered at $\mathrm{C}$ touches the circles centered at $\mathrm{B}$ and $\mathrm{D}$. It is easy to see that $A B C D$ is a rhombus with side length 2. Let $\mathrm{O... | 4or3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,526 |
6. There are 10 red and 10 yellow apples with a total weight of 1 kg. It is known that the weights of any two apples of the same color differ by no more than 40 g. All apples were sequentially paired "red with yellow," with red apples chosen in non-decreasing order and yellow apples in non-increasing order (i.e., the h... | 6. Answer: 136 g. Solution Estimate. Note that the weights of the pairs also cannot differ by more than 40 g. Therefore, if the weight of the heaviest pair is more than 136 g, then the weight of each of the other nine pairs will be more than 96 g, but then the total weight of all apples will be more than 1 kg, which is... | 136 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 15,527 |
Problem 1. Parabolas $y=x^{2}+a x+b$ and $y=x^{2}+c x+d$ intersect the $O x$ axis at the point $(2019 ; 0)$. Prove that if the points of their secondary intersection with the $O x$ axis are symmetric with respect to the origin, then the points of their intersection with the $O y$ axis are also symmetric with respect to... | Solution. Let the first parabola intersect the $O x$ axis again at point $r$, and the second at point $-r$. Then, by Vieta's theorem, $b=2019 r, d=-2019 r$, i.e., $b=-d$. But these parabolas intersect the $O y$ axis at points $(0 ; b)$ and $(0 ; d)$, from which the required follows.
## Criteria
4 p. A complete and ju... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,528 |
Problem 2. Is it possible to color all natural numbers in red and blue so that any two numbers differing by 5 are of different colors, and any two numbers differing by a factor of two are of different colors?
Answer: no, it is not possible. | Solution. Let's look at the numbers 10, 15, and 20. From the condition, it follows that their colors must be pairwise different, which is impossible since there are only two colors.
## Criteria
4 p. A complete and justified solution is provided.
0 p. The correct answer is provided, but there is no justification.
0 ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,529 |
Problem 3. On a plane, there are a square and an equilateral triangle such that the area of each of these two figures is numerically equal to the perimeter of the other. Find the side length of the given square. | Answer: $2 \sqrt[3]{2} \sqrt{3}$.
Solution. Let the side of the equilateral triangle be $a$, and the side of the square be $b$. Then $3 a=b^{2}$ and $4 b=\frac{\sqrt{3}}{4} a^{2}$. From this, $b^{4}=9 a^{2}=48 \sqrt{3} b$. It follows that $b^{3}=48 \sqrt{3}$, from which $b=2 \sqrt[3]{2} \sqrt{3}$.
## Criteria
4 6. A ... | 2\sqrt[3]{2}\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,530 |
Problem 4. Little Boy and Karlson have a long chocolate bar $15 \times 100$. They take turns eating square pieces of any size from it (pieces can only be eaten along the grid lines). Karlson starts. The one who cannot make a move loses. Who wins with correct play? | Solution. Karlson eats a $14 \times 14$ square out of the middle of the chocolate bar, so that the axis of symmetry of the square and the axis of symmetry of the rectangle coincide. Then, relative to the common axis of symmetry, the remaining part of the chocolate bar is divided into two identical parts. From this poin... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,531 |
Problem 5. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$. Point $P$ is the midpoint of edge $A A_{1}$, point $Q$ is the midpoint of edge $C D$, and point $R$ is the midpoint of edge $B_{1} C_{1}$. Prove that $\angle P B_{1} Q < \angle P R Q$. | Solution. Let the side of the cube be $2a$. Then, by the Pythagorean theorem,
$$
\begin{gathered}
P B_{1}=\sqrt{4 a^{2}+a^{2}}=\sqrt{5} a \\
Q B_{1}=\sqrt{4 a^{2}+4 a^{2}+a^{2}}=3 a \\
P R=R Q=P Q=\sqrt{4 a^{2}+a^{2}+a^{2}}=\sqrt{6} a
\end{gathered}
$$
Then triangle $P Q R$ is equilateral, so $\angle P R Q=60^{\circ}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,532 |
Problem 6. For positive numbers $a, b$ and $c$, prove the inequality
$$
\frac{1+b c}{a}+\frac{1+c a}{b}+\frac{1+a b}{c}>\sqrt{a^{2}+2}+\sqrt{b^{2}+2}+\sqrt{c^{2}+2}
$$ | Solution. Note that
$$
\frac{b c}{a}+\frac{c a}{b}=\frac{c\left(a^{2}+b^{2}\right)}{a b} \geqslant \frac{2 a b c}{a b}=2 c
$$
Similarly, $\frac{c a}{b}+\frac{a b}{c} \geqslant 2 a$ and $\frac{a b}{c}+\frac{b c}{a} \geqslant 2 b$. Adding these three inequalities, we get
$$
\frac{2 b c}{a}+\frac{2 c a}{b}+\frac{2 a b}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,533 |
# 1. CONDITION
According to expert forecasts, apartment prices in Moscow will fall by $20 \%$ in rubles and by $40 \%$ in euros within a year. In Sochi, apartment prices will fall by $10 \%$ in rubles within a year. By how much will apartment prices in Sochi fall in euros? It is assumed that the exchange rate of the e... | Solution. Consider equally priced apartments in Moscow and Sochi, let the cost of either be $a$ rubles, which corresponds to $b$ euros. After a year, the cost of the apartment in Moscow will be 0.8 rubles, which is equal to $0.6 b$ euros. From this, $a$ rubles after a year will correspond to $3 b / 4$ euros. In Sochi, ... | 32.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,535 |
# 3. CONDITION
Inside an isosceles triangle ($A B=B C$), points $M$, $N$, and $K$ are marked (point $N$ is the closest to side $A C$) such that $M N\|B C$ and $N K\| A B$. Prove that $A M+K C>M N+N K$. | Solution. Extend the segment $M N$ beyond point $N$ to intersect side $A C$ at point $F$, and draw $M E$ parallel to $B A$ (point $E$ lies on side $A C$ - see the figure). In triangle $A M E$, angle $A E M$ is obtuse, so the side opposite it is the largest side of the triangle. Then $A M > E M = M F > M N$. Similarly, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,536 |
# 4. CONDITION
Can a 3 x 3 x 3 cube be sawn into 9 corners? A corner is a figure consisting of three 1 x 1 x 1 cubes, as shown in the figure; in the cube, a corner can be oriented parallel to any of the faces. Justify your answer.
.
# | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,537 |
# 5. CONDITION
When sequentially calculating the sum of a hundred addends on a calculator: $20.12+20.12+20.12+\ldots+20.12$, Petya made several mistakes by shifting the decimal point one place in some of the addends—right in some, left in others. Could his result have been exactly twice the correct one? Justify your a... | Solution. First method. Let $x$ be the number of terms where Petya moved the decimal point to the right, $y$ - to the left, and $z$ - where he did not make a mistake. Then $x+y+z=100$ and, if the answer to the question is positive, then $201.2 x + 2.012 y + 20.12 z = 2012 \cdot 2$. The second equation can be simplified... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,538 |
# 6. CONDITION
In a row from left to right, 13 weights with masses of 1 g, 2 g, 3 g, ..., 13 g were placed. Only seven of them, standing in a row (their order has not changed), remain, while the other six weights have been lost. Can the masses of the remaining weights be determined in two weighings on a balance scale?... | Solution. Let the weight of the leftmost of the remaining weights be x. It is sufficient to determine what x is. In the first weighing, we compare the three leftmost weights and the two immediately following them. If the weights are equal, then we have $x+x+1+x+2$ $=\mathrm{x}+3+\mathrm{x}+4$, i.e., $\mathrm{x}=4$ and ... | itispossible | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,539 |
9.1. There are seven cards on the table. In one move, it is allowed to flip any five cards. What is the minimum number of moves required to flip all the cards? | 9.1. Answer. 3 moves.
Obviously, one move is not enough. After two moves, there will be at least three cards that have been flipped twice, which means these cards will be in their original position.
Let's provide an example of flipping all the cards in three moves. Number the cards from 1 to 7 and flip cards numbered... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,541 |
9.2. The roots of the quadratic equation $a x^{2}+b x+c=0$ are 2007 times the roots of the quadratic equation $c x^{2}+d x+a=0$. Prove that $b^{2}=d^{2}$. | 9.2. Let $x_{1}, x_{2}$ be the roots of the second equation, $n=2007$, then by Vieta's theorem $x_{1}+x_{2}=$ $=-\frac{d}{c}, x_{1} x_{2}=\frac{a}{c}, n x_{1}+n x_{2}=-\frac{b}{a}, n x_{1} n x_{2}=\frac{c}{a}$, i.e., $\frac{d n}{c}=-n\left(x_{1}+x_{2}\right)=\frac{b}{a}$, $\frac{n^{2} a}{c}=n^{2} x_{1} x_{2}=\frac{c}{a... | b^{2}=^{2} | Algebra | proof | Yes | Yes | olympiads | false | 15,542 |
9.3. On the side $AC$ of triangle $ABC$, a point $B_{1}$ is taken. Let $I$ be the center of the inscribed circle of the triangle. The circumcircle of triangle $A B_{1} I$ intersects the side $AB$ again at point $C_{1}$. The circumcircle of triangle $C B_{1} I$ intersects the side $BC$ again at point $A_{1}$. Prove that... | 9.3. Note that $A I$ is the bisector of angle $B A C$. From the equality of angles $C_{1} A I$ and $B_{1} A I$, it follows that the chords $C_{1} I$ and $B_{1} I$ are equal. Similarly, $B_{1} I = A_{1} I$ (see Fig. 4). Therefore, point $I$ is the center of the circumscribed circle for all triangles $A_{1} B_{1} C_{1}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,543 |
9.4. In the country, there are 20 cities. An airline wants to organize two-way flights between them so that from any city, it is possible to reach any other city with no more than $k$ transfers. At the same time, the number of air routes from any city should not exceed four. What is the smallest $k$ for which this is p... | 9.4. Answer. $k=2$.
Note that at least two transfers will be required. Indeed, from an arbitrary city $A$ without a transfer, one can reach no more than 4 cities, and with exactly one transfer - no more than $4 \cdot 3=12$ cities (since one of the flights from each of these cities leads back to $A$). Therefore, if usi... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,544 |
9.5. In a set of five sticks, no three sticks can form a triangle. Could it be that, by breaking one of the sticks into two, we obtain six sticks, from which two isosceles triangles can be formed? | 9.5. Answer. It could.
For example, a set of sticks with lengths $1,1,2,3,5$ would work. By breaking the stick of length 5 into two sticks of lengths 2 and 3, we can form two isosceles triangles with sides $1,2,2$ and $1,3,3$. | Itcould | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,545 |
9.7. Let each of the natural numbers $n, n+1, n+2$ be divisible by the square of any of its prime divisors. Prove that the number $n$ is divisible by the cube of some of its prime divisors. | 9.7. Suppose the opposite: $n$ is divisible exactly by the second power of each of its prime divisors. Then $n=k^{2}$, where $k$ is the product of all prime divisors $p$. Therefore, it is sufficient to prove that $n$ cannot be a perfect square.
Suppose the opposite: $n=m^{2}$, where $m$ is a natural number. If $n$ is ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,547 |
9.8. A circle is circumscribed around triangle $A B C$. Let $A_{0}$ and $C_{0}$ be the midpoints of its arcs $B C$ and $A B$, not containing vertices $A$ and $C$. It turns out that the segment $A_{0} C_{0}$ is tangent to the circle inscribed in triangle $A B C$. Find the angle $B$. | 9.8. Let the incircle of triangle $ABC$ be denoted by $\omega$, its center by $I$, and the points of tangency of $\omega$ with $AC$ and $A_0 C_0$ by $K$ and $L$ respectively. Then, in triangles $IKC$ and $ILA_0$, angles $IKC$ and $ILA_0$ are right angles (as angles between a radius and a tangent), and $IK = IL$, since ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,548 |
1. Solve the equation: $\cos ^{2} x+\cos ^{2} 2 x=1$. | Answer: $\frac{\pi}{2}+\pi k, \pm \frac{\pi}{6}+\pi k$ (for integer $k$).
## Solution:
Since $\cos 2 x=2 \cos ^{2} x-1$, by substituting $y=\cos ^{2} x$, the equation reduces to $y+(2 y-1)^{2}=1$. Expanding and simplifying, we get $4 y^{2}-3 y=0$, from which $y=0$ or $y=\frac{3}{4}$.
The first of these values leads ... | \frac{\pi}{2}+\pik,\\frac{\pi}{6}+\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,549 |
3. Triangle $A B C$ is equilateral. On side $A C$, point $M$ is marked, and on side $B C$, point $N$, such that $M C=B N=2 A M$. Segments $M B$ and $A N$ intersect at point $Q$. Find the angle $C Q B$.
# | # Answer: $90^{\circ}$.
## Solution:
Since $M C=B N$ and $A C=B C$, then $A M=A C-M C=B C-B N=N C$.
From the fact that $A M=N C, A B=A C$ and $\angle B A M=\angle C A N=60^{\circ}$, it follows that triangles $B A M$ and $A C N$ are congruent, so $\angle N A C=\angle M B A=\alpha$.
From triangle $A B M: \angle A M B... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,551 |
4. In a math club, there are 6 schoolchildren, each of whom has at least 3 friends in this club (friendship is mutual). If someone gets a book, they read it and then give it to one of their friends who hasn't read the book yet. Prove that the teacher can give a new book to one of the schoolchildren so that it will even... | # Solution:
## First Method.
We will build a chain of friends. Take one student (hereafter referred to as Student 1), choose any of his friends (denoted as Student 2), now choose a friend of Student 2, but not Student 1 (this can be done because Student 2 has at least three friends) - this is Student 3. For Student 3... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,552 |
5. Find all pairs of natural numbers ( $m, n$ ), for which $\operatorname{LCM}(m, n)=3 m+2 n+1$. (Recall that LCM is the least common multiple of the numbers.) | Answer: $(3,10)$ and $(4,9)$.
## Solution:
## First method.
Note that $m$ does not divide $n$ (otherwise $LCM(m, n)=m$). In this case, $3 m+1=7$, which is not equal to $2 n$, since it is odd.
(2) Let $m>n$ (i.e., $n \leq m-1$). Then $2 n+1=k m$, but $2 n+1 \leq 2(m-1)+11$, so $LCM(m, n)$ is divisible by $d$, but $L... | (3,10)(4,9) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,553 |
1. How to draw six lines on a plane so that six intersection points are formed (an intersection point is a point through which at least two lines pass)?
# | # Solution.
The figure on the right shows one of the possible examples.
When checking, it is important to consider that the intersection point may not be drawn, and it is important to under... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,554 |
3. Vasya claims that he has two pentagons and a triangle, from which he can form both a $4 \times 6$ rectangle and a $3 \times 8$ rectangle. Is he right? | Answer: Vasya is right. Solution. The example is shown in the figure.
There are many similar examples. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,555 |
4. In the record of three two-digit numbers, there are no zeros, and in each of them, both digits are different. Their sum is 41. What could be their sum if the digits in them are swapped? | Answer: 113.
Solution. In all numbers, the tens digit is 1. Otherwise, the largest is at least 21, and the other two are at least 12. Their sum is no less than $12+12+21=44$, which is not equal to 41. The sum of the units digits is 11 (zeros are not allowed). Therefore, the sum with the digits rearranged is 113.
Ther... | 113 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,556 |
1. Each of the 5 brothers owns a plot of land. One day, they pooled their money, bought a neighbor's garden, and divided the new land equally among themselves. As a result, Andrey's plot increased by $10 \%$, Boris's plot - by $\frac{1}{15}$, Vladimir's plot - by $5 \%$, Grigory's plot - by $4 \%$, and Dmitry's plot - ... | Answer: By $5 \%$.
Solution. Let A, B, V, G, D be the areas of the plots of each brother, respectively. Then, according to the problem, $\frac{1}{10} \mathrm{~A}=\frac{1}{15} \mathrm{~B}=\frac{1}{20} \mathrm{~V}=\frac{1}{25} \mathrm{~G}=\frac{1}{30} \mathrm{~D}$ (*). Denoting $\mathrm{A}=x$, we find: $\mathrm{B}=1.5 x... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,557 |
2. The participants of the Olympiad left 9 pens in the office. Among any four pens, at least two belong to the same owner. And among any five pens, no more than three belong to the same owner. How many students forgot their pens, and how many pens does each student have? | Answer. There are three students, each owning three pens.
Solution. No student owned more than three pens, as otherwise the condition "among any five pens, no more than three belonged to one owner" would not be met. There are a total of 9 pens, so there are no fewer than 3 students. On the other hand, among any four p... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,558 |
3. In triangle $A B C$, the altitudes $B B_{1}, C C_{1}$ and the median $A A_{1}$ are drawn. Prove that triangle $A_{1} B_{1} C_{1}$ is equilateral if $\angle C+\angle B=120^{\circ}$. | Solution. In a right-angled triangle, the median is equal to half the hypotenuse. Then from $\triangle C B C_{1}$ and $\triangle C B B_{1}$, we get $C_{1} A_{1}=B A_{1}=A_{1} C$ and $B_{1} A_{1}=B A_{1}=A_{1} C$, which means $C_{1} A_{1}=B_{1} A_{1}$. Notice that $\angle C_{1} A_{1} B_{1}=180-\angle C_{1} A_{1} B-\angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,559 |
4. What is the minimum number of cells that need to be colored on a $6 \times 6$ board so that, for any placement (including rotations and flips) of a figure consisting of 4 cells in the shape of the letter $\Gamma$, there is at least one colored cell? | Answer: 12.
Solution: Consider a $2 \times 3$ rectangle. It is obvious that at least 2 cells need to be colored in it. Divide the $6 \times 6$ board into 6 rectangles of $2 \times 3$. In each, at least 2 cells need to be colored, so in total, at least 12 cells need to be colored. An example with 12 cells is shown on t... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,560 |
5. Find all natural numbers $N<10000$ such that $N=26 \cdot S(N)$, where $S(N)$ denotes the sum of the digits of the number $N$. | Answer: 234, 468.
Solution. Note that the number $N$ is at most a three-digit number, otherwise $N \geq 1000 > 936 = 26 \cdot 4 \cdot 9 \geq 26 \cdot S(N)$. It remains to consider numbers less than 1000. As is known, a number and the sum of its digits give the same remainder when divided by 9. Consider the equation $N... | 234,468 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,561 |
1. The roots of the quadratic trinomial $a x^{2}+b x+c$ are $\sin 42^{\circ}$ and $\sin 48^{\circ}$. Prove that $b^{2}=a^{2}+$ 2ac. | Solution. Let $x_{1}=\sin 42^{\circ}$ and $x_{2}=\sin 48^{\circ}$. Since $\sin 48^{\circ}=\cos 42^{\circ}$, then $x_{1}^{2}+x_{2}^{2}=1$.
By Vieta's formulas, $x_{1}+x_{2}=-\frac{b}{a} ; x_{1} \cdot x_{2}=\frac{c}{a}$.
Then $b^{2}=\left(-\left(x_{1}+x_{2}\right) \cdot a\right)^{2}=\left(x_{1}^{2}+x_{2}^{2}+2 x_{1} x_... | b^{2}=^{2}+2 | Algebra | proof | Yes | Yes | olympiads | false | 15,562 |
2. Along a circular highway, 30 houses of heights $1, 2, 3, \ldots, 30$ floors (exactly one house of each height) have been built. We will call a house interesting if it is taller than one of its neighboring houses but shorter than the other. It turned out that exactly 10 of these houses are interesting. Prove that the... | Solution. A one-story house cannot be interesting.
Then the smallest total height of interesting houses is $2+3+\ldots+11=65>64$.
That is, it cannot be equal to 64. | 65>64 | Combinatorics | proof | Yes | Yes | olympiads | false | 15,563 |
3. Given an increasing positive geometric progression $\mathrm{b}_{\mathrm{n}}$.
It is known that $b_{4}+b_{3}-b_{2}-b_{1}=5$. Prove that $b_{6}+b_{5} \geq 20$. | Solution. Let $b_{2}=b_{1} q$, since the progression is increasing, then $q>1$.
Notice that $b_{3}+b_{4}=b_{1} q^{2}+b_{2} q^{2}=\left(b_{1}+b_{2}\right) q^{2}$ and $b_{5}+b_{6}=b_{3} q^{2}+b_{4} q^{2}=\left(b_{3}+b_{4}\right) q^{2}$
That is, the numbers $\left(b_{1}+b_{2}\right),\left(b_{3}+b_{4}\right),\left(b_{5}+... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,564 |
4. In triangle $A B C$, points $M$ and $N$ are the midpoints of sides $A C$ and $B C$ respectively. It is known that the point of intersection of the medians of triangle $A M N$ is the point of intersection of the altitudes of triangle $A B C$. Find the angle $A B C$. | Answer: $45^{\circ}$.
Solution. Triangles $E T N$ and $A T B$ are similar, therefore, $T N: T B=T E: T A=E N: A B=1: 4$. Therefore, $C T=1 / 2 B T$. Since $H$ is the intersection point of the medians of triangle $A M N$, $E H=1 / 3 A E=E T$. Therefore, $H T=1 / 2 A T$.
Thus, right triangles $C T H$ and $B T A$ are si... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,565 |
5. Prove that the number $\left[\frac{n}{1}\right]+\left[\frac{n}{2}\right]+\ldots+\left[\frac{n}{n}\right]+[\sqrt{n}]$ is even for any natural $n$. ([x] - the integer part of $x$, that is, the greatest integer not exceeding $x$.) | Solution. Let $S(n)=\left[\frac{n}{1}\right]+\left[\frac{n}{2}\right]+\ldots+\left[\frac{n}{n}\right]$ and $F(n)=\left[\frac{n}{1}\right]+\left[\frac{n}{2}\right]+\ldots+\left[\frac{n}{n}\right]+[\sqrt{n}]$. We need to find the difference $S(n+1)-S(n)$. For this, we will look at the difference $\left[\frac{n+1}{k}\righ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,566 |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,569 |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,570 |
Problem 5.4. Arrange the digits from 1 to 6 (each must be used exactly once) so that the sum of the three numbers located on each of the 7 lines is equal to 15. In your answer, indicate which digits should be placed at positions $A-F$.
![](https://cdn.mathpix.com/cropped/2024_05_06_e7f9c87b6a37ffba3564g-09.jpg?height=... | Answer: $A=4, B=1, C=2, D=5, E=6, F=3$.
Solution. According to the condition, $A, D, E$ are distinct digits, not exceeding 6, the sum of which is 15. If these digits are taken to be the maximum possible, then their sum is $4+5+6=15$. Therefore, $A, D, E$ are $4,5,6$ in some order (if at least one of the digits is no m... | A=4,B=1,C=2,D=5,E=6,F=3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,571 |
Problem 5.8. There are 7 completely identical cubes, on which 3 points are marked on one face, 2 points on two faces, and 1 point on the rest. These cubes were glued together to form a figure in the shape of the letter "P", as shown in the image, with the number of points on any two touching faces being the same.
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,574 |
Problem 6.7. Eight paper squares $2 \times 2$ were laid out on the table one after another until a large square $4 \times 4$ was formed. The last square placed on the table was $E$. The image shows how the squares are visible: square $E$ is fully visible, and the other squares are partially visible. Which square was pl... | Answer: $G$.
Solution. It is clear that different squares were placed in different positions (otherwise, the one placed last would completely cover the one placed earlier, but we can see at least one cell from each square).
The positions of squares $A, C, G$, and $F$ can be determined immediately, as they contain cor... | G | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,575 |
Problem 7.2. Denis divided a triangle into nine smaller triangles, as shown in the figure, and placed numbers in them, with the numbers in the white triangles being equal to the sums of the numbers in the adjacent (by sides) gray triangles. After that, Lesha erased the numbers 1, 2, 3, 4, 5, and 6 and wrote the letters... | Answer: a1 b3 c2 d5 e6 f4.
Solution. Note that the number 6 can be uniquely represented as the sum of three numbers from the set of numbers from 1 to 6, which is $6=1+2+3$ (or the same numbers in a different order).
Now let's look at the numbers $B, D$, and $E$. The maximum value of the sum $D+E$ is the sum $5+6=11$,... | a1b3c2d5e6f4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,576 |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
. Therefore, it must contain the num... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,580 |
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,581 |
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
 | Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =... | 103 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,583 |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,584 |
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers located inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
We get 8 rectangles $1 \t... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,585 |
Problem 10.6. The graph of the quadratic trinomial $y=\frac{2}{\sqrt{3}} x^{2}+b x+c$ intersects the coordinate axes at three points $K, L$, and $M$, as shown in the figure below. It turns out that $K L=K M$ and $\angle L K M=120^{\circ}$. Find the roots of the given trinomial.
. According to the problem, triangle $O M K$ is a right triangle with a $30^{\circ}$ angle at vertex $M$, so $O M=\sqrt{3} K O=\sqrt{3} p$ and $K M=2 K O=2 p$. Al... | 0.51.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,586 |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,587 |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,588 |
1. Variant 1.
Find the sum:
$$
(-2021)+(-2020)+(-2019)+\ldots+2023+2024
$$ | Answer: 6069.
Solution. By pairing numbers that differ in sign, we get that in each such pair the sum is 0, and without pairs, the numbers left are $0, 2022, 2023, 2024$. | 6069 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,589 |
# 2. Option 1.
In a certain three-digit number $N$, the last two digits were swapped and the resulting number was added to the original. The sum turned out to be a four-digit number starting with 173. Find the largest possible sum of the digits of the number $N$. | Answer: 20.
Solution: Let the original number be $\overline{a b c}$, and the last digit of the sum $\overline{a b c}+\overline{a c b}$ be $x$. Then, $100 a+10 b+c+100 a+10 c+b=200 a+11(c+b)=1730+x$. If $a1730+x$.
Therefore, $a=8$. Thus, $(800+10 b+c)+(800+10 c+b)=1730+x$, i.e., $11(b+c)=130+x$. From this, it follows ... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,590 |
# 3. Option 1.
In the village, 7 people live. Some of them are liars who always lie, while the rest are knights (who always tell the truth). Each resident said whether each of the others is a knight or a liar. A total of 42 answers were received, and in 24 cases, a resident called a fellow villager a liar. What is the... | Answer: 3.
Solution: The phrase "He is a knight" would be said by a knight about a knight and by a liar about a liar, while the phrase "He is a liar" would be said by a knight about a liar and by a liar about a knight. Therefore, in each pair of "knight-liar," the phrase "He is a liar" will be said twice. Since this p... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,591 |
4. Variant 1.
Given an equilateral triangle $A B C$ with an area of 210. Inside it, points for which vertex $A$ is neither the closest nor the farthest are painted red. What is the area of the painted part of the triangle? | Answer: 41.
Solution.
Consider a point $D$ for which vertex $B$ is the nearest, and vertex $C$ is the farthest. Let $M, N, K$ be the midpoints of sides $AB, BC$, and $AC$ respectively. $O$ i... | 41 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,592 |
Variant 2.
What is the smallest sum that nine consecutive natural numbers can have if this sum ends in 2050306? | Answer: 22050306.
Option 3.
What is the smallest sum that nine consecutive natural numbers can have if this sum ends in $1020156$?
Answer: 31020156. | 31020156 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,594 |
# 6. Variant 1.
What is the largest root that the equation
$$
(x-a)(x-b)=(x-c)(x-d)
$$
can have if it is known that $a+d=b+c=2022$, and the numbers $a$ and $c$ are different? | Answer: 1011.
Solution. By expanding the brackets and combining like terms, we find $x=\frac{c d-a b}{c+d-a-b}$. Note that $c+d-a-b \neq 0$, because otherwise $c+d=a+b$, and considering the equality $a+d=b+c$, we would get $a=c$.
Considering that $d=2022-a, c=2022-b, x=\frac{2022^{2}-2022 a-2022 b}{2(2022-a-b)}=\frac... | 1011 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,595 |
7. Variant 1.
In a $101 \times 101$ square, an $88 \times 88$ corner square is painted red. What is the maximum number of non-attacking queens that can be placed on the board without placing the figures on the red cells? A queen attacks along the horizontal, vertical, and diagonals of the square. Attacking through the... | Answer: 26.
Solution. Consider 13 horizontal and 13 vertical strips of size $1 \times 101$, not containing any red cells. Each of the queens stands on at least one of these strips, and no strip can contain more than one queen, so there are no more than 26 queens. To construct an example, consider two rectangles of siz... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,596 |
8. Variant 1.
A line parallel to the leg $A C$ of the right triangle $A B C$ intersects the leg $B C$ at point $K$, and the hypotenuse $A B$ at point $N$. On the leg $A C$, a point $M$ is chosen such that $M K=M N$. Find the ratio $\frac{A M}{M C}$, if $\frac{B K}{B C}=14$. | Answer: 7.
Solution.
Drop the altitude $M H$ from point $M$ in the isosceles triangle $M N K$. Then $M H K C$ is a rectangle and $M C=K H=H N$. Let $M C=K H=H N=y$. Let $K B=x$, then $C K=C ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,597 |
3. It is known that the sum of the squares of two natural numbers is divisible by 7. Is it true that their product is divisible by 49? Justify your answer.
# | # Solution
Consider the set of remainders of the division of a natural number by 7: $\{0,1,2,3,4,5,6\}$. Therefore, the set of remainders of the division of the square of a natural number by 7 consists of the numbers $\{0,1,2,4\}$. Based on the obtained set of remainders, the sum of the squares of natural numbers will... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,599 |
4. Perpendiculars AP and AK are dropped from vertex A to the bisectors of the external angles ABC and ACB of triangle ABC. Find the length of segment PK, if the perimeter of triangle ABC is p.
# | # Solution
In triangle $\mathrm{ABC}$, we draw the external angle bisectors at vertices B and C. Drop perpendiculars from vertex A to these bisectors. Extend the constructed perpendiculars ... | \frac{p}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,600 |
5. How many integer solutions $(x, y)$ does the equation $|x|+|y|=n$ have?
# | # Solution
Graphical interpretation of the equation - the boundaries of a square with sides on the lines $x+y=n, -x+y=n, x-y=n, -x-y=n$. The side lying on the line $x+y=n$ corresponds to $\mathrm{n}+1$ points with integer coordinates, the side lying on the line $-x+y=n$ corresponds to $\mathrm{n}+1$ points with intege... | 4n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,601 |
1. There are 100 fish in two aquariums combined. When 30 fish were moved from the first aquarium and 40 from the second, the number of fish left in each aquarium was equal. How many fish were originally in each aquarium? | Answer: 45 and 55 fish respectively.
Solution: After the relocation, there were 100-30-40=30 fish left in the aquariums. This means 15 in each. Therefore, there were initially $15+30=45$ fish in the first one, and $15+40=55$ fish in the second one. | 4555 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,602 |
2. Vasya multiplied twelve fours, and Petya - twenty-five twos. Whose number turned out to be greater? Justify your answer.
# | # Answer. At Petya.
Solution. Since two twos in the product give 4, the product of 12 Vasya's fours is the same as the product of 24 twos. And since Petya multiplied 25 twos (25>24), the result he got was larger (by a factor of two). | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,603 |
3. Can the rectangle shown below be cut into five squares? (The squares do not have to be the same size, there should be no leftover pieces)
 | Answer. For example, like this:
 | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,604 |
4. Three gnomes, Pili, Eli, and Sleepy, found a diamond, a topaz, and a copper basin in a cave. Eli has a red hood and a longer beard than Pili. The one who found the basin has the longest beard and a blue hood. The gnome with the shortest beard found the diamond. Who found what? Explain your answer. | Answer: Spali found the copper basin, Pili found the diamond, and Eli found the topaz.
Solution: Since the gnome with the longest beard has a blue hood, Eli does not have the longest beard. Pili also does not have the longest beard (since it is shorter than Eli's). Therefore, the longest beard belongs to Spali, the me... | Spalifoundthecopperbasin,Pilifoundthediamond,Elifoundthetopaz | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,605 |
5. Write 7 consecutive natural numbers such that among the digits in their notation there are exactly 16 twos. (Consecutive numbers differ by 1.) | Answer. Any of the following two sequences will do:
2229, 2230, 2231, 2232, 2233, 2234, 2235
2215, 2216, 2217, 2218, 2219, 2220, 2221 | 2229,2230,2231,2232,2233,2234,2235or2215,2216,2217,2218,2219,2220,2221 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,606 |
Task No. 1.1
## Condition:
Five friends - Masha, Nastya, Irina, Olya, and Anya - meet in the park every day after buying ice cream from the shop around the corner. One day, the girls had a conversation.
Irina: I was the first in line!
Olya: No one was after me.
Anya: Only one person was after me.
Masha: There wer... | Answer: 3
Exact match of the answer - 1 point
## Solution.
From the statements of Irina and Olya, it is clear that they were first and last, respectively. Since there was only one person after Anya, it was Olya. Nastya stood next to Irina, but she could not have stood in front of her, so Nastya was second. This mean... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,607 |
# Task No. 1.2
## Condition:
Five friends - Katya, Polina, Alyona, Lena, and Svetlana - meet in the park every day after buying ice cream from the shop around the corner. One day, the girls had a conversation.
Polina: I stood next to Alyona.
Alyona: I was the first in line!
Lena: No one was after me.
Katya: There... | # Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task №1.1
# | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,608 |
# Task № 1.3
## Condition:
Five friends - Sasha, Yulia, Rita, Alina, and Natasha - meet in the park every day after buying ice cream from the little shop around the corner. One day, the girls had a conversation.
Sasha: There were five people in front of me.
Alina: There was no one after me.
Rita: I was the first i... | Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task №1.1
# | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,609 |
# Task No. 1.4
## Condition:
Five friends - Kristina, Nadya, Marina, Liza, and Galia - meet in the park every day after buying ice cream from the shop around the corner. One day, the girls had a conversation.
Kristina: There were five people in front of me.
Marina: I was the first in line!
Liza: No one was behind ... | # Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task №1.1
# | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,610 |
# Task № 2.1
## Condition:
Karlson and Little together weigh 10 kg more than Freken Bok, and Little and Freken Bok weigh 30 kg more than Karlson. How much does Little weigh? Give your answer in kilograms. | Answer: 20
Exact match of the answer -1 point
Solution.
Let's denote Carlson's mass by K, Freken Bok's mass by F, and Little One's mass by M. Then from the condition, it follows that $K+M=F+10$ and $M+F=K+30$. Adding these equations term by term, we get $K+2M+F=F+K+40$. From this, $M=20$. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,611 |
# Task No. 3.1
## Condition:
Polina makes jewelry on order for a jewelry store. Each piece of jewelry consists of a chain, a stone, and a pendant. The chain can be silver, gold, or iron. Polina has stones - a diamond, an emerald, and a quartz - and pendants in the shape of a star, a sun, and a moon. Polina is only sa... | # Answer: 24
Exact match of the answer - 1 point
## Solution.
Notice that the iron ornament with the sun is definitely in the middle, as it must be between the gold and silver ornaments. For the first position, we can choose a silver or gold chain. After choosing a chain for the first position, the chain for the thi... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,615 |
# Task № 3.2
## Condition:
Artyom makes watches for a jewelry store on order. Each watch consists of a bracelet and a dial. The bracelet can be leather, metal, or nylon. Artyom has round, square, and oval dials. Watches can be mechanical, quartz, or electronic.
Artyom is only satisfied when the watches are arranged ... | Answer: 12
Exact match of the answer -1 point
Solution by analogy with task №3.1
# | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,616 |
# Task № 3.3
## Condition:
Alina makes phone cases on order for a tech store. Each case has a pattern and a charm.
The case can be silicone, leather, or plastic. Alina has charms in the shapes of a bear, a dinosaur, a raccoon, and a fairy, and she can draw the moon, the sun, and clouds on the case.
Alina is only sa... | Answer: 36
Exact match of the answer - 1 point
Solution by analogy with task №3.1
# | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,617 |
# Task № 3.4
## Condition:
Anton makes watches for a jewelry store on order. Each watch consists of a bracelet, a precious stone, and a clasp.
The bracelet can be silver, gold, or steel. Anton has precious stones: zircon, emerald, quartz, diamond, and agate, and clasps: classic, butterfly, and buckle. Anton is only ... | Answer: 48
Exact match of the answer -1 point
Solution by analogy with task №3.1
# | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,618 |
# Task № 5.1
## Condition:
Polina thought of a natural number. Her friends asked one question each:
Masha: Is it divisible by 11?
Irina: Is it divisible by 13?
Anya: Is it less than 15?
Olya: Is it divisible by 143?
Polina answered affirmatively to only two of the four questions. What numbers could Polina have t... | Solution.
If Olya were right, the number would be divisible by 143, but in that case, it would also be divisible by 11 and 13, which means there would be at least three affirmative answers. Therefore, the number is not divisible by 143. If both Masha and Irina were right, the number would be divisible by both 11 and 1... | 1113 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,623 |
Task No. 5.3
## Condition:
Kristina thought of a natural number. Her friends each asked one question:
Yulia: Is it divisible by 17?
Nastya: Is it divisible by 19?
Vika: Is it less than 20?
Dasha: Is it divisible by 323?
Kristina answered affirmatively to only two of the four questions. What numbers could Kristin... | Solution by analogy with task №5.1
# | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,625 | |
# Task № 6.1
## Condition:
Yasha and Grisha are playing a game: first, they take turns naming a number from 1 to 105 (Grisha names the number first, the numbers must be different). Then each counts the number of different rectangles with integer sides, the perimeter of which is equal to the named number. The one with... | # Answer: 104
## Exact match of the answer -1 point
## Solution.
We will show that by naming the number 104, Grisha will win. Consider a rectangle with sides a and b. Its perimeter \( P = 2(a + b) \Rightarrow (a + b) = P / 2 \), then the length of the smaller side can take integer values from 1 to the integer part o... | 104 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,627 |
# Task № 6.2
## Condition:
Sasha and Misha are playing a game: first, they take turns naming a number from 1 to 213 (Misha names the first number, the numbers must be different). Then each counts the number of different rectangles with integer sides, the perimeter of which is equal to the named number. The winner is ... | Answer: 212
Exact match of the answer -1 point
Solution by analogy with task №6.1
## Condition:
Oksana and Seryozha are playing a game: they take turns naming a number from 1 to 165 (Oksana goes first, the numbers must be different). Then each counts the number of different rectangles with integer sides whose perim... | 164 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,628 |
# Task № 6.4
## Condition:
Dima and Vlad are playing a game: first, they take turns naming a number from 1 to 97 (Dima names the first number, the numbers must be different). Then each of them counts the number of different rectangles with integer sides, the perimeter of which is equal to the named number. The winner... | Answer: 96
Exact match of the answer -1 point
Solution by analogy with task №6.1
# | 96 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,629 |
# Assignment No. 7.1
## Condition:
Professor Severus Snape has prepared three potions, each with a volume of 300 ml. The first potion makes the person who drinks it smart, the second makes them beautiful, and the third makes them strong. To ensure the potion works, one needs to drink at least 30 ml of it. Severus Sna... | # Answer: 80
## Exact match of the answer -1 point
## Solution.
First, Hermione drinks $300 / 2=150$ ml of the wisdom potion, and then she pours 150 ml into the second jug, after which the second jug contains 150 ml of the wisdom potion and 300 ml of the beauty potion. When she drinks half of the contents of this ju... | 80 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,630 |
# Task No. 7.2
## Condition:
Professor Severus Snape has prepared three potions, each with a volume of 600 ml. The first potion makes the person who drinks it smart, the second makes them beautiful, and the third makes them strong. To ensure the potion works, one needs to drink at least 30 ml of it. Severus Snape was... | Answer: 40
Exact match of the answer -1 point
Solution by analogy with task №7.1
# | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,631 |
# Task № 7.3
## Condition:
Potions teacher Severus Snape has prepared three potions, each with a volume of 400 ml. The first potion makes the person who drinks it smart, the second makes them beautiful, and the third makes them strong. To ensure the potion works, one needs to drink at least 30 ml of it. Severus Snape... | Answer: 60
Exact match of the answer -1 point
Solution by analogy with task №7.1
# | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,632 |
# Task No. 7.4
## Condition:
Professor Severus Snape has prepared three potions, each with a volume of 480 ml. The first potion makes the person who drinks it smart, the second makes them beautiful, and the third makes them strong. To ensure the potion works, one needs to drink at least 30 ml of it. Severus Snape was... | Answer: 50
Exact match of the answer -1 point
Solution by analogy with task №7.1
# | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,633 |
# Task № 8.1
## Condition:
A board of $2022 \times 2022$ is given. Olya and Masha take turns coloring $2 \times 2$ squares on it red and blue, with the sides of the squares aligned with the grid lines. The girls agreed that each cell can be painted no more than once in blue and no more than once in red. Cells that ar... | # Solution.
Let the number of purple cells be denoted by \( x \). Each move colors a \( 2 \times 2 \) square, meaning the number of "colorings" is \( 4k \), where \( k \) is the number of moves. On the other hand, let \( x \) be the number of purple cells, which are the cells that the girls painted twice, so \( 2x \) ... | 2022\cdot2020 | Combinatorics | MCQ | Yes | Yes | olympiads | false | 15,634 |
# Task № 8.2
## Condition:
A board of size $2022 \times 2022$ is given. Maxim and Andrey take turns coloring $2 \times 2$ squares on it in yellow and red, with the sides of the squares aligned with the grid lines. The boys agreed that each cell can be painted no more than once in blue and no more than once in red. Ce... | Solution by analogy with task №8. 1
# | 2021\cdot20202022\cdot2020 | Combinatorics | MCQ | Yes | Yes | olympiads | false | 15,635 |
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