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9.4. The number 12 is written on the board. Every minute, the number is either multiplied or divided by 2 or 3, and the result is written on the board in place of the original number. Prove that the number that will be written on the board exactly one hour later will not be equal to 54. | 9.4. Solution. Note that $12=2^{2} \cdot 3$, and $54=2 \cdot 3^{3}$. Each minute, one of the exponents changes by one, i.e., the sum of the exponents changes parity. From this, it follows that after an hour, the parity of the sum of the exponents will be the same as that of the initial number. However, initially, the s... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,606 |
9.5. The numbers $1,2,3, \ldots, 2017,2018,2019$ are written on a board. It is allowed to erase any two numbers and replace them with the absolute value of their difference. In the end, one number will remain on the board. Can it be zero? Explain. | 9.5. Solution. The sum of all numbers written on the board will be odd (the sum of 1010 odd numbers and 1009 even numbers is an odd number). When erasing 2 numbers, there can be the following 3 cases:
a) Two even numbers are erased, then the absolute difference will be an even number, and the new sum will be an odd nu... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,607 |
7.1. The average height of 11 football players on the team is 182 cm. During the match, the referee sent off one player, and the average height of the remaining players became 181 cm. What is the height of the player who was sent off? | Answer: 192 cm. Solution. Let $S$ be the sum that results from adding the heights of the 10 remaining football players. Then $\frac{S}{10}=181$ and $\frac{S+x}{11}=182$, where $x$ is the height of the removed football player. From this, $S=1810$ and $x=182 \cdot 11-S=2002-1810=192$. | 192 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,608 |
7.2. Petya wrote down all natural numbers from 1 to $n$ in a row on the board and counted the total number of digits written. It turned out to be 777. What is $n$? | Answer: 295. Solution: Since a total of 777 digits have been written, $n$ must be a three-digit number: indeed, in the case of a two-digit $n$, no more than $9+2 \cdot 90=189$ digits would have been written, and in the case of a four-digit (or more) number, more than $9+2 \cdot 90+3 \cdot 900=2889$ digits would have be... | 295 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,609 |
7.3. In a bag for a bingo game, there are 90 barrels with numbers from 1 to 90. What is the minimum number of barrels that need to be drawn at random from the bag to guarantee getting a barrel with a number divisible by 3 or 5 (or both 3 and 5)? | Answer: 49. Solution. We will call numbers that are divisible by 3 or 5 desirable. The number of numbers from 1 to 90 that are divisible by 3 is 30 ( =90: 3 ), and those divisible by 5 is 18 ( =90:5). If we add $30+18=48$, then all numbers that are divisible only by 3 and only by 5 will be counted once, and numbers tha... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,610 |
7.4. How many six-digit natural numbers exist, each of which has adjacent digits with different parity | Answer: 28125. Solution. If the first (most significant) digit is even, then it can be chosen in four ways $(2,4,6,8)$, and all subsequent ones can be chosen in five ways (possible candidates for the second, fourth, and sixth digits are $1,3,5,7,9$, and for the third and fifth - $0,2,4,6,8$). In the end, by the rule of... | 28125 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,611 |
7.5. Kolya drew $n$ segments and marked all their intersection points in red. Could it happen that on each segment there are exactly three red points, if a) $n=11$; b) $n=100 ?$ | Answer: a) could; b) could. Solution. a) See the example in the figure. The set of segments in this example consists of two parts: 6 segments on the left, and 5 on the right. b) If we place 20 copies of the right part of the example from part a), we will get the desired arrangement of 100 segments.
 was calculated. Could five consecutive differences (out of seven) be equal to the numbers $2,1,6,1,2$? | Answer: Could not.
Solution: Subtract the smallest of the given numbers from all of them. This operation does not change the differences between the numbers. Thus, we can assume that the numbers $A=0, B=1, C=2, D=3, E=4, F=5, G=6$ are arranged in a circle. The difference 6 is the maximum possible difference, and it ca... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,613 |
7.2. Points $A, B, C, D$ are located at the vertices of rectangle $ABCD$, its sides and diagonals $AC$ and $BD$ are roads. The first car traveled the route $B \rightarrow$ $C \rightarrow A \rightarrow D$ in one hour, and the second car traveled the route $D \rightarrow B \rightarrow C \rightarrow A$ in one hour. After ... | Answer: In 40 minutes.
Solution: The diagonals of a rectangle have the same length and are longer than any of its sides. In one hour, together the two cars would travel three times the length of side $B C$ and three times the diagonal, since one car travels two sides equal to $B C$ and one diagonal in an hour, while t... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,614 |
7.3. Is it possible to arrange 100 numbers 1 and one number -1 in a circle so that the product of any three consecutive numbers is positive? | Answer: No.
Solution: Suppose it is possible. Since there are a total of $100+101=201=3 \cdot 67$ numbers, we can divide them all into 67 consecutive triplets. In each triplet, the product of the numbers is positive, so the product of all numbers is also positive. However, the product of 100 numbers 1 and 101 numbers ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,615 |
7.4. Find all solutions to the puzzle COW + COW = MILK. Different letters correspond to different digits, the same letters correspond to the same digits
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7.4. Find all solutions to the puzzle COW + COW = MILK. Different letters correspond to different digits, the same letters correspond to the same ... | Answer. Two solutions $(302015+302015=604030$ and $304015+304015=608030)$.
Solution. The equality in the hundreds place could only have occurred in two cases: $0+0=0$, that is, $\mathrm{O}=0$, which could only happen if there were no carries from the tens to the hundreds; and if $\mathrm{O}=9$ (in the case of a carry ... | 302015+302015=604030304015+304015=608030 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,616 |
7.5. In the hall, there are liars and knights. Liars always lie, while knights always tell the truth. Each person pointed to one of those present and said: "He is a liar." It turned out that someone said this about each person in the hall. Could there have been exactly 101 people in the hall? | Answer: It could not.
Solution: Suppose there could have been exactly 101 people in the hall. Since each person pointed to one of those present, and someone pointed to each of those present, exactly one of those present pointed to each person. Note that a knight could only point to a liar, and a liar could only point ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,617 |
1. A straight road passes not far from Mount Fuji. The driver of a car noticed the mountain 60 km to the north, and an hour later - 45 km to the west. What is the shortest distance the car passed from Mount Fuji? | Answer: 36 km.
Solution: The mountain and the observation points are at the vertices of a right-angled triangle with legs of 60 and 45. The hypotenuse in this case is 75. By equating the areas of the triangle calculated in different ways, we get that the product of the legs is equal to the product of the hypotenuse an... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,618 |
2. This year, the son and daughter are so many years old that the product of their ages is 7 times less than the father's age. And in three years, the product of their ages will already be equal to the father's age. Find the father's age. | Answer: 21 years
Solution. Let $n$ be the son's age, $m$ be the daughter's age, then the father's age is $7 m n$. In three years, the son's age will be $n+3$, the daughter's age will be $m+3$, and the father's age will be $7 \mathrm{mn}+3$, and according to the condition, the equation $(n+3)(m+3)=7 m n+3$ will be sati... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,619 |
4. A right triangle $A B C$ (cathetus $B C$ is greater than cathetus $A C$) is inscribed in a circle. A point $D$ is chosen on side $B C$ such that $B D=A C$, and point $M$ is the midpoint of the arc $A C B$. Find the angle $C D M$. | Answer: $45^{\circ}$.
Solution. Note that $A B$ is the diameter of the circumscribed circle. Connect point $M$ with points $A, B, C$, and $D$. Since arcs $A M$ and $B M$ are equal, the chords $A M$ and $B M$ that subtend them are also equal. Segments $B D$ and $A C$ are equal by the condition. Finally, angles $M B C$ ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,621 |
5. Given 700 different natural numbers, each of which does not exceed 2017. Prove that some two of them differ by 3, 4, or 7. | Solution. Let $S_{1}=\left\{n_{1}, n_{2}, \ldots, n_{700}\right\}$ be the set of given numbers. Denote $S_{2}=\left\{n_{1}+3, n_{2}+3, \ldots, n_{700}+3\right\}, S_{3}=\left\{n_{1}+7, n_{2}+7, \ldots, n_{700}+7\right\}$. Let $S=S_{1} \cup S_{2} \cup S_{3}$. The numbers in the set $S$ do not exceed 2024. If we assume th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,622 |
9.1. Igor added ten consecutive natural numbers, then divided the resulting sum by the sum of the next ten natural numbers. Could he have gotten the number 0.8? | Answer: It could not.
Solution. Suppose that the division resulted in 0.8. Let the smallest number of the first sum be denoted by $n$. Then this sum is: $n+(n+1)+\ldots+(n+9)=10 n+45$. Each term of the second sum is 10 more than the corresponding term of the first sum, so the second sum is 100 more than the first.
Th... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,623 |
9.2. On the coordinate plane, the graphs of a linear and a quadratic function are plotted (see figure). The equation of the linear function has the form $y=c x+2 c$ for some number $c$. Using the same parameter $c$, write the equation of the quadratic function and explain your solution. | Answer: $y=0.5 c(x+2)^{2}$.
Solution. Let's find the coordinates of the points of intersection of the graph of the linear function with the coordinate axes: ( $0 ; 2 \mathrm{c}$ ) and (-2; 0). The graph of the quadratic function (parabola) touches the $O X$ axis at the point ( $-2 ; 0$ ), hence its formula has the for... | 0.5(x+2)^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,624 |
9.3. In triangle $A B C$, the altitude $B H$ and the medians $A M$ and $C K$ are drawn. Prove that triangles $K H M$ and $A B C$ are similar. | Solution. From the condition of the problem, it follows that $M K$ is the midline of triangle $A B C$, which means $M K \| A C$ and $M K=0.5 A C$ (see Fig. 9.3). Since $H M$ is the median of the right triangle $B H C$, then $H M=0.5 B C$. Moreover, since $H M=M C$, then $\angle A C B=\angle M H C=\angle H M K$. Therefo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,625 |
9.4. We will call a natural number interesting if it can be factored into natural factors, each of which is less than 30. Prove that from 10000 interesting numbers, one can always choose two such that their product is a perfect square. | Solution. Consider the prime factorization of an interesting number, then each of these factors is also less than 30. There are exactly 10 prime numbers less than thirty: 2, 3, 5, 7, 11,
13, $17,19,23,29$. This means that each interesting number can be represented as: $2^{n_{1}} \cdot 3^{n_{2}} \ldots . .29^{n_{10}}$,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,626 |
9.5. Given an isosceles trapezoid $A B C D$. Points $Q$ and $P$ are considered on the lateral sides $A B$ and $C D$ respectively, for which $C P=A Q$. Prove that the midpoints of all such segments $P Q$ lie on one straight line. | Solution. We will prove that the midpoint of an arbitrary segment $P Q$ satisfying the condition lies on the midline $M N$ of the trapezoid $A B C D$.
First method. Draw segments $Q F$ and $P E$, parallel to the bases of the trapezoid (see Fig. 9.5a). Then, in trapezoids $E B C P$ and $A Q F D$, the angles at the base... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,627 |
9.6. A team of workers is repairing an apartment. To avoid damaging the floor in the room (a checkered square of size $4 \times 4$), they laid 13 two-cell rugs along the grid lines. Suddenly, it turned out that one rug was needed in another room. Prove that the workers can choose it in such a way that the repair in the... | Solution: Suppose it is impossible to remove a mat without exposing a piece of the floor. Then, under each mat, there must be a cell covered only once (otherwise, a mat, under which each cell is covered at least twice, can be removed). This means that the number of cells covered only once is no less than 13. Therefore,... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,628 |
8.2. A boy always lies on Tuesdays, and tells the truth only on Thursdays and Fridays. One day, he was asked his name for six days in a row. The answers were as follows: Kolya, Petya, Kolya, Petya, Vasya, Petya. What is the boy's name? | Answer: Kolya. Since among the boy's answers there are no two identical ones in a row, he either started answering on Friday or finished on Thursday. But the latter is impossible, because the boy cannot give the same answers on Tuesday and Thursday. Therefore, he started answering on Friday and told the truth. | Kolya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,629 |
8.3. In trapezoid $A B C D$ with base $A D$, it is known that $A B=B C, A C=C D$ and $B C+C D=A D$. Find the angles of the trapezoid. | Answer: $\quad \angle A=72^{\circ}, \quad \angle B=108^{\circ}$, $\angle C=144^{\circ}, \angle D=36^{\circ}$.
Mark a point $K$ on the base $A D$ of the trapezoid such that $A K=B C$. Then $K D=C D$ and $A B C K-$ is a rhombus. Let the angle $\angle C A K$ be $\alpha$. We get, $\angle C A K=$
, there are exactly 7 dominoes whose sum of numbers is even (and the sum on the remaining dominoes is odd) | Answer: Yes, it is possible. We will paint the cells of the square in a checkerboard pattern and initially write ones in all black cells, and twos in all white cells. Then, in any domino, the sum of the numbers will be 3, since one cell in it is white and the other is black. Now, replace the twos with ones in 7 white c... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,632 |
10.5. Oleg drew an empty $50 \times 50$ table and wrote a non-zero number above each column and to the left of each row. It turned out that all 100 written numbers are distinct, with 50 of them being rational and the other 50 being irrational. Then, in each cell of the table, he wrote the product of the numbers written... | Answer: 1250 works.
Solution. First, we show that there are at least 1250 irrational numbers in the table. Suppose along the left side of the table, there are $x$ irrational and $50-x$ rational numbers. Then along the top side, there are $50-x$ irrational and $x$ rational numbers. Since the product of a non-zero ratio... | 1250 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,633 |
10.6. Initially, several (more than one) natural numbers are written on the board. Then, every minute, a number equal to the sum of the squares of all the numbers already written on the board is added (so, if the numbers 1, 2, 2 were initially written on the board, then on the first minute the number \(1^2 + 2^2 + 2^2\... | Solution. Let $S_{1}, \ldots, S_{100}$ be the numbers written on the board in the first 100 minutes. Suppose that before writing the number $S_{i}$, the numbers on the board were $a_{1}, \ldots, a_{k}$. Then $S_{i}=a_{1}^{2}+a_{2}^{2}+\ldots+a_{k}^{2}$, and the next number written is $S_{i+1}=a_{1}^{2}+a_{2}^{2}+\ldots... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,634 |
10.7. A convex polygon is cut by non-intersecting diagonals into isosceles triangles. Prove that this polygon contains two equal sides.
(A. Hrybalko) | Solution. We will prove the statement by induction on the number of sides $n$ of the polygon. The base case $n=3$ is obvious. Now we will derive the statement for a $k$-gon $(k \geqslant 4)$, assuming that it is true for polygons with fewer sides than $k$.
Thus, let a convex $k$-gon $P$ be divided by non-intersecting ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,635 |
10.8. Circle $\omega$ is circumscribed around an acute-angled triangle $ABC$. A point $D$ is chosen on side $AB$, and a point $E$ is chosen on side $BC$ such that $AC \parallel DE$. Points $P$ and $Q$ on the smaller arc $AC$ of circle $\omega$ are such that $DP \parallel EQ$. The rays $QA$ and $PC$ intersect line $DE$ ... | Solution. Since quadrilateral $A B C Q$ is inscribed and $A C \| D E$, we have $\angle B E X=\angle B C A=\angle B Q A=\angle B Q X$. Therefore, quadrilateral $X B E Q$ is inscribed, from which $\angle X B Q = \angle X E Q = \angle D E Q$. Similarly, quadrilateral $Y B D P$ is inscribed, and $\angle P B Y = \angle P D ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,636 |
11.1. The number $x$ is such that $\sin x + \tan x$ and $\cos x + \cot x$ are rational numbers. Prove that $\sin 2x$ is a root of a quadratic equation with integer coefficients.
(N. Agakhanov) | Solution. Let $a=\sin x+\tan x$ and $b=\cos x+\cot x$. Introduce the notations: $u=\sin x+\cos x$ and $v=\sin x \cdot \cos x$. By the condition, the numbers $c=a+b=u+\frac{1}{v}$ and $d=a \cdot b=v+u+1$ are rational. Hence, $k=d-c=v+1-\frac{1}{v}$. Therefore, $t=\sin 2 x=2 v-$ is a root of the quadratic equation $t^{2}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,637 |
11.2. Each of 100 schoolchildren has a stack of 101 cards numbered from 0 to 100. The first schoolchild shuffles the stack, then takes one card from the top of the resulting stack, and each time a card is taken (including the first time), writes on the board the arithmetic mean of the numbers on all the cards taken by ... | Solution. On the 1st step, each of the 100 people was given one of the numbers from the set $A_{1}=\{0,1,2, \ldots, 100\}$.
On the 2nd step, one of the numbers from the set $A_{2}=$ $=\left\{\frac{1}{2}, \frac{2}{2}, \frac{3}{2}, \ldots, \frac{199}{2}\right\}$.
On the 100th step, one of the numbers from the set $A_{1... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,638 |
11.3. In each row of the $100 \times n$ table, the numbers from 1 to 100 are arranged in some order, with no repetitions within a row (the table has $n$ rows and 100 columns). It is allowed to swap two numbers in a row that differ by 1, provided they are not adjacent. It turns out that it is impossible to obtain two id... | Answer: $2^{99}$.
Solution. We will associate a string $x_{1}, x_{2}, \ldots, x_{100}$ of numbers from 1 to 100 with a sequence of 99 signs $$ according to how the adjacent numbers are ordered. That is, if $x_{k} < x_{k+1}$, the $k$-th sign is $<$, and if $x_{k} > x_{k+1}$, the $k$-th sign is $>$. Note that the allowe... | 2^{99} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,639 |
11.4. The circle $\omega$ is circumscribed around triangle $ABC$, where $AB < AC$. The angle bisectors of triangle $ABC$ intersect at point $I$. From the midpoint $M$ of side $BC$, a perpendicular $MH$ is dropped to the line $AI$. The lines $MH$, $BI$, and $AB$ form triangle $T_{b}$, while the lines $MH$, $CI$, and $AC... | Solution. Let the points of intersection of the line $M H$ with the lines $A B, A C, B I$ and $C I$ be denoted by $P, Q, X$ and $Y$ respectively (see Fig. 6). Let the lines $A I, B I$ and $C I$ intersect the circle $\omega$ again at points $A_{1}, B_{1}$ and $C_{1}$ respectively. Denote $\angle B A I = \angle C A I = \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,640 |
7.1. Does there exist a four-digit natural number with distinct non-zero digits that has the following property: if this number is added to the same number written in reverse order, the result is divisible by $101 ?$ | Answer. It exists.
Solution. For example, the number 1234 works. Indeed, $1234+4321=5555=101 \cdot 55$.
Remark. A number $\overline{a b c d}$ with distinct non-zero digits satisfies the condition if and only if $a+d=b+c$.
Comment. Any correct example with verification that it fits - 7 points.
Any correct example wi... | 1234 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,641 |
7.2. There are 9 cards with numbers $1,2,3,4,5,6,7,8$ and 9. What is the maximum number of these cards that can be laid out in some order in a row so that on any two adjacent cards, one of the numbers is divisible by the other? | Answer: 8.
Solution: Note that it is impossible to lay out all 9 cards in a row as required. This follows from the fact that each of the cards with numbers 5 and 7 can only have one neighbor card with the number 1. Therefore, both cards 5 and 7 must be at the edges, and the card with the number 1 must be adjacent to e... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,642 |
7.3. Petya bought one cupcake, two muffins, and three bagels, Anya bought three cupcakes and a bagel, and Kolya bought six muffins. They all paid the same amount of money for their purchases. Lena bought two cupcakes and two bagels. How many muffins could she have bought for the same amount she spent? | Answer: 5 cupcakes.
Solution: The total cost of Petya and Anya's purchases is equal to the cost of two of Kolya's purchases. If we denote P, K, and B as the costs of a cake, a cupcake, and a bagel, respectively, we get the equation: $(P + 2K + 3L) + (3P + 5) = 12K$, from which it follows that $4P + 4B = 10K$, or $2P +... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,643 |
7.4. In the class, there are 26 students. They agreed that each of them would either be a liar (liars always lie) or a knight (knights always tell the truth). When they came to the class and sat at their desks, each of them said: “I am sitting next to a liar.” Then some students moved to other desks. Could it be that a... | Answer: Could not.
Solution. Note that the phrase "I sit next to a liar" could only have been said in the case where a liar and a knight sit at the same desk. This means that in the class of liars and knights, there are an equal number - 13 each. The phrase "I sit next to a knight" could only have been said in the cas... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,644 |
7.5. What is the minimum number of 3-cell corners that need to be painted in a $6 \times 6$ square so that no more corners can be painted? (Painted corners must not overlap.) | Answer: 6.
Solution. Let the cells of a $6 \times 6$ square be painted in such a way that no more corners can be painted. Then, in each $2 \times 2$ square, at least 2 cells are painted, otherwise, a corner in this square can still be painted. By dividing the $6 \times 6$ square into 9 $2 \times 2$ squares, we get tha... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,645 |
1. Which two-digit numbers are more numerous: those with both digits of the same parity, or those with digits of different parity?
# | # Answer: There are an equal number of numbers.
Solution. All two-digit numbers are divided into nine tens. In each ten, with the same tens digit, there are five numbers of each type. Therefore, in the end, there are an equal number of these types of numbers. Criteria. Only answer: 0 points. | There\\an\equal\\of\ | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,646 |
3. Is it possible to arrange balls of eight colors in five boxes standing in a circle so that each box contains three balls of different colors, and no two balls of the same color are in adjacent boxes? | Answer: possible.
Instructions. Among 15 balls distributed in boxes, there cannot be three balls of the same color, since balls of the same color do not lie next to each other, and no fewer than six boxes are required. Therefore, we have seven colors of balls, two of each, and one ball of an eighth color. For convenien... | possible | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,648 |
4. In a room, 13 chairs are arranged in a row. Three jewelers sit on three adjacent chairs, and the middle one hides a diamond in the chair he is sitting on. The inspector has six detectors that show whether someone has sat on a chair or not. Can he place the detectors on the chairs before the jewelers arrive so that t... | Answer: can.
Instruction. Layout: s s d d d s s d d d s s (s - chair without detector, d - chair with detector). Since there are no three empty (without detectors) chairs in a row, at least one of the detectors will go off. If one goes off, then they were sitting on the chair with the triggered detector and on the two... | can | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,649 |
1. The numbers $a, b$ and $\sqrt{a}+\sqrt[3]{b}$ are rational ( $a \geqslant 0$ ). Prove that $\sqrt[3]{b}$ is rational. | Solution. Let $\sqrt{a}+\sqrt[3]{b}=r$, then $\sqrt[3]{b}=r-\sqrt{a}$, after cubing we get $b=r^{3}-3 r^{2} \sqrt{a}+3 r a-a \sqrt{a},\left(a+3 r^{2}\right) \sqrt{a}=r^{3}+3 r a-b$. If the number $a+3 r^{2}$ is zero, then $a=0$ and $r=0$, so $b=0, \sqrt[3]{b}$ is rational. In all other cases $a+3 r^{2}>0$, so $\sqrt{a}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,650 |
2. There are 7 indistinguishable weights. Boris knows the mass of each weight, while Arkady only knows that this set of weights contains all integer masses from 1 to 7 grams. How can Boris, using no more than three weighings on a balance scale, confirm the weight of each weight for Arkady? | Solution. Let us denote by $a_{m}$ the weight with mass $m$, and the mass itself. Boris can, for example, conduct the following three weighings: $\left.a_{1}+a_{2}+a_{3}a_{2}+a_{4}\right.$.
We will prove that these weighings are sufficient. The total mass of three weights is at least $1+2+3=6$, so the inequality $a_{1... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,651 |
3. Each natural number is painted either blue or red. For any coloring, will there always exist three distinct monochromatic natural numbers $x, y$ and $z$ such that $x y=z^{2}$? | Answer: Yes.
Solution. Suppose there is a coloring in which such numbers do not exist, and we will arrive at a contradiction. Let's see how the even powers of two are colored. If starting from $2^{2}$ they are colored alternately, then we can take a monochromatic triple $x=2^{2}, y=2^{10}, z=2^{6}$. Therefore, there w... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,652 |
4. In a non-convex quadrilateral $A B C D$, each of the angles $B, C$, and $D$ is equal to $30^{\circ}$ (the angle at vertex $A$ is greater than a straight angle). Points $N$ and $T$ are the midpoints of sides $B C$ and $C D$ respectively. Prove that triangle $A N T$ is equilateral. | First solution. Let the line $AB$ intersect $CD$ at point $E$, and the line $AD$ intersect $BC$ at point $F$. $\angle CEB = 180^\circ - \angle ECB - \angle CBE = 120^\circ$, similarly $\angle CFD = 120^\circ$, so $\angle EAF = 360^\circ - 2 \cdot 120^\circ - 30^\circ = 90^\circ$. Triangles $CBE$ and $CDF$ are isosceles... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,653 |
1. Find the smallest number that contains only the digits 2 and 3 in equal quantities, and is divisible by 2 and 3. | Answer: 223332.
Sketch of the solution. If a number is divisible by 3, then the sum of its digits is divisible by 3, which means the number of twos is a multiple of three, and thus, there must be at least three twos. Therefore, the desired number is a six-digit number. Since it is divisible by 2, it ends in 2. The num... | 223332 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,655 |
2. A rectangle with sides $6 \mathrm{~cm}$ and $3 \mathrm{~cm}$ was cut into three rectangles of equal perimeter. What can the perimeter of these rectangles be? Find all possible answers. | Answer: 14 cm, 10 cm, 10.5 cm.
Sketch of the solution.
Let's cut a rectangle with sides $a$ and $b$. Cases 1 and 2. Cut the rectangle into three equal rectangles with two cuts parallel to side $a$.
^{25}=2^{100}$. Therefore, their sum is greater than $2^{100}$. Such numbers do not exist.
Criteria. Only answer... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,657 |
4. Two chess players played 100 games against each other. 11 points were awarded for a win, $x$ points for a draw, and no points for a loss. Find all possible values of $\mathrm{x}$, if the chess players scored a total of 800 points, and $x$ is a natural number. | Answer: 3 and 4.
Sketch of the solution. Let $n$ games be decisive, then the number of draws is $100-n$, and the players scored $11 n+2 x(100-n)$. Since a total of 800 points were scored, we get the equation $11 n+2 x(100-n) = 800. 11 n-2 n x=800-200 x . n(11-2 x)=200(4-x) 3 n=200(4-x)-2 n(4-x)$. From which $3 n=2(4-x... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,658 |
5. There are 50 trees of 25 species, with two trees of each species. Can they be planted in a row so that between any two trees of the same species, there is one or three trees? | Answer: It is impossible.
Sketch of the solution. Suppose the trees are planted in the required manner. Let's number the trees in the order of their arrangement in a row from 1 to 50. Note that trees, between which one tree or three trees grow, have either both even numbers or both odd numbers in the numbering. Since ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,659 |
6. Several cells on a $14 \times 14$ board are marked. It is known that no two of the marked cells are in the same row and the same column, and also that a knight can, starting from any marked cell, reach any other marked cell via marked cells. What is the maximum possible number of marked cells? | Answer: 14.
Sketch of the solution.
Estimate. In each row, there is no more than one marked cell (field), and therefore no more than 14 cells are marked.
Example in the image.
 | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,660 |
8.1. Compare the numbers: $A=2011 \cdot 20122012 \cdot 201320132013$ and $B=2013 \cdot 20112011 \times$ $\times 201220122012$. | Answer: $A=B$.
$A=2011 \cdot 20122012 \cdot 201320132013=2011 \cdot 2012 \cdot 10001 \cdot 2013 \cdot 100010001$;
$B=2013 \cdot 20112011 \cdot 201220122012=2013 \cdot 2011 \cdot 10001 \cdot 2012 \cdot 100010001$.
Notice that each of the products equals 8146492851648619183716.
+ full justified solution (including di... | B | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,661 |
8.2. In the formula of the linear function $y=k x+b$, instead of the letters $k$ and $b$, insert the numbers from 1 to 20 (each one only once) so that 10 functions are obtained, the graphs of which pass through the same point. | Answer: for example, $y=1 x+20, y=2 x+19, \ldots, y=10 x+11$ or $y=1 x+2, y=3 x+$ $+4, \ldots, y=19 x+20$.
In the first case, the graph of each function will pass through the point $(1 ; 21)$, and in the second case - through the point $(-1 ; 1)$.
Other examples are possible.
+ the correct answer is provided and the... | 1x+20,2x+19,\ldots,10x+11passingthrough(1;21)1x+2,3x+4,\ldots,19x+20passingthrough(-1;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,662 |
8.5. Ten football teams each played one game against each of the others. As a result, each team ended up with exactly $x$ points. What is the greatest possible value of $x$? (Win - 3 points, draw - 1 point, loss - 0 points.) | Answer: 13.
Evaluation. Let's prove that $x$ cannot be greater than 13. Indeed, in each match, either 3 points are awarded (if one of the teams wins) or 2 points (if there is a draw). In total, $\frac{10 \cdot 9}{2}=45$ matches were played, meaning no more than 135 points were awarded, that is, the total points scored... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,664 |
1. In three identical boxes, there are two balls each: one box contains two white balls, another contains two black balls, and the third contains one white and one black ball. Each box has a label: one shows two white balls, another shows two black balls, and the third shows one white and one black ball. It is known th... | 1. In the box labeled BC, both balls have the same color. We take one ball out of it. Suppose it is white. Then there are two white balls in this box. In the box labeled CC, there are a white and a black ball, as there are no other options. Finally, in the box labeled BB, there are two black balls.
The case when the b... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,665 |
2. Given a $6 \times 3$ table. Each cell is filled with a number 0 or 1. If at the intersection of any two rows and any two columns all four numbers are the same, then the table is considered to be filled unsuccessfully. Is it possible to fill the table in such a way to avoid this phenomenon? | 2. Answer: Yes, it is possible.
Consider the following table filling:
| 0 | 0 | 1 |
| :--- | :--- | :--- |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 0 |
| 1 | 0 | 1 |
| 0 | 1 | 1 |
In each row, there are two identical elements. It is easy to check that in any other row, these elements are already in different positions.... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,666 |
3. From points $A$ and $B$, two cars set off towards each other simultaneously with constant speeds. One hour before the first car arrived at $B$ and four hours before the second car arrived at $A$, they met. Find the ratio of the speeds of the cars. | 3. Answer: the speed of the first car is twice as high.
Let the speed of one car be $k$ times the speed of the other. Since they started at the same time, one car will travel a distance $k$ times greater. After this, the segments will switch, and the slower car, whose speed is $k$ times less, will have to cover a dist... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,667 |
4. In a convex quadrilateral, the bisectors of all its angles are drawn. Can each side of the quadrilateral intersect any bisector at a point other than a vertex? | 4. Answer: No, it cannot.
Consider quadrilateral $ABCD$ and draw its diagonals. Let them intersect at point $O$. Consider the side farthest from it. Without loss of generality, we can assume that this is side $AB$. Since the distance from point $O$ to line $AB$ is not less than the distance from $O$ to line $AD$, the ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,668 |
5. Can a $5 \times 5$ square be cut into two types of rectangles: $1 \times 4$ and $1 \times 3$, so that exactly 8 rectangles are obtained?
7 points are awarded for a complete solution to each problem
The maximum total score is 35 | 5. Answer: no, it cannot.
In the square, there are 25 cells, while 8 rectangles of $1 \times 3$ in total have 24 cells. Therefore, among the 8 rectangles, exactly one must be of the form $1 \times 4$. We will show that these rectangles cannot cover the square.
Reasoning by contradiction, let's temporarily remove the ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,669 |
1. Solve the equation $1+2^{n}+3^{n}+5^{n}=2 k$ in integers. | # Solution
For $n \geq 1$, the left side becomes an odd number, while $2k$ is even, so there are no solutions.
For $n=0$, we get $k=2$.
For $n=-1$, we get an equation that has no integer solution for $k$.
For $n \leq -2$, the expression on the left takes values greater than 1 but less than 2. Therefore, there is no... | n=0,k=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,670 |
2. Prove that there exists a number of the form $20212021 \ldots 2021 \ldots 20210 \ldots 0$, which is divisible by 2022.
# | # Solution
Let's write down the numbers 2021, 20212021, 202120212021,....,2021....2021. In the last number, the sequence 2021 repeats 2022 times. Consider the remainders when each of these numbers is divided by 2022.
None of the written numbers are divisible by 2022, as they are all odd. The number of written numbers... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,671 |
3. There is 250 ml of a $70\%$ acetic acid solution and 500 ml of a $5\%$ acetic acid solution. Find the maximum volume of a $p\%$ acetic acid solution that can be obtained from the available solutions (no additional water can be added).
# | # Solution
Let's take x ml of the 70% solution, and V ml - the volume of the resulting p% solution. Then we have the equation
$$
0.7 x + 0.05(V - x) = \frac{p}{100} V
$$
from which we find
$$
\frac{x}{V} = \frac{p - 5}{65}
$$
That is, we take $(p - 5)$ parts of the 70% solution and $(70 - p)$ parts of the 5% solut... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,672 | |
4. In a right-angled triangle with legs $a$ and $b$ ( $a>b$ ), two identical circles are placed. The circles touch each other externally and each touches the hypotenuse and one of the legs. Find the radius of such a circle. | Solution
1) Let $\mathrm{O}$ and $\mathrm{Q}$ be the centers of these circles. $\mathrm{OQ}$ is parallel to the hypotenuse, since the centers of the circles are at the same distance from it... | \frac{1}{2}\sqrt{^2+b^2}-\frac{1}{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,673 |
5. Solve the inequality $\sqrt{\frac{x}{1-x}}-2 \sqrt[4]{\frac{1-x}{x}} \leq 1-\frac{1}{x}$. | # Solution
Let $\sqrt[4]{\frac{1-x}{x}}=t$. Then the original inequality transforms into the system of inequalities
$$
\left\{\frac{\left(t^{3}-1\right)^{2}}{t^{2}} \leq 0\right.
$$
the solution to which is only $t=1$.
Solving the equation $\sqrt[4]{\frac{1-x}{x}}=1$, we find $x=\frac{1}{2}$.
## Grading Criteria
... | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 14,674 |
1. The numbers $p$ and $b$ are roots of the quadratic equation $x^{2}+2020 a x+c=0, a \neq 0$. Find the sum of the roots of the quadratic equations $a x^{2}+b x+d=0$ and $a x^{2}+p x+q=0$, if each of them has 2 distinct real roots. | Solution. Since $p$ and $b$ are the roots of the quadratic equation $x^{2}+2020 a x+c=0$, by Vieta's theorem, $p+b=-2020 a$. Let $x_{1}$ and $x_{2}$ be the roots of the equation $a x^{2}+b x+d=0$, and $x_{3}$ and $x_{4}$ be the roots of the equation $a x^{2}+p x+q=0$. Then, by Vieta's theorem, $x_{1}+x_{2}=-\frac{b}{a}... | 2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,675 |
3. Triangle $A B C$ is inscribed in a circle. Points $M$ and $N$ are such that segment $A M$ is a diameter, and segment $A N$ is perpendicular to side $B C$. Prove that $B N = C M$.
 | Solution 1. Chords $C B$ and $A N$ are perpendicular by the condition, and chords $M N$ and $A N$ are also perpendicular because triangle $A N M$ is a right triangle with hypotenuse $A M$. Thus, we have two parallel chords $C B$ and $M N$, between which the arcs $C N$ and $B M$ are equal. Therefore, chords $B N$ and $C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,676 |
4. On his birthday, Piglet baked a large cake weighing 10 kg and invited 100 guests. Among them was Winnie-the-Pooh, who is fond of sweets. The birthday boy announced the rule for dividing the cake: the first guest cuts a piece of the cake that is $1 \%$, the second guest cuts a piece of the cake that is $2 \%$ of the ... | Solution. The first guests in the queue receive increasingly larger pieces of the pie because the remaining part of the pie is large at the initial stages of division. However, since the remaining part of the pie decreases, there will come a point when guests start receiving smaller pieces of the pie. At which guest wi... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,677 |
5. On the website of the football club "Rostov," a poll is being conducted to determine which of the $m$ football players the website visitors consider the best at the end of the season. Each visitor votes once for one player. The website displays the rating of each player, which is the percentage of votes cast for the... | Solution. Let $a$ be the greatest loss of a percentage point due to rounding when determining a footballer's rating. Then, according to rounding rules, $aa m \geqslant 5$, which means $0.5 m>5$ or $m>10$.
We will show that a solution exists when $m=11$. For example, let $m=11$ and 73 visitors voted, with 33 of them vo... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,678 |
2. Since $a_{k}=\frac{k \cdot(k+1)}{2}=\frac{1}{2} \cdot\left(k^{2}+k\right)$, then $a_{1}=\frac{1}{2} \cdot\left(1^{2}+1\right), a_{2}=\frac{1}{2} \cdot\left(2^{2}+2\right)$, $\ldots ., a_{n}=\frac{1}{2} \cdot\left(n^{2}+n\right)$
$$
\begin{aligned}
& \quad S=\frac{1}{2}\left(1^{2}+2^{2}+\cdots+n^{2}\right)+\frac{1}{... | Answer: $2017 \cdot 1009 \cdot 673$. | 2017\cdot1009\cdot673 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,681 |
# Problem 10.1
For all real $x$ and $y$, the equality $f\left(x^{2}+y\right)=f(x)+f\left(y^{2}\right)$ holds. Find $f(-1)$.
## Number of points 7 | Answer 0.
Solution
Substituting $x=0, y=0$, we get $f(0)=f(0)+f(0)$, that is, $f(0)=0$.
Substituting $x=0, y=-1$, we get $f(-1)=f(0)+f(1)$, that is, $f(-1)=f(1)$.
Substituting $x=-1, y=-1$, we get $f(0)=f(-1)+f(1)$. Therefore, $2 f(-1)=0$.
# | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,683 |
# Problem 10.2
Prove that if the numbers $x, y, z$ are solutions to the system for some values of $p$ and $q$:
$y=x^{2}+p x+q, z=y^{2}+p y+q, x=z^{2}+p z+q$,
then the inequality $x^{2} y+y^{2} z+z^{2} x \geq x^{2} z+y^{2} x+z^{2} y$ holds.
Points: 7 | Solution
Multiply the first, second, and third equations of the system by $y$, $z$, and $x$ respectively, and add them together, we get:
$x^{2}+y^{2}+z^{2}=x^{2} y+y^{2} z+z^{2} x+p(x y+x z+y z)+q(x+y+z)$.
On the other hand, multiplying the first equation of the original system by $z$, the second by $x$, and the thi... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 14,684 |
# Problem 10.3
Non-zero numbers $a$ and $b$ satisfy the equation $a^{2} b^{2}\left(a^{2} b^{2}+4\right)=2\left(a^{6}+b^{6}\right)$. Prove that at least one of them is irrational.
## Number of points 7 | Solution
Rewriting the equation as $\left(a^{4}-2 b^{2}\right)\left(b^{4}-2 a^{2}\right)=0$, we get that either $\left(\frac{a^{2}}{b}\right)^{2}=2$ or $\left(\frac{b^{2}}{a}\right)^{2}=2$. As is known, both equations are impossible for rational $a$ and $b$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,685 |
# Task 10.4
In triangle $A B C$, the altitude $B H$ and medians $A M$ and $C K$ are drawn. Prove that triangles $K H M$ and $A B C$ are similar.
Number of points 7 | Solution
Figure
From the problem statement, it follows that $M K$ is the midline of triangle $A B C$, hence $M K \| A C$ and $M K=0.5 A C$ (see fig.). Since $H M$ is the median of the right... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,686 |
# Task 10.5
A team of workers is repairing an apartment. To avoid damaging the floor in a room (a checkered square of size $4 \times 4$), they laid 13 two-cell rugs along the grid lines. Suddenly, it turned out that one rug was needed in another room. Prove that the workers can choose it in such a way that the repair ... | # Solution
Assume that it is impossible to remove a mat without exposing a piece of the floor. Then, under each mat, there must be a cell covered only once (otherwise, a mat, under which each cell is covered at least twice, can be removed). This means that the number of cells covered only once is no less than 13. Ther... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,687 |
1. A positive number $x$ was increased by $69 \%$. By what percentage did the number $\sqrt{\frac{x}{5}}$ increase? | # Solution.
The increased number will be $1.69 x$, then from the proportion
$\sqrt{\frac{x}{5}}-100 \%$, we determine $y=\frac{1.3 \sqrt{\frac{x}{5}} \cdot 100}{\sqrt{\frac{x}{5}}}=130$ (\%), so the difference will be
equal to $30 \%$.
Answer: by $30 \%$.
Instructions. Only the answer - 0 points; the answer obtain... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,688 |
2. Prove that if $a(a+b+c)<0$, then the equation $a x^{2}+b x+c=0$ has 2 real roots.
# | # Solution.
Proof. Since $a(a+b+c)<0$, \\ $a+b+c<0$ if $a>0$, and $a+b+c>0$ if $a<0$.
Consider the first case. Since $a>0$, the branches of the parabola defined by the formula $y=a x^{2}+b x+c$ are directed upwards. And since $y(1)=a+b+c<0$, the parabola intersects the $O X$ axis at 2 points. Therefore, the equation ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,689 |
3. Given a quadrilateral $A B C D$. Lines $D A$ and $C B$ intersect at point $E$, and lines $A B$ and $D C$ intersect at point $F$. It is known that the bisectors of angles $B E A$ and $B F C$ are perpendicular. Prove that a circle can be circumscribed around $A B C D$. | # Solution.
Proof.
Let $K$ and $N$ be the points of intersection of the angle bisector of $\angle AEB$ with the lines $AF$ and $DF$. In triangle $KFN$, the bisector is also an altitude. Ther... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,690 |
4. Is the number $\operatorname{tg} \sqrt{5 \pi}-1$ positive or negative?
# | # Solution.
$$
\text { We will prove that } \frac{5 \pi}{4}0$.
Answer: positive.
Instructions. Only the answer - 0 points; if the inequality (1) for $\pi$ uses an approximate value of 3.14 (after squaring) - 4 points; the square root is found approximately, for example, using a calculator - 0 points.
$$ | positive | Calculus | math-word-problem | Yes | Yes | olympiads | false | 14,691 |
5. In each cell of a $10 \times 10$ board, there is a grasshopper. At the whistle, each grasshopper jumps over one cell diagonally (not to the adjacent diagonal cell, but to the next one). As a result, some cells may end up with more than one grasshopper, while some cells will be unoccupied. Prove that in this case, th... | # Solution.
We will paint the cells of the board black and white, as shown in the figure. As a result, 60 cells will be painted black, and 40 cells will be painted white. Notice that from a black cell, the grasshopper can only jump to a white cell, and from a white cell, it can only jump to a black cell. Consequently,... | 20 | Combinatorics | proof | Yes | Yes | olympiads | false | 14,692 |
9.1. Ninety-nine natural numbers are arranged in a circle. It is known that any two adjacent numbers differ either by 1, by 2, or one is twice the other. Prove that at least one of these numbers is divisible by 3. (S. Berlov) | Solution. Suppose none of the numbers is divisible by 3. Then each number gives a remainder of 1 or 2 when divided by 3. But numbers giving the same non-zero remainder when divided by 3 cannot differ by 1 or 2; they cannot differ by a factor of two either. Therefore, adjacent numbers give different remainders when divi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,693 |
9.2. Seryozha chose two different natural numbers $a$ and $b$. He wrote down four numbers in his notebook: $a, a+2, b$ and $b+2$. Then he wrote on the board all six pairwise products of the numbers from the notebook. What is the maximum number of perfect squares that can be among the numbers on the board?
(S. Berlov)
... | # Answer. Two.
Solution. Note that no two squares of natural numbers differ by 1, because $x^{2}-y^{2}=(x-y)(x+y)$, where the second bracket is greater than one. Therefore, the numbers $a(a+2)=(a+1)^{2}-1$ and $b(b+2)=(b+1)^{2}-1$ are not squares. Moreover, the numbers $ab$ and $a(b+2)$ cannot both be squares, otherwi... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,694 |
10.1. Let's call a natural number good if it has exactly two prime numbers among its divisors. Can 18 consecutive natural numbers be good?
(O. Podlipsky) | Answer: No.
Solution: Suppose there are 18 consecutive good numbers. Among them, there will be three numbers divisible by 6. Let these numbers be $6n, 6(n+1)$, and $6(n+2)$. Since these numbers are good, and the prime factorization of each includes a 2 and a 3, they cannot have any other prime factors.
Next, only one... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,696 |
10.2. Given a function $f$ defined on the set of real numbers and taking real values. It is known that for any $x$ and $y$ such that $x>y$, the inequality $(f(x))^{2} \leqslant$ $\leqslant f(y)$ holds. Prove that the set of values of the function is contained in the interval $[0,1]$.
(A. Khryabrov) | Solution. By the condition $f(y) \geqslant(f(y+1))^{2} \geqslant 0$ for any $y$, so all values of the function are non-negative.
Now let $f\left(x_{0}\right)=1+a>1$ for some $x_{0}$. We will prove by induction on $n$ that for any $y1+2^{n} a$. For $n=1$ we have $f(y) \geqslant\left(f\left(x_{0}\right)\right)^{2}=1+2 a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 14,697 |
10.3. A safe has $n$ cells numbered from 1 to $n$. Initially, each cell contained a card with its number. Vasya rearranged the cards in some order so that the $i$-th cell now contains a card with the number $a_{i}$. Petya can swap any two cards with numbers $x$ and $y$, paying $2|x-y|$ rubles for it. Prove that Petya c... | The first solution. Let $\left(b_{1}, \ldots, b_{n}\right)$ be an arbitrary arrangement of cards (here $b_{i}$ is the number on the card in the $i$-th cell). We call its cost the number $\left|b_{1}-1\right|+\left|b_{2}-2\right|+\ldots+\left|b_{n}-n\right|$.
Lemma. For any arrangement $\left(b_{1}, \ldots, b_{n}\right... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,698 |
10.4. Triangle $ABC (AB > BC)$ is inscribed in the circle $\Omega$. Points $M$ and $N$ are chosen on sides $AB$ and $BC$ respectively such that $AM = CN$. Lines $MN$ and $AC$ intersect at point $K$. Let $P$ be the incenter of triangle $AMK$, and $Q$ be the excenter of triangle $CNK$ that touches side $CN$. Prove that t... | The first solution. We will use the following well-known lemma about the trident.
Lemma. Let $L$ be the midpoint of the arc $YZ$ (not containing point $X$) of the circumcircle of triangle $XYZ$. Let $I$ be the incenter of triangle $XYZ$, and $I_x$ be the excenter of this triangle, touching side $YZ$. Then $LY = LZ = L... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,699 |
11.1. Does there exist a positive number $a$ such that for all real $x$ the inequality
$$
|\cos x|+|\cos a x|>\sin x+\sin a x ?
$$
holds? | Answer. Does not exist.
Solution. Suppose that $01$, then, denoting $a x=t$ and $b=1 / a$, we will transform
the inequality from the condition into $|\cos b t|+|\cos t|>\sin b t+\sin t$, reducing the problem to the previous case. | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 14,700 |
11.2. Petya and Vasya are playing a game on an $n \times n$ checkered board. Initially, the entire board is white, except for the corner cell, which is black and contains a rook. The players take turns. Each turn, a player moves the rook horizontally or vertically, and all the cells the rook passes through (including t... | # Answer. Petya.
Solution. One of the winning strategies for Petya is to make the longest possible vertical move on each of his turns (for example, his first move will be to go vertically to the opposite corner of the board). We will show that, acting according to this strategy, he will win.
We will call a white cell... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,701 |
11.3. Positive rational numbers $a$ and $b$ are written as decimal fractions, each of which has a minimal period consisting of 30 digits. The decimal representation of the number $a-b$ has a minimal period length of 15. For what smallest natural $k$ can the minimal period length of the decimal representation of the num... | # Answer. $k=6$.
Solution. By multiplying, if necessary, the numbers $a$ and $b$ by a suitable power of ten, we can assume that the decimal representations of the numbers $a, b, a-b$, and $a+k b$ are purely periodic (i.e., the periods start immediately after the decimal point).
We will use the following known fact: t... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,702 |
11.4. Triangle $ABC (AB > BC)$ is inscribed in the circle $\Omega$. Points $M$ and $N$ are chosen on sides $AB$ and $BC$ respectively such that $AM = CN$. Lines $MN$ and $AC$ intersect at point $K$. Let $P$ be the incenter of triangle $AMK$, and $Q$ be the excenter of triangle $CNK$ that touches side $CN$. Prove that t... | The first solution. We will use the following well-known lemma about the trident.
Lemma. Let $L$ be the midpoint of the arc $YZ$ (not containing point $X$) of the circumcircle of triangle $XYZ$. Let $I$ be the incenter of triangle $XYZ$, and $I_x$ be the excenter of this triangle, touching side $YZ$. Then $LY = LZ = L... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,703 |
1. On a piece of paper, the number 5000 is written. Little Boy and Karlson take turns dividing this number by any of the following numbers: $2, 5, 10$. The one who cannot divide the remaining number evenly on their turn loses. In which position should Little Boy play and how should he play to win? | 1. Factorize the number into prime factors: $5000=2^{3} 5^{4}$. To win, Little One must play first: first, he divides the number by 2, and then repeats the actions of Karlson, then Little One will always be able to divide evenly. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,704 |
4. The King of Beasts, the Lion, announced a volleyball tournament. Several volleyball teams consisting of different animals have gathered. In the tournament, each team plays against each other. Prove that after each game, there will be two teams that have played the same number of matches by this point. | 4. Let the number of all teams be n. The number of matches played by a given team at some point in time can range from 0 to n-1, a total of n options. If one team has played with all n-1 teams, then no other team could have played 0 matches. Then for all n-1 teams, there are n-2 options left for the number of matches: ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,707 |
5. There are 5 visually identical weights, of which three weights weigh the same, the fourth weight is heavier than each of the three, and the fifth weight is lighter than each of the three. Determine the heaviest and the lightest weights in three weighings. | 5. Let's number the weights as $1,2,3,4,5$. (We will use equalities or inequalities to show the comparisons of the weights of the weights)
1) Weigh weights 1 and 2.
a) $1=2$.
2) Weigh 3 and 4. The weights cannot be equal.
Let $34$ be symmetric)
3) Weigh 1 and 5.
a) $1=5$, so 3 is light, 4 is heavy.
b) $1<5$, so 5 i... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,708 | |
1. For negative numbers $a, b$ and $c$, the equalities $\frac{c}{a+b}=2$ and $\frac{c}{b-a}=3$ hold. Which is greater: $c$ or $a$? | Answer: $c<a$.
Solution. From the given equalities, it follows that:
$$
2(a+b)=c=3(b-a) \Rightarrow b=5a \Rightarrow c=12a
$$
Since the number $a$ is negative, multiplying it by 12 makes it smaller. Therefore, $c<a$.
Comment. In the absence of a complete solution, if all transformations are correctly performed and ... | < | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,709 |
2. Thirty beads (blue and green) were laid out in a circle. For 26 beads, the neighboring one was blue, and for 20 beads, the neighboring one was green. How many blue beads were there? | Answer: 18 blue beads.
Solution. Let's find the number of beads that have both a blue and a green bead next to them: $26+20-30=$ 16. The number of beads that have only blue beads next to them is $26-16=10$. The number of blue beads is $\frac{10 \cdot 2+16}{2}=18$. Here is an example of such an arrangement. Let's denot... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,710 |
3. At the Olympiad, 64 participants arrived. It turned out that each of them had no more than 8 acquaintances among the participants. Prove that there will be 8 participants who are pairwise unfamiliar with each other. | Solution. Let's number the people from 1 to 64. Take the first person, who is acquainted with no more than 8 others, so there remain at least $64-8-1=55$ people with whom he is not acquainted. Take the second person from these 55, who is acquainted with no more than 8 others, so there remain at least $55-8-1=46$ people... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,711 |
4. Cut an arbitrary triangle with side $a$ into 3 parts, from which a triangle can be formed that has a side of length $2a$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The note about translation is not part of the t... | Solution. Consider an arbitrary triangle ABC. Let the side of length $a$ be $AB$. In triangle $\mathrm{ABC}$, draw the midline $MN \parallel AB$, and on the line $MN$, mark the segment $NP$ such that $NP = AB$. On the line $AB$ beyond point $B$, mark points $R$ and $Q$ such that: $BR = MN = RQ = AB / 2$.
Then, since $... | Logic and Puzzles | other | Yes | Yes | olympiads | false | 14,712 | |
5. Vasya and Petya are playing the following game. Vasya names a non-zero digit, and Petya inserts it in place of one of the asterisks in the product $* * * \times * * * \times * * * \times * * *$, and Vasya can see exactly where. Can Petya arrange the digits so that the resulting product is divisible by 9 regardless o... | Answer: Yes, he will be able to.
Solution. Petya will act as follows: digits that give a remainder of 0 when divided by 3, he will place in the first number; digits that give a remainder of 1 when divided by 3, he will place in the second number; and digits that give a remainder of 2 when divided by 3, he will place i... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,713 |
# 1. Clone 1
The figure is divided into 7 equal squares and several rectangles. The perimeter of rectangle A is 112 cm. What is the perimeter of rectangle B? Express your answer in centimeters.
![](https://cdn.mathpix.com/cropped/2024_05_06_1f50971f08007f421b41g-01.jpg?height=485&width=483&top_left_y=1008&top_left_x=... | # Answer: 168
## Solution
1st method. Rectangles A and B are composed of identical squares: Rectangle A consists of three, and Rectangle B consists of four. Let's call the side of such a square a "stick" and count the perimeter in sticks. The perimeter of Rectangle B is four sticks more than the perimeter of Rectangl... | 168 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,714 |
# 3. Clone 1
A rope was divided into 19 equal parts and arranged in a snake-like pattern. After that, a cut was made along the dotted line. The rope split into 20 pieces: the longest of them is 8 meters, and the shortest is 2 meters. What was the length of the rope before it was cut? Express your answer in meters.
![... | Answer: 114
Solution
Let's find the length of each of the 19 equal parts into which the rope was divided. To do this, we need to add the shortest piece to half of the longest piece obtained after the cut. We get $8 / 2 + 2 = 6$. Then the total length of the rope is $19 \cdot 6 = 114$ meters.
## Clone 2
A rope was d... | 152 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,716 |
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