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8. Egor wrote a number on the board and encrypted it according to the rules of letter puzzles (different letters correspond to different digits, the same letters correspond to the same digits). The result was the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible ... | Solution. The letter A must equal 0. The remaining 6 letters are non-zero digits with a sum that is a multiple of 3. Note that each remainder when divided by 3 appears three times. By enumeration, we find all sets of remainders whose sum is a multiple of three: 000111, 000222, 111222, 001122. We count the 6-element sub... | 21600 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,671 |
1. Can a two-digit number be divisible by five other two-digit numbers? | Solution. Yes: for example, 84 is a multiple of $12,14,21,28,42$. Other examples are 60, 90 and 96. | 84,60,90,96 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,672 |
2. The pond has a square shape. On the first frosty day, the part of the pond within 10 meters of the nearest shore froze. On the second day, the part within 20 meters froze, on the third day, the part within 30 meters, and so on. On the first day, the area of open water decreased by $19 \%$. How long will it take for ... | Solution. It is not hard to understand that a pond of $200 \times 200$ fits, for which the answer is - in 10 days (since each day the side decreases by 20 meters). There are no other options, as the larger the side of the pond, the smaller the percentage that will freeze on the first day.
More rigorously: let the side... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,673 |
3. How many ways are there to cut a $10 \times 10$ square into several rectangles along the grid lines such that the sum of their perimeters is 398? Ways that can be matched by rotation or flipping are considered different. | Solution: 180 ways.
If the entire square is cut into 100 unit squares, the sum of the perimeters will be $4 \times 100=400$. Therefore, we need to reduce this sum by 2, which is achieved by keeping one internal partition intact (in other words, the square is cut into 98 squares and 1 domino). There are a total of 180 ... | 180 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,674 |
4. A rectangle $11 \times 12$ is cut into several strips $1 \times 6$ and $1 \times 7$. What is the minimum total number of strips? | Solution. Answer: 20. The example is shown in the figure.
Evaluation: we will paint every seventh diagonal so that 20 cells are shaded (see figure). Each strip contains no more than one cell, so there are no fewer than 20 strips.
. However, it is forbidden to place a package inside... | Solution. 8. Example: ((6)(2)) ((3)(4)) ((1)4) (there are other examples).
There cannot be more than 8 packages. Indeed, then the sum of the number of candies in the packages (or rather, the number of incidences of candies to packages) is not less than $1+2+\ldots+9=45$. But there are 20 candies, so at least one of th... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,676 |
1. There is a rectangular sheet, white on one side and gray on the other. It was folded as shown in the picture. The perimeter of the first rectangle is 20 more than the perimeter of the second rectangle. And the perimeter of the second rectangle is 16 more than the perimeter of the third rectangle. Find the area of th... | Solution. From the figure, it can be seen that when folding, the perimeter of the rectangle decreases by twice the short side, so the short side of rectangle-1 is $20 / 2=10$, the short side of rectangle-2 is $16 / 2=8$. Therefore, the long side of rectangle-1 is 18, and the long side of the original sheet is 28. Thus,... | 504 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,677 |
3. Egor wrote a number on the board and encrypted it according to the rules of letter puzzles (different letters correspond to different digits, the same letters correspond to the same digits). The result was the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible ... | Solution. For a number to be divisible by 8, "АЛА" must be divisible by 8, with "А" -
the expression in parentheses is clearly divisible by 8, so it is sufficient to require that ("А" + 2 * "... | 67200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,679 |
4. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same... | Solution. $A$ and $C$ meet every 7 minutes, while $A$ and $B$ meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is $7 \cdot 53 = 371$ minutes. At the same time, $B$ and $D$ also meet every 7 minutes, so on the 371st minute, car $D$ will be at the same point as th... | 371 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,680 |
5. In a row, the squares of the first 2022 natural numbers are written: $1, 4, 9, \ldots, 4088484$. For each written number, except the first and the last, the arithmetic mean of its left and right neighbors was calculated and written below it (for example, under the number 4, $\left.\frac{1+9}{2}=5\right)$ was written... | Solution. Let's look at an arbitrary number in the row $x^{2}$. Under it, it will be written
$$
\frac{(x-1)^{2}+(x+1)^{2}}{2}=\frac{x^{2}-2 x+1+x^{2}+2 x+1}{2}=\frac{2 x^{2}+2}{2}=x^{2}+1
$$
Thus, each time the number increases by one. Initially, there are 2022 numbers, and each time their count decreases by 2, so th... | 10231311025154 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,681 |
7. Let's call a natural number useful if it does not contain any zeros or identical digits in its decimal representation, and the product of all its digits is divisible by their sum. Do there exist two consecutive three-digit useful numbers? | Solution. Yes, for example 578 and 579; 875 and 876.
Criteria. Correct answer without verification - 5 points. | 578579;875876 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,683 |
8. On the island of knights, who always tell the truth, and liars, who always lie, a school was opened. All $2 N$ students of different heights lined up in pairs (i.e., in two columns). The first two people said: "I am taller than two people: my partner and the person behind me." The last two said: "I am also taller th... | Solution. A) In each pair, there is no more than one knight, so there are no more than $N$ knights (the example is achieved by placing $N$ taller students in a checkerboard pattern). B) Since all students are of different heights, the tallest of them is definitely taller than their neighbors, so he is a knight, that is... | N | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,684 |
1. The pond has a rectangular shape. On the first frosty day, the part of the pond within 10 meters of the nearest shore froze. On the second day, the part within 20 meters froze, on the third day, the part within 30 meters, and so on. On the first day, the area of open water decreased by 20.2%, and on the second day, ... | Solution. First method. Let the sides of the pond be $a$ and $b$ meters, then
$(a-20)(b-20)=(1-0.202) a b, (a-40)(b-40)=(1-0.388) a b$,
from which $20(a+b)-400=0.202 a b, 40(a+b)-1600=0.388 a b$, that is, $800=0.016 a b, a b=5000$ and further $a+b=525$. It turns out that the sides are 400 and 125 meters.
Answer: on ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,685 |
3. In rhombus $A B C D$, points $E$ and $F$ are the midpoints of sides $A B$ and $B C$ respectively. Point $P$ is such that $P A=P F, P E=P C$. Prove that point $P$ lies on line $B D$. | Solution. Note that point $P$ lies at the intersection of the perpendicular bisectors of $A F$ and $E C$. Consider point $Q$, which is symmetric to $P$ with respect to line $B D$. It has the same properties as point $P$, that is, $Q A=Q F, Q E=Q C$. However, this means that point $Q$ also lies at the intersection of th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,686 |
5. Petya and Vasya are playing a game. They have a strip of 10 cells. Each turn, a player writes any digit in any free cell. However, they do not take turns in the usual way. First, Petya makes as many moves as he wants (but less than 10); then he asks Vasya to make one move; after that, Petya makes all the remaining m... | Solution. Petya has a winning strategy. For example, such a strategy: he writes 04 in the last two cells. Note that if a number ends in 02 or 52, then its square ends in 04. We will prove that for any of Vasya's moves, Petya will be able to find a perfect square.
Suppose Vasya moves to the hundreds place. Consider the... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,687 |
1. Come up with five different natural numbers whose product is 1000. | Solution. Note that $1000=10 \cdot 10 \cdot 10=2 \cdot 5 \cdot 2 \cdot 5 \cdot 2 \cdot 5=1 \cdot 2 \cdot 5 \cdot(2 \cdot 2) \cdot(5 \cdot 5)=1 \cdot 2 \cdot 5 \cdot 4 \cdot 25$. Answer: $1,2,4,5,25$. | 1,2,4,5,25 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,688 |
3. Here is a problem from S. A. Rachinsky's problem book (late 19th century): "How many boards 6 arshins long and 6 vershoks wide are needed to cover the floor of a square room with a side of 12 arshins?" The answer to the problem is: 64 boards. Determine from these data how many vershoks are in an arshin. | Solution. The area of the room is $12 \cdot 12=144$ square arshins. Therefore, the area of each board is $144 / 64=2.25$ square arshins. Since the length of the board is 6 arshins, its width should be $2.25 / 6=3 / 8=6 / 16$ arshins. Thus, 6 vershoks make up $6 / 16$ arshins, meaning 1 vershok is $1 / 16$ arshin.
Anot... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,690 |
4. On a glade, two firs, each 30 meters tall, grow 20 meters apart from each other. The branches of the firs grow very densely, and among them are some that are directed straight towards each other, and the length of each branch is half the distance from it to the top. A spider can crawl up or down the trunk (strictly ... | Solution. From the diagram, it can be seen that the branches of the firs intersect at a height of no more than 10 meters from the ground. Indeed, at this height, the distance to the treetop is 20 meters, so the length of each branch is $20 / 2 = 10$ meters, and the total length of the branches of the two firs is equal ... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,691 |
5. Nikita has a magic jar. If you put $n$ candies in the jar and close it for an hour, the number of candies inside will increase by the sum of the digits of $n$. For example, if there were 137 candies, it would become $137+1+3+7=148$. What is the maximum number of candies Nikita can get in 20 hours and 16 minutes, if ... | Solution. We need to strive to have as many candies as possible at the end of each hour. However, this does not mean that we should always put all the candies in the jar. The greatest sum of digits (i.e., the greatest increase in the number of candies) is achieved with a number where all digits (except the first) are n... | 267 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,692 |
1. The sum of the digits of a natural number is 2017. Moreover, any ten consecutive digits of the number are all different. Find the first 10 digits of the smallest and largest of such numbers. Justify your answer. | Solution. Among every ten consecutive digits, all digits from 0 to 9 appear exactly once, meaning the sum of every 10 consecutive digits is 45. Therefore, the first digit is the same as the eleventh, the second is the same as the twelfth, and so on, which means the number consists of repeating blocks of 10 digits (the ... | 25678901349765432108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,693 |
2. Find all pairs of real numbers $x$ and $y$ for which
$$
25^{x^{4}-y^{2}}+25^{y^{4}-x^{2}}=\frac{2}{\sqrt{5}}
$$ | Solution. Note that
$$
\frac{25^{x^{4}-y^{2}}+25^{y^{4}-x^{2}}}{2} \geqslant \sqrt{25^{x^{4}-y^{2}} \cdot 25^{y^{4}-x^{2}}}
$$
(inequality between the arithmetic mean and the geometric mean). The right-hand side is equal to
$$
5^{x^{4}-x^{2}+y^{4}-y^{2}}=5^{\left(x^{2}-1 / 2\right)^{2}+\left(y^{2}-1 / 2\right)^{2}-1... | (-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}});(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}});(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}});(\frac{1}{\sqrt{2}},\frac{1}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,694 |
4. Prove that for any natural $n \leqslant 2017$ a rectangle $1 \times n$ can be cut into 50 pieces and these pieces can be used to form a square. | Solution. Cut off as many rectangles of size $1 \times \sqrt{n}$ from the original rectangle as needed so that the remaining part has a length between $\sqrt{n}$ and $2 \sqrt{n}$. Since $\sqrt{n}<45$, fewer than 45 rectangles will be cut off. Cut the remaining part into three figures (a large triangle, a small triangle... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 11,696 |
5. The angle between the diagonals of a trapezoid is $60^{\circ}$. Prove that the sum of the lengths of the non-parallel sides is not less than the length of the longer base. | Solution. Let the bases of the trapezoid be $AD$ and $BC$, and the diagonals intersect at point $O$. First, consider the more complex case where $\angle COD = 60^\circ$.
Lemma. Suppose a regular triangle $ABK$ is constructed outside a side of an arbitrary triangle $ABC$. Then for any point $P$, the inequality $PA + PB... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,697 |
1. Tom and Jerry are playing on an infinite field of hexagonal cells. Initially, only 4 cells are black, and the rest are white (see the figure). Players take turns repainting cells, starting with Tom: Tom, on his turn, repaints one black cell to white, and Jerry -
. Jerry can form a "triangle" from three cells (i.e., color one of the two hexagons adjacent to both $A$ and $B$ - let's call it $C$). No matter which of these three cells Tom paints (let's say $A$), t... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 11,698 |
2. There are 49 identical squares. Form two rectangles from them so that their perimeters differ by a factor of 2. No squares should be left over.
(L. S. Koreschkova) | Answer: $7 \times 4$ and $1 \times 21$ or $1 \times 33$ and $1 \times 16$.
Solution. The sum of their areas is 49 ($28+21$ in the first case and $33+16$ in the second), and their perimeters indeed differ by a factor of 2: in the first case $2 \cdot(7+4)=22$, $2 \cdot(1+21)=44$; in the second $2 \cdot(1+16)=34, 2 \cdot... | 7\times41\times21or1\times331\times16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,699 |
3. In the cells of a $5 \times 5$ square, natural numbers from 1 to 5 are placed such that in each column, each row, and each of the two main diagonals, all numbers are distinct. Can the sum of the numbers in the cells shaded in the figure be equal to 20?
(L. S. Korechkova)
![](https://cdn.mathpix.com/cropped/2024_05... | Answer: No.
Solution. Suppose it can. Then, for the sum of the numbers in the shaded cells to be 20, all the numbers in them must be "5". Then only one more "5" can be placed, but both diagonals of this number still do not contain it, so it must be placed in the center of the board. However, this would result in two "... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,700 |
4. Ira has two identical squares and two identical triangles, from which she formed three figures as shown in the picture, and then calculated the perimeters of these figures. The perimeter of the first figure turned out to be 26, the second - 32, and the third - 30. Find the lengths of the sides of the triangle.
(S. ... | Answer: $3,4,5$.
Solution. Let's denote the sides of the triangle: the shortest side as $k$, the middle side as $c$, and the longest side as $-d$. The perimeter of figure №2 differs from figure №1 by two short sides of the triangle, so $k=(32-26): 2=3$. Now we can find the middle side of the triangle from figure №3 - ... | 3,4,5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,701 |
5. Participants of the spring math camp on Pi Day (March 14) decided to give each other squares if they were just acquaintances, and circles if they were friends. Andrey noticed that each boy received 3 circles and 8 squares, while each girl received 2 squares and 9 circles. And Katya calculated that a total of 4046 fi... | Solution. Note that each camper was given exactly $3+8=2+9=11$ figurines. Then the total number of figurines is the product of 11 and the number of participants, but 4046 is not divisible by 11.
Criteria. The idea of an equal number of figurines per participant - 3 points.
Reasoning about the number of boys and/or gi... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 11,702 |
6. How many numbers from 1 to 999 without the digit "0" are written in the Roman numeral system exactly one symbol longer than in the decimal system?
(P. D. Mulyenko)
Reference. To write a number in Roman numerals, you need to break it down into place value addends, write each place value addend according to the tabl... | Solution. Note that, regardless of the digit place and other digits of the number, a decimal digit $a$ is written as:
- one Roman numeral when $a=1$ and $a=5$,
- two Roman numerals when $a$ is $2,4,6,9$,
- three Roman numerals when $a=3$ and $a=7$,
- four Roman numerals when $a=8$.
Thus, in suitable numbers, only the... | 68 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,703 |
1. Each cell of a $1000 \times 1000$ board is painted blue or white. We will call a cell balanced if among its neighbors there are an equal number of blue and white cells. Is it possible to color the board so that there are more than 600000 blue balanced cells? (Cells are considered neighboring if they share a side.) | Solution. It is possible. The figure shows how to make approximately $2 / 3$ of the cells on the board blue and balanced.
A more precise calculation: in each row (except the first and last), ... | 998\cdot664>600000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,704 |
3. In three-dimensional space, a standard coordinate system is given. Find the area of the set of points satisfying the following conditions: $x^{2}+y^{2}=5,|x-y|<1,|y-z|<1$. | Solution. We will construct the set of points in the $xy$-plane satisfying the first two conditions (which do not depend on $z$). The first condition defines a circle, the second - a strip, their intersection is two arcs. The length of each arc is $\sqrt{5}(\operatorname{arctg} 2-\operatorname{arctg} 1 / 2)$.
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,706 |
1. Igor and Pasha are playing a game, taking turns placing natural numbers at the vertices of a regular hexagon (each can choose any free vertex and place any natural number in it). After six moves, when the game ends, the judge writes the product of the numbers at the two ends of each side of the hexagon on each side.... | Answer: Pasha.
Solution. Let's number the vertices in a circle from 1 to 6 and mentally divide them into 3 pairs of diametrically opposite vertices (1-4, 2-5, 3-6).
To win, Pasha should place the same number that Igor placed on his previous move in the opposite vertex. Then, by the end of the game, all numbers in opp... | Pasha | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,709 |
2. There are 81 squares of the same size. Arrange them into two rectangles so that their perimeters are the same. No squares should be left over.
(L. S. Koreschkova) | Answer: $3 \times 11$ and $6 \times 8$.
Solution. The sum of their areas is $33+48=81$, and their perimeters are indeed equal: $2 \cdot(3+11)=2 \cdot(6+8)=28$.
Note. There are no other suitable pairs of rectangles.
Criteria. A correct example is provided without calculations showing that it fits - 5 points. At least... | 3\times116\times8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,710 |
3. Eight boys (Vasya, Dima, Egor, Ilya, Kolya, Petya, Temya, and Fyodor) stood in a line in some order, after which they counted from 1 to 8, with the following conditions:
- Dima's number was three times Ilya's number;
- Fyodor stood somewhere after the third boy but before Kolya;
- Vasya's number was half of Petya's... | Answer: Egor, Ilya, Tema, Vasya, Fedy, Dima, Kolya, Petya.
Solution. According to the fourth condition, Tema stands in the third place. Since Dima's number is divisible by 3 according to the first condition, and the third number is occupied, Dima must have the number 6, and Ilya - number 2. From the third condition, i... | Egor,Ilya,Tema,Vasya,Fedy,Dima,Kolya,Petya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,711 |
4. Ira has two identical squares and two identical triangles, from which she formed three figures as shown in the picture, and then calculated the perimeters of these figures. The perimeter of the first figure turned out to be 74, the second - 84, and the third - 82. Find the lengths of the sides of the triangle.
(S. ... | Answer: $5,12,13$.
Solution. Let's denote the sides of the triangle: the shortest side as $k$, the middle side as $c$, and the longest side as $d$. The perimeter of figure №2 differs from figure №1 by two short sides of the triangle, so $k=(84-74): 2=5$. Now we can find the middle side of the triangle from figure №3 -... | 5,12,13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,712 |
5. In a spring math camp, between 50 and 70 children arrived. On Pi Day (March 14), they decided to give each other squares if they were just acquaintances, and circles if they were friends. Andrey calculated that each boy received 3 circles and 8 squares, while each girl received 2 squares and 9 circles. And Katya not... | Answer: 60.
Solution. Let the number of boys be $m$, and the number of girls be $-d$. Then $3 m + 9 d = 8 m + 2 d$ (the number of circles equals the number of squares). Transforming, we get $5 m = 7 d$, which means the number of boys and girls are in the ratio $7: 5$. Therefore, the total number of children is divisib... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,713 |
6. How many numbers from 1 to 999 are written in the Roman numeral system with the same number of symbols as in the decimal system?
(P. D. Mulyenko)
Reference. To write a number in Roman numerals, you need to break it down into place value addends, write each place value addend according to the table, and then write ... | Solution. Note that, regardless of the digit place and other digits of the number, a decimal digit a is written as:
- zero Roman numerals when $a=0$,
- one Roman numeral when $a=1$ and $a=5$,
- two Roman numerals when $a$ is $2,4,6,9$,
- three Roman numerals when $a=3$ and $a=7$,
- four Roman numerals when $a=8$.
Thu... | 52 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,714 |
1. Milla and Zhena came up with a number each and wrote down all the natural divisors of their numbers on the board. Milla wrote down 10 numbers, Zhena wrote down 9 numbers, and the largest number written on the board twice is 50. How many different numbers are written on the board? | Solution. Note that a number written twice is a common divisor of the original numbers; the largest such number is their GCD. Therefore, all numbers written twice are divisors of the number 50, that is, the numbers $1,2,5,10,25,50$. Thus, among the listed numbers, exactly 6 are repeated, and the number of different num... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,715 |
2. On graph paper, a polygon with a perimeter of 36 is drawn, with its sides running along the grid lines. What is the maximum area it can have? | Solution. Consider the extreme verticals and horizontals. Moving from them inward does not allow reducing the perimeter, but it decreases the area. Therefore, the rectangle has the largest area. If A and B are the lengths of its sides, then A + B = 18.
By trying different rectangles with a perimeter of 36, such as (1,... | 81 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,716 |
3. In a circle, three equal chords are drawn passing through one point. Prove that these chords are diameters.
# | # Solution.
## Method 1.
Let AB, CD, EF be three chords, and M be their point of intersection. We need to prove that the figure consisting of segments AB and CD is symmetric with respect to one of the diameters.
Indeed, since chords AB and CD are equal, the arcs ACB and CAD, on which they subtend, are also equal; si... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,717 |
4. The brothers found a treasure of gold and silver. They divided it so that each received 100 kg. The eldest received the most gold - 25 kg - and one eighth of all the silver. How much gold was in the treasure? | Solution. 1) The elder brother received 75 kg of silver, which is one-eighth of the total amount; therefore, the total mass of silver is 600 kg.
2) The others received more silver than the elder brother, i.e., each received more than 75 kg. If there are at least eight brothers, then in total they would receive more th... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,718 |
5. Three people want to travel from city $A$ to city $B$, which is 45 kilometers away from $A$. They have two bicycles. The speed of a cyclist is 15 km/h, and the speed of a pedestrian is 5 km/h. What is the minimum time they can take to reach $B$, given that the bicycles cannot be left unattended on the road? | Solution. Two people ride 15 kilometers, then one sets off on foot, while the second stays to guard both bicycles until the third arrives.
It is easy to verify that among similar options, which differ only in the choice of the stopping point, the best one is when it is exactly in the middle. Complications involving on... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,719 |
6. Lev took two natural numbers, added their sum to their product, and as a result got 1000. What numbers could Lev have taken? Find all options. | Solution. If we denote the numbers of Lev as a and b, we get: a + b + ab = 1000. Adding one to both sides: 1 + a + b + ab = 1001, or (1 + a)(1 + b) = 7 · 11 · 13. Since a and b are natural numbers, 1 + a > 1 and 1 + b > 1. Therefore, one of the six variants must be true:
a) 1 + a = 7, 1 + b = 11 · 13, from which a = 6... | 6142,1090,1276 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,720 |
1. Let's call a natural number useful if it does not contain any zeros or identical digits in its decimal representation, and the product of all its digits is divisible by their sum. Find the two largest consecutive (i.e., differing by 1) useful numbers. | Solution. The numbers 9875213 and 9875214 are useful. We will prove that there do not exist larger consecutive useful numbers. The sums of the digits of consecutive numbers are also consecutive (otherwise, there would be a transition across a place value, meaning a 0 at the end). However, the maximum possible sums of t... | 98752139875214 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,721 |
2. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same... | Solution. $A$ and $C$ meet every 7 minutes, while $A$ and $B$ meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is $7 \cdot 53 = 371$ minutes. At the same time, $B$ and $D$ also meet every 7 minutes, so on the 371st minute, car $D$ will be at the same point as th... | 371 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,722 |
3. Prove that a regular pentagon can be cut into 4 pieces, from which a rectangle can be formed without gaps or overlaps. | Solution. By cutting and rearranging the triangle, a trapezoid can be obtained (this follows from the marked angles and the equality of sides). Next, by cutting off two right triangles from the trapezoid, we can rotate them and obtain a rectangle (see the figure).
 that a point $M$ is convenient for a circle with center $O$ if and only if $O M$ is twice the radius.
1
^{1 / 3}= \\
=3^{a^{2}+b+c+1+b^{2}+a+c+1+c^{2}+a+b+1}=3^{(a+1)^{2}+(b+1)^{2}+(c+1)^{2}} \geqslant 1
\end{gathered}
$$... | =b==-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,725 |
6. The park is a $10 \times 10$ grid of cells. A lamp can be placed in any cell (but no more than one lamp per cell).
a) The park is called illuminated if, no matter which cell a visitor is in, there is a $3 \times 3$ square of 9 cells that contains both the visitor and at least one lamp. What is the minimum number of... | Solution. a) 4. Divide the park into 4 quarters (squares $5 \times 5$), then there must be at least one lamp in each quarter (to illuminate, for example, the corner cells). By placing one lamp in the center of each quarter, we get an example.
b) 10.
Estimate. In each corner square $3 \times 3$ there must be at least t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,726 |
7. Let's call the efficiency of a natural number $n$ the fraction of all natural numbers from 1 to $n$ inclusive that have a common divisor with $n$ greater than 1. For example, the efficiency of the number 6 is $\frac{2}{3}$.
a) Does there exist a number with an efficiency greater than $80\%$? If so, find the smalles... | Solution. Let's move on to studying inefficiency (1 minus efficiency). From the formula for Euler's function, it follows that it is equal to $\frac{p_{1}-1}{p_{1}} \cdot \ldots \frac{p_{k}-1}{p_{k}}$, where $p_{1}, \ldots, p_{k}$ are all distinct prime divisors of $n$. Then, by adding a new prime factor, we can increas... | 30030 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,727 |
8. A certain continuous function $f$ is such that $f(f(f(f(f(0)))))=0$. Prove that the equation $f(f(x))=x$ has at least one root. | Solution. We will prove that $f(x)=x$ has a solution. Indeed, if this is not the case, then (due to the continuity of the function $f(x)-x$), $f(x)$ is either always greater or always less than $x$, meaning that applying $f$ always changes the result in the same direction. However, this would contradict the condition o... | proof | Algebra | proof | Yes | Yes | olympiads | false | 11,728 |
1. The sum of the digits of a natural number is 2017. Moreover, any ten consecutive digits of the number are distinct. How many digits can the number have? List all possible answers and prove that there are no others. | Solution. Among every ten consecutive digits, all digits from 0 to 9 appear exactly once, meaning the sum of every 10 consecutive digits is 45.
Notice that $2017=44 \cdot 45+37$. This means the number consists of 44 blocks of 10 digits each, and at the end, there are a few more digits with a sum of 37. In other words,... | 446to449 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,729 |
5. On an $8 \times 8$ board, several dominoes (i.e., $2 \times 1$ rectangles) can be placed without overlapping each other. Let $N$ be the number of ways to place 32 dominoes, and $S$ be the number of ways to place 16 dominoes. Which is greater, $N$ or $S$? Ways that can be transformed into each other by rotation or re... | Solution. 16 dominoes can be arranged in more ways than 32. To prove this, consider a certain correspondence (not one-to-one) between 32-partitions and 16-arrangements (i.e., ways to arrange 32 dominoes and ways to arrange 16 dominoes).
If there are 16 horizontal dominoes, we leave only them. The vertical dominoes are... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,733 |
1. Since a syllable consists of two different letters, identical letters can only appear at the junction of syllables.
First, let's find the number of combinations of two syllables with a matching letter at the junction. Such syllables (in terms of the arrangement of vowels and consonants) are either AMMO $(3 \cdot 8 ... | Answer: $2 \cdot 264 \cdot 48-2 \cdot 24^{2}=24192$ funny words. | 24192 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,734 |
3. Two players play a game. They take turns naming three-digit numbers that have no zeros in their notation, and the sum of the digits is divisible by 9. At the same time, each subsequent number must start with the same digit that the previous number ends with, for example: $351-189$ - 936 - $621 \ldots$... Repeating n... | Solution. The first player wins. One possible strategy is as follows. He calls the number 999, and then in response to any number $\overline{A B C}$ called by the second player, he calls the number $\overline{C B A}$ (the same number "backwards"). Note that after this, the second player will again have to call a number... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,737 |
4. Flint has five sailors and 60 gold coins. He wants to distribute them into wallets, and then give the wallets to the sailors so that each gets an equal number of coins. But he doesn't know how many sailors will be alive by the time of the distribution. Therefore, he wants to arrange the coins so that they can be eve... | Solution. Yes. For example, with the following number of coins in the wallets: $12,12,8,7,6,5$, $4,3,3$.
Note that a smaller number of wallets is insufficient (see the solution to problem 4 for grade 6). | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,738 |
5. On an $8 \times 8$ board, several dominoes (i.e., $2 \times 1$ rectangles) can be placed without overlapping each other. Let $N$ be the number of ways to place 32 dominoes, and $T$ be the number of ways to place 24 dominoes. Which is greater, $N$ or $T$? Ways that can be transformed into each other by rotation or re... | Solution. There are more ways to arrange 24 dominoes. We will prove this.
Consider any partition of the board into 32 dominoes. Remove the four dominoes that contain the corner cells. In addition, from each side of the board, remove one domino that contains one of the middle cells of that side. (An example is shown in... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,739 |
2. Is it possible to choose natural coefficients $A, B$, and $C$ in the expression $A \cdot 5^{n}+B \cdot 3^{n-1}+C$ so that none of them is divisible by 8, but the result is divisible by 8 for any natural $n$?
(L. S. Korechkova) | Solution. Yes. Note that $5^{n}$ gives a remainder of 1 when divided by 4, and $3^{n-1}$ is always odd. Therefore, we can take $A=2, B=4, C=2$.
Criteria. Answer "yes" without explanation - 0 points. No more than one point if one of the coefficients is a multiple of 8.
Correct coefficients given without explanation - ... | A=2,B=4,C=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,740 |
3. Once Valera left home, walked to the cottage, painted 11 fence boards there, and returned home 2 hours after leaving. Another time, Valera and Olga went to the cottage together, painted 8 fence boards (without helping or hindering each other), and returned home together 3 hours after leaving. How many boards will Ol... | Solution. This is a more complex version of problem 5 for 5th grade.
The strange result (that the characters accomplished less work in more time when working together) is explained by the different times spent walking, since the speed of "joint" walking is equal to the lower of the two walkers' speeds. The second time... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,741 |
4. In a rectangular grid $20210 \times 1505$, two diagonals were drawn and all cells through which they passed were painted. How many cells were painted?
(O. A. Pyayve, A. A. Tessler) | Solution. This is a more complex version of problem 3 for 5th grade.
First, let's determine how many cells one diagonal crosses. Note that $20210=215$. $94,1505=215 \cdot 7$. Therefore, the ... | 42986 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,742 |
5. On a plane, an equilateral triangle and three circles with centers at its vertices are drawn, and the radius of each circle is less than the height of the triangle. A point on the plane is painted yellow if it lies inside exactly one of the circles; green if inside exactly two; and blue if inside all three. It turns... | Solution. The sum of the areas of the three circles is $1000+2 \cdot 100+3 \cdot 1=1203$; the sum of the areas of the three "lenses" is $100+3 \cdot 1=103$ (a "lens" is the intersection of two circles).
The area of the triangle is $S_{1}-S_{2}+S_{3}$, where
$S_{1}=1203 / 6-$ the sum of the areas of the three 60-degre... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,743 |
2. How many five-digit numbers are roots of the equation $x=[\sqrt{x}+1][\sqrt{x}]$? The symbol $[a]$ denotes the integer part of the number $a$, that is, the greatest integer not exceeding $a$.
(0. A. Pyayve) | Solution. Let $n=[\sqrt{x}]$, then $[\sqrt{x}+1]=[\sqrt{x}]+1=n+1$, which means $x=n(n+1)$.
All numbers of the form $x=n(n+1)$ are suitable, since for them $n<\sqrt{x}<n+1$, meaning $[\sqrt{x}]$ is indeed equal to $n$.
It remains to count the five-digit numbers of this form. Note that 99$\cdot$100 $<10000<100 \cdot 1... | 216 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,744 |
3. In parallelogram $ABCD (AB \neq BC)$, two heights, $BH$ and $BK$, are drawn from the obtuse angle $B$ (the bases of the heights lie on the sides of the parallelogram and do not coincide with its vertices). Triangle $BHK$ turned out to be isosceles. Indicate all possible values of angle $BAD$.
(L. S. Korechkova) | Solution. Since the sides of the parallelogram are not equal, the heights are not equal either (for example, from the area formula), that is, $B H \neq B K$. This means that in triangle $B H K$, side $K H$ is equal to one of the other two. For definiteness, let's assume that $K H = B K$, and $H \in A D, K \in C D$.
Le... | anyangleintherange(45;90),except60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,745 |
6. Given a rectangle of size $2021 \times 4300$. Inside it, there is a billiard ball. It is launched in a straight line, forming a $45^{\circ}$ angle with the sides of the rectangle. Upon reaching a side, the ball reflects at a $45^{\circ}$ angle; if the ball hits a corner, it exits along the same line it entered. (An ... | # Solution.
We will replicate the rectangle multiple times by reflecting it relative to its sides. In adjacent columns (rows), copies of the rectangle will be oriented differently, but when shifted by an even number of columns and an even number of rows, the orientation will match the initial one. Now, we can consider... | 294 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,746 |
5. In the country, there are 100 cities, and several non-stop air routes are in operation between them, such that one can travel from any city to any other, possibly with layovers. For each pair of cities, the minimum number of flights required to travel from one to the other was calculated. The transportation difficul... | Solution. First, we prove that the maximum transportation difficulty occurs when the cities are connected "in a chain."
Indeed, we can consider the graph as a tree (otherwise, we can remove some edges to leave a tree - the difficulty will increase). Choose the longest path in the tree. Suppose there are vertices not i... | 8332500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,751 |
1. Three lines divide a circle into 7 parts. Is it possible to distribute the numbers from 1 to 7, one in each area, so that the sums of the numbers on either side of each line are equal? | Solution. Yes, for example, like this:
 | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,752 |
2. Marina needs to buy a notebook, a pen, a ruler, a pencil, and an eraser to participate in the Olympiad. If she buys a notebook, a pencil, and an eraser, she will spend 47 tugriks. If she buys a notebook, a ruler, and a pen, she will spend 58 tugriks. How much money will she need for the entire set, if the notebook c... | Solution. If Marina buys two sets from the condition, she will spend $47+58=105$ tugriks, but she will buy an extra notebook, so the full set of school supplies costs $105-15=90$ tugriks.
Criteria. Only the answer without explanation - 1 point. If in the solution they try to determine the cost of the pen and pencil (a... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,753 |
3. An accident has occurred in the reactor of a research spacecraft, and toxic substances are leaking from it. All corridors between rooms are equipped with airtight doors, but there is no time to close individual doors. However, the captain can still give the command "Close $N$ doors," after which the ship's artificia... | Solution. There are a total of 23 corridors on the spaceship. If no more than 21 doors are closed, then the corridors between the reactor and the right engine, and between the right engine and the lounge, may remain open, which means the crew will be in danger. Therefore, at least 22 doors must be closed.
Criteria. Co... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,754 |
4. On an island of knights, who always tell the truth, and liars, who always lie, a school was opened. All $2 N$ students of different heights stood in a circle, after which each one said: "I am taller than the student standing opposite me!" How many knights are studying at the school? | Solution. In each pair, one of the two students is indeed taller than their opposite neighbor, so they are telling the truth and are a knight. In total, there are $N$ knights.
Criteria. If the fact that there is definitely one knight in a pair is not justified - 1 point. If a specific example is used as justification ... | N | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,755 |
5. Egor wrote a number on the board and encrypted it according to the rules of letter puzzles (different letters correspond to different digits, the same letters correspond to the same digits). He got the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible by 5? | Solution. The number must be divisible by 5, so the letter "A" is equal to 0 or 5. If it is equal to 0, then for the other letters ("G", "V", "T", "E", "M", "L") there are $A_{9}^{6}=9!/ 3$! options; if "A" is equal to 5, then for the other letters there are $8 \cdot A_{8}^{5}=8$!/3! options, since "G" cannot be zero. ... | 114240 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,756 |
7. Three cars $A, B$ and $C$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ drives counterclockwise. All cars move at constant (but pairwise distinct) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time. After another 46 minutes, ... | Solution. $A$ and $C$ meet every 7 minutes, and $A$ and $B$ meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is $7 \cdot 53=371$ minutes.
Criteria. If 53 minutes is replaced with 46 - 3 points. Solved by trial with lengths and speeds -1 point. Only the answer w... | 371 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,757 |
8. There is a rectangular sheet, white on one side and gray on the other. It was folded as shown in the picture. The perimeter of the first rectangle is 20 more than the perimeter of the second rectangle. And the perimeter of the second rectangle is 16 more than the perimeter of the third rectangle. Find the perimeter ... | Solution. From the figure, it can be seen that when folding, the perimeter of the rectangle decreases by twice the short side, so the short side of rectangle-1 is $20 / 2=10$, the short side of rectangle-2 is $16 / 2=8$. Therefore, the long side of rectangle-1 is 18, and the long side of the original sheet is 28. Thus,... | 92 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,758 |
2. Katya decided to calculate the sum of the cubes of all natural divisors of some natural number, and she got the result $M A T H$. But then she discovered that she had forgotten one of the divisors. Adding its cube, she got the correct result - MASS. Find the smallest possible value of the number $M A T H$. (MATH and... | Solution. Answer: 2017. The original natural number is $12 ; 12^{3}+6^{3}+4^{3}+2^{3}+1^{3}=$ 2017; if you add $3^{3}$, you get 2044.
We will prove that there are no smaller suitable numbers.
1) For any number less than 10, the sum of the cubes of all divisors, as is easily verified, is less than a thousand.
2) $10^{... | 2017 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,760 |
4. Two players play a game like this. They take turns naming eight-digit numbers that have no zeros in their notation, and the sum of the digits is divisible by 9. At the same time, each subsequent number must start with the same digit that the previous number ends with. Repeating numbers is not allowed. The player who... | Solution. The first player wins. One of his possible strategies is described below.
We will divide the suitable numbers into pairs by matching each number with its reverse. This pairing does not apply to palindromes (i.e., numbers that read the same in both directions).
Notice that the number of palindromes starting ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,762 |
5. On an $8 \times 8$ board, several dominoes (i.e., $2 \times 1$ rectangles) can be placed without overlapping each other. Let $N$ be the number of ways to place 32 dominoes, and $F$ be the number of ways to place 16 tokens on this board (no more than one token per cell). Which is greater, $N$ or $F$? Ways that can be... | Solution. We will prove that the number of arrangements of 16 chips is greater than the number of partitions into 32 dominoes. To do this, we will devise a rule that associates each partition into 32 dominoes with an arrangement of 16 chips, such that different partitions correspond to different arrangements. In other ... | N<F | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,763 |
6. Several plants and zombies (no more than 20 creatures in total) attended the "Plants VS Zombies" gathering, and it turned out that all creatures were of different heights. Plants always tell the truth to those who are shorter than them and lie to those who are taller. Zombies, on the contrary, lie to shorter creatur... | Solution. Let the total number of beings be $n$, and exactly $z$ of them are zombies. When plants greet, they say to everyone “I am taller than you”, and zombies say to everyone “I am shorter”. Each zombie said this phrase to everyone except themselves, so we get $z(n-1)=20$. Given that $n-1<20$, the possible cases are... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,765 |
1. There are 28 students in the class. On March 8th, each boy gave each girl one flower - a tulip, a rose, or a daffodil. How many roses were given if it is known that there were 4 times as many roses as daffodils, but 10 times fewer than tulips? (A. A. Tesler) | Solution. This is a more complex version of problem 4 for 5th grade.
Let the number of narcissus be $x$, then the number of roses is $4x$, and the number of tulips is $40x$, so the total number of flowers is $45x$. The number of flowers is the product of the number of boys and the number of girls. If there are $m$ boy... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,767 |
3. On a plane, an equilateral triangle and three circles with centers at its vertices are drawn. A point on the plane is painted yellow if it lies inside exactly one of the circles; green if inside exactly two; and blue if inside all three. Can the yellow area be 100, the green area 10, and the blue area 1? (P. D. Muly... | Solution. Yes. For example, let's choose the radii of the circles so that their areas are 1, 11, and 111; and make the side of the triangle \(a\) sufficiently small (for example, \(a \leqslant \frac{1}{2}\)). We will prove that the smaller circle always lies inside the larger one (from this it will follow that the area... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,768 |
4. On the plane, there are four points $A, B, C, D$ such that $A B=B C=C D, B D=D A=A C$. Find the angles of the quadrilateral with vertices at these points. | Solution. Angles $\angle A B C$ and $\angle D C B$ are equal due to the congruence of triangles $A B C$ and $D C B$.
a) If vertices $A$ and $D$ are on opposite sides of line $B C$, then these angles are alternate interior angles. Therefore, segments $A B$ and $C D$ are parallel and equal, making $A B D C$ a parallelog... | 72,72,108,108 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,769 |
5. The magical clock, in addition to the usual pair of hands, has a second pair that is symmetrical to the first at every moment relative to the vertical axis. It is impossible to determine which hands are real from a photograph of the clock. Furthermore, just like with ordinary clocks, it is impossible to distinguish ... | Solution. See the solution to problem 6 for 10th grade.
Instead of the last paragraph of the solution to problem 6 for 10th grade, it is sufficient to provide an example with exactly three undefined photographs and prove that there are exactly three. For example, let the first three photos be taken at 5:50, 11:40, 17:... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,770 |
6. Find all real solutions of the system of equations
$$
\left\{\begin{array}{l}
\frac{1}{x}=\frac{32}{y^{5}}+\frac{48}{y^{3}}+\frac{17}{y}-15 \\
\frac{1}{y}=\frac{32}{z^{5}}+\frac{48}{z^{3}}+\frac{17}{z}-15 \\
\frac{1}{z}=\frac{32}{x^{5}}+\frac{48}{x^{3}}+\frac{17}{x}-15
\end{array}\right.
$$
(A. B. Vladimirov) | Solution. Let $F(t)=32 t^{5}+48 t^{3}+17 t-15$. Then the system has the form $F\left(\frac{1}{y}\right)=\frac{1}{x}, F\left(\frac{1}{z}\right)=\frac{1}{y}$, $F\left(\frac{1}{x}\right)=\frac{1}{z}$. From this, it follows that $F\left(F\left(F\left(\frac{1}{x}\right)\right)\right)=\frac{1}{x}$. Note that $F(0.5)=0.5$ and... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,771 |
1. Milla and Zhena came up with a number each and wrote down all the natural divisors of their numbers on the board. Milla wrote 10 numbers, Zhena - 9, and the number 6 was written twice. How many different numbers are on the board? | Solution. Since the number 6 is written twice, both original numbers (denote them as a and b) are divisible by 6.
If Vera's number has 10 divisors, then its factorization is either $p^{9}$ or $p^{1} \cdot q^{4}$ (where p and q are some prime numbers); the first is impossible since it is divisible by 6. Valya's number ... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,772 |
3. The brothers found a treasure of gold and silver. They divided it so that each received 100 kg. The elder brother received $1 / 5$ of all the gold and $1 / 7$ of all the silver, while the younger brother received $1 / 7$ of all the gold. What share of the silver did the younger brother receive? | # Solution.
1) Let the mass of gold in the treasure be $z$ kg, and the mass of silver be $s$ kg. The eldest brother received $z/5 + s/7$ kg; this is less than $(z+s)/5$, but more than $(z+s)/7$. From this, it follows that the number of brothers is more than five but less than seven, meaning there are six of them.
2) N... | \frac{9}{49} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,774 |
4. Three people want to travel from city A to city B, which is 45 kilometers away from A. They have two bicycles. The speed of a cyclist is 15 km/h, and the speed of a pedestrian is 5 km/h. What is the minimum time they can reach B, if the bicycle can be left unattended on the road? | Solution. Two people ride a bicycle for 10 kilometers, then one of them leaves the bicycle by the road and walks the next 10 kilometers, while the other continues for the next 10 kilometers and also leaves the bicycle (which the first one should pick up later), the third one walks the first 10 kilometers and then rides... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,775 |
6. On graph paper, a polygon with a perimeter of 2014 is drawn, with its sides running along the grid lines. What is the maximum area it can have? | Solution. First, consider the extreme verticals and horizontals. Moving from them inward does not allow reducing the perimeter, but it decreases the area. Therefore, the rectangle has the largest area. If A and B are the lengths of its sides, then $A + B = 1007$.
Now, among different rectangles with a perimeter of 201... | 253512 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,777 |
2. It is known that on Monday, the painter painted twice as slowly as on Tuesday, Wednesday, and Thursday, and on Friday - twice as fast as on these three days, but worked 6 hours instead of 8. On Friday, he painted 300 meters more of the fence than on Monday. How many meters of the fence did the painter paint from Mon... | Solution. Let's take $100 \%$ of the fence length that the painter painted on Tuesday, Wednesday, and Thursday. Then Monday accounts for $50 \%$, and Friday for $-150 \%$. Therefore, 300 meters of the fence correspond to $150 \% - 50 \% = 100 \%$. For the entire week (from Monday to Friday), it is $500 \%$, i.e., 1500 ... | 1500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,779 |
3. Find the number of four-digit numbers in which all digits are different, the first digit is divisible by 2, and the sum of the first and last digits is divisible by 3. | Solution. The first digit can be 2, 4, 6, or 8. If the first is 2, the last can be 1, 4, 7; if the first is 4, the last can be 2, 5, 8; if the first is 6, the last can be 0, 3, (6 does not fit), 9; if the first is 8, the last can be 1, 4, 7. In total, there are 3 + 3 + 3 + 3 = 12 options for the first and last digits. ... | 672 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,780 |
4. In the Olympionov family, it is a tradition to especially celebrate the day when a person turns as many years old as the sum of the digits of their birth year. Kolya Olympionov had such a celebration in 2013, and Tolya Olympionov had one in 2014. Who is older and by how many years? | Solution. Let's determine in which year a person could be born if adding the sum of the digits of their birth year results in 2013 or 2014. It is clear that the year of such a birth cannot be later than 2014. Since the sum of the digits of each number from 1 to 2014 does not exceed \(1+9+9+9=28\), the birth year cannot... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,781 |
5. Karlson bought several pancakes (25 rubles each) and several jars of honey (340 rubles each) at the buffet. When he told Little Man how much he spent at the buffet, Little Man was able to determine, based on this information alone, how many jars of honey and how many pancakes Karlson bought. Could this amount have e... | Solution. Could. For example, let's say Karlson spent $4 \cdot 340 + 25 \cdot 40 = 2360$ rubles. Suppose Karlson can make up this amount in some other way; for this, he should spend x rubles less on pancakes and x rubles more on honey (or vice versa). But then, for x rubles, he can buy both a whole number of pancakes a... | 2360 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,782 |
6. The brothers found a treasure of gold and silver. They divided it so that each received 100 kg. The eldest received the most gold - 30 kg - and one fifth of all the silver. How much gold was in the treasure | Solution. 1) The elder brother received 70 kg of silver, which is one fifth of the total amount; therefore, the total mass of silver is 350 kg.
2) The others received more silver than the elder brother, i.e., each received more than 70 kg. If there are at least five brothers, then in total they would receive more than... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,783 |
3. A fly is sitting at vertex $A$ of a triangular room $A B C$ ( $\angle B=60^{\circ}, \angle C=45^{\circ}, A C=5$ m). At some point, it flies out in a random direction, and each time it reaches a wall, it turns $60^{\circ}$ and continues flying in a straight line (see figure). Can it happen that after some time, the f... | Solution. Let the fly take off at an angle of 60 degrees to the line $A C$. Consider the equilateral triangle $A K C$ with side $A C$. Note that its sides $A K$ and $K C$ can be divided into parts (into infinitely many parts) such that each part equals the next segment of the fly's trajectory. The sum of these parts is... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,785 |
4. Three students wrote a two-digit number on the board, each of which is a perfect square. It turned out that if you "glue" them into one six-digit number, it is also a square of a natural number. Find all such six-digit numbers. | Solution. Let the two-digit squares be $x^{2}, y^{2}$, and $z^{2}$, then $x, y, z \in\{4,5,6,7,8,9\}$. The six-digit number is denoted by $t^{2}(t>0)$. Then
$$
t^{2}=10000 x^{2}+100 y^{2}+z^{2}=(100 x)^{2}+(10 y)^{2}+z^{2}
$$
Let $t=100 x+k$. Obviously, $k \in \mathbb{N}$, since $t^{2}>(100 x)^{2}$. Then
$$
\begin{a... | 166464=408^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,786 |
5. Does there exist a natural number $x$ such that among the statements " $x+1$ is divisible by 2019 ", " $x+2$ is divisible by 2018 ", " $x+3$ is divisible by 2017 ", \ldots " $x+2017$ is divisible by 3 ", " $x+2018$ is divisible by 2 ", exactly half are true? | Solution. Note that the conditions can be replaced by the following: «$x+2020$ is divisible by 2019», «$x+2020$ is divisible by 2018», «$x+2020$ is divisible by 2017», … «$x+2020$ is divisible by 3», «$x+2020$ is divisible by 2». Thus, $x+2020$ must be divisible by half of the numbers $2,3, \ldots, 2019$. For example, ... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,787 |
1. Pasha and Igor are flipping a coin. If it lands on heads, Pasha wins; if tails, Igor wins. The first time the loser pays the winner 1 ruble, the second time - 2 rubles, then - 4, and so on (each time the loser pays twice as much as the previous time). After 12 games, Pasha is 2023 rubles richer than he was initially... | Answer: 9 (all except 4, 8, and 1024).
Solution. We need to place the signs in the equation $\pm 1 \pm 2 \pm 2^{2} \pm 2^{3} \pm \ldots \pm 2^{9} \pm 2^{10} \pm 2^{11}=2023$. If we choose all plus signs, the sum will be $2^{0}+\ldots+2^{11}=2^{12}-1=4095$, so we need to replace plus signs with minus signs before the n... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,788 |
2. Somewhere in the ocean, there is an island of Misfortune, on which several cities are located, connected by roads in such a way that a random tourist can get from any city to any other. It turned out that if any two cities are placed under quarantine and all roads leading to them are blocked, it is still possible to... | Solution. Not always. We will prove that for the example shown in the figure (vertical edges are chosen), this is not the case.
1) No matter which two roads we close, the graph will remain connected (after closing one road, it takes one of the two forms shown below; it is obvious that in each case, closing another roa... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,789 |
3. Solve the equation: $[20 x+23]=20+23 x$. Recall that $[a]$ denotes the integer part of the number, that is, the greatest integer not exceeding $a$. (S. S. Koresheva) | Solution. Let both sides of the equation be denoted by $n$, which is an integer. Then $x=\frac{n-20}{23}$ and $n \leqslant 20 x+23<n+1$. This means that
$$
n \leqslant \frac{20(n-20)}{23}+23<n+1
$$
which is
$$
n \leqslant 43<n+7 \frac{2}{3}
$$
It follows that $x=1-\frac{k}{23}$ for any integer $0 \leqslant k \leqsl... | \frac{16}{23},\frac{17}{23},\frac{18}{23},\frac{19}{23},\frac{20}{23},\frac{21}{23},\frac{22}{23},1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,790 |
4. Given a right triangle $ABC$ with a right angle at $A$. On the leg $AC$, a point $D$ is marked such that $AD: DC = 1: 3$. Circles $\Gamma_{1}$ and $\Gamma_{2}$ are then constructed with centers at $A$ and $C$ respectively, passing through point $D$. $\Gamma_{2}$ intersects the hypotenuse at point $E$. Circle $\Gamma... | Answer: 13.
Solution. Let $AC = x$. Then $AD = x / 4, DC = CE = 3x / 4, BE = BC - CE = \sqrt{x^2 + 25} - 3x / 4$. According to the problem, $\angle AFB = 90^\circ$, so $AF^2 + FB^2 = AB^2$, which means $AD^2 + BE^2 = 25$. Expressing everything in terms of $x$ and simplifying, we get $13x = 12 \sqrt{x^2 + 25}$. Squarin... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,791 |
5. Given six cards, on which the digits $1,2,4,5,8$ and a comma are written. From them, all possible numbers are formed (each card must be used exactly once, the comma cannot stand at the beginning or at the end of the number). What is the arithmetic mean of all such numbers?
(M. V. Karluкова) | Solution. First, let's assume there is no comma, that is, we find the sum of integers from the given digits. There are 5! = 120 ways to order the digits. When summing them, each of the five possible digits will appear 4! times in each place value, so the sum of the digits in the units place is $4! \cdot (1+2+4+5+8) = 2... | 1234.4321 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,792 |
6. On the coordinate plane, points $A(0,0)$ and $B(1000,0)$ were marked, as well as points $C_{1}(1,1)$, $C_{2}(2,1), \ldots, C_{999}(999,1)$. Then all possible lines $A C_{i}$ and $B C_{i}(1 \leqslant i \leqslant 999)$ were drawn. How many integer-coordinate intersection points do all these lines have? (An integer-coo... | Solution. Let $a_{n}$ and $b_{n}$ denote the lines passing through $A$ and $B$ respectively, as well as through a point on $l$ with an abscissa that is $n$ greater than the abscissa of $A$ (where $1 \leqslant n \leqslant 999$). The lines $a_{n}$ and $a_{m}$ for $n \neq m$ intersect at a non-integer point (between $A B$... | 2326 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,793 |
3. Find the number of five-digit numbers in which all digits are different, the first digit is divisible by 2, and the sum of the first and last digits is divisible by 3. | Solution. The first digit can be 2, 4, 6, or 8. If the first is 2, the last can be $1, 4, 7$; if the first is 4, the last can be $2, 5, 8$; if the first is 6, the last can be $0, 3, (6$ does not fit), 9; if the first is 8, the last can be $1, 4, 7$. In total, there are $3+3+3+3=13$ options for the first and last digits... | 4032 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,796 |
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