problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
6. Prove that for any selection of 68 different natural numbers, less than 100, among the selected numbers, there will be one that is equal to the sum of some three other selected numbers. | Solution. Let $\mathrm{a}<\mathrm{b}$ be the smallest two of 68 numbers, and $\mathrm{x}_{1}<\mathrm{x}_{2}<\ldots<\mathrm{x}_{66}$ be the rest. Since all numbers are distinct, then $\mathrm{b}<34$. Consider 66 numbers of the form $\mathrm{a}+\mathrm{b}+\mathrm{x}_{\mathrm{k}}$, as well as all 68 numbers $\mathrm{a}, \... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,439 |
7. In each cell of a $10 \times 10$ board, there sits a rabbit. Between rabbits in adjacent cells, there are partitions that can be removed. What is the minimum number of partitions that need to be removed so that any rabbit can visit any other rabbit, traveling through no more than 17 cells (not counting the starting ... | Solution. An example for 100 partitions is given in the figure. The correctness of the example follows from the fact that, as can be easily seen, any rabbit can reach the central 4-cell square by moving no more than 8 cells. From the central square to any cell, it can also be reached in no more than 8 moves. No more th... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,440 |
1. A checkered snake with a coloring pattern of "green cell - red - green - blue - green - red - green - blue and so on" has fallen onto a 2018 × 2018 checkered board. Two red cells, adjacent diagonally, were found. Prove that the snake turns in one of the green cells. | # Solution:
Assume that the snake only turns in red and blue cells. Let one of the two red cells mentioned in the problem have coordinates \((x, y)\). Then, the next blue cell (in the direction of the snake's movement) has one coordinate the same, while the other differs by 2, meaning the parity of each coordinate is ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,441 |
2. On the base $AE$ of the trapezoid $ABCE$, a point $D$ is chosen such that $S_{ABCD} = S_{CDE}$. It is known that $ABCD$ is a parallelogram, and its diagonals intersect at point $O$. A point $T$ is chosen on the segment $DE$. Prove that if $OT \| BE$, then $OD \| CT$. | Solution:
1) Since $A B C D$ is a parallelogram, then $A O=O C, B O=O D, A D=B C$.
2) From the formulas for the area of a parallelogram and a triangle, it follows that $D E=A D+B C$, hence $D E=2 A D$.
3) $B O=O D, O T \| B E$, therefore $D T=T E$ by Thales' theorem. Thus, $T$ is the midpoint of $D E$, and $D T=A D$.
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,442 |
4. Vasya, who you know from the first round, came up with $n$ consecutive natural numbers, for each he wrote down the sum of the digits, and as a result, he also got $n$ consecutive numbers (possibly not in order). For what maximum $n$ is this possible?
## SOLUTION: | Answer: when $n=18$.
Example: the numbers $392,393, \ldots 399,400,401, \ldots 409$ have digit sums of $14,15, \ldots, 21,4,5, \ldots, 13$, that is, all from 4 to 21.
Estimation. Let $n \geqslant 19$.
a) If all numbers are within the same hundred (...00-...99), then the difference in digit sums of 18 is only achieve... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,443 |
2. In the company, several employees have a total monthly salary of 10000 dollars. A kind manager proposes to triple the salary for everyone earning up to 500 dollars, and increase the salary by 1000 dollars for the rest, so the total salary will become 24000 dollars. A mean manager proposes to reduce the salary to 500... | Solution. Note that the increase proposed by the kind manager is twice as large as the salary proposed by the evil manager (this is true for both poor and rich employees). The increase according to the kind manager's proposal is 14000, so the salary according to the evil manager's proposal is 7000. | 7000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,444 |
5. On a notebook sheet, it is allowed to draw lines passing through at least two grid nodes, but not coinciding with the grid lines. Is it possible to draw three lines so that they form a triangle with an area of $1 / 3$ of a cell? | Solution. Yes, see the figure (in the figure, each "original" cell is divided into 9 smaller cells). The area of the triangle can be found by subtracting the area of everything else from the area of the rectangle.
}{2}$.
For even $i$, the sum $\frac{i(i+1)}{2}$ is divisible by $i+1$, which means 2017 must also be divisible by $i+1$. Since 2017 is a prime number, this is only possible when $i=2016$... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,448 |
7. Prove that a lattice polygon with an area of 300 and a perimeter of 300 has a side longer than 1. (The polygon does not have holes, that is, its boundary is a closed broken line without self-intersections.) | Solution. Let's calculate the number of grid segments located inside the figure. Note that each cell has 4 sides, so the quadrupled number of cells equals twice the number of internal segments plus the perimeter of the figure (if we count 4 sides for each cell, then the internal segments are counted twice, and the peri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,449 |
4. In the game of "new cups and balls," three different balls are placed under three small glasses, and then the glasses with the balls are somehow swapped so that no ball remains in its place. Initially, the balls were arranged as follows: red, blue, white. Can they end up in reverse order after the hundredth round?
... | # Solution:
In such a game, only three combinations are possible: "red, blue, white", "blue, white, red", and "white, red, blue". Any permutation allowed by the condition of the problem leads to one of these combinations. Therefore, the permutation "white, blue, red" cannot be obtained either after the hundredth round... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,452 |
6. There are 30 people in the bar. The bartender knows that among them, there are 10 knights (who always tell the truth), 10 liars (who always lie), and 10 troublemakers. The bartender can ask person $X$ about person $Y$: "Is it true that $Y$ is a troublemaker?" If $X$ is not a troublemaker, they will answer the questi... | Answer: 19. Indeed, with the first question, a peaceful client may be identified, so the bartender cannot guarantee to keep all peaceful clients.
Let's show how the bartender can leave 19 peaceful clients.
Solution 1. First, he asks everyone about client $A$ until someone throws him out of the bar. Let's say $B$ thre... | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,453 |
7. In a $100 \times 100$ square, 10000 cuts were made along the grid lines (each cut one cell long), and it split into 2500 four-cell figures. How many of them are $2 \times 2$ squares? | Solution:
Answer: 2300.
Solution. Note that the perimeter of a $2 \times 2$ square is 8, while for the other four-cell figures (rectangle $1 \times 4$, T-shape, L-shape, or S-shape) it is 10. Let the number of squares be $x$, and the number of other figures be $2500-x$. Then their total perimeter is $8 x + 10 \cdot (... | 2300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,454 |
10. The metro of city $\mathrm{N}$ consists of three lines, its scheme is shown in the figure. Initially, the trains are at the marked stations and start moving in the direction indicated by the arrow. Every minute, each train travels exactly one section between stations. Upon arriving at a terminal station, the train ... | 10. Note that the train takes exactly 7 minutes to travel the entire red branch, 8 minutes for the blue branch, and 9 minutes for the green branch. This means the trains return to their initial positions on the red branch every 14 minutes, on the blue branch every 16 minutes, and on the green branch every 9 minutes. Si... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,456 |
1. In a house, there are 300 apartments. In apartments whose numbers are divisible by 5, cats live, and in the rest of the apartments, there are no cats. If the sum of the digits of the apartment number is divisible by 5, then a dog definitely lives in such an apartment, and in the rest of the apartments, there are no ... | Solution. Let the apartment number be of the form $\overline{a b c}$, where $a, b, c$ are the digits of the apartment number (some of which may be equal to 0). The apartment number where the cat and the dog live must satisfy three conditions: | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,460 |
2. Does there exist such a grid figure from which exactly 7 cells can be cut out so that the remaining part does not fall apart into two pieces, and it is possible to cut out a single cell in seven different ways so that the remaining part falls apart? If two pieces touch only at a corner, they fall apart. | Solution. It exists. Here are examples:
There are other variants as well. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,461 |
3. Milla was writing the letters M and L in uppercase. At the end, she counted that the letter matched the previous one 59 times, and did not match 40 times. Determine the maximum number of letters M that Milla could have written, and prove that it is indeed the maximum.
| Solution. Answer: 80 letters.
For each letter except the first, it is known whether it matches the previous letter or not. Therefore, Milla wrote $1+59+40=100$ letters.
Divide all the written letters into groups of consecutive identical letters. Then, groups of M (M-groups) will alternate with groups of L (L-groups).... | 80 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,462 |
4. Is it possible to arrange the integers from 2 to 17 in a $4 \times 4$ table so that the sums in all rows are equal and no row contains two numbers where one divides the other? | Solution. No. Suppose we managed to arrange the numbers in the table according to the condition. Then the sum of all numbers in the table is $2+3+\ldots+17=152$. Therefore, the sum of the numbers in one row is $152: 4=38$. Consider the row in which the number 2 is located. In this row, there cannot be any other even nu... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,463 |
5. There are 20 chocolate candies on the table. Masha and Bear are playing a game according to the following rules. Players take turns. In one move, a player can take one or several candies from the table and eat them. Masha goes first, but on this move, she cannot take all the candies. In all other moves, players cann... | Solution. Masha will win if she eats 4 candies on her first move.
Let's arrange the candies in a row, number them, and assume that the players take candies in a row from left to right.
Suppose Masha eats one candy on her first move. In this case, she will lose, as until the end of the game, the players will take one ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,464 |
6. Nikita writes down numbers one after another according to the following rule: first, he writes down three natural numbers. Then he repeats the same procedure: he adds the last three numbers and appends the resulting sum at the end. Can Nikita write down 9 prime numbers in a row, acting this way?
| Solution. This is possible, and in a unique way:
$$
5,3,3,11,17,31,59,107,197 .
$$ | 5,3,3,11,17,31,59,107,197 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,465 |
7. Petya thought of a 9-digit number obtained by rearranging the digits of the number 123456789. Vitya is trying to guess it. For this, he chooses any 9-digit number (possibly with repeated digits and zeros) and tells it to Petya, who then responds with how many digits of this number match the ones in his thought numbe... | Solution. Vitya's first move is to name the number 122222 222. Then Petya can only respond with 0, 1, or 2.
If Petya answers 0, then neither the one nor the two hit their places. This means the two must be in the first position. There's no need to ask further.
If Petya answers 2, then both the one and the two hit the... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,466 |
3.1. Prove that among the numbers $1,3,9$ there are monochromatic ones. | 1. Prove that among the numbers $1,3,9$ there are monochromatic ones.
Let the numbers 1, 3, and 9 be of three colors $A, B$, and $C$ respectively.
Among the numbers 1, 2, and 3, there are two of the same color, so 2 is color $A$ or $B$. However, $1+1=2$, so 1 and 2 are of different colors. Therefore, 2 is color $B$.
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,467 |
3.4. Immortal Mikhail comes up with a new charming coloring every day. Prove that there is a charming coloring that he will never come up with. | 3. Provide an example of a charming coloring in which the numbers 20 and 30 are the same color.
Let's color all numbers that are congruent to 1 and 4 modulo 5 in color $A$, and those congruent to 2 and 3 modulo 5 in color $B$. Consider the equation $a+b=c$. It is easy to verify that if exactly one of the numbers is di... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,469 |
1.2. Let there be no such four troglodytes $A, B, C$ and $D$ that $A$ is friends with $B, C$ and $D$, while $B, C$ and $D$ are not friends with each other. Das found a sociable set of size $k$. Prove that Shakti can find a strange sociable set of size no more than $k$.
# | # Solution.
Consider a sociable set $A$ of the smallest size. Note that we can choose $A$ such that for each troglodyte $a$ in $A$, there exists a troglodyte $v_{a}$ whose only friend in the set $A$ is the troglodyte $a$. Indeed, otherwise, if $a$ is friends with someone in $A$, then it can simply be removed, and if i... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,471 |
1.3. Show that a strange sociable set of no more than 81 troglodytes may not exist.
# | # Solution.
Let there be 10 troglodytes who are friends with each other, denoted as $K$; and each of $K$ is also friends with nine other different remaining troglodytes. There are no other friendships. Any strange sociable set $A$ contains at most one troglodyte from $K$ (say, Vasya), and is forced to include all trog... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,472 |
1.4. Prove that Shakti will always be able to find a strange sociable set of no more than 82 troglodytes.
# | # Solution.
Consider a communicative set $A$ of the smallest size. Note that we can choose $A$ such that for each troglodyte $a$ in $A$, there exists a troglodyte $v_{a}$ whose only friend in the set $A$ is the troglodyte $a$. Indeed, otherwise, if $a$ is friends with someone in $A$, it can simply be removed, and if i... | 82 | Combinatorics | proof | Yes | Yes | olympiads | false | 21,473 |
3.1. Solve the inequality in natural numbers:
$$
n \leqslant n! - 4^n \leqslant 4n
$$ | # Solution.
We will prove by induction that $n!>4^{n}+4 n$. The base case $n=10$ is verified; the inductive step:
$$
(n+1)!>(n+1)\left(4^{n}+4 n\right)=4^{n+1}+16 n+(n-3)\left(4^{n}+4 n\right)>4^{n+1}+4(n+1)
$$
Thus, there are no solutions. | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,478 |
3.2. Find all natural $n$ and composite $k$ such that
$$
n \leqslant n!-k^{n} \leqslant k n
$$ | # Solution.
Notice that $n^{n / 2}+n^{3 / 2}9$, so it makes sense to look for solutions with $k$ in the range from $\sqrt{n}$ to $n / 2$.
Furthermore, if $k \leqslant n / 3$, then among the factors of $n!$ there will be $k, 2k$, and $3k$, meaning that $n!$ will be divisible by $k^{3}$. It is not hard to notice that t... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,479 |
3.4. Solve the inequality in natural numbers:
$$
n \leqslant n!-k^{n} \leqslant k n
$$ | # Solution.
Since the case of composite (in particular, even) $k$ has already been discussed in point 2, it remains, as in point 3, to consider the case $n!=k^{n}+k^{2}$. Notice that the left-hand side of the equation is divisible by $k-1$, while the right-hand side is congruent to 2 modulo $k-1$ (and $k>3$). This is ... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,481 |
1. Along the shore of a circular island, there are 5 huts: A, B, V, G, D (in that order). The shortest distances along the shore between some of them are as follows: between A and V - 16 km, between B and G - 18 km, between V and D - 22 km, between G and A - 23 km, between D and B - 18 km. Provide an example where the ... | Solution. There are only two examples:
| | AB | BC | CD | DA | AC |
| :---: | :---: | :---: | :---: | :---: | :---: |
| I | 5 km | 11 km | 7 km | 15 km | 13 km |
| II | 12 km | 4 km | 14 km | 17 km | 6 km | | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,482 |
3. An Indian engineer thought of a natural number, listed all its proper natural divisors, and then increased each of the listed numbers by 1. It turned out that the new numbers are all the proper divisors of some other natural number. What number could the engineer have thought of? Provide all options and prove that t... | Solution. Let the original number be $n$, and the "some other number" be $m$. The number $m$ is odd, because an even number has a divisor 2, which could not have resulted. Therefore, all its divisors are odd. This means all divisors of $n$ are even. Thus, $n$ is a power of two.
$n=4$ and $n=8$ work, corresponding to $... | n=4n=8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,484 |
4. The number 100 is represented as the sum of several two-digit numbers, and in each addend, the digits are swapped. What is the largest number that could result from the new sum? | Solution. If the digits in the two-digit number $\overline{b a}$ are swapped, the number increases by $9(a-b)$. Therefore, the new sum is $S=100+9 U-9 D$, where $D$ is the sum of the tens digits, and $U$ is the sum of the units digits in the original addends. Since $10 D+U=100$, then $S=1000-99 D$. Thus, we need to min... | 406 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,485 |
6. Given a string of 2021 letters A and B. Consider the longest palindromic substring. What is its minimum possible length? A palindrome is a string that reads the same from right to left and from left to right. | Solution. The minimum possible length of the maximum palindrome is 4.
We will prove that it cannot be less than 4. Consider the 5 letters in the center of the string. If these are alternating letters, then it is a palindrome of length 5. Suppose among these five letters there are two identical letters standing next to... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,487 |
7. At a round table, 99 gnomes are sitting. The hobbit Bilbo knows all the gnomes, but he cannot see how they are seated because his eyes are blindfolded. Bilbo can name any two gnomes, and all the gnomes will answer in unison how many gnomes are sitting between these two gnomes (along the shortest arc). Can Bilbo find... | Solution. Bilbo asks the first 49 questions about some dwarf $A$ and any 49 other dwarfs. If he gets an answer of "0" to at least one of the questions, his goal is achieved. Otherwise, he receives answers ranging from 1 to 48 (with each answer appearing no more than twice). Each time after receiving an answer, Bilbo cr... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,488 |
3. On the bus route, there are only four stops - "Initial", "First", "Final", and "Last". At the first two stops, passengers only got on, and at the remaining stops, they only got off. It turned out that 30 passengers got on at the "Initial" stop, and 14 passengers got off at the "Last" stop. At the "First" stop, three... | Answer: Those traveling from "First" to "Final" are six more.
Let $x$ be the number of people who got on at "First". Then, $3x$ people got off at "Final". Since the number of people getting on equals the number of people getting off, we have $30 + x = 3x + 14$, from which $x = 8$. Let $y$ be the number of people trave... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,489 |
4. Baron Munchausen placed a horse in some cells of an $N \times N$ board. He claims that no one will find two different $4 \times 4$ squares on this board (with sides along the grid lines) with the same number of horses. For what largest $N$ can his words be true?
# | # Solution:
Answer: $N=7$.
The number of knights in a $4 \times 4$ square can range from 0 to 16, i.e., there are 17 possible variants. The number of $4 \times 4$ squares on an $N \times N$ board is $(N-3)^{2}$ (since the top-left cell of the square can occupy positions from the far left to the fourth from the right ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,490 |
5. Is it possible to divide the numbers from 1 to 80 into fours such that in each four, the largest number equals the sum of the other three?
# | # Solution:
Answer: No, it is not possible.
Method 1. Suppose it is possible. Note that there are only 20 groups. Consider the sum of the largest numbers in all groups. This sum is no more than the sum of the 20 largest numbers, i.e., $61+62+\ldots+79+80=(61+80)+(62+79)+\ldots+(70+71)=141 \cdot 10=1410$. This is less... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,491 |
6. Pompous Vova has an iPhone XXX, and on that iPhone there is a voice-command calculator: "Multiply my number by two and subtract two from the result," "Be so kind as to multiply my number by three and then add four," and finally, "Add seven to my number!" The iPhone knows that Vova initially had the number 1. How man... | Answer: 9000 (or 18000, if negative numbers are considered).
We will prove that we can obtain all four-digit numbers. Note that the command +7 allows us to obtain from the current number all larger numbers with the same remainder when divided by seven. Therefore, it is sufficient to use the first two buttons to obtain... | 9000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,492 |
7. In the detachment, there are one hundred people, and each has three friends in the detachment. For duty, it is required to appoint groups of three people, each of whom is friends with each other. For 99 consecutive days, it was possible to appoint such groups of three without repeating them. Prove that this will als... | Solution:
Method 1. Suppose it is not possible, i.e., all possible triplets have already taken shifts. Consider any duty triplet $A, B, C$. Suppose someone from them has taken a shift with someone not from this triplet (denote them as $A$ and $D$). Since each person in the group has only three friends, $A$ has no frie... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,493 |
1. Winnie-the-Pooh decided to give Eeyore a pot of honey. On the way to Eeyore, he tried the honey from the pot several times. When he tried the honey for the first time, the pot became half as heavy. And after the second time, the pot became half as heavy again. And after the third! And after the fourth! Indeed, after... | Solution. $\quad$ Answer: 3000 g. Indeed, $3000=(((200 \cdot 2) \cdot 2) \cdot 2) \cdot 2)-200$. | 3000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,494 |
2. Petya has several 5-ruble coins and several 2-ruble coins. Vanya has as many 5-ruble coins as Petya has 2-ruble coins, and as many 2-ruble coins as Petya has 5-ruble coins. Petya has 60 rubles more than Vanya. Which coins does Petya have more of - 5-ruble or 2-ruble? By how many? | Solution. Answer: Petya has 20 more 5-ruble coins.
Suppose Petya has more 2-ruble coins than 5-ruble coins. Let Petya remove one 2-ruble coin, and Vanya remove one 5-ruble coin. Then the difference in the sums of Petya's and Vanya's coins will increase by 3. If they repeat this operation until Petya has an equal numbe... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,495 |
3. Draw a $4 \times 4$ square grid in your notebook. In the cells of this grid, draw 8 diagonals such that no cell contains more than one diagonal, the diagonals do not share endpoints, and no more diagonals can be added while maintaining these rules. | Solution. For example, like this:
 | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,496 |
5. Vasya wrote down several different numbers in his notebook, and then started writing down some of their pairwise sums. When he finished, it turned out that each originally written number was included in exactly 13 sums. Could the sum of all the written numbers be equal to $533$? | Solution. Each initially written number appears in the final sum 14 times: once on its own and 13 times as part of a pairwise sum. Therefore, 533 should be divisible by 14, but it is not. Therefore, the sum of 533 could not have been obtained. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,498 |
7. There are several (more than two) stacks of paper with different numbers of sheets. If there is a pair of stacks with a different number of sheets, then a sheet of paper can be removed from any other stack. Is it possible to make it so that all stacks have the same number of sheets? | Solution. Suppose this is possible. Then, with the last move, we removed a sheet from some stack, and the number of sheets in all stacks became equal. This means that before we removed the last sheet, the number of sheets in all stacks, except for the stack from which the sheet was removed, was equal. However, in order... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,500 |
1. The Ministry of Truth announced that employment in Oceania fell by $15\%$ from the previous level in January, while unemployment increased by $10\%$ from the previous level. What is the current unemployment rate in Oceania, according to the Ministry's statement? (Employment is the proportion of the working-age popul... | Solution. Let unemployment be $x$ percent, then employment is $100-x$ percent. According to the statement, these shares have turned into $1.1 \cdot x$ and $0.85 \cdot(100-x)$ respectively, and still sum up to 100 percent. Then we get the equation:
$$ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,501 | |
2. The sum of the factorials of three consecutive natural numbers is divisible by 61. Prove that the last of these numbers is no less than 61. (The factorial of a number $n$ is the product of all numbers from 1 to $n$ inclusive.) | Solution. Let the last of the numbers be $k$. Then the sum of the factorials can be rewritten as follows:
$$
(k-2)!+(k-1)!+k!=(k-2)!\cdot(1+(k-1)+(k-1) \cdot k)=(k-2)!\cdot k^{2}=1 \cdot 2 \cdot \ldots \cdot(k-2) \cdot k^{2}
$$
Since 61 is a prime number, one of the numbers $1,2, \ldots, k-2, k$ must be divisible by ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,502 |
4. Prince George has 100 coins, some of which may be counterfeit (possibly all or none). George can show the expert between 10 and 20 coins, and the expert will tell him how many of them are counterfeit. The problem is that the only expert in the area is Baron Münchhausen, and he exaggerates: the result given by the ba... | Solution. It will work. We will submit $X$ random coins for examination, and then the same $X$ plus 1 more. If the baron says the same number both times, the added coin is genuine; otherwise, it is counterfeit.
Thus, we can divide the coins into groups of 10, and check each group for counterfeits in 11 queries: first,... | 110<120 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,504 |
5. Two players take turns placing non-overlapping dominoes on a $2020 \times 2020$ board, each covering two cells. The second player's goal is to cover the entire board with dominoes, while the first player's goal is to prevent this. Who can ensure a win? | Solution. The first player will win. For this, he can play, for example, as follows. On his first move, he places a domino on cells a2 and a3. The second player does not want to get a domino b1-b2 or b1-c1 on the board, as both would isolate cell a1. He cannot cover cells b2 and c1 simultaneously, so he must cover b1, ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,505 |
6. In the Lemon Kingdom, there are 2020 villages. Some pairs of villages are directly connected by paved roads. The road network is arranged in such a way that there is exactly one way to travel from any village to any other without passing through the same road twice. Agent Orange wants to fly over as many villages as... | Solution. It will not be possible to fly over all villages if some village is connected by roads to all others (
However, it is possible to fly over all villages except one. Let's form a flight plan: choose the first village randomly, and each subsequent one randomly from those that can be flown to from the previous o... | 2019 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,506 |
7. Misha came up with two composite numbers: \(a\) and \(b\). On the board, in the left column, he listed all the proper natural divisors of the number \(a\), and in the right column, all the proper natural divisors of the number \(b\). There were no identical numbers on the board. Misha wants the number \(a + b\) to n... | Solution. For each divisor $c$ of the number $a$, there is a divisor $a / c$. If $c \neq a / c$, we will strike out the smallest of them from the corresponding column. Since no more than one number is struck out from each pair of the form $c, a / c$, a total of no more than half of the numbers in the column are struck ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,507 |
2. Aseya divided Vasya's favorite number by her favorite number, and Buseya divided Vasya's favorite number by her favorite number. Then both girls wrote on the board the divisor, the quotient, and the remainder. The five numbers on the board are 2020, 2020, 2021, 2021, 2021. Can the sixth number be determined uniquely... | Solution. No. Let's consider two options for what the favorite numbers of the kids could have been:
| | Vasya | Asya | Buseya |
| :---: | :---: | :---: | :---: |
| Option 1 | $2021 \cdot 2021+2020$ | 2021 | 2021 |
| Option 2 | $2021 \cdot 2021+2020=2022 \cdot 2020+2021$ | 2021 | 2022 |
It is clear that in the first ... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,508 |
3. The Elector George has 100 coins, some of which are counterfeit (possibly all or none). George can show the expert from 10 to 20 coins, and the expert will tell him how many of them are counterfeit. The problem is that the only expert in the entire region is Baron Münchhausen, and he exaggerates: the result given by... | Solution. It will work. We will submit $X$ random coins for expertise, and then the same $X$ plus 1 more. If the baron says the same number both times, the added coin is genuine; otherwise, it is counterfeit. This way, we can divide the coins into groups of 10, and check each group for counterfeits in 11 queries: first... | 110 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,509 |
4. There are 50 boys and 50 girls standing in a row in some order. In this row, there is exactly one group of 30 children standing consecutively, with an equal number of boys and girls. Prove that there will be a group of 70 children standing consecutively, in which the number of boys and girls is also equal. | Solution. We will arrange the children in a circle. Now let's look at all possible groups of 30 children standing in a row (we will call such groups squads). It is sufficient to prove that there are at least two of them with an equal number of boys and girls. Indeed, then when breaking the cycle back into a row, one of... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,510 |
5. In triangle $A B C$, angle $B$ is right. On side $B C$, the midpoint $M$ is marked, and on the hypotenuse, a point $K$ is found such that $A B=A K$ and $\angle B K M=45^{\circ}$. Additionally, on sides $A B$ and $A C$, points $N$ and $L$ are found respectively, such that $B C=C L$ and $\angle B L N=45^{\circ}$. In w... | Solution. Let $\angle B A C=\alpha$. Then, since $\triangle A B K$ is isosceles with base $B K$, $\angle A B K=\angle A K B=90^{\circ}-\frac{\alpha}{2}$. Due to the right-angled nature of $\triangle A B C$, the angle $\angle A C B$ is $90^{\circ}-\alpha$, and since $\triangle C B L$ is isosceles with base $B L$, $\angl... | AN:BN=1:2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,511 |
6. Two players take turns placing non-overlapping dominoes on a $2021 \times 2021$ board, each covering two cells. The second player's goal is to cover the entire board with dominoes, except for one cell, while the first player's goal is to prevent this. Who can ensure a win? | Solution. The first player will win.
First, let's show how to capture one cell. For this, the first player can play, for example, as follows. The first move is to place a domino on cells a2 and a3. The second player does not want to get a domino on b1-b2 or b1-c1, as both would isolate cell a1. The second player canno... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,512 |
1. At each meeting of the Numismatists Club, four coins are considered and the most expensive and the cheapest of them are determined. Alice brought five ancient coins of different values to the Club. How can the coin of medium value among these coins be determined in three meetings of the Club? | # Solution:
## Method 1.
Notice that the coin of median value is neither the most expensive nor the cheapest in any quartet.
1st session: consider any quartet of coins. This way, we identify 2 coins that are definitely not the median.
2nd session: replace the most expensive coin from the first quartet with a previo... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,513 |
3. A program for each four-digit number prints the product of its digits. Which numbers will be printed by the program exactly once?
## SOLUTION: | Answer: $1,5^{4}(625), 7^{4}(2401), 8^{4}(4096), 9^{4}(6561)$.
It is clear that 0 will be printed several times. Note that the permutation of digits in a number does not change its product, so only once can be printed the products of the digits of numbers with the same digits. It is not hard to see that the products o... | 1,625,2401,4096,6561 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,514 |
5. Arrange the numbers from 1 to 202 in a row so that the following condition is met: any two numbers, between which there are at least 100 other numbers, differ by no more than 100.
# | # Solution:
Arrange the numbers as follows:
$(102 ; 103105107 \ldots 199201 ; 1009896 \ldots 64 ; 1 ; 2 ; 357 \ldots 9799 ; 202200198 \ldots 106104 ; 101)$
or differently | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,515 |
7. The board had the number 1 written on it. The restless Sasha gets a new natural number every minute, trying to subtract $2^{100}$ from the last number written. If it turns out that the result has already been written before, or that the current number is less than $2^{100}$, then Sasha adds $3^{100}$. Will the numbe... | # Solution:
Answer: Yes (as with any other natural number).
We will track the remainder of the current number when divided by $2^{100}$. If we get a number with the same remainder as $5^{100}$, but greater than $5^{100}$, then the number $5^{100}$ will also appear.
Since if a number $a$ appears in the sequence, then... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,516 |
2. Through the center $O$ of the circle circumscribed around the regular pentagon $A B C D E$, lines parallel to the sides $A B$ and $B C$ are drawn. The points of intersection of these lines with $B C$ and $C D$ are denoted as $F$ and $G$ respectively. What is the ratio of the areas of quadrilateral $O F C G$ and pent... | # Solution.
Draw other lines through point $O$ parallel to the sides of the pentagon. We see that $A B C D E$ consists of five quadrilaterals equal to $O F C G$. | \frac{1}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,518 |
3. In each cell of a $70 \times 70$ board, there is a knight or a liar. Knights always tell the truth, and liars always lie. Each of them said: "There are as many knights in my row as there are in my column." Can there be exactly 2021 knights on the board? | # Solution.
Notice that the problem does not change with the permutation of rows and columns. We will assume that the first $n$ cells in the top row are knights, and the rest are liars. Similarly, we will assume that the first $n$ cells in the first column are knights (since the top-left corner is a knight), and the r... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,519 |
4. In the country, each city is connected to three other cities by bus routes. Initially, all routes were state-owned. Some routes were transferred to two private companies. Now, to travel any closed route, services of both private companies are required. Prove that there will be a city from which it is impossible to l... | # Solution.
Consider a graph: vertices - cities, edges - routes. Let there be $n$ cities, then there are 3n/2 routes. Remove the routes of the first company, we get that in the remaining graph there are no cycles, which means there are no more than $n-1$ edges. Therefore, the first company has no less than $n / 2+1$ e... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 21,520 |
5. Let $a_{1}, a_{2}, \ldots, a_{2021}, b_{1}, b_{2}, \ldots, b_{2021}$ be pairwise distinct natural numbers. Consider the graphs of functions of the form
$$
y=\frac{a_{i}}{x+b_{i}}
$$
(2021 functions in total). Can it happen that the abscissas of all intersection points of these graphs are integers? | # Solution.
It can. Consider some functions of the specified form with pairwise distinct coefficients. Find the abscissa of the intersection point. Solve the equation $\frac{a_{i}}{x+b_{i}}=\frac{a_{j}}{x+b_{j}}$, we get $x=\frac{b_{i} a_{j}-a_{i} b_{j}}{a_{i}-a_{j}}$. Let $N-$ be the LCM of all numbers $a_{i}-a_{j}$.... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,521 |
6. In triangle $A B C$, points $X$ and $Y$ are marked such that rays $A X, C Y$ intersect on the extensions of segments $A X$ and $C Y$ and are perpendicular to lines $B Y$ and $B X$ respectively. The sum of the distances from $X$ and $Y$ to line $A C$ is less than the height $B H$. Prove that $A X + C Y < A C$. | # Solution.
The condition on the sum of distances can be reformulated as follows: the midpoint $M$ of segment $X Y$ hangs at a height less than $h / 2$, where $h=B H$. This means that if we mark a point $N$ such that $X B Y N$ is a parallelogram, then $X N$ and $Y N$ will intersect $A C$. If $X N$ and $Y N$ intersect ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,522 |
7. On a circle, 100 different points are chosen. Peter and Ekaterina are playing a game. In the first move, Peter chooses three triangles with vertices at the chosen points, and then each player takes turns choosing one such triangle. At any point, all the chosen triangles must have a common interior point, and triangl... | # Solution.
Notice that for any point lying inside the hundred-gon formed by the selected points, but not on the diagonals, there exists an even number of triangles with vertices at the vertices of the hundred-gon that contain this point. Indeed, let's move such a point inside the hundred-gon. When we cross some diago... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,523 |
2. Which numbers are more among the first trillion natural numbers: valid or invalid?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 2. Note that among the first trillion natural numbers $10^{12}-9^{12}$ contain the digit nine in their representation, of which no more than $10^{12} / 9$ are divisible by 9. Therefore, at least $\mathrm{N}=10^{12}-9^{12}-10^{12} / 9$ contain the digit nine and are not divisible by it, i.e., are defective. Note that $\... | Number Theory | proof | Yes | Yes | olympiads | false | 21,524 | |
3. Prove that among any 20 consecutive numbers, there will be both a suitable and an unsuitable one. | 3. Among twenty numbers, two end in 9; neither of them can be divisible by 9, so one of them is invalid.
Why can't all 20 numbers be invalid? Note that among the 20 numbers, there is a complete set of ten $(\ldots 0, \ldots 1, \ldots, \ldots 9)$. Let $k$ be the highest digit in the first of these numbers (...0); then ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,525 |
4. There are several cards. On each of them, on each of the two sides, a circle is drawn: red, blue, or yellow. Among any 30 cards, there is a card with a red circle, among any 40 cards, there is a yellow circle, and among any 50, there is a blue one. In total, there are 20 cards with circles of different colors. Prove... | # Solution:
There are no more than 29 cards without red circles, no more than 39 cards without yellow, and no more than 49 cards without blue. Adding all these numbers, we get no more than 117. In this process, single-colored cards have been counted twice, and two-colored cards have been counted once. Therefore, the d... | 48 | Combinatorics | proof | Yes | Yes | olympiads | false | 21,527 |
7. Kopyatych took some natural number, raised it to the 1st, 2nd, 3rd, 4th, and 5th powers. Then he encrypted the numbers by replacing the same digits with the same letters and different digits with different letters. He wrote each encrypted number on a separate piece of paper. But Nyusha left only a fragment of each p... | # Solution:
It is clear that the letters К, Ё, Ж, И are of the same parity, and among them there are no 0 or 5.
If К, Ё, Ж, И are even, then all powers except the first one must be divisible by 4, and then О, Ш are also even digits, which is not good.
The last digit of the number in the first and fifth power is the ... | 189 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,528 |
2. A circle is divided into 1000 sectors, all of which are white. Every minute, some 500 consecutive sectors are repainted - white sectors become red, and red sectors become white. At a certain repainting, the number of white sectors did not change. Prove that for one of the neighboring (previous or next) repaintings -... | # Solution.
Consider any pair of diametrically opposite sectors. On each move, exactly one sector from this pair is repainted, so after each odd repainting they are of different colors, and after each even - the same color. In particular, after each odd move, there is exactly one white sector in each such pair, i.e., ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,530 |
3. We remind you that the sum of the angles in a triangle equals 180 degrees. In triangle $A B C$, angle $A$ is a right angle. Let $B M$ be the median of the triangle, and $D$ be the midpoint of $B M$. It turns out that $\angle A B D = \angle A C D$. What are these angles? | # Solution.
Draw $A D$. Since triangle $A B M$ is a right triangle, its median $A D$ equals half the hypotenuse $B M$. Therefore, $\angle A B D = \angle B A D = \alpha$, and $\angle A D M$ equals $2 \alpha$, as the exterior angle of triangle $A B D$. However, since $D M = A D$, $\angle D A M = \angle D M A = 90^{\circ... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,531 |
4. Joe has 99 coins of three types: light coins weighing 1 gram, medium coins weighing 2 grams, and heavy coins weighing 3 grams. Each type of coin is stored in a box labeled, respectively, LIGHT, MEDIUM, and HEAVY. One night, Amy relabeled the boxes so that now none of the labels match the contents of the boxes. Show ... | # Solution.
Each pan should receive $99 \cdot 3$ grams. Let $a$ LIGHT coins, $b$ MEDIUM, and $c$ HEAVY coins land on the left pan. Then
$$
\begin{aligned}
2 a+3 b+c & =3 a+b+2 c=297 \\
2(99-a)+3(99-b)+(99-c) & =3(99-a)+(99-b)+2(99-c)=297
\end{aligned}
$$
Any solution to this system will work, for example, $a=82, b=4... | =82,b=43,=4 | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 21,532 |
5. Find all natural numbers whose proper divisors can be divided into pairs such that in each pair the numbers differ by 545. A proper divisor of a natural number is a natural divisor other than 1 and the number itself.
# | # Solution.
Let $n$ be such a number. In each pair of divisors, there is one of different parity, meaning there is an even one, so there is a divisor 2. Then there is a divisor 547. On the other hand, if $d$ is the greatest proper divisor, which is $n / 2$, then $d-545$ does not exceed $n / 3$. We get that
$$
\frac{n... | 2\cdot547 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,533 |
6. Prove that on a $533 \times 533$ board, colored in a checkerboard pattern, there exist monochromatic cells A and B with the following property: "the number of ways to tile the board without cell B with dominoes is not equal to the number of ways to tile the board without cell A with dominoes." Ways that differ by ro... | # Solution.
Let's introduce notation similar to chess notation. Notice that we can remove both $a 1$ and $b 2$. Then, in the tiling without $b 2$, the domino covering $a 1$ can be rotated to now cover $b 2$. Thus, it is clear that different tilings without $b 2$ correspond to different tilings without $a 1$, so there ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,534 |
1. The figure shows one side of a five-story building, in some of the windows the light is on (they are drawn in white). In this building everything is as usual: all apartments are one-story, the layout of apartments on each floor is the same, each apartment has at least one window on this side, windows from one apartm... | # Solution.
(a) 25. This number of apartments is obtained when each apartment has exactly one window facing the given side of the house. There cannot be more apartments than windows.
(b) No, it cannot, because the number of apartments must be divisible by the number of floors.
(c) If the dark window on the first flo... | No | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,535 |
2. In this problem, we consider figures drawn on graph paper. Each such figure consists of whole cells glued together along their sides, forming a single piece. If some cells are cut out of the figure, it may fall apart into several pieces (when two pieces touch only at a corner, they fall apart).
(a) Provide an examp... | # Solution.
(a) For example, a $2 \times 2$ square. Or a rectangle $n \times m$, where $n \geqslant 2$ and $m \geqslant 2$.
(b) Here are some examples:
There are other variants as well.
(... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,536 |
3. In the Magic and Wizardry club, all first and second-year students wear red robes, third-year students wear blue, and fourth-year students wear black.
Last year, at the general assembly of students, there were 15 red, 7 blue, and several black robes, while this year - blue and black robes are equal in number, and r... | # Solution.
(a) 7 mantles. Since 7 blue mantles from last year will turn black this year and there are an equal number of blue and black mantles this year, there will also be 7 blue mantles this year, which will turn black next year.
(b) 6 students. This year, there are 7 blue, 7 black, and $14=2 \times 7$ red mantle... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,537 |
4. Let in a harmonious country there be $n$ schools, each of which is connected by non-stop routes to exactly $d$ others. Prove that $d < 2 \sqrt[3]{n}$.
# | # Solution:
Notice that if we hang the graph from an arbitrary vertex $v$, the number of edges from the second to the third level will be $d(d-1)$, since the graph cannot contain triangles. Now notice that if three such edges lead from vertices $u_{1}, u_{2}, u_{3}$ to vertex $w$, then the medians of vertices $u_{1}, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,538 |
2. Can the sequence $(-3,-1,1,3)$ be transformed into $(-3,-1,-3,3)$ (numbers in exactly this order)
## SOLUTION:
Translating the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Answer: No.
Let some sequence of polynomials transform $(-3,-1,1,3)$ into $(-3,-1,-3,3)$. It is not hard to see that a single such cubic polynomial does not exist (this can be understood by substituting all conditions and solving a linear system of equations, or by noting that such a polynomial must have the form $x+a... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 21,539 |
2. The Elector George has 100 coins, some of which may be counterfeit (possibly all or none). George can show the expert between 10 and 20 coins, and the expert will tell him how many of them are counterfeit. The problem is that the only expert in the area is Baron Munchausen, and he exaggerates: the result given by th... | Solution. It will work. We will submit 10 random coins for examination, and then the same 10 and one more. If the baron says the same number both times, the added coin is genuine; otherwise, it is counterfeit. Thus, in two questions, we can check any coin for being counterfeit.
Criteria. Full solution - 4 points. 1 po... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,541 |
3. The stubborn robot "Inverter" stands on an infinite plane and faces east. This robot understands only two commands: STEP and LEFT. When the robot sees the command STEP, it moves forward exactly 1 meter. When the robot sees the command LEFT, it turns left exactly $90^{\circ}$ while remaining in place. The robot is ca... | Solution. Example program for the stubborn robot: Nnnnshhnn. There are many other examples.
Criteria. 4 points if the correct algorithm is provided. 0 points in all other cases. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,542 |
4. Each of five friends multiplied several consecutive numbers starting from 1. It turned out that one of the products is equal to the sum of the other four. Find all possible values of this product and show that there are no other values. | Solution. The product of all consecutive numbers from 1 to $x$ is called the factorial of the number $x$ and is denoted by $x!$. Thus, we need to solve the equation $x!=a!+b!+c!+d!$. Let the numbers $a, b, c, d$ be ordered in ascending order. Then $x>d$, i.e., $x! \geqslant x \cdot d!$. But we know that $x! \leqslant 4... | 24 | Number Theory | proof | Yes | Yes | olympiads | false | 21,543 |
5. At a round table, 8 gnomes are sitting, each with three diamonds. Every minute, the gnomes simultaneously do the following: they divide all their diamonds into two piles (possibly one or both piles are empty), then give one pile to their left neighbor and the other to their right neighbor. At some point, all the dia... | Solution. Let's number the gnomes in order. Notice that gnomes with even numbers always share with gnomes with odd numbers, and vice versa. Since all distributions happen simultaneously, all the diamonds of the gnomes with even numbers will go to the gnomes with odd numbers and vice versa. That is, the gnomes with even... | 512 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,544 |
6. In a $4 \times 4$ square, cells are colored in several colors such that in any $1 \times 3$ rectangle, there are two cells of the same color. What is the maximum number of colors that can be used? | Solution. Maximum 9 colors. See example in the picture. We will prove that more is not possible. Any row (row or column) gives a maximum of three colors, so the first row + first column will give a maximum of $3+3-1=5$ colors.
We will prove that the remaining $3 \times 3$ square will give a maximum of 4 colors. Indeed... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,545 |
7. In a row, there are 50 boys and 50 girls standing in some order. In this row, there is exactly one group of 30 children standing in a row, in which there are an equal number of boys and girls. Prove that there will be a group of 70 children standing in a row, in which there are also an equal number of boys and girls... | Solution. We will arrange the children in a circle. Now let's look at all possible groups of 30 children standing in a row (we will call such groups squads). It is sufficient to prove that there are at least two of them with an equal number of boys and girls. Indeed, then when breaking the cycle back, one of these two ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,546 |
1. Captain Billy the pirate plundered 1010 gold doubloons and set sail on his ship to a deserted island to bury them as treasure. Every evening of the voyage, he paid each of his pirates one doubloon. On the eighth day of the voyage, the pirates plundered a Spanish caravel, and Billy's treasure doubled, while the numbe... | Solution. Answer: 30
Before the pirates looted the caravel, Billy managed to pay the pirates their daily wages 7 times. After this, his fortune doubled. This is equivalent to Billy having 2020 doubloons before the voyage, and he paid the wages 14 times. After this, Billy paid the wages 40 times to half of the remainin... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,547 |
2. Is it possible to write down a sequence of 10 integers such that the product of any three consecutive numbers is divisible by 6, while the product of any two consecutive numbers is not divisible by 6? | Solution: No.
Firstly, none of the listed numbers can be divisible by 6, because the product of this number and any of its neighbors will be divisible by 6. Suppose the first number is not divisible by 2 or 3. Then the product of the first three numbers will not be divisible by 6, since the product of the second and t... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,548 |
3. Krosch, Yozhik, Nyusha, and Barash ate a bag of candies. Then Sovunya asked: “Well, how many candies did each of you eat?” To which the kids replied as follows:
Krosch: “Yozhik, Nyusha, and I together ate only 120 candies.”
Yozhik: “Nyusha and Barash ate 103 candies (I've been watching them for a while).”
Nyusha:... | Solution. In all three statements, each of the boys is mentioned exactly twice. Therefore, the sum $120+103+152$ should be even. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,549 |
4. The length of the escalator is 200 steps. When Petya walks down the escalator, he manages to count 50 steps. How many steps will he count if he runs twice as fast? | Solution. 80 steps. Let's call the step of the escalator from which Petya begins his descent the first step. When Petya walks down, he passes 50 steps. During this time, the first step manages to descend $200-50=150$ steps. Therefore, the escalator moves three times faster than Petya walks. When Petya runs, the speed r... | 80 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,550 |
7. In the company, there are elves, fairies, and gnomes. Each elf is friends with all fairies except for three, and each fairy is friends with twice as many elves. Each elf is friends with exactly three gnomes, and each fairy is friends with all gnomes. Each gnome is friends with exactly half of the elves and fairies c... | Solution. Answer: 12.
Let $n$ be the number of elves, $m$ be the number of fairies, and $k$ be the number of gnomes. Then the number of friendly pairs "elf-fairy" is $n(m-3)$, and "fairy-elf" is $-m \cdot 2(m-3)$. But these are the same pairs, therefore,
$$
n(m-3)=2 m(m-3)
$$
From which, $n=2 m$. Counting the friend... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,553 |
3. The set contained weights of 5, 24, and 43 grams, with an equal number of each type. All the available weights were weighed, and their total mass was found to be $606060 \ldots 60$ grams. Prove that more than 10 weights are lost. | Solution. If no weight has been lost, the total weight should be divisible by 24, but it has a remainder of 12 when divided by 24 (since it is divisible by 3 and 4, but not by 8). Therefore, the total weight of the lost weights also has a remainder of 12 when divided by 24. We need to understand how to achieve a remain... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,555 |
4. In a convex quadrilateral $ABCD$, angle $A$ is $40^{\circ}$, angle $D$ is $45^{\circ}$, and the bisector of angle $B$ bisects $AD$. Prove that $AB > BC$. | Solution. Consider a point $T$ on the extension of side $A B$ beyond point $B$ such that $B T = B C$. Let $M$ be the midpoint of $A D$. Then $\angle B T C = \angle T C B = \frac{1}{2} \angle A B C$, which means $B M$ is parallel to $T C$. Denote by $Q$ the intersection point of lines $T C$ and $A D$; we will show that ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,556 |
6. In opposite corners of a chessboard stand knights. Two players take turns cutting out free squares from the board. The player who, after their move, leaves one knight unable to reach the other across the board loses. Which player can always win, regardless of the other's moves? | Solution. Note that at the moment before the last move, we will have only the cells forming a path between the knights, otherwise, we could make more than one move. Since the cell under the knight changes color when the knight moves, all paths between the knights consist of an odd number of cells. Therefore, before the... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,557 |
7. Find all primes $p$ and natural numbers $n$ that satisfy the equation $p^{2}+n^{2}=3 p n+1$. | Solution. Obviously, $n^{2}-1=p(3 n-p)$, that is, $n$ is equal to $\pm 1$ modulo $p$. Substitute $n=a p \pm 1$ into the equation, after transformations we get
$$
\left(a^{2}+1-3 a\right) p^{2}= \pm(3-2 a) p
$$
Thus, $(3-2 a) \vdots p$.
If $a=0$, we get the solution $p=3, n=1$.
Otherwise, $a \geqslant 3$. On the oth... | p=3,n=1orp=3,n=8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,558 |
10. The metro of city $\mathrm{N}$ consists of three lines, its scheme is shown in the figure. Initially, the trains are at the marked stations and start moving in the direction indicated by the arrow. Every minute, each train travels exactly one section between stations. Upon arriving at a terminal station, the train ... | 10. Note that the train takes exactly 7 minutes to travel the entire red branch, 8 minutes for the blue branch, and 9 minutes for the green branch. This means the trains return to their initial positions on the red branch every 14 minutes, on the blue branch every 16 minutes, and on the green branch every 9 minutes. Si... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,560 |
1. Today's date is written as: 22.11.2015. Name the last past date that is written with the same set of digits. | Solution. In 2015, there are no earlier such dates. The previous suitable year is 2012, the last month in it - 12, the largest possible number - 15. Answer: 15.12.2012. | 15.12.2012 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,565 |
2. A number is written on the board. In one move, you can either increase or decrease any of its digits by three (if it results in a digit), or swap two adjacent digits. Show how to transform the number 123456 into 654321 in 11 moves. | Solution. In the first 5 moves, we move 6 to the beginning: 612345. Then we swap 1 and 2, 4 and 5, getting 621354. Now we increase 2 and 1 by three, and decrease 5 and 4 by three - resulting in 654321. | 654321 | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 21,566 |
3. A group of toddlers in a kindergarten has 90 teeth in total. Any two toddlers together do not have more than 9 teeth. What is the minimum number of toddlers that can be in the group? | Solution. If all children have fewer than 5 teeth, then there are no fewer than $90 / 4$, i.e., no fewer than 23 children.
If one child has exactly 5 teeth, then the others have no more than 4, and there are no fewer than $1+85 / 4$, i.e., also no fewer than 23 children.
If any of the children have between 6 and 9 te... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,567 |
4. Given nine natural numbers. The first is written only with ones, the second - only with twos, ..., the ninth - only with nines. Can one of these numbers be equal to the sum of all the others? | Solution. No, because in this case the sum of all numbers should be even, but it is odd. Another possibility is to use a brute force method by the last digit.
## Grade 5. Lecture Hall. Solutions | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,568 |
6. At a masquerade ball, there were 20 people. In each dance, two people participated - a boy and a girl. It turned out that ten of them danced with three partners, two (Sasha and Zhenya) - with five, and the remaining eight - with six. Prove that Sasha and Zhenya are of different genders. | Solution. Note that the number of dances performed by everyone except Sasha and Zhenya is a multiple of three. Suppose that Sasha and Zhenya are of the same gender (for example, both are boys). Then the number of all dances performed by girls is divisible by 3, while the number of dances performed by boys is not. Since... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,570 |
7. A chessboard $(8 \times 8)$ was cut into several equal parts in such a way that all white cells remained uncut, while each black cell was cut. How many parts could have been obtained? | Solution. Note that there are 32 white cells, and each part contains an integer number of white cells, so the answer must be a divisor of 32.
Obviously, it cannot be 1. The answers $2, 4, 8, 16$, and 32 are possible. To achieve them,
![](https://cdn.mathpix.com/cropped/2024_05_06_b5ab07e644997c69f8fdg-1.jpg?height=50... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,571 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.