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17.17. In each cell of a rectangular sheet of squared paper, a number is written, and in each cell that is not on the edge, the arithmetic mean of the numbers in the adjacent cells is written. All the numbers written are different. Prove that the largest number is on the edge (i.e., at least one of the adjacent cells i... | 17.17. Suppose the largest number $a$ is not at the edge. Then it has all four neighboring numbers $a_{1}, a_{2}, a_{3}, a_{4}$ in the table, and at the same time, $a=\left(a_{1}+a_{2}+a_{3}+a_{4}\right) / 4$. But $a>a_{1}, a>a_{2}, a>a_{3}, a>a_{4}$. Therefore, $a>\left(a_{1}+a_{2}+a_{3}+a_{4}\right) / 4$. This leads ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,622 |
17.18. There are 13 weights, each weighing an integer number of grams. It is known that any 12 of them can be distributed on 2 scales, 6 weights on each, so that equilibrium is achieved. Prove that all weights are of the same weight. | 17.18. First, let's prove that all weights are either simultaneously even or odd. Suppose 12 weights are distributed on two scales, 6 on each, such that equilibrium is achieved. Assume a weight of \(a\) grams is set aside, and one of the weights on the scales weighs \(b\) grams, where \(a\) and \(b\) have different par... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,623 |
17.19. a) From the numbers $1,2,3,4,5,6,7, \ldots, 199,200$, 101 numbers are chosen arbitrarily. Prove that among the chosen numbers, there will be two such that one divides the other.
b) From the two hundred numbers: $1,2,3, \ldots, 199,200$, one number less than 16 and 99 other numbers are chosen. Prove that among t... | 17.19. a) Let's consider the greatest odd divisors of the selected numbers. Among the numbers from 1 to 200, there are exactly 100 different greatest odd divisors (the numbers $1, 3, \ldots, 199$). Therefore, two of the selected numbers have the same greatest odd divisor. This means that the two selected numbers differ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,624 |
17.20. From one bacterium, $n$ bacteria were obtained as follows. First, the bacterium divided into two, then one of the two resulting bacteria divided into two, then one of the three resulting bacteria divided into two, and so on. Prove that for any natural number $a \leqslant n / 2$ there was a moment when there exis... | 17.20. Let a certain bacterium, from which $m$ bacteria eventually resulted, divide into two bacteria. From these two "sisters," $m'$ and $m''$ bacteria eventually resulted, and $m' + m'' = m$. Therefore, one of the numbers $m'$ and $m''$ is not less than $m / 2$. Each time, we will choose the bacterium that has no few... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,625 |
17.21. We took three numbers $x, y, z$ and computed the absolute values of their pairwise differences $x_{1}=|x-y|, y_{1}=|y-z|$, $z_{1}=|z-x|$. In the same way, from the numbers $x_{1}, y_{1}, z_{1}$ we constructed the numbers $x_{2}, y_{2}, z_{2}$ and so on. It turned out that for some $n$ we got $x_{n}=x, y_{n}=y, z... | 17.21. Answer: $y=z=0$. The numbers $x_{n}, y_{n}, z_{n}$ are non-negative, so the numbers $x, y, z$ are also non-negative. If all the numbers $x$, $y, z$ were positive, then the largest of the numbers $x_{1}, y_{1}, z_{1}$ would be strictly less than the largest of the numbers $x, y, z$, and then the largest of the nu... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,626 |
17.22. A set of numbers $a_{1}, a_{2}, \ldots, a_{n}$ is associated with the set $b_{1}=\frac{a_{1}+a_{2}}{2}, b_{2}=\frac{a_{2}+a_{3}}{2}, \ldots, b_{n-1}=\frac{a_{n-1}+a_{n}}{2}, b_{n}=\frac{a_{n}+a_{1}}{2}$. Then the same operation is performed on the resulting set of numbers, and so on. Prove that if the resulting ... | 17.22. Suppose not all numbers $x_{1}, x_{2}, \ldots, x_{n}$ are equal. Let the largest of these numbers be $N$, and it appears exactly $k$ times among them. Then in the new sequence, there are no numbers greater than $N$, and the number $N$ appears no more than $k-1$ times. Therefore, after several operations, the lar... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,627 |
18.1. On 44 trees planted in a circle, 44 chizhik birds were sitting (one on each tree). From time to time, some two chizhiks simultaneously fly to the neighboring trees in opposite directions (one - clockwise, the other - counterclockwise). Will the chizhiks ever be able to gather on one tree? | 18.1. Answer: they will not be able to. Let's number the trees clockwise. Suppose at some moment in time, there are $n_{k}$ chizhi sitting on the $k$-th tree. Consider the sum $n_{1}+2 n_{2}+\ldots+44 n_{44}$. If one chizh flies to the next tree clockwise, this sum either increases by 1 or decreases by 43, and if it fl... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,628 |
18.2. With a triplet of numbers $(a, b, c)$, it is allowed to perform the following operation: increase one number by 2, and simultaneously decrease the other two by 1. Is it possible to obtain a triplet with two zeros from the triplet $(13,15,17)$ using such operations? | 18.2. Answer: no. In the specified operation, the first two numbers ($a, b$) are replaced either with ($a-1, b-1$), or with ($a+2, b-1$), or with ($a-1, b+2$). In all three cases, the remainder of the division of the number $a-b$ by 3 does not change. The numbers 13 and 15 give different remainders when divided by 3. T... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,629 |
18.3. In the sequence $1,0,1,0,1,0, \ldots$ each term starting from the seventh is equal to the last digit of the sum of the six preceding terms. Prove that in this sequence, the six numbers $0,1,0,1,0,1$ will never appear consecutively.
## 18.2. Semi-invariants | 18.3. We will replace the sextet of numbers $\left(x_{1}, \ldots, x_{6}\right)$ with the sextet of numbers $\left(x_{2}, x_{3}, \ldots, x_{6}, x_{7}\right)$, where $x_{7}$ is the last digit of the number $x_{1}+\ldots+x_{6}$. We will show that under such an operation, the last digit of the number $S\left(x_{1}, \ldots,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,630 |
18.4. Let $a_{1} \leqslant a_{2} \leqslant \ldots \leqslant a_{n}, b_{1} \leqslant b_{2} \leqslant \ldots \leqslant b_{n}$ and $\sigma(1), \ldots$ $\ldots, \sigma(n)$ be any permutation of the numbers $1, \ldots, n$. Prove that
$$
\sum a_{i} b_{n-i} \leqslant \sum a_{i} b_{\sigma(i)} \leqslant \sum a_{i} b_{i}
$$ | 18.4. Suppose first that $\sigma$ and $\tau$ are two permutations of the numbers $1, \ldots, n$, for which $\sigma(k)=\tau(k+1)$, $\sigma(k+1)=\tau(k)$, and $\sigma(i)=\tau(i)$ for $i \neq k, k+1$. Then if $\sigma(k+1)>\sigma(k)$, we have $\sum a_{i} b_{\sigma(i)}-\sum a_{i} b_{\tau(i)}=\left(a_{k+1}-a_{k}\right)\left(... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,631 |
18.5. $N$ soldiers are lined up in a single row (shoulder to shoulder). On the command "left," all turn $90^{\circ}$ simultaneously, but some turn left while others turn right. Exactly one second later, each soldier who is now face to face with a neighbor turns around - by $180^{\circ}$. Another second later, each sold... | 18.5. a) Let $m$ be the number of all soldiers who have turned to face a given soldier. Then this soldier will turn "around" no more than $m$ times. Indeed, at the moments when the soldier does not turn, the number $m$ does not change, and at the moments when the soldier turns, the number $m$ decreases by 1. It remains... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 38,632 |
18.6. Integers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, whose sum is positive, are arranged in a circle. If $x_{i}<0$, then the numbers $x_{i-1}, x_{i}, x_{i+1}$ can be replaced by $x_{i-1}+x_{i}, -x_{i}$, $x_{i+1}+x_{i}$ (it is assumed that $x_{i+5}=x_{i}$). Prove that after a finite number of such operations, all numbers... | 18.6. First, note that the sum of the numbers does not change with the specified operations. For each set $X=\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)$, let $F(x)=\sum_{i=1}^{5}\left(x_{i}-x_{i+2}\right)^{2}$. Suppose $x_{i}0
$$
i.e., $F(Y)<F(X)$. This means that after each operation, the integer non-negative num... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,633 |
18.7. Let $\sigma(1), \sigma(2), \ldots, \sigma(n)$ be some permutation of the numbers $1,2, \ldots, n$. Suppose that this permutation can be obtained in two different ways by a series of transpositions: once using $m_{1}$ transpositions, and another time using $m_{2}$ transpositions. Prove that $m_{1}$ and $m_{2}$ are... | 18.7. Let us take arbitrary pairwise distinct numbers $x_{1}, \ldots, \ldots, x_{n}$ and consider the number $\prod_{i>j} \frac{x_{i}-x_{j}}{x_{\sigma(i)}-x_{\sigma(j)}}= \pm 1$. We will show that if this number equals 1, then the numbers $m_{1}$ and $m_{2}$ are even, and if it equals -1, then the numbers $m_{1}$ and $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,634 |
18.8. The game "15" is as follows. In a square with a side of 4, there are 15 tiles - squares with a side of 1. The tiles are numbered from 1 to 15:
| 1 | 2 | 3 | 4 |
| :---: | :---: | :---: | :---: |
| 5 | 6 | 7 | 8 |
| 9 | 10 | 11 | 12 |
| 13 | 14 | 15 | |
Any tile that borders (by side) with the empty cell can b... | 18.8. Answer: no, it cannot. Let's assign the number 16 to the free cell. Then, after each move, we get some permutation of the numbers $1, \ldots, 16$. Each move represents a transposition. We want to end up with the transposition $14 \leftrightarrow 15$.
We will show that the free cell can return to its initial posi... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,635 |
18.9. Prove that any permutation of the numbers $1, \ldots, n$ can be obtained from the set $1, \ldots, n$ by performing several transpositions.
A permutation of the numbers $1, \ldots, n$ is called even if it can be obtained by performing an even number of transpositions; otherwise, it is called odd. Problem 18.7 sho... | 18.9. It is sufficient to prove that from any permutation $\sigma(1), \ldots, \ldots, \sigma(n)$ one can obtain the set $1, \ldots, n$ by performing several transpositions (these transpositions can be performed in reverse order). If $\sigma(1) \neq 1$, then we swap 1 and $\sigma(1)$. Then, in the new set of numbers, we... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,636 |
18.10. Let $\sigma$ be some permutation of the numbers $1, \ldots, n$. We call any pair of numbers $i < j$ such that $i > \sigma(j)$ an inversion. Prove that the parity of the permutation coincides with the parity of the number of all inversions.
Each permutation $\sigma$ of the numbers $1, \ldots, n$ can be represent... | 18.10. It is sufficient to prove that after each transposition, the parity of the number of inversions changes. Let the numbers $\sigma(i)$ and $\sigma(j)$ swap places during the transposition. Then an inversion can only appear or disappear for the pairs $(i, k)$ and $(j, k)$. It is easy to see that the appearance or d... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,637 |
18.11. a) Let the permutation $\sigma$ of the numbers $1, \ldots, n$ be a product of $m$ cycles. After this permutation, we perform a transposition of any two numbers. Prove that the resulting permutation is a product of $m \pm 1$ cycles.
b) Prove that a permutation is even if and only if the number $n-m$ is even. | 18.11. a) Let the numbers $a$ and $b$ be transposed. Consider two cases.
1. The numbers $a$ and $b$ are in the same cycle ( $a, a_{1}, \ldots, a_{p}, b, b_{1}, \ldots, b_{q}$ ). After the transposition, this cycle splits into two cycles ( $a, a_{1}, \ldots, a_{p}$ ) and $\left(b, b_{1}, \ldots, b_{q}\right)$, since $a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,638 |
19.1. In three boxes lie balls - black, white, and green (one ball in each box). On the first box, it says "white," on the second - "black," on the third "white or green." It is known that none of the inscriptions are true. Determine which balls are in which boxes. | 19.1. The third box can contain only a black ball. A white ball can lie only in the second box. The green ball lies in the first box. | not\found | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,639 |
19.2. A traveler arrived on an island, part of whose inhabitants always tell the truth, while the others always lie. What question should the traveler ask an inhabitant of the island to find out whether he always tells the truth or always lies? | 19.2. The question to be asked is: "If you always told the truth, how would you answer the question: 'Are you a liar?'?" Someone who always tells the truth would answer this question with: "no", while someone who always lies would answer: "yes".
保留了原文的换行和格式，但请注意，最后一句的翻译在英文中通常不需要换行，除非有特定的格式要求。如果需要严格保持格式，可以这样写：
19.2. T... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,640 |
19.3. One person always tells the truth, while the other person always lies. What question should you ask them so that they answer it the same way? | 19.3. "Do you always tell the truth?" | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,641 |
19.5. Given 8 objects, one of which is marked. It is required to ask 3 questions, to which only "yes" and "no" answers are given, and find out which object is marked. | 19.5. Let's number the items from 0 to 7. We will ask questions about whether the selected item belongs to the groups $\{0,2,4,6\},\{0,1,4,5\}$, and $\{0,1,2,3\}$. Let $\varepsilon_{i}=0$ if the answer to the $i$-th question is "yes", and $\varepsilon_{i}=1$ otherwise. Then the selected item has the number $\varepsilon... | \varepsilon_{1}+2\varepsilon_{2}+4\varepsilon_{3} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,643 |
19.6. In the institute, there are truth-lovers, who always tell the truth, and liars, who always lie. One day, each employee made two statements.
1) There are not even ten people in the institute who work more than I do.
2) At least a hundred people in the institute earn more than I do.
It is known that the workload ... | 19.6. Answer: 110. From the first statement, for the truth-teller with the least workload, it follows that there are no more than 10 truth-tellers in the institute, and for the liar with the greatest workload, it follows that there are no fewer than 10 truth-tellers in the institute. From the second statement, for the ... | 110 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,644 |
19.7. Three wise men are sitting on chairs back to back, so the one sitting in front cannot see those sitting behind. They know there are 3 white and 2 black hats. The wise men close their eyes and hats are placed on their heads, after which the remaining hats are removed. The wise men open their eyes, and the one sitt... | 19.7. Answer: the one sitting in front knows that he has a white hat. If the two sages in front had black hats, then the sage sitting at the back would know that he has a white hat. Therefore, at least one of them has a white hat. So, if the sage sitting in front had a black hat, the one sitting in the middle would rea... | the\one\sitting\in\front\has\\white\hat | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,645 |
19.8. In a train carriage, several sages are traveling. When the train passed through a tunnel, dust got into the windows. The conductor came in and said: "Some of you have dirty faces. Unfortunately, there is no water on the train. But there will be major stops soon, so you can get off the train and wash your face." O... | 19.8. We will prove that if $n$ sages are dirty, then all of them will go to wash their faces at the $n$-th stop. For $n=1$, this is obvious: the dirty sage sees that all the others are not dirty, and he knows that someone is dirty. Suppose that the required statement is proved if no more than $n$ sages are dirty. Cons... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,646 |
19.12. Prove that for any given truth table using the connectives \&, ᄀ and $\vee$, a statement can be constructed that has the given truth table, in the case of: a) one variable; b) two variables; c) $n$ variables. | 19.12. a) It is easy to see that the statements $p, \neg p, p \& \neg p$ and $p \vee \neg p$ correspond to all possible truth tables.
b) The statements $p \& q, p \&(\neg q), \neg p \& q$ and $\neg p \&(\neg q)$ take the value T (True) for exactly one of the four possible sets of values of the variables $p$ and $q$. T... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 38,650 |
19.13. Part of the island's inhabitants always tell the truth, while the others always lie. A traveler wants to find out which of the two roads leads to village $A$, by asking one question to a local resident he meets. Will he be able to do this? | 19.13. Let $p$ be the statement that the road leads to $A$, and $q$ be the statement that the local resident you meet is truthful. We obtain the following table of desired answers and, accordingly, the truth table of the required statement (naturally, it is constructed based on the fact that a truthful person tells the... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,651 |
20.1. The transporter needs to ferry a wolf, a goat, and some cabbage across a river. The boat can only take one of these items at a time. Additionally, the cabbage cannot be left alone with the goat, and the goat cannot be left alone with the wolf. How can the transportation be carried out? | 20.1. First, the goat should be transported across the river. After that, the ferryman returns, and there are two further options. One needs to transport the cabbage (wolf), return with the goat, leave the goat on the shore, transport the wolf (cabbage), and then transport the goat. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,652 |
20.2. Three soldiers and three bandits need to cross a river. They found a boat that can only hold two people. Soldiers cannot be left on either bank if there are fewer of them than bandits. How can they all cross the river? | 20.2. We will indicate in parentheses who is in the boat. Crossing the river can be done as follows: PPCC(PC), PPCC(P)C, PCC(PP)C, PCC(P)PC, CC(PP)PC, CC(C)PPP, C(CC)PPP, C(P)PPCC, (PC)PPCC. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,653 |
20.3. The family approached the bridge at night. Dad can cross it in 1 minute, Mom in 2, the son in 5, and grandma in 10. They have one flashlight. The bridge can only hold two people at a time. How can they cross the bridge in 17 minutes? (If two people are crossing the bridge, they move at the slower of their speeds.... | 20.3. The main idea is that the grandmother should cross the bridge together with the grandson. First, the father and mother go, then the father returns with the flashlight, after that the grandmother and grandson go, then the mother returns with the flashlight, finally, the father and mother cross the bridge. In total... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,654 |
20.4. Two players play the following game. There are 9 cards with numbers $1,2, \ldots, 9$. The players take turns picking one card for themselves. The winner is the one who has three cards with numbers that sum to 15. Prove that this game is equivalent to a game of "tic-tac-toe" on a $3 \times 3$ board.
## 20.2. Pour... | 20.4. It is easy to check that there are exactly 8 triples of cards that sum to $15: (1,5,9), (1,6,8), (2,4,9), (2,5,8), (2,6,7)$, $(3,4,8), (3,5,7)$, and $(4,5,6)$. These 8 triples of numbers are precisely the triples of numbers lying on a straight line in the table
| 2 | 9 | 4 |
| :--- | :--- | :--- |
| 7 | 5 | 3 |
... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 38,655 |
20.5. There is a full jug of milk with a capacity of 8 liters and two empty jugs of 5 liters and 3 liters. How to divide the milk into two equal parts | 20.5. First solution. If 8-liter, 5-liter, and 3-liter jugs are filled with \(a\), \(b\), and \(c\) liters of milk, respectively, we will denote this as \((a, b, c)\). The following sequence of pourings can be applied: \((8,0,0) \rightarrow (3,5,0) \rightarrow (3,2,3) \rightarrow (6,2,0) \rightarrow (6,0,2) \rightarrow... | (4,4,0) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,656 |
20.6. There is a full jug of milk with a capacity of 12 liters and two empty jugs of 8 liters and 5 liters. How to divide the milk into two equal parts | 20.6. The following sequence of pourings can be applied:
$$
\begin{aligned}
(12,0,0) \rightarrow(4,8,0) \rightarrow(0,8,4) & \rightarrow(8,0,4) \rightarrow \\
\rightarrow(8,4,0) \rightarrow(3,4,5) & \rightarrow(3,8,1) \rightarrow(11,0,1) \rightarrow \\
& \rightarrow(11,1,0) \rightarrow(6,1,5) \rightarrow(6,6,0)
\end{a... | (6,6,0) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,657 |
20.7. There are four barrels. The first one holds 24 buckets, the second 13, the third 11, and the fourth 5. The first barrel is filled with water, while the other barrels are empty. How can you divide the water into three equal parts?
## 20.3. Tournaments
When solving problems about tournaments, you need to know the... | 20.7. The following pourings can be applied:
$$
\begin{aligned}
& (24,0,0,0) \rightarrow(19,0,0,5) \rightarrow(8,0,11,5) \rightarrow \\
& \quad \rightarrow(8,11,0,5) \rightarrow(0,11,8,5) \rightarrow(0,13,8,3) \rightarrow(8,13,0,3) \rightarrow \\
& \quad \rightarrow(8,13,3,0) \rightarrow(8,8,3,5) \rightarrow(8,8,8,0)
... | (8,8,8,0) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,658 |
20.9. Ali, Borya, and Vasya played several games of chess, and each played the same number of games. Could it have turned out that Ali had the fewest losses, Borya had the most wins, and Vasya had the most points? | 20.9. Answer: yes, it could. To construct a corresponding example, consider three types of tournaments with the following tables:
| | Wins | Draws | Losses | Points |
| :---: | :---: | :---: | :---: | :---: |
| Alik | 0 | 4 | 0 | 2 |
| Borya | 1 | 2 | 1 | 2 |
| Vasya | 1 | 2 | 1 | 2 |
| | Wins | Draws | Losses | P... | =5,b=3,=1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,660 |
20.10. a) In a chess tournament, two students from grade 7 and a certain number of students from grade 8 participated. The two seventh-graders scored 8 points, and each of the eighth-graders scored the same number of points. How many eighth-graders participated in the tournament? Find all possible solutions.
b) In a c... | 20.10. a) Answer: 7 or 14. Let $x$ be the number of eighth-graders, and $y$ be the number of points scored by each eighth-grader. By calculating the total points scored by all participants in the tournament in two ways, we arrive at the equation
$$
x y + 8 = \frac{(x+2)(x+1)}{2}
$$
i.e.,
$$
2 y = \frac{(x+2)(x+1) - ... | 7or14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,661 |
20.11. In the competitions, $2^{n}$ boxers participate. Each day, $2^{n-1}$ pairs of boxers fight (each boxer has exactly one fight). All boxers have different strengths, and in each fight, the stronger one wins. Prove that in $n(n+1) / 2$ days, the place of each boxer can be determined. (The schedule for each day is s... | 20.11. We will use induction on $n$. For $n=1$, the statement is obvious. Suppose that $2^{n-1}$ boxers can be ordered in $n(n-1) / 2$ days. We will prove that then $2^{n}$ boxers can be ordered in $n(n+1) / 2$ days. We will divide the boxers into two groups $X$ and $Y$ of $2^{n-1}$ people each. First, fights will only... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,662 |
20.12. There are several bags, each containing a sufficient number of coins. In one of the bags, the coins are counterfeit, while in all the others, they are genuine. The weight of a genuine coin is known, and it is known that a counterfeit coin is 1 gram lighter than a genuine one. How can you identify the bag with th... | 20.12. Let's number the bags and take from each bag as many coins as its number. We will weigh all these coins. Their weight is less than the weight of the same number of real coins by as many grams as the number of the bag with the fake coins. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,663 |
20.13. Among 26 identical-looking coins, there is one counterfeit coin that is lighter than the others. Prove that the counterfeit coin can be found in three weighings using a balance scale (without pointers or weights). | 20.13. Let's take 18 coins and place half of them on one scale pan, and the other half on the other pan. If one group of coins turns out to be lighter, then the counterfeit coin is in this group. If the scales balance, then the counterfeit coin is among the eight remaining coins.
After the first weighing, we are left ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 38,664 |
20.14. Among $2 n+1$ coins, there are $2 k$ counterfeit ones ( $k \leqslant n$ ), and the weight of a counterfeit coin differs from that of a genuine coin by 1 gram. How can one determine in a single weighing on a balance scale with a pointer whether a chosen coin is counterfeit or not? | 20.14. Let's set aside the chosen coin and place half of the remaining coins on one pan of the scales, and the other half on the other pan. Let there be $k_{1}$ counterfeit coins among the first $n$ coins, and $k_{2}$ counterfeit coins among the other $n$ coins. Then, the weight on the first pan is $n a \pm k_{1}$, whe... | if\the\chosen\\is\counterfeit,\the\pointer\reading\will\be\an\odd\,\\if\the\chosen\\is\genuine,\the\pointer\reading\will\be\an\even\ | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,665 |
20.15. There are $n$ bags, each containing a sufficient number of coins. There are several different types of coins. Coins of different types have different weights, and their weights differ by an integer number of grams. The weight of coins of each type is known. In each bag, there are coins of one type, but the numbe... | 20.15. Let the weight of the lightest coin be $a$ grams, and the weight of the heaviest coin be $a+d-1$ grams. Number the bags with the numbers $0,1,2, \ldots, n-1$ and take exactly $d^{i}$ coins from the bag with number $i$. Then, if the coins in the bag with number $i$ weigh $a+m_{i}$, the weight of the selected coin... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,666 |
20.16. Some of the 20 metal cubes, identical in size and appearance, are aluminum, the rest* are duralumin (heavier). How can you determine the number of duralumin cubes using 11 weighings on a balance with two pans and no weights? | 20.16. Let's put one cube on each pan of the balance. There are two possible cases.
Case 1. One of the cubes turned out to be heavier in the first weighing.
In this case, one of the selected cubes is aluminum, and the other is duralumin. Place the selected cubes on one pan and compare them with the remaining cubes. S... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,667 |
20.17. a) Given $\frac{1}{2}\left(3^{n}-3\right)$ coins $(n \geqslant 2)$, among which there is one counterfeit coin that differs in weight from the genuine ones. Using $n$ weighings on a balance scale without weights, find the counterfeit coin and determine whether it is heavier or lighter than the genuine ones.
b) D... | 20.17. a) Consider all $n$-digit numbers in the ternary system, except for the numbers $00 \ldots 0, 11 \ldots 1$, and $22 \ldots 2$. We will pair these numbers such that the sum of the numbers in each pair is equal to $22 \ldots 2$. Each coin will be associated with one of these pairs. We will call the right marker of... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,668 |
20.18. a) In a rectangular table composed of positive numbers, the product of the sum of the numbers in any column and the sum of the numbers in any row equals the number at their intersection. Prove that the sum of all numbers in the table is equal to one.
b) In a rectangular table, the product of the sum of the numb... | 20.18. a) First solution. Let $x_{1}, \ldots, x_{n}$ be the sums of the numbers in the rows, and $y_{1}, \ldots, y_{m}$ be the sums of the numbers in the columns. At the intersection of the $i$-th row and the $j$-th column, the number $x_{i} y_{j}$ is placed. Therefore, the sum of the numbers in the $i$-th row is $x_{i... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,669 |
20.19. A square table with $n^2$ cells is filled with numbers from 1 to $n$ such that each row and each column contains all these numbers. Prove that if $n$ is odd and the table is symmetric with respect to the diagonal running from the top left corner to the bottom right, then all the numbers $1, 2, 3, \ldots, n$ will... | 20.19. The table is symmetric with respect to the diagonal, so each number located outside the diagonal corresponds to an equal number in the symmetric position. Therefore, there is an even number of ones, an even number of twos, and so on, located outside the diagonal. By the condition, each row contains all numbers f... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,670 |
20.20. Non-negative numbers are written in the cells of an $8 \times 8$ table, the sum of which is 1956.
a) The sum of the numbers on one of the diagonals is 112. The numbers symmetrically located relative to this diagonal are equal. Prove that the sum of the numbers in any column is less than 1035.
b) The sum of the... | 20.20. a) Suppose the sum of the numbers in some column is $S \geqslant 1035$. Consider the row symmetric to this column with respect to the highlighted diagonal. The sum of the numbers in this row is also $S$, and the sum of all numbers in this column and this row is $2 S-s$, where $s$ is the number at their intersect... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,671 |
20.21. The numbers $1,2, \ldots, k^{2}$ are arranged in a square table:
$$
\begin{array}{cccc}
1, & 2, & \ldots, & k \\
k+1, & k+2, & \ldots, & 2 k \\
\ldots & \ldots & \ldots & \ldots \\
(k-1) k+1, & \ldots, & \ldots, & k^{2} .
\end{array}
$$
We will write down an arbitrary number from this table, and then cross out... | 20.21. Answer: $\frac{k\left(k^{2}+1\right)}{2}$.
Let's write the given table as
$$
\begin{array}{cccc}
k \cdot 0+1, & k \cdot 0+2, & \ldots, & k \cdot 0+k \\
k \cdot 1+1, & k \cdot 1+2, & \ldots, & k \cdot 1+k \\
\ldots & \ldots & \ldots & \ldots \\
(k-1) k+1, & (k-1) k+2, & \ldots, & (k-1) k+k
\end{array}
$$
Each ... | \frac{k(k^{2}+1)}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,672 |
20.22. In a table of size $1987 \times 1987$, real numbers are written, the absolute value of each of which does not exceed 1. It is known that the sum of the four numbers at the intersection of any two rows and two columns is zero. Prove that the sum of all numbers does not exceed 1987. | 20.22. Consider a $3 \times 3$ table and cut out its top-left corner. We will show that if such a figure is placed in this table, the sum of the 8 numbers in it does not exceed 2. Indeed, consider two $2 \times 2$ squares located in the bottom-left and top-right corners of the $3 \times 3$ table, and take the sum of th... | 1987 | Combinatorics | proof | Yes | Yes | olympiads | false | 38,673 |
21.1. Find all three-digit numbers, any natural power of which ends in three digits that form the original number (in the same order). | 21.1. Answer: 376 and 625. Let $N$ be the desired number. Then $N^{2}-N=N(N-1)$ is divisible by 1000. The numbers $N$ and $N-1$ are coprime, so one of them is divisible by 8, and the other by 125. Let's first assume $N=125 k$. Then $k \leqslant 8$. Among the numbers $125 k-1, k=1, \ldots, 8$, only the number 624 is div... | 376625 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,674 |
21.2. a) Prove that for any natural number $n$ there exist exactly two natural $n$-digit* numbers $a_{n}$ and $b_{n}$ (different from $00 \ldots 0$ and $0 \ldots 01$), for which $a_{n}^{2}$ ends in $a_{n}$, and $b_{n}^{2}$ ends in $b_{n}$.
b) Prove that $a_{n}+b_{n}=10^{n}+1$.
c) Suppose the numbers $a_{n}$ and $a_{n... | 21.2. a) The number $a_{n}^{2}-a_{n}=a_{n}\left(a_{n}-1\right)$ must be divisible by $10^{n}=2^{n} 5^{n}$. The numbers $a_{n}$ and $a_{n}-1$ are coprime and $a_{n} \neq 0$ and 1, so either $a_{n}$ is divisible by $2^{n}$ and $a_{n}-1$ is divisible by $5^{n}$, or $a_{n}$ is divisible by $5^{n}$ and $a_{n}-1$ is divisibl... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,675 |
21.3. The numbers $2^{n}$ and $5^{n}$ start with the digit $a$. What is $a$? | 21.3. Answer: 3. By the condition $a \cdot 10^{p}<2^{n}<(a+1) 10^{p}$ and $a \cdot 10^{q}<5^{n}<$ $<(a+1) 10^{q}$. Therefore, $a^{2} 10^{p+q}<10^{n}<(a+1)^{2} 10^{p+q}$, i.e., $a^{2}<10^{n-p-q}<$ $<(a+1)^{2}$. At the same time, $(a+1)^{2} \leqslant 100$. Thus, $a^{2}<10<(a+1)^{2}$, i.e., $a=3$. The number 3 is the lead... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,676 |
21.4. a) Prove that the number $2^{n}$ can start with any set of digits.[^3]b) Prove that the number $0.12481632 \ldots$ (powers of two are written consecutively) is irrational. | 21.4. a) Let $A$ be a given natural number. We will show that a natural number $n$ can be chosen such that $10^{m} A<2^{n}<10^{m}(A+1)$, i.e., $m+\lg A<n \lg 2<m+\lg (A+1)$. An equivalent condition is that there exist natural numbers $m$ and $n$ such that $\lg A < n \lg 2 - m < \lg (A+1)$. The number $\lg 2$ is irratio... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,677 |
21.6. Prove that the second to last digit of the number $3^{n}$ is even for any $n>2$. | 21.6. Let $N=a_{0}+a_{1} \cdot 10+a_{2} \cdot 100+\ldots$ Then $3 N=3 a_{0}+30 a_{1}+$ $+300 a_{2}+\ldots$ Therefore, if the number $a_{1}$ is even, then the second-to-last digit of the number $3 N$ has the same parity as the second-to-last digit of the number $3 a_{0}$. The equality $3^{4}=81$ shows that the last digi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,679 |
21.7. Let $a$ be the last digit of the number $2^{n}, b=2^{n}-a$. Prove that if $n>3$, then $a b$ is divisible by 6. | 21.7. Since $2^{4}=16$, the number $2^{4 k}$ ends in 6. Accordingly, the numbers $2^{4 k+1}, 2^{4 k+2}, 2^{4 k+3}$ end in $2,4,8$. For numbers of the form $2^{4 k}$, the required statement is obvious, since $a=6$. Note also that the number $a$ is always even. Therefore, it is enough to check that the numbers $2^{4 k+1}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,680 |
21.8. Let \(a_{1}, a_{2}, \ldots\) be distinct natural numbers, in the decimal representation of which the digit 1 does not appear. Prove that among the numbers \(a_{n} / n\) there are arbitrarily large numbers.
## 21.4. Sum of digits | 21.8. The number of natural numbers that do not exceed $10^{k}$ and in whose decimal representation the digit 1 does not appear, does not exceed $9^{k}-1$. Indeed, on each of the $k$ digits stands one of 9 digits, and all these digits cannot be zeros simultaneously. Therefore, $a_{n} \geqslant 10^{k}$ for some $n \leqs... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,681 |
21.9. Let $a \geqslant 2$ be a natural number that is not divisible by 2 or 5. Prove that the sum of the digits of the number $a^{m}$ can be arbitrarily large for sufficiently large $m$. | 21.9. Suppose the sum of the digits of numbers of the form $a^{m}$ is bounded. Then the sum of the digits of the number $a^{m}$ does not exceed the sum of the digits of the number $a^{N}$ for some fixed $N$. Let $a^{N}<10^{k}$. According to problem 17.2, there exists a natural number $n$ for which $a^{n}-1$ is divisibl... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,682 |
21.11. Let $P(x)$ be a polynomial with natural coefficients, and $a_{n}$ be the sum of the digits of the number $P(n)$. Prove that some number appears in the sequence $a_{1}, a_{2}$, $a_{3}, \ldots$ infinitely many times.
## 21.5. Various problems about decimal representation | 21.11. Let $P(x)=b_{m} x^{m}+b_{m-1} x^{m-1}+\ldots+b_{0}$. Choose a natural number $k$ such that the number $10^{k}$ is greater than any of the numbers $b_{0}, b_{1}, \ldots, b_{m}$. Then in the decimal representation of the number $P\left(10^{k}\right)$, the digits of the number $b_{m}$ come first, followed by zeros,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,684 |
21.12. Find a four-digit number that is a perfect square and has the first two digits the same and the last two digits also the same. | 21.12. Let $a$ be the first and second digits, $b$ be the third and fourth. Then the given number is equal to $11(b+100 a)$, so $b+100 a=11 x^{2}$ for some natural number $x$. Moreover, $100 \leqslant b+100 a \leqslant$ $\leqslant 908$, hence $4 \leqslant x \leqslant 9$. By calculating the squares of the numbers $44, \... | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,685 |
21.13. The numbers $2^{n}$ and $5^{n}$ (in decimal notation) are written one after the other. How many digits does the resulting number have? | 21.13. Answer: $n+1$. Suppose the number $2^{n}$ has $p$ digits, and the number $5^{n}$ has $q$ digits. Then $10^{p-1}<2^{n}<10^{p}$ and $10^{q-1}<5^{n}<10^{q}$. Multiplying these inequalities, we get $10^{p+q-2}<10^{n}<10^{p+q}$. Therefore, $p+q=n+1$. | n+1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,686 |
21.14. Prove that any positive number $a$ can be represented as the sum of nine numbers, the decimal representations of which contain only the digits 0 and $k$, where $k$ is a fixed digit different from zero. | 21.14. Let's represent the number $a / k$ as the sum of nine numbers, the decimal representations of which contain only the digits 0 and 1. Using the fact that $a=k(a / k)$, we will obtain the required representation. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,687 |
21.15. Find all three-digit numbers that are equal to the sum of the factorials of their digits. | 21.15. Answer: 145. Let \( N = 100x + 10y + z \) be the desired number, for which \( N = x! + y! + z! \). The number \( 7! = 5040 \) is a four-digit number, so no digit of \( N \) exceeds 6. Therefore, the number \( N \) is less than 700. But then no digit of \( N \) exceeds 5, since \( 6! = 720 \). The inequality \( 3... | 145 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,688 |
21.16. All integers are written in a row, starting from one. What digit stands at the 206,788th place?
## 21.6. Periods of Decimal Fractions and Repunits
| Let $p$ be a prime number different from 2 and 5. The length of the period of the number $p$ is the number of digits in the period of the decimal representation o... | 21.16. Answer: the digit 7. There are exactly 9 single-digit numbers, 99-9=90 two-digit numbers, 999-99-9=900 three-digit numbers, 9000 four-digit numbers, and so on. Single-digit numbers occupy the first 9 positions in the written sequence, two-digit numbers occupy $90 \cdot 2=180$ positions, three-digit numbers occup... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,689 |
21.17. Let $n$ be a natural number not exceeding $p-1$. Prove that the number of digits in the period of the decimal representation of the number $n / p$ is equal to the length of the period of the number $p$.
保留源文本的换行和格式，直接输出翻译结果。 | 21.17. The length of the period of the decimal representation of the fraction $n / p$ is the smallest natural number $d$ for which $n\left(10^{d}-1\right)$ is divisible by $p$. Since the numbers $n$ and $p$ are coprime, $n\left(10^{d}-1\right)$ is divisible by $p$ if and only if $10^{d}-1$ is divisible by $p$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,690 |
21.18. Prove that the length of the period of the number $p$ is a divisor of the number $p-1$. | 21.18. The period length of a number $p$ is the smallest natural number $d$ for which $10^{d}-1$ is divisible by $p$. According to Fermat's Little Theorem (problem 31.1), $10^{p-1}-1$ is divisible by $p$. Let $p-1=a d+r$, where $0 \leqslant r0$. Then $10^{p-1}=\left(10^{d}\right)^{a} 10^{r} \equiv 10^{r}(\bmod p)$, hen... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,691 |
21.19. The periods of regular fractions with denominator 7 are obtained from each other by cyclic permutation:
$$
\begin{array}{lll}
1 / 7=0,(142857), & 3 / 7=0,(428571), & 2 / 7=0,(285714) \\
6 / 7=0,(857142), & 4 / 7=0,(571428), & 5 / 7=0,(714285)
\end{array}
$$
Prove that all prime numbers $p$ for which the period... | 21.19. We will divide 1 by $p$ in a column. As a result, some remainders will periodically repeat. For example, when dividing 1 by 7, the remainders $1,3,2,6,4,5$ periodically repeat. In the case where the period length is $p-1$, this sequence contains all possible remainders $1,2, \ldots, p-1$.
Let's return to the ex... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,692 |
21.20. The period of the fraction $1 / 7=0,(142857)$ has the following property: $142+857=999$. Prove that the period of the fraction $1 / p$ has a similar property for any prime number $p$, for which the length of the period is $p-1$.
$$
\% \quad \% \quad \%
$$
A repunit is a natural number of the form $111 \ldots 1... | 21.20. The number $p-1$ is even; let's write it as $p-1=2 k$. The period of the fraction $1 / p$, written as a natural number, has the form $a \cdot 10^{k}+b$. We need to prove that $a+b=10^{k}-1$.
The number $10^{2 k}-1=\left(10^{k}-1\right)\left(10^{k}+1\right)$ is divisible by $p$. At the same time, $10^{k}-1$ is n... | 10^{k}-1 | Number Theory | proof | Yes | Yes | olympiads | false | 38,693 |
21.21. Prove that the length of the period of a prime number $p \neq 3$ is equal to the number of ones in the smallest repunit divisible by $p$. | 21.21. Let the period length of a prime number $p$ be $d$. This means that $d$ is the smallest natural number for which $10^{d}-1$ is divisible by $p$. Clearly, $10^{d}-1=9 R_{d}$, where $R_{d}$ is a repunit containing $d$ ones. If $p$ is a prime number different from 3, then $9 R_{d}$ is divisible by $p$ if and only i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,694 |
21.22. Let $p \geqslant 7$ be a prime number. Prove that the number $\underbrace{111 \ldots 11}_{p-1}$ is divisible by $p$.
## 21.7. Definition of the $d$-ary representation of a number | 21.22. If $p$ is a prime number different from 2 and 5, then the number $10^{p-1}-1$ is divisible by $p$ (Problem 31.1). This number has the form $\underbrace{99 \ldots 9}_{p-1}$. Therefore, if $p \neq 3$, the number $\underbrace{11 \ldots 1}_{p-1}$ is also divisible by $p$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,695 |
21.23. Let $d>1$ be a natural number. Prove that any natural number $n$ can be uniquely represented in the form $n=a_{0}+a_{1} d+a_{2} d^{2}+\ldots+a_{k} d^{k}$, where $0 \leqslant a_{i} \leqslant d-1$ are integers, and $a_{k} \neq 0$.
The expression $n=a_{0}+a_{1} d+a_{2} d^{2}+\ldots+a_{k} d^{k}$ is called the $d$-a... | 21.23. Choose $k$ such that $d^{k} \leqslant n<d^{k+1}$. Let $a_{k}=\left[n / d^{k}\right]$. Then $1 \leqslant a_{k} \leqslant d-1$ and $n=a_{k} d^{k}+n^{\prime}$, where $0 \leqslant n^{\prime}<d^{k}$. If $n^{\prime} \neq 0$, then choose $l<k$ such that $d^{l} \leqslant n^{\prime}<d^{l+1}$, and let $a_{l}=\left[n^{\pri... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,696 |
21.24. Let $n$ be a natural number, $a_{0}$ be the remainder of the division of $n$ by 2. Let $n_{1}=\left(n-a_{0}\right) / 2$, $a_{1}$ be the remainder of the division of $n_{1}$ by 2, and so on until we get $n_{m}=1$ and $a_{m}=1$. Prove that $a_{m} 2^{m}+a_{m-1} 2^{n-1}+\ldots+a_{1} \cdot 2+a_{0}-$ is the binary rep... | 21.24. Let $n=b_{k} 2^{k}+b_{k-1} 2^{k-1}+\ldots+b_{1} \cdot 2+b_{0}$. Then $a_{0}=b_{0}$ and $n_{1}=b_{k} 2^{k-1}+b_{k-1} 2^{k-2}+\ldots+b_{1}$. Therefore, $a_{1}=b_{1}$ and $n_{2}=b_{k} 2^{k-2}+$ $+b_{k-1} 2^{k-3}+\ldots+b_{2}$ and so on. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,697 |
21.25. Prove that $(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots\left(1+x^{2^{n}}\right)=$ $=1+x+x^{2}+x^{3}+\ldots+x^{2^{n+1}-1}$. | 21.25. Each number from 0 to $2^{n+1}-1$ can be uniquely represented as the sum of different numbers $0,1,2,2^{2}, \ldots, 2^{n}$. Therefore, the term $x^{m}$, where $0 \leqslant m \leqslant 2^{n+1}-1$, appears exactly once when expanding the brackets in the given product. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,698 |
21.26. Prove that the number of odd coefficients in the polynomial $(1+x)^{n}$ is $2^{d}$, where $d$ is the sum of the digits in the binary representation of the number $n$ (i.e., the number of ones in the binary representation of the number $n$). | 21.26. It is clear that $(1+x)^{2} \equiv 1+x^{2}(\bmod 2)$. Therefore, induction on $m$ shows that $(1+x)^{2^{m}} \equiv 1+x^{2^{m}}(\bmod 2)$. Let $n=2^{m_{1}}+$ $+2^{m_{2}}+\ldots+2^{m_{d}}$, where $0 \leqslant m_{1}<m_{2}<\ldots<m_{d}$. Then
$$
\begin{aligned}
&(1+x)^{n}=(1+x)^{2^{m_{1}}}(1+x)^{2^{m_{2}}} \ldots(1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,699 |
21.27. (Nim game.) There are three piles of stones. Two players take turns taking any number of stones from any one pile, but only from one. The player who takes the last stone wins.
Let's write the number of stones in the piles in binary: $a_{0}+a_{1} \cdot 2+a_{2} \cdot 2^{2}+\ldots, b_{0}+b_{1} \cdot 2+b_{2} \cdot ... | 21.27. a) Let $d_{k}$ be the odd number with the largest index $k$. Then one of the numbers $a_{k}, b_{k}, c_{k}$ is equal to 1, for example, $a_{k}=1$. The beginner takes stones from the first pile so that the numbers $a_{k+1}, a_{k+2}, \ldots$ do not change, while each of the numbers $d_{k}, d_{k-1}, d_{k-2}, \ldots,... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,700 |
21.28. Prove that for any natural number $n$ the number
$$
n-\left[\frac{n}{2}\right]-\left[\frac{n}{2^{2}}\right]-\left[\frac{n}{2^{3}}\right]-\ldots
$$
is equal to the sum of the digits in the binary representation of $n$.
See also problems $19.5,33.5$.
## 21.9. Other number systems | 21.28. Let $n=a_{0}+2 a_{1}+2^{2} a_{2}+\ldots+2^{k} a_{k}$ be the binary representation of the number $n$. Then
$$
\begin{aligned}
& {\left[\frac{n}{2}\right]=a_{1}+2 a_{2}+2^{2} a_{3}+\ldots+2^{k-1} a_{k}} \\
& {\left[\frac{n}{2^{2}}\right]=a_{2}+2 a_{3}+\ldots+2^{k-2} a_{k}} \\
& {\left[\frac{n}{2^{3}}\right]=a_{3}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,701 |
21.29. Let $a_{1}, a_{2}, \ldots$ be distinct natural numbers, in the decimal representation of which 100 consecutive ones do not appear. Prove that among the numbers $a_{n} / n$ there are arbitrarily large numbers. | 21.29. In the $d$-ary numeral system, where $d=10^{100}$, we obtain a sequence of natural numbers that do not contain the digit $c=\underbrace{11 \ldots 1}_{100}$. As in problem 21.8, we get $a_{n} \geqslant d^{k}$ for some $n \leqslant(d-1)^{k}$. In this case, $a_{n} / n \geqslant\left(1+\frac{1}{d-1}\right)^{k}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,702 |
21.30. Let's write a natural number $n$ in the $p$-ary numeral system: $n=a_{0}+a_{1} p+a_{2} p^{2}+\ldots+a_{m} p^{m}$. Prove that the number $\left[\frac{n}{p}\right]+\left[\frac{n}{p^{2}}\right]+\left[\frac{n}{p^{3}}\right]+\ldots$ equals $\frac{n-\left(a_{0}+a_{1}+\ldots+a_{m}\right)}{p-1}$. | 21.30. It is easy to see that $\left[\frac{n}{p^{k}}\right]=0$ for $k>m$ and
$$
\begin{aligned}
{\left[\frac{n}{p^{m}}\right] } & =a_{m} \\
{\left[\frac{n}{p^{m-1}}\right] } & =a_{m} p+a_{m-1}
\end{aligned}
$$
$$
\left[\frac{n}{p}\right]=a_{m} p^{m-1}+a_{m-1} p^{m-2}+\ldots+a_{1}
$$
Therefore,
$$
\begin{aligned}
{\... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,703 |
21.31. Let $p-$ be a prime number. Prove that if $p^{k}-$ is the highest power of $p$ dividing $C_{n+m}^{m}$, then $k$ is the number of carries when adding the numbers $m$ and $n$ in the $p$-ary numeral system. | 21.31. The highest power of a prime number $p$ that divides $n!$ is $\left[\frac{n}{p}\right]+\left[\frac{n}{p^{2}}\right]+\left[\frac{n}{p^{3}}\right]+\ldots$ Therefore, according to problem 21.30, the highest power of the number $p$ that divides $C_{n+m}^{m}=$ $=\frac{(m+n)!}{m!n!}$, is
$$
\frac{m+n-s(m+n)}{p-1}-\fr... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,704 |
21.32. Let $p$ be a prime number. Write the natural numbers $a$ and $b$ in base $p$: $a=a_{0}+a_{1} p+\ldots+a_{m} p^{m}$, $b=b_{0}+b_{1} p+\ldots+b_{m} p^{m}$. Prove that
$$
C_{b}^{a} \equiv C_{b_{0}}^{a_{0}} C_{b_{1}}^{a_{1}} \ldots C_{b_{m}}^{a_{m}}(\bmod p)
$$ | 21.32. According to problem $14.30(1+x)^{p} \equiv 1+x^{p}(\bmod p)$. By induction on $k$ we get $(1+x)^{p^{k}} \equiv 1+x^{p^{k}}(\bmod p)$. Therefore,
$$
\begin{aligned}
(1+x)^{b}= & (1+x)^{b_{0}+b_{1} p+\ldots+b_{m} p^{m}}=(1+x)^{b_{0}}(1+x)^{b_{1} p} \ldots(1+x)^{b_{m} p^{m}} \equiv \\
& \equiv(1+x)^{b_{0}}\left(1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,705 |
21.33. If we consider digits in different positions as different, then in a $d$-ary numeral system, $n d$ digits allow us to write $d^{n}$ numbers (from 0 to $d^{n}-1$). Which numeral system is the most economical in this respect, i.e., allows recording the largest number of numbers using a given number of digits? (Whe... | 21.33. Answer: system with base 3. Using $m=d n$ digits, we can write $d^{m / d}$ numbers. Therefore, we need to prove that $3^{m / 3} \geqslant d^{m / d}$, i.e., $3^{d} \geqslant d^{3}$ for any natural $d$. This inequality is proven in the solution to problem 13.11. | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,706 |
21.34. Prove that any natural number $n$ can be uniquely represented in the form $n=a_{1} \cdot 1!+a_{2} \cdot 2!+\ldots$ $\ldots+a_{k} \cdot k!$, where $a_{i}$ are integers satisfying the inequalities $0 \leqslant a_{i} \leqslant i$, and $a_{k} \neq 0$. | 21.34. According to problem 13.5, the inequality $1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \ldots + k \cdot k! < (k+1)!$ holds, so for the given $n$, the number $k$ is uniquely determined. Specifically, the inequalities $k! \leqslant n < (k+1)!$ must be satisfied. Then, choose $a_{k}$ such that the inequalities $a_{k} \c... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,707 |
21.35. Prove that any rational number $p / q \neq 0$ can be uniquely represented in the form
$$
\frac{p}{q}=x_{1}+\frac{x_{2}}{2!}+\frac{x_{3}}{3!}+\ldots+\frac{x_{n}}{n!}
$$
where $x_{1}, \ldots, x_{n}$ are integers, and $0 \leqslant x_{k}<k$ for $k \geqslant 2$ and $x_{n} \neq 0$.
See also problem 15.17. | 21.35. Suppose the required equality holds. Multiplying both sides of the equality by \( n! \), we get an equality of the form \(\frac{n! \cdot p}{q} = n X + x_{n}\), where \( X \) is an integer. Thus, \(\frac{n! \cdot p}{q}\) is an integer and \( x_{n} \) is the remainder of this number when divided by \( n \). Note t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,708 |
22.1. Prove that among any six people, there will be either three who are pairwise acquainted, or three who are pairwise unacquainted. (In other words, in any graph with six vertices, there are either three vertices that are pairwise connected by edges, or three vertices that are pairwise not connected by edges.) | 22.1. Let $v-$ be an arbitrary vertex of a graph with six vertices. Among the remaining five vertices, there are either three vertices connected by edges to $v$, or three vertices not connected by edges to $v$. Let $v_{1}, v_{2}, v_{3}$ be the vertices connected by edges to $v$. If the vertices $v_{1}, v_{2}, v_{3}$ ar... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,709 |
22.2. Prove that in any graph, the number of vertices from which an odd number of edges emanate is even. | 22.2. Let's calculate in two ways the number of pairs consisting of an edge and one of its ends. On the one hand, this number is equal to twice the number of edges; in particular, it is even. On the other hand, it is equal to the sum of the numbers of edges leaving all vertices. This sum is even, so it includes an even... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,710 |
22.3. Given a segment $O A$. From the end of the segment $A$, 5 segments $A B_{1}, A B_{2}, A B_{3}, A B_{4}, A B_{5}$ extend. From each point $B_{i}$, five new segments can extend, or no new segments, and so on. Can the number of free ends of the constructed segments be equal to $1001$? | 22.3. Answer: yes, it can. When five segments are drawn from the end of a segment, 5 new free ends appear and one old end disappears. As a result, the number of free ends increases by 4. Therefore, if sets of five segments are drawn $k$ times, the number of free ends is $4 k+1$. For $k=250$, we get the required number ... | 1001 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,711 |
22.4. a) Prove that a graph in which an even number of edges come out of each vertex can be represented as a union of non-intersecting cycles.
b) Prove that a graph in which the number of vertices from which an odd number of edges come out is equal to \(2n\), can be represented as a union of non-intersecting cycles an... | 22.4. a) We apply induction on the number of edges in the graph. For a graph with two edges, the statement is obvious. Now, take an arbitrary graph and traverse its edges, starting from some vertex and not passing through the same edge twice. Since an even number of edges emanate from each vertex, we can continue the t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,712 |
22.6. Prove that if a graph is connected and the number of vertices from which an odd number of edges emerge does not exceed two, then there exists an Eulerian path in this graph.
## 22.2. Directed Graphs
- A graph is called directed if a direction of movement is specified on each of its edges, i.e., an edge exits on... | 22.6. According to problem 22.4, this graph can be represented as a union of non-intersecting cycles and, possibly, one non-self-intersecting path. If the graph consists only of a non-self-intersecting path or only of a cycle, then everything is clear. If there is a non-self-intersecting path, we will take it, and if t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,714 |
22.7. Prove that any connected graph with an even number of edges can be oriented so that an even number of edges leave each vertex.
## 22.3. Matchings
A matching is a set of edges in a graph with no common vertices. A matching is called maximum if it contains the largest possible number of edges (i.e., any matching ... | 22.7. First, orient the graph arbitrarily. Consider all vertices from which an odd number of edges emanate. Their number is even. Indeed, the sum of the number of edges emanating from all vertices is equal to the number of edges, so it is even. Thus, if there is at least one vertex $v_{1}$ from which an odd number of e... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,715 |
22.8. Prove that a matching is maximum if and only if the graph has no path that alternates between edges of the graph and has the following properties: 1) of any two consecutive edges of the path, one belongs to the matching and the other does not; 2) no edges of the matching leave the initial and final vertices of th... | 22.8. Suppose first that the graph has a path with the specified properties. From the initial and final points of this path, there are edges that do not belong to the matching, so the path consists of an odd number of edges \(v_{1} v_{2}, v_{2} v_{3}, \ldots, v_{2 n-1} v_{2 n}\). We construct a new matching by removing... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,716 |
22.9. Let the vertices of a graph be partitioned into two non-intersecting sets $X$ and $Y$, such that all edges connect vertices from different sets. Prove that a matching including all vertices of set $X$ exists if and only if for any subset of selected vertices from set $X$, the number of vertices connected to the s... | 22.9. One direction of the statement is obvious: if a matching exists, then the number of vertices from the set $Y$ connected to the selected vertices even only by the edges of the matching is already equal to the number of selected vertices.
Suppose now that the specified condition is satisfied, but the required matc... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,717 |
22.10. At a dance, several young men and several young women arrived, and each young woman is acquainted with exactly $k$ young men and each young man is acquainted with exactly $k$ young women, where $k \geqslant 1$. Prove that they can be paired up so that in each pair, the young man and the young woman are acquainte... | 22.10. First, let's show that the number of boys is exactly the same as the number of girls. Let the number of boys be $a$, the number of girls be $b$, and the number of all pairs of boys and girls who know each other be $n$. Then $k a = n = k b$, so $a = b$.
Choose an arbitrary group of $a_{1}$ boys. Let the number o... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,718 |
23.2. Let \(a\) and \(b\) be complex numbers. Prove that
\[
|a+b|^{2}-|a-b|^{2}=4 \operatorname{Re}(a \bar{b})
\] | 23.2. It is clear that $|a \pm b|^{2}=(a \pm b)(\bar{a} \pm \bar{b})=|a|^{2}+|b|^{2} \pm(a \bar{b}+\bar{a} b)$. Further, the number $a \bar{b}+\bar{a} b=a \bar{b}+\bar{a} \bar{b}$ is real, so it equals $\operatorname{Re}(a \bar{b}+\bar{a} b)=2 \operatorname{Re}(a \bar{b})$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,720 |
23.3. Let \( z \) and \( w \) be complex numbers. Prove that
\[
|z+w|^{2}+|z-w|^{2}=2|z|^{2}+2|w|^{2}
\] | 23.3. Let $z=a+i b$ and $w=c+i d$, where $a, b, c, d$ are real numbers. Then
$$
|z \pm w|^{2}=(a \pm c)^{2}+(b \pm d)^{2}=a^{2} \pm 2 a c+c^{2}+b^{2} \pm 2 b d+d^{2}
$$
Therefore, $|z+w|^{2}+|z-w|^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right)=2|z|^{2}+2|w|^{2}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,721 |
23.4. Let $a, b$ and $c$ be complex numbers. Prove that the following inequalities are equivalent:
1) $\operatorname{Re}[(a-c)(\bar{c}-\bar{b})] \geqslant 0$
2) $\left|c-\frac{a+b}{2}\right| \leqslant \frac{1}{2}|a-b|$.
## 23.2. De Moivre's Formula | 23.4. It is clear that
$$
\operatorname{Re}[(a-c)(\bar{c}-\bar{b})]=-\operatorname{Re}(a \bar{b})-|c|^{2}+\operatorname{Re}(a \bar{c}+\bar{b} c)
$$
and
$$
\begin{array}{r}
\frac{1}{4}|a-b|^{2}-\left|c-\frac{a+b}{2}\right|^{2}=\frac{1}{4}\left(|a-b|^{2}-|a+b|^{2}\right)-|c|^{2}+\operatorname{Re}[c(\bar{a}+\bar{b})]= ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,722 |
23.5. Prove that $(\cos \varphi+i \sin \varphi)^{n}=\cos n \varphi+i \sin n \varphi$ for any natural $n$ (de Moivre's formula). | 23.5. The modulus of the number $z=\cos \varphi+i \sin \varphi$ is 1, and its argument is $\varphi$. Therefore, the modulus of the number $z^{n}$ is 1, and its argument is $n \varphi$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,723 |
23.6. a) Prove that the numbers $\sin ^{2} \frac{\pi}{2 n+1}, \sin ^{2} \frac{2 \pi}{2 n+1}, \ldots$ $\ldots, \sin ^{2} \frac{n \pi}{2 n+1}$ are the roots of the polynomial
$$
C_{2 n+1}^{1}(1-x)^{n}-C_{2 n+1}^{3}(1-x)^{n-1} x+C_{2 n+1}^{5}(1-x)^{n-2} x^{2}-\ldots
$$
$$
\ldots+(-1)^{n} x^{n}
$$
b) Prove that the numb... | 23.6. a) According to de Moivre's formula $\cos (2 n+1) \alpha+i \sin (2 n+1) \alpha=$ $=(\cos \alpha+i \sin \alpha)^{2 n+1}$. Therefore,
$$
\begin{aligned}
& \sin (2 n+1) \alpha=C_{2 n+1}^{1} \cos ^{2 n} \alpha \sin \alpha- \\
& \quad-C_{2 n+1}^{3} \cos ^{2 n-2} \alpha \sin ^{3} \alpha+\ldots+(-1)^{n} \sin ^{2 n+1} \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,724 |
23.7. Using the result of problem 23.6, calculate the following sums and products:
a) $\operatorname{ctg}^{2} \frac{\pi}{2 n+1}+\operatorname{ctg}^{2} \frac{2 \pi}{2 n+1}+\ldots+\operatorname{ctg}^{2} \frac{n \pi}{2 n+1}$;
b) $\frac{1}{\sin ^{2} \frac{\pi}{2 n+1}}+\frac{1}{\sin ^{2} \frac{2 \pi}{2 n+1}}+\ldots+\frac{... | 23.7. а) The considered sum is equal to
$$
\frac{C_{2 n+1}^{3}}{C_{2 n+1}^{1}}=\frac{n(2 n-1)}{3}
$$
б) The identity $\frac{1}{\sin ^{2} \alpha}=1+\operatorname{ctg}^{2} \alpha$ shows that the considered sum is equal to $\frac{n(2 n-1)}{3}+n=\frac{2 n(n+1)}{3}$.
в) Answer: $\frac{\sqrt{2 n+1}}{2^{n}}$. The square of... | \frac{\sqrt{2n+1}}{2^{n}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,725 |
23.8. a) Prove that
$$
\begin{array}{r}
x^{2 n}-1=\left(x^{2}-1\right)\left(x^{2}-2 x \cos \frac{\pi}{n}+1\right)\left(x^{2}-2 x \cos \frac{2 \pi}{n}+1\right) \times \ldots \\
\ldots \times\left(x^{2}-2 x \cos \frac{(n-1) \pi}{n}+1\right)
\end{array}
$$
b) Prove that
$$
\sin \frac{\pi}{2 n} \sin \frac{2 \pi}{2 n} \l... | 23.8. a) The roots of unity of degree $2 n$, different from $\pm 1$, have the form $\cos \frac{k \pi}{n} \pm i \sin \frac{k \pi}{n}$, where $k=1,2, \ldots, n-1$. It remains to note that
$$
\left(x-\cos \frac{k \pi}{n}-i \sin \frac{k \pi}{n}\right)\left(x-\cos \frac{k \pi}{n}+i \sin \frac{k \pi}{n}\right)=x^{2}-2 x \co... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,726 |
23.10. Prove that the primitive roots of unity of degree $n$ are the numbers $\cos \frac{2 m \pi}{n}+i \sin \frac{2 m \pi}{n}$, where the number $m$ is coprime with $n$. | 23.10. For any integer $m$, the number $z=\cos \frac{2 m \pi}{n}+i \sin \frac{2 m \pi}{n}$ is an $n$-th root of unity. In this case, $z^{k}=\cos \frac{2 k m \pi}{n}+i \sin \frac{2 k m \pi}{n}$. For the number $z$ to be a primitive root of unity, it is necessary that all numbers $m, 2 m, \ldots, (n-1) m$ do not divide $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,728 |
23.11. Let $\varepsilon$ be a primitive $n$-th root of unity. Prove that
$$
1+\varepsilon^{k}+\varepsilon^{2 k}+\ldots+\varepsilon^{(n-1) k}= \begin{cases}0 & \text { if } 1 \leqslant k \leqslant n-1 \\ n & \text { if } k=n\end{cases}
$$ | 23.11. If $1 \leqslant k \leqslant n-1$, then $\varepsilon^{k} \neq 1$, therefore $1+\varepsilon^{k}+\varepsilon^{2 k}+\ldots$ $\ldots+\varepsilon^{(n-1) k}=\frac{\varepsilon^{n k}-1}{\varepsilon^{k}-1}=\frac{1-1}{\varepsilon^{k}-1}=0$. For $k=n$ we get the sum of $n$ terms, each of which is equal to 1. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,729 |
23.12. Let $z_{1}, \ldots, z_{n}$ be the vertices of a regular $n$-gon in the complex plane, and $z_{0}$ be its center. Prove that if $P(z)$ is a polynomial of degree not higher than $n-1$, then $P\left(z_{1}\right)+\ldots+P\left(z_{n}\right)=n P\left(z_{0}\right)$. | 23.12. Let $\varepsilon$ be a primitive $n$-th root of unity. Then $z_{m}=z_{0}+a \varepsilon^{m}$ for $m=1, \ldots, n$. Therefore,
$$
z_{m}^{k}=\left(z_{0}+a \varepsilon^{m}\right)^{k}=z_{0}^{k}+C_{m}^{1} z_{0}^{k-1} a \varepsilon^{m}+C_{m}^{2} z_{0}^{k-2} a^{2} \varepsilon^{2 m}+\ldots+a^{k} \varepsilon^{k m}
$$
Bu... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,730 |
23.13. Let $\varepsilon$ be a primitive $n$-th root of unity. Prove that $(1-\varepsilon)\left(1-\varepsilon^{2}\right)\left(1-\varepsilon^{3}\right) \ldots\left(1-\varepsilon^{n-1}\right)=n$. | 23.13. It is clear that $1+x+x^{2}+\ldots+x^{n-1}=\frac{x^{n}-1}{x-1}=(x-\varepsilon)\left(x-\varepsilon^{2}\right) \ldots$ $\ldots\left(x-\varepsilon^{n-1}\right)$. The equality $(x-\varepsilon)\left(x-\varepsilon^{2}\right) \ldots\left(x-\varepsilon^{n-1}\right)=1+x+x^{2}+\ldots$ $\ldots+x^{n-1}$ holds for all $x$, i... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,731 |
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