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values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
26.18*. (PRC, 81). The cells of an $n \times n$ chessboard, where $n$ is an even number greater than 2, are colored with $n^{2} / 2$ colors such that each color is used to color exactly two cells. Prove that it is possible to place $n$ rooks on the board such that they stand on cells of different colors and do not atta... | 26.18. Note that on a board of size $m \times m$, one can place $m$ non-attacking rooks in exactly $m!$ ways, since on the first row a rook can be placed in $m$ ways, then on the second row in $(m-1)$ ways, and so on, finally on the last row in only one way. Suppose the statement of the problem is false. Then, if the b... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,256 |
26.19*. (PRC, 79; Australia, 82). In each cell of a table of size $n \times n(n \geqslant 2)$, some letter is placed. It is known that all rows of the table are different. Prove that there is a column in it, after erasing which the remaining table will also not have identical rows. | 26.19. Suppose the statement of the problem is false. Then for any value of $i=1,2, \ldots, n$ there exists a pair of rows that differ only in the $i$-th element. Fix one such pair for each $i$ and construct the following graph: each row of the table is called a vertex of the graph, and pairs of vertices corresponding ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,257 |
26.20*. (Jury, Luxembourg, 83). In a space with a rectangular coordinate system, consider the set $E$ of points with integer coordinates taking values from 0 to 1982. Each of these points is colored either red or blue. How many colorings exist that have the following property: the number of red vertices of any parallel... | 26.20. First, we prove that the coloring of points in the set \(E\) satisfies the condition of the problem if and only if any rectangle with vertices from \(E\) and sides parallel to the axes has an even number of red vertices. Suppose the number of red vertices of some rectangle \(\Gamma_{0}\) is odd and, for example,... | 2^{5947} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,258 |
27.1. (Belgium, 80). Each of the two urns contains white and black balls, and the total number of balls in both urns is 25. One ball is randomly drawn from each urn. Knowing that the probability that both drawn balls will be white is 0.54, find the probability that both drawn balls will be black. | 27.1. Let the total number of balls in the first and second urns be \( m_{\mathrm{i}} \) and \( m_{2} \) respectively (for definiteness, assume \( m_{1} \leqslant m_{2} \)), and the number of white balls in these urns be \( k_{1} \) and \( k_{2} \) respectively. Then the probability that both drawn balls are white is \... | 0.04 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,259 |
27.2. (England, 73). A teacher and students of a certain class are asked questions. The probability that the teacher's answer will be correct is $\alpha$, and the probability of a correct answer from a student is $\beta$ or $\gamma$ depending on whether a boy or a girl is answering, respectively. The probability that t... | 27.2. Let $\delta$ denote the probability that a randomly chosen student will give the correct answer. The probability of the answer of a randomly chosen student coinciding with the teacher's answer is the sum of the probability $\alpha \delta$ of both giving the correct answer and the probability $(1-\alpha)(1-\delta)... | \frac{\frac{1}{2}-\gamma}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,260 |
27.3. (СІІІА, 83). From the vertices of a regular $n$-gon $(n \geqslant 6)$, two different triples of points are chosen at random. What is the probability that the two triangles, whose vertices are the chosen triples, do not intersect? | 27.3. Let's divide all possible pairs of triples of vertices into $C_{n}^{\natural}$ groups, collecting in one group those and only those pairs of triples that form the same set of six vertices. On the one hand, each such group contains as many elements as there are ways to split a fixed set of six vertices into two tr... | 0.3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,261 |
27.4. (GDR, 78). A circle is circumscribed around a regular $2 n$-gon. A triplet of its distinct vertices is called one-sided if there exists a semicircle on which these vertices lie (the ends of the semicircle belong to it). What is the probability that a randomly chosen triplet of vertices will be one-sided? | 27.4. If two vertices out of three have already been chosen, the number of ways to choose the third vertex so that the resulting triplet is one-sided depends on the angular distance 1 between the two chosen vertices. (The angular distance between vertices $A$ and $B$ is defined as $l=\widehat{A O B} \cdot(n / \pi)$, wh... | \frac{3n}{2(2n-1)} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,262 |
27.5. (Jury, Brazil, 82; Australia, 83). An urn contains $n$ white and $m$ black balls, and next to the urn is a box with a sufficiently large number of black balls. The following operation is performed: a pair of balls is randomly drawn from the urn; if they are of the same color, a black ball from the box is moved to... | 27.5. Since the parity of the number of white balls contained in the urn does not change after each operation, the last ball will be white if and only if the number $n$ is odd. Therefore, the desired probability is either 1 (if $n$ is odd), or 0 (if $n$ is even). | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,263 |
27.6. (Australia, 82). Player $A$ tosses a coin $n+1$ times, while player $B$ tosses it $n$ times. What is the probability that in the end, player $A$ will have more "heads" than player $B$? | 27.6. Let players $A$ and $B$ get $m$ and $k$ "heads" respectively. Then the desired probability $p$ of the event $m>\boldsymbol{k}$ is equal to the probability $q$ of the event $(n+1)-m>n-k$, i.e., the probability that player $A$ gets more "tails" than player $B$ (since "heads" and "tails" appear with equal probabilit... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,264 |
27.7. (New York, 76). The sequence ( $i_{1} ; i_{2} ; \ldots ; i_{n}$ ) is formed from the first $n>3$ natural numbers, arranged in a random order. What is the probability that for all $k=1,2, \ldots, n$ the inequality $i_{k} \geqslant k-3$ holds? | 27.7. Let's find the number of sequences ( $i_{1} ; \ldots ; i_{n}$ ), for which $i_{k} \geqslant k-3$ for $k=1,2, \ldots, n$. The number $i_{n}$ can take four values: $n, n-1, n-2, n-3$. The number $i_{n-1}$ can take five values: $f, n-1, n-2, n-3, n-4$, except for the value already occupied by the number $i_{n}$. Thu... | \frac{4^{n-3}\cdot3!}{n!} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,265 |
27.8. (USA, 75). A deck of $n$ distinct playing cards, arranged in a random order, contains three aces. The top cards of the deck are removed one by one until the second ace is removed. Prove that the average number of cards removed is $(n+1) / 2$. | 27.8. Let's divide all the possible card distributions into pairs of mutually "inverse" decks: if in some deck the cards are arranged in a certain order, then we will pair it with a deck of cards going in the reverse order. Suppose now that in some deck the second ace from the top has the number $k$. Then, to draw the ... | (n+1)/2 | Combinatorics | proof | Yes | Yes | olympiads | false | 29,266 |
27.9. (New York, 81). The sequence ( $i_{1} ; i_{2} ; \ldots ; i_{n}$ ) is composed of the first $n$ natural numbers arranged in a random order. Find the average number of inversions (disorders) in this sequence, where an inversion is defined as each pair of numbers $i_{j}>i_{k}$, for which $j<k$. | 27.9. Let's divide all rows into pairs of mutually "inverse" rows: $\left(i_{1} ; \ldots ; i_{n}\right)$ and $\left(i_{n} ; \ldots ; i_{1}\right)$. Any two natural numbers $m>l$, not exceeding $n$, form an inversion in exactly one of the rows in each such pair. Consequently, in each such pair of rows, there are exactly... | \frac{n(n-1)}{4} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,267 |
27.10. (Belgium, 81). Two players $A$ and $B$ observe a boy who continuously flips a coin. The results of the flips are recorded sequentially using letters: on the $k$-th position in the sequence, the letter $O$ or the letter $P$ is placed depending on whether "heads" or "tails" appears on the $k$-th flip, respectively... | 27.10. After the first letter $O$ (from the moment the boy is observed, with probability $I$ the letter $O$ will appear at least once), with equal probability of $1 / 4$, one of 364 combinations can follow:
$$
P O, \quad O O, \quad P P, \quad O P
$$
In the first case, player $B$ wins, in the second case, player $A$ w... | B | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,268 |
27.11. (Belgium, 77). Three shooters $A, B, C$ decided to duel simultaneously. They positioned themselves at the vertices of an equilateral triangle and agreed on the following: the first shot is taken by $A$, the second by $B$, the third by $C$, and so on in a circle; if one of the shooters is eliminated, the duel con... | 27.11. Let's consider three events that may occur after the first shot of shooter $A$.
1) $C$ is hit. Then with probability 1, shooter $A$ will be hit by the first shot of $B$.
2) $B$ is hit. Then: either with probability 0.5, shooter $C$ will hit $A$ with his first shot, or with probability $0.5 \cdot 0.3$, shooter $... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 29,269 |
27.12*. (Belgium, 78). A point moves along the edges of the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$. From any vertex, it can move to one of the three edges (emanating from this vertex) at random with equal probability, equal to $1 / 3$. The vertices $B^{\prime}$ and $C^{\prime}$ are special in that, ... | 27.12. Note that from any vertex of the cube, the point will land in one of the vertices $B^{\prime}, C^{\prime}$ with a probability not less than $1 / 9$ (since for any vertex of the cube, different from $B^{\prime}$ and $C^{\prime}$, there exists a route passing through no more than two edges and connecting it to $B^... | \frac{4}{7},\frac{3}{7} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,270 |
27.13*. (Belgium, 82). On the inner wrapper of each chocolate bar in the "Great Mathematician" series, one of the $n$ outstanding mathematicians is depicted, and the portrait of each of them appears with equal probability, equal to $1 / n$. On average, how many chocolate bars need to be bought to collect the complete s... | 27.13. Let's call a chocolate bar special if it features a portrait that has not been encountered before. Suppose among the purchased chocolate bars there are exactly $k$ different ones. Denote by $M_{k}$ the average number of chocolate bars that need to be bought after this, so that the last one turns out to be specia... | n(1+\frac{1}{2}+\ldots+\frac{1}{n-1}+\frac{1}{n}) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,271 |
1. 200 students are arranged in a rectangle with 10 people in each transverse row and 20 people in each longitudinal row. In each transverse row, the shortest student is selected, and then among the 20 selected, the tallest is chosen; on the other hand, from the same 200 students, in each longitudinal row, the tallest ... | 1. Let $A$ be the first of the two selected students, and $B$ be the second. If $A$ and $B$ stand in the same transverse row, then $B$ is taller than $A$, because $A$ is the shortest student in their transverse row; if $A$ and $B$ stand in the same longitudinal row, then $B$ is also taller than $A$, because $B$ is the ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 29,272 |
2. Each person who has ever lived on Earth has exchanged a certain number of handshakes with others. Prove that the number of people who have exchanged an odd number of handshakes is even. | 2. Let's form the total sum of the number of handshakes made by each person. This sum is necessarily even because each handshake between two individuals $A$ and $B$ increases the number of handshakes made by $A$ by 1 and the number of handshakes made by $B$ by 1, and thus contributes a term of 2 to the total sum. On th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,273 |
3. Prove that among any six people, there will be either three who are pairwise acquainted, or three who are pairwise unacquainted. | 3. Let $A$ be one of the given individuals; it is clear that either $1^{\circ} A$ has among the present three acquaintances $B_{1}, B_{2}$, and $B_{3}$; or $2^{\circ}$ there are three individuals $C_{1}, C_{2}$, and $C_{3}$, with none of whom $A$ is acquainted (among the 5 individuals other than $A$, he has either thre... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,275 |
4. Several people are present at the meeting (obviously, more than one person - otherwise, what kind of "meeting" would it be?).
a) Is it possible that no two of those present have the same number of acquaintances?
b) Prove that for any number of meeting participants, it can happen that no three of those present have... | 4. a) Each of the $N$ present individuals can have $0,1,2, \ldots, N-1$ acquaintances. Here we have $N$ possible values for the number of acquaintances; however, if someone has 0 acquaintances, then no one has $N-1$ acquaintances (and vice versa—if someone has $N-1$ acquaintances, then no one has 0 acquaintances). From... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,276 |
5. At a certain meeting, there were $2 n$ people, each of whom was acquainted with at least $n$ of those present. Prove that four of those present can be seated at a round table for 4 people so that each person is seated between their acquaintances. | 5. If all the present people are acquainted with each other, then the possibility of seating 4 people in this manner raises no doubts. Now let $A$ and $B$ be unacquainted with each other. Each of them has among the other $2 n-2$ present people no fewer than $n$ acquaintances; since $n+n=2 n=(2 n-2)+2$, $A$ and $B$ have... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,277 |
6. A large number of scientists arrived at the congress; some of them were previously acquainted with each other, others -[^1]not. It turned out that no two scientists with the same number of acquaintances had any common acquaintances. Prove that among the scientists present at the congress, there is a scientist who is... | 6. Consider a scientist $A$, who has the largest number $n$ of acquaintances among those present (or one of such scientists if there are several); at this point, $n>0$, since we assumed that some of the participants are acquainted with each other. All acquaintances of $A$ have a different number of acquaintances (since... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,278 |
7. A thousand delegates from different countries arrived at the congress. It is known that any three of them can communicate with each other without the help of others (however, it is possible that one of the three will have to serve as a translator for the other two). Prove that all the delegates of the congress can b... | 7. Let's choose three delegates of the congress; among them, there will be two who know one language - we will place them in the same room of the hotel. From the remaining 998 delegates, we will again select three, among whom there will again be two who can be placed in the same room - and so on, until we are left with... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,279 |
8. At an international conference, there are 17 scientists. Each pair of them converses in a certain language; in total, all 17 people know three languages. Prove that among the participants, there will be three who speak to each other in the same language. | 8. Let $A$ be one of the scientists present at the conference; with each of the 16 others, he speaks in one of three (or fewer, if he does not know all three) languages. It is not hard to see that there is a language (let's call it "Tarabar") on which $A$ speaks with at least 6 scientists: indeed, otherwise, the number... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,280 |
9. At a certain meeting, there were $n$ people. It is known that any two acquaintances among them do not have any common acquaintances, and any two people who are not acquainted with each other have exactly two common acquaintances.
a) Prove that all attendees have the same number of acquaintances.
b) For which $n$ i... | 9. a) Let's choose one of the attendees $A$; all the people acquainted with him we will denote as $A_{1}, A_{2}, \ldots, A_{\text {k }}$. It is clear that no two of these people are acquainted with each other and any two people, say $A_{i}$ and $A_{j}$, have two common acquaintances: $A$ and some person $A_{i j}$; here... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,281 |
10. In the city of "Diversity," 10,000 residents live, and any two of them are either enemies or friends. Each day, no more than one resident can "start a new life," which means breaking up with all their friends and befriending all their enemies; meanwhile, any three residents can become friends with each other. Prove... | 10. Let $A, B$ and $C$ be three residents of the city. It is clear that it is possible for all of them to be friends with each other; it is also possible that one of them (say, $A$) is not friends with either $B$ or $C$, while $B$ and $C$ are friends with each other: in this case, for $A, B$, and $C$ to all become frie... | 5000 | Combinatorics | proof | Yes | Yes | olympiads | false | 29,282 |
11*. In the country of Oz, there are several castles, from each of which three roads lead. A wandering knight set out from his ancestral castle and embarked on a journey through the country. The knight loves variety; therefore, upon reaching each subsequent castle, he turns left if he turned right the previous time, an... | 11. It is natural to call a "road" a segment of the path between two castles; all the castles in the country of Oz are connected by some finite number \( n \) of roads. If a knight travels through the country for a long time, he will travel a large number of roads; if this number is \(\geqslant 4 n+1\), then he will tr... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 29,283 |
12*. At King Arthur's court, $2 n$ knights have gathered, each of whom has no more than $n-1$ enemies among those present. Prove that Merlin, Arthur's advisor, can seat the knights around a round table in such a way that no one sits next to their enemy. | 12. Let's agree to call "friends" any two knights who are not enemies; further, let's start by seating all the knights around a round table arbitrarily. Suppose somewhere at the table
a)
!... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,284 |
13. a) It is known that among 80 coins there is one counterfeit coin, which is lighter than the others, which all have the same weight. Using four weighings on a balance scale without weights, find the counterfeit coin.
b) It is known that among $n$ coins there is one counterfeit coin, which is lighter than the others... | 13. a) Let's divide our coins into three groups: two groups of 27 coins each and one of 26 coins. During the first weighing, we will place the groups of 27 coins on the scales. If the scales do not balance, the counterfeit coin is on the "lighter" side. If the scales balance, the counterfeit coin is in the group of 26 ... | [\log_{3}(n-\frac{1}{2})+1] | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 29,285 |
14. 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 no more than 11 weighings on a balance scale without weights?
Note. The problem assumes that all cubes can be aluminum, but they cannot all be dura... | 14. Let's place one cube on each pan of the scales (first weighing). In this case, two different scenarios may occur.
$1^{\circ}$. During the first weighing, one pan of the scales tips. In this case, one of the two weighed cubes is definitely aluminum, and the other is duralumin. Next, we place these two cubes on one ... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 29,286 |
15*. Among 12 coins, there is one counterfeit coin. It is known that the counterfeit coin differs in weight from the genuine ones, but it is not known whether it is lighter or heavier. The genuine coins all weigh the same. Using three weighings on a balance scale without weights, identify the counterfeit coin and deter... | 15. Let's divide our coins into three groups of four coins each. During the first weighing, we will place one group of four coins on each pan of the scales. There are two possible scenarios:
$1^{\circ}$. The scales balance.
$2^{\circ}$. One of the pans outweighs the other.
Let's consider both scenarios separately.
... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 29,287 |
16. a) One day, a guest came to the hotel owner, K, without money but with a silver chain consisting of seven links. The owner agreed to keep the guest for a week on the condition that the guest would give him one of the chain links as payment each day. What is the minimum number of links that need to be cut so that th... | 16. a) It is sufficient to saw off one third link; in this case, the chain will break into two parts, containing 2 and 4 links respectively, and one separate (sawed) link. On the first day, the guest will give this link; on the second day, he will take it back and give in exchange the part of the chain consisting of tw... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 29,288 |
17. In the city of Lisse, all metro stations are connected to each other, i.e., from any station, you can travel to any other station (possibly with transfers). Prove that in this case, one metro station can be closed (without the possibility of passing through it) so that from any of the remaining stations, you can st... | 17. Let $S$ be some metro station, and $T$ be the farthest station from $S$, i.e., such that the shortest path from $S$ to $T$ passes through a greater (or at least not fewer) number of stations than the shortest path from $S$ to any other station. Now, let's close station $T$. With this, we can still travel from $S$ t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,289 |
18*. In all the streets of the city of Zurbagan, there was two-way traffic. However, when the city authorities needed to carry out repairs on all roads, they were forced to temporarily introduce one-way traffic on some streets, leaving traffic two-way on the remaining streets; after this, on the contrary, one-way traff... | 18. We will prove the statement of the problem by mathematical induction on the number of city intersections (from which more than two roads originate). If in Zurbagan there are only $d a$ intersections $A$ and $B$, then the statement of the problem is obvious: at least two roads lead from $A$ to $B$ (if there were onl... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 29,290 |
19*. In the country of Dolphinia, there are $n$ cities, each pair of which is connected by a highway, and movement along all highways is one-way. Prove that if $n \neq 2$ or 4, the directions of movement along the highways can be chosen such that from any city to any other city, one can travel with no more than one int... | 19. It is clear that if a country has 2 cities connected by a one-way highway, it will be impossible to travel from one city to the other. If there are 4 cities, we can consider them as the vertices of a quadrilateral \(A_{1} A_{2} A_{3} A_{4}\); in this case, either the movement along the perimeter of the quadrilatera... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,291 |
21. Can a knight's move in chess get from the bottom left corner of the board to the top right, visiting each square exactly once? | 21. To traverse all 64 squares of a chessboard, visiting each square once, a knight must make 63 moves. Since with each move the knight transitions from a white square to a black one or from a black square to a white one, after even-numbered moves the knight will land on squares of the same color as the starting square... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 29,293 |
22. The problem of the suicidal king. On a chessboard of size $1000 \times 1000$ cells, there is a white king and 499 black rooks. Prove that, regardless of the initial position of the pieces (and regardless of the play of the black pieces), the king can "play for a loss," i.e., in a series of moves, place itself on a ... | 22. The King can adopt the following "suicidal" (but suicide is his goal!) strategy: first, he will move to the lower left corner of the board, and from there, he will move diagonally up to the right. After the King makes his 1st diagonal move, he will end up on the starred field marked on Fig. 10, a; at this point, if... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 29,294 |
23. 12 fields are arranged in a circle, and on four adjacent fields stand four differently colored chips: red, yellow, green, and blue.
One move allows you to move any chip from the field it is on, over four other fields, to the fifth (if it is free) in either of the two possible directions. After several moves, the c... | 23. Let's change the order of the fields around the circle, specifically, arranging them in such a way that one can move from one field to an adjacent one. In other words, after field $l$, we will place 6 (since according to the problem, one can move from field 1 to 6), after 6 we will place field 11 (since one can mov... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 29,295 |
24. Among the students admitted to the university, there are exactly 50 who know English, exactly 50 who know French, and exactly 50 who know German (of course, there may be students who know two or even all three languages, so the total number of students who know at least one language is, in general, less than \(3 \c... | 24. First of all, let's prove that if in some group of students, each of the three languages - a (English), f (French), and n (German) is known by exactly $n$ people, where $n \geqslant 2$, then it is possible to form a group of students in which each language is known by exactly 2 people. It is clear that from the for... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,296 |
25. a) At a sports competition, 20 athletes are participating; the competition is judged by 9 judges. Each judge ranks the competitors from 1st to 20th according to their assessment of the performances, and it turned out later that the ratings of all judges did not differ significantly: no athlete received places from ... | 25. a) It is clear that the smallest value of the "average position" is 1 - this value is achieved if all judges assign the 1st place to the same athlete. On the other hand, 5 or more athletes cannot receive (from different judges) the first place: indeed, these $n \geqslant 5$ athletes would then collectively receive ... | 2048 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 29,297 |
26*. The Spartakiad lasted $n$ days; during which $N$ sets of medals were awarded; on the 1st day, 1 set of medals and $1 / 7$ of the remaining quantity were awarded; on the 2nd day - 2 sets of medals and $1 / 7$ of the remaining quantity; ...; on the penultimate, $(n-1)$-th day - $(n-1)$ sets of medals and $1 / 7$ of ... | 26. Let the number of medal sets remaining to be awarded at the beginning of the $i$-th day of the competition be denoted by $N_{i}$, where $i=1,2, \ldots, \ldots, n$; it can, however, be assumed that the quantity $N_{i}$ makes sense even for $i>n$, becoming 0 in these cases. From the conditions of the problem (and our... | n=6,N=36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,298 |
27. Five friends, one of whom had a monkey, once bought a bag of nuts, which they intended to divide among themselves the next morning. However, one of the friends woke up at night and wanted nuts; he divided all the nuts in the bag into five equal parts, with one extra nut left over, which he gave to the monkey, and t... | 27. First solution. Let $n$ be the number of nuts each friend received in the morning; in this case, $5n + 1$ is the number of nuts that were in the bag in the morning. The last of those who woke up at night, obviously, took $\frac{5n + 1}{4}$ for himself, and before that, there were $5 \frac{5n + 1}{4} + 1 = \frac{25n... | 15621 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,299 |
28. Two brothers sold a flock of sheep that belonged to both of them, taking as many rubles for each sheep as there were sheep in the flock. The money received was divided as follows: first, the elder brother took ten rubles from the total amount, then the younger brother took ten rubles, after that the elder brother t... | 28. Let the number of sheep in the flock be denoted by $n$; in this case, the brothers received $n$ rubles for each sheep, and thus the total amount they received is $N=n \cdot n=n^{2}$ rubles. Let $d$ be the number of whole tens in the number $n$, and $e$ be the number of units; then $n=10 d+e$ and
$$
N=(10 d+e)^{2}=... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 29,300 |
30. Which integers decrease by an integer factor when the last digit is erased? | 30. It is easy to see that when the last digit of an integer is crossed out, the number decreases by at least 10 times. Exactly 10 times the numbers decrease when the last digit is crossed out in numbers ending in zero; therefore, all these numbers satisfy the condition of the problem.
Now suppose that an integer $x$ ... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,302 |
31. a) Find all integers that start with the digit 6 and decrease by 25 times when this digit is erased.
b) Prove that there do not exist integers that decrease by 35 times when the first digit is erased. | 31. a) Let the desired number have $k+1$ digits; in this case, it has the form $6 \cdot 10^{k} + y$, where $y$ is a $k$-digit number (which may start with one or several zeros). According to the problem, we have:
$$
6 \cdot 10^{k} + y = 25 \cdot y
$$
from which it follows that
$$
y = \frac{6 \cdot 10^{k}}{24}
$$
Th... | 6250\ldots0(n=0,1,2,\ldots) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,303 |
32*. An integer decreases by 9 times when a certain digit is crossed out; at the same time, the resulting number is also divisible by 9.
a) Prove that to divide the resulting number by 9, it is also sufficient to cross out one digit in it.
b) Find all integers that satisfy the condition of the problem. | 32. a) First of all, let us show that no number \( N \) can decrease by 9 times when a digit, other than the second one, is erased. Indeed, if this were not the case, denoting the digits of the number \( N \) by \( a_{0}, a_{1}, \ldots, a_{n} \), i.e.,
\[
a_{0} \cdot 10^{n} + a_{1} \cdot 10^{n-1} + \ldots + a_{n} = N
... | 10125,2025,30375,405,50625,6075,70875 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,304 |
33. a) Find all numbers that, when the third digit is erased, decrease by an integer factor.
b)* Find all numbers that, when the second digit is erased, decrease by an integer factor. | 33. a) Similarly to the solution of problem 32 a), assuming that the integer $N$ is reduced by a factor of $m$ when the third digit is erased, we have:
$$
\begin{gathered}
N=a_{0} \cdot 10^{n}+a_{1} \cdot 10^{n-1}+a_{2} \cdot 10^{n-2}+\ldots+a_{n} \\
10 \cdot \frac{N}{m}=a_{0} \cdot 10^{n}+a_{1} \cdot 10^{n-1}+a_{3} \... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,305 |
34. a) Find the smallest integer starting with the digit 1 such that if this digit is moved to the end, the number triples. Find all such numbers.
b) What digits can non-zero integers start with if they triple when the first digit is moved to the end? Find all such numbers. | 34. a) First solution. Let's denote the (m-digit) number obtained by removing the first digit 1 from the desired number as \(X\). In this case, according to the problem, we have:
\[
\left(1 \cdot 10^{m} + X\right) \cdot 3 = 10 X + 1
\]
from which
\[
X = \frac{3 \cdot 10^{m} - 1}{7}
\]
From the last equation, it is ... | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,306 |
35. Find the smallest natural number ending in the digit 6 that quadruples when its last digit is moved to the beginning of the number. | 35. First solution. Let the number $X$ satisfy the conditions of the problem:
$$
X=\widehat{a_{1} a_{2} \ldots a_{n-1} 6} \quad \text { and } \quad 4 X=\widehat{6 a_{1} a_{2} \ldots a_{n-1}},
$$
where $a_{1}, a_{2}, \ldots, a_{n-1}, 6$ are the digits of the number $X$. Since the last digit of $X$ is 6, the last digit... | 153846 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,307 |
36. Prove that there are no non-zero integers which, when the first digit is moved to the end, increase by 5 times, 6 times, or 8 times. | 36. If when a number is increased by 5 times, the number of its digits does not change, then the sought number must start with the digit 1. By moving this digit to the end, we get a number ending with the digit 1. But such a number cannot be divisible by 5.
Similarly, it is proven that there are no numbers that increa... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,308 |
37. Prove that there are no non-zero integers that double when the first digit is moved to the end. | 37. First solution. Since the product of the desired number by 2 has the same number of digits, the first digit of the number cannot be greater than 4. Since by moving the first digit to the end we should get an even number (the doubled original number), the first digit must be even. Therefore, it can only be 2 or 4.
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,309 |
38. a) Prove that there are no non-zero integers which, when the first digit is moved to the end, increase by 7 or 9 times.
b) Prove that there are no non-zero integers which, when the first digit is moved to the end, increase by 4 times. | 38. a) First solution. A number that increases by 7 times when the first digit is moved to the end must start with the digit 1 (otherwise, the number, which is 7 times larger, would have more digits than the original). Denoting the number obtained from the original by removing the first digit as $X$, we will have (comp... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,310 |
39. Find the smallest integer starting with the digit 7 that decreases threefold when this digit is moved to the end. Find all such numbers. | 39. First solution. Let the unknown digits of the desired number be denoted by $x, y, \ldots, z, t$. In this case, using the notation from the second solution of problem 34 a), we have:
$$
\text { - } \overline{7 x y \ldots z t} \cdot \frac{1}{3}=\overline{x y \ldots z t 7}
$$
or
$$
\overline{x y \ldots z t 7} \cdot... | 7241379310344827586206896551 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,311 |
41. a) Find a six-digit number that increases by 6 times when the last three digits of the number, without changing their order, are moved to the beginning of the number.
b) Prove that there does not exist an eight-digit number that increases by 6 times when the last four digits are moved to the first four positions w... | 41. a) Let the number formed by the first three digits of the desired number $N$ be denoted by $p$, and the number formed by the last three digits of $N$ be denoted by $q$. In this case, the condition of the problem gives
$$
1000 q + p = 6(1000 p + q) \quad(=6 N)
$$
from which we get
$$
(1000 q + p) - (1000 p + q) =... | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,313 |
42. Find a six-digit number, the products of which when multiplied by 2, 3, 4, 5, and 6 are written with the same digits as the number itself, but in a different order. | 42. Let $x$ be the number satisfying this condition. Since $6x$, like $x$, is a six-digit number, the first digit of the decimal representation of the number $x$ is 1. Therefore:
1) the first digits of the decimal representation of the numbers $x, 2x, 3x, 4x, 5x$, and $6x$ are all different, so they form a complete set... | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,314 |
43. Find all three-digit numbers that are equal to the arithmetic mean of all numbers obtained from the given number by all possible permutations of its digits (including, of course, the "identity permutation" that leaves all the digits of the number in place). | 43. Let \( N = \overline{xyz} = 100x + 10y + z \) be the desired number, where \( x, y, z \) are the digits of the number. By permuting the digits of \( N \), we obtain the numbers \( N_1 = \overline{yxz} = 100y + 10x + z, \ldots, N_5 = \overline{zyx} = 100z + 10y + x \). The sum \( N + N_1 + \ldots + N_5 \) of all the... | 111,222,333,444,555,666,777,888,999,407,518,629,370,481,592 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,315 |
44. Let $A$ be some positive integer, and $A^{\prime}$ be the number formed by the same digits but in a different order. Prove that if $A + A^{\prime} = 10^{10}$, then $A$ is divisible by 10. | 44. It is clear that the numbers $A$ and $A^{\prime}$ must be 10-digit numbers; let $A=\overline{a_{11} a_{9} a_{8} \ldots a_{1}}$ and $A^{\prime}=\overline{a_{10}^{\prime} a_{9}^{\prime} a_{8}^{\prime} \ldots a_{1}^{\prime}}\left(a_{10}, a_{9}, a_{8}, \ldots, a_{1}\right.$ - the consecutive digits of the number $A$, a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,316 |
45. Let $M$ be some 17-digit number, and $N$ be the number obtained by "reversing" $M$, i.e., written with the same digits but in reverse order. Prove that at least one digit in the decimal representation of the number $M+N$ is even.
## 3. PROBLEMS ON DIVISIBILITY OF NUMBERS
The topics initiated by the problems in th... | 45. Let us imagine the numbers $M$ and $N$ written one under the other and the process of adding these numbers "column by column"; suppose that all the digits of the resulting sum $M+N$ are odd. It follows that the sum of the digits in the last column is odd; and then the sum of the digits in the first column, which di... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,317 |
46. Prove that for any integer $n$.
a) $n^{3}-n$ is divisible by 3;
b) $n^{5}-n$ is divisible by 5;
c) $n^{7}-n$ is divisible by 7;
d) $n^{11}-n$ is divisible by 11;
e) $n^{13}-n$ is divisible by 13.
Note. Note that $n^{9}-n$ is not necessarily divisible by 9 (for example, $2^{9}-2=510$ is not divisible by 9).
Pr... | 46. a) $n^{3}-n=(n-1) n(n+1)$, and among three consecutive integers, one is necessarily divisible by 3.
b) $n^{5}-n=n(n-1)(n+1)\left(n^{2}+1\right)$. If an integer ends in one of the digits $0,1,4,5,6$ or 9, then one of the first three factors is divisible by 5. If it ends in one of the digits $2,3,7$ or 8, then $n^{2}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,318 |
47. Prove that for any integer $n$
a) $3^{6 n}-2^{6 n}$ is divisible by 35 (here $n \geqslant 0$ );
b) $n^{5}-5 n^{3}+4 n$ is divisible by 120;
c) $n^{2}+3 n+5$ is not divisible by 121. | 47. a) The difference of identical even powers is divisible by the sum of the bases; therefore, $3^{6 n}-2^{6 n}=27^{2 n}-8^{2 n}$ is divisible by $27+8=35$.
b) It is not difficult to verify that
$n^{5}-5 n+4 n=n\left(n^{2}-1\right)\left(n^{2}-4\right)=(n-2)(n-1) n(n+1)(n+2)$.
But among five consecutive integers, on... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,319 |
48. Prove that for all integers $m$ and $n$
a) ${ }^{*} m n\left(m^{60}-n^{60}\right)$ is divisible by 56786730;
b) $m^{5}+3 m^{4} n-5 m^{3} n^{2}-15 m^{2} n^{3}+4 m n^{4}+12 n^{5}$ is not equal to 33. | 48. a) It is not difficult to verify that
$$
56786730=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 31 \cdot 61
$$
thus, it remains to prove that the given expression is divisible by each of these prime divisors. If $m$ and $n$ are both odd, then $m^{80}-n^{80}$ is even; therefore, $m n\left(m^{60}-n^{60}\right)$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,320 |
49. For which positive integers $n$ is the number $20^{n}+$ $+16^{n}-3^{n}-1$ divisible by 323? | 49. First of all, note that $323=17 \cdot 19$; thus, we need to determine when the number $N$ given in the problem's condition is divisible by both 17 and 19. Let's first consider the case where the number $n=2 k$ is even. It is clear that $20^{n}-3^{n}$ is divisible by $20-3=17$ for all $n$; on the other hand, $16^{n}... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,321 |
51. What remainders can the hundredth power of an integer give when divided by 125? | 51. Any integer either is divisible by 5 or can be represented in one of the following four forms: $5k+1, 5k+2, 5k-2$, or $5k-1$. If a number is divisible by 5, then its hundredth power is clearly divisible by $5^3 = 125$. Further, using the binomial theorem, we get:
$$
(5k \pm 1)^{100} = (5k)^{100} \pm \ldots + \frac... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,323 |
52. Prove that if an integer $N$ is coprime with 10, then the 101st power of the number $N$ ends with the same three digits as $N$ (for example, $1233^{101}$ ends with the digits 233, and $37^{101}$ ends with the digits 037). | 52. We need to prove that if $N$ is coprime with 10, then $N^{101}-N=N\left(N^{100}-1\right)$ is divisible by 1000, i.e., that $N^{100}-1$ is divisible by 1000. First of all, it is clear that if $N$ is odd, then $N^{100}-1=\left(N^{50}+1\right)\left(N^{25}+1\right)\left(N^{25}-1\right)$ is divisible by 8. Furthermore, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,324 |
53. Find a three-digit number, every integer power of which ends in the three digits that make up the original number. | 53. Let \( N \) be the desired number; then, in particular, \( N^2 - N \) ends with three zeros, i.e., it is divisible by 1000. Since \( N^2 - N = N(N-1) \), and \( N \) and \( N-1 \) are coprime numbers, this is only possible if one of these numbers is divisible by 8, and the other by 125 (neither of these numbers can... | 625376 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,325 |
54*. Let $N$ be an even number not divisible by 10. What will be the tens digit of the number $N^{20}$? What will be the hundreds digit of the number $N^{200}$? | 54. Let's find the last two digits of the number \(N^{20}\). The number \(N^{20}\) is divisible by 4 (since \(N\) is even). Further, the number \(N\) is not divisible by 5 (otherwise it would be divisible by 10) and, therefore, can be represented in the form \(5k \pm 1\) or in the form \(5k \pm 2\) (see the solution to... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,326 |
55. Prove that the sum
$$
1^{k}+2^{k}+3^{k}+\ldots+n^{k}
$$
where $n$ is any positive integer, and $k$ is odd, is divisible by $1+2+3+\ldots+n$. | 55. The sum $1+2+3+\ldots+n$ equals $\frac{n(n+1)}{2}$; therefore, we need to prove that if $k$ is odd, then $S_{k}=1^{k}+2^{k}+3^{k}+\ldots+$ $+n^{k}$ is divisible by $\frac{n(n+1)}{2}$.
First, note that for odd $k$, $a^{k}+b^{k}$ is divisible by $a+b$. Now let's consider two cases separately:
A. The number $n$ is e... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,327 |
56. To state the divisibility rule for numbers by 11. | 56. Let
$$
N=a_{n} \cdot 10^{n}+a_{n-1} \cdot 10^{n-1}+a_{n-2} \cdot 10^{n-2}+\ldots+a_{1} \cdot 10+a_{0}
$$
- be the given number (where $a_{n}, a_{n-1}, a_{n-2}, \ldots, a_{i}, a_{0}$ are the digits of the number, which can take values $0,1,2, \ldots, 9$).
Subtract from $N$ the number
$$
M=a_{0}-a_{1}+a_{2}-a_{3}... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,328 |
57. The number $123456789(10)(11)(12)(13)(14)$ is written in the base-15 numeral system, i.e., this number is equal to
(14) $+(13) \cdot 15+(12) \cdot 15^{2}+(11) \cdot 15^{3}+\ldots+2 \cdot 15^{12}+15^{13}$. What remainder does it give when divided by 7? | 57. The number 15 gives a remainder of 1 when divided by 7. Therefore, it follows that
$$
15^{2}=(7 \cdot 2+1) \cdot(7 \cdot 2+1)=7 n_{1}+1
$$
gives a remainder of 1 when divided by 7,
$$
15^{3}=15^{2} \cdot 15=\left(7 n_{1}+1\right) \cdot(7 \cdot 2+1)=7 n_{2}+1
$$
gives a remainder of 1 when divided by 7, and gene... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,329 |
60. Let $a$ be the last digit of the even number $N=2^{\text {h }}$, $A$ be the number that $N$ becomes if this last digit is erased. Prove that for all $k>3$ the number $aA$ is divisible by 6. | 60. Since $a$, obviously, is an even number, we only need to prove that the product $aA$ is divisible by 3. Since the last digit of the number $2^{k+1}=2N$ is the same as that of the number $2a$ (where $a$ is the last digit of $N$), then, sequentially multiplying the powers of two by 2 (i.e., sequentially increasing th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,332 |
62. Prove that $11^{10}-1$ is divisible by 100. | 62. It is not difficult to verify that
\[
\begin{aligned}
& 11^{10}-1^{10}= \\
& \quad=(11-1)\left(11^{9}+11^{8}+11^{7}+11^{6}+11^{5}+11^{4}+11^{3}+11^{2}+11+1\right) .
\end{aligned}
\]
It is easy to see that the second factor on the right-hand side is divisible by 10, since it is the sum of 10 terms, each of which e... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,333 |
63. Prove that $2222^{5555}+5555^{2222}$ is divisible by 7. | 63. $2222^{5555}+5555^{2222}=$
$$
=\left(2222^{2555}+4^{55555}\right)+\left(5555^{2222}-4^{22222}\right)-\left(4^{5555}-4^{22222}\right) .
$$
Of the three expressions in parentheses, the first is divisible by $2222+4=2226=7 \cdot 318$ (the sum of odd powers is divisible by the sum of the bases) and, consequently, is ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,334 |
64. Prove that a number composed of $3^{\text {n }}$ identical digits is divisible by $3^{n}$ (for example, the number 222 is divisible by 3, the number 777777777 is divisible by 9, etc.). | 64. Let's use the method of mathematical induction. The number $\overline{a a a}$, composed of three identical digits $a$ (the bar above is placed to avoid confusion with the product $a \cdot a \cdot a$), is divisible by 3 (since the sum of the digits of this number, which is $3a$, is divisible by 3). Next, assume that... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,335 |
65. Find the remainder when the number
$$
10^{10}+10^{\left(10^{0}\right)}+\ldots+10^{\left(10^{10}\right)}
$$
is divided by 7. | 65. First of all, note that \(10^{6}-1=999999\) is divisible by 7 (since \(999999=7 \cdot 142857\)). From this, it easily follows that \(10^{\mathrm{N}}\), where \(N\) is any integer, gives the same remainder when divided by 7 as \(10^{r}\), where \(r\) is the remainder from dividing \(N\) by 6. Indeed, if \(N=6k+r\), ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,336 |
68. a) What is the last digit of the number
$$
\left(\ldots\left(\left(\left(7^{7}\right)^{7}\right)^{7}\right)^{\ldots 7}\right)
$$
(raising to the power of 7 is repeated 1000 times)? What are the last two digits of this number?
b) What is the last digit of the number
$$
7\left(7^{\left(.7^{\left(7^{7}\right)}\righ... | 68. a) If you multiply two numbers, one of which ends in the digit $a$, and the second in the digit $b$, then their product will end in the same digit as the product $a b$. This observation allows us to easily solve the given problem. We will sequentially raise to powers, keeping track only of the last digit of the num... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,339 |
69*. Determine the five last digits of the number
$$
\left.N=9^{\left(9^{(} \cdot 9^{\left(9^{9}\right)}\right)} \ldots\right),
$$
written using 1001 nines in a similar manner to the number in problem 68). | 69. Let's consider the sequence of numbers:
$$
\begin{aligned}
& 1^{\circ} \cdot Z_{1}=9 \\
& \begin{aligned}
2^{\circ} \cdot Z_{2} & =9^{Z_{1}}=(10-1)^{Z_{1}}= \\
& =10^{Z_{1}}-C_{Z_{1}}^{1} \cdot 10^{Z_{1}-1}+\ldots+C_{Z_{1}}^{1} \cdot 10-1
\end{aligned}
\end{aligned}
$$
where the omitted terms in the expansion are... | 45289 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,340 |
70. For which natural numbers $n$ is the sum $5^{n}+n^{5}$ divisible by 13? What is the smallest $n$ that satisfies this condition? | 70. First, let's find the remainders of the division by 13 of the numbers \(5^n\) and \(n^5\) for the first few values of \(n = 0, 1, 2, \ldots\). It is more convenient to start with the numbers \(5^n\):
\(n\)
| 0 | 1 | 2 | 3 | 4 | . | . | . |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 1 | 5 | 25 | 1... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,341 |
71. What are the last two digits of the number
$$
n^{a}+(n+1)^{a}+(n+2)^{a}+\ldots+(n+99)^{a}
$$
a) $a=4$
b) $a=8$ ? | 71. It is clear that the last two digits of the numbers $n, n^{2}, n^{3}, \ldots$, where $n$ is a non-negative integer, depend only on the last two digits of the number $n$ - this follows from the rule of multiplying multi-digit numbers "in a column". On the other hand, the last digits of 100 consecutive non-negative i... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,342 |
72*. Find the last 1000 digits of the number
$$
N=1+50+50^{2}+50^{3}+\ldots+50^{999}
$$ | 72. According to the formula for the sum of the terms of a geometric progression
$$
N=\frac{50^{1000}-1}{50-1}=\frac{80^{1000}-1}{49}
$$
But $\frac{1}{49}$ turns into a pure periodic decimal fraction, the period of which, consisting of 42 digits, is not difficult to find by simple division:
$$
\frac{1}{49}=0.(020408... | p\underbrace{PP\ldotsP}_{23\text{times}} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,343 |
73. A natural number $M$ is divisible by 7; prove that if the number of digits of the number is divisible by 6, then by moving the last digit to the front, we get a number $N$, which is also divisible by 7. | 73. Let the original number be $M=10 A+a$, where $a$ is the last digit of $M$; then, the number $N$ obtained from $M$ in the specified manner will, obviously, be equal to $a \cdot 10^{6n-1} + A$, where $6n$ is the number of digits in $M$. We now consider the expression
$$
M-3 N=(10 A+a)-\left(3 \cdot 10^{6 n-1} a+3 A\... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,344 |
74. How many zeros does the product of all integers from 1 to 100 inclusive end with?
In the future, we will use the following notation:
$$
1 \cdot 2 \cdot 3 \cdot 4 \ldots(n-1) \cdot n=n!
$$
(read as: $n$ factorial). Thus, the previous problem can be more succinctly stated as: how many zeros does the number 100! en... | 74. The number of bullets at the end of a number indicates how many times 10 is a factor in that number. The number 10 is equal to the product of $2 \cdot 5$; in the product of all integers from 1 to 100, the factor 2 appears in a higher power than the factor 5. Therefore, the product $1 \cdot 2 \cdot 3 \ldots 100$ is ... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,345 |
76. Is the number of combinations of 1000 elements taken 500 at a time divisible by 7? | 76. The number of combinations of 1000 elements taken 500 at a time is $\frac{1000!}{(500!)^{2}}$: Since 7 is a prime number, the highest power of 7 that divides 1000! (see the second solution to problem 756) is $\left[\frac{1000}{7}\right]+\left[\frac{1000}{49}\right]+ \left[\frac{1000}{343}\right]=142+20+2=164$. The ... | notdivisible7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,347 |
77. a) Find all numbers $n$ between 1 and 100 such that $(n-1)!$ does not divide $n$.
b) Find all numbers $n$ between 1 and 100 such that $(n-1)!$ does not divide $n^2$. | 77. a) The number ( $n-1$ )! does not divide $n$ only in the case when $n$ is a prime number or if $n=4$. Indeed, if $n$ is a composite number that can be represented as the product of two different factors $a$ and $b$, then both $a$ and $b$ are less than $n-1$ and, consequently, are included in ( $n-1$ )!; hence, ( $n... | )2,3,4,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89, | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,348 |
78*. Find all integers $n$ that are divisible by all integers not exceeding $\sqrt{n}$. | 78. Suppose that the number $n$ is divisible by all numbers $m$ less than or equal to $\sqrt{n}$. Let's form the least common multiple $K$ of all such numbers $m$. Clearly, this will include all prime numbers less than $\sqrt{n}$, and each prime number $p$ will be raised to a power $k$ such that $p^k \leq \sqrt{n}$, bu... | 24,12,8,6,4,2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,349 |
79. a) Prove that the sum of the squares of five consecutive integers cannot be a perfect square of an integer.
b) Prove that the sum of the same even powers of three consecutive integers cannot equal an even power of an integer.
c) Prove that the sum of the same even powers of nine consecutive integers cannot equal ... | 79. a) Let's denote five consecutive integers as
$$
n-2, n-1, n, n+1, n+2
$$
Then
$$
\begin{aligned}
&(n-2)^{2}+(n-1)^{2}+n^{2}+(n+1)^{2}+(n+2)^{2}= \\
&=5 n^{2}+10=5\left(n^{2}+2\right)
\end{aligned}
$$
If the number $5\left(n^{2}+2\right)$ were a perfect square, it would be divisible by 5, hence $n^{2}+2$ would b... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,350 |
80. a) Let $A$ and $B$ be two different seven-digit numbers, each composed of all the digits from 1 to 7. Prove that $A$ does not divide $B$.
b) From all the digits from 1 to 9, form three three-digit numbers that are in the ratio $1: 2: 3$. | 80. a) The sum of the digits of each of the numbers $A$ and $B$ is
$$
1+2+3+4+5+6+7=28
$$
hence, both numbers give a remainder of 1 when divided by 9 (each number gives the same remainder when divided by 9 as the sum of its digits). But if $\frac{A}{B}=n$, or equivalently, $A=n B$, where $n$ is an integer different f... | 219,438,657 | Number Theory | proof | Yes | Yes | olympiads | false | 29,351 |
81. The square of an integer ends with four identical digits. Which ones? | 81. A perfect square can only end in the digits 0, 1, 4, 9, 6, and 5. Furthermore, the square of any even number is clearly divisible by 4, while the square of any odd number gives a remainder of 1 when divided by 4 \(\left((2 k)^{2}=4 k^{2},(2 k+1)^{2}=4\left(k^{2}+k\right)+1\right)\); therefore, the square of no numb... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,352 |
82. Prove that if both sides of a rectangle and its diagonal are expressed in integers, then the area of the rectangle is divisible by 12. | 82. Let the sides of the rectangle be denoted by \(x\) and \(y\), and the diagonal by \(z\); then, by the Pythagorean theorem, we have:
\[
x^{2} + y^{2} = z^{2}
\]
We need to prove that the product \(xy\) is divisible by 12. First, we will show that this product is divisible by 3, and then that it is divisible by 4. ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,353 |
83. Prove that if all coefficients of the quadratic equation
$$
a x^{2}+b x+c=0
$$
- are odd integers, then the roots of the equation cannot be rational. | 83. According to the formula for solving a quadratic equation, we have:
$$
x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}
$$
Therefore, for the roots of the equation to be rational, the expression $b^{2}-4 a c$ must be a perfect square. Let $b=2 n+1, a=2 p+1, c=2 q+1$; in this case, we will have:
$$
\begin{aligned}
& b^{2}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,354 |
84. Prove that if the sum of simple fractions
$$
\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}
$$
( $n$ - an integer) is converted to a decimal, the resulting fraction will be a mixed periodic decimal. | 84. We have:
$$
\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}=\frac{3 n^{2}+6 n+2}{n(n+1)(n+2)}
$$
The numerator of our fraction is not divisible by 3, while the denominator is, as it is the product of three consecutive integers. Therefore, the denominator has factors other than 2 and 5, which means the decimal expansion w... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,355 |
85. Prove that the numbers
a) $M=\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}$
b) $N=\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+m}$
c) $K=\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{2 n+1}$
( $n, m$-positive integers) cannot be integers. | 85. a), b). Let's bring all the terms in the sum
$$
M=\frac{1}{2}+\ldots+\frac{1}{n}\left(\text { n } N=\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+m}\right)
$$
to a common denominator. Among all the fractions in our sum, consider the one whose denominator contains the highest power of two (such a fract... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,356 |
86. a) Prove that the fraction $\frac{a^{3}+2 a}{a^{4}+3 a^{2}+1}$ is irreducible for any integer value of $a$.
b) Indicate all (natural) numbers by which the fraction $\frac{5 n+6}{8 n+7}$ can be reducible for integer $n$. | 86. The fraction $\frac{a^{3}+2 a}{a^{4}+3 a^{2}+1}$ is reducible or irreducible simultaneously with the fraction $\frac{a^{4}+3 a^{2}+1}{a^{3}+2 a} \Rightarrow a+\frac{a^{2}+1}{a^{3}+2 a}$, or - simultaneously with the fraction $\frac{a^{2}+1}{a^{3}+2 a}$. The fraction $\frac{a^{2}+1}{a^{3}+2 a}$ is reducible or irred... | 13 | Number Theory | proof | Yes | Yes | olympiads | false | 29,357 |
88. Prove that there exists a number divisible by $5^{1000}$, the decimal representation of which does not contain any zeros. | 88. It is obvious that the decimal representation of all numbers $5^n$, where $n$ is a natural number, ends with the digit 5; therefore, the representation of the number $a=5^{1000}$ also ends with the digit 5 (not zero!). Now, suppose that the representation of the number $a$ contains zeros; let the first of them (cou... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,359 |
89. Prove that all numbers of the form 10001, 100010001, 1000100010001, ... are composite. | 89. The numbers of our sequence have the form $1+10^{4}+10^{8}+\ldots+10^{4 k}$. Consider alongside these numbers the numbers $1+10^{2}+10^{4}+10^{6}+\ldots$ $\ldots+10^{2 k}$. It is not difficult to verify by direct inspection that
$$
\begin{aligned}
& 10^{4 k+4}-1=\left(10^{4}-1\right) \cdot\left(1+10^{4}+10^{8}+\ld... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,360 |
90. Prove that any two numbers in the sequence
$2+1, \quad 2^{2}+1, \quad 2^{4}+1, \quad 2^{8}+1, \quad 2^{16}+1, \ldots, 2^{2^{n}}+1, \ldots$ are pairwise coprime.
Note. From the result of this problem, it follows, in particular, that there are infinitely many prime numbers (see also problems 234 and 349). Indeed, i... | 90. Note that
\[
\begin{aligned}
22^{n}-1 & =\left(2^{2} 2^{n-1}+1\right)\left(2^{2^{n-1}}-1\right)= \\
& =\left(2^{n-1}+1\right)\left(2^{n-2}+1\right)\left(2^{n-2}-1\right)=\ldots \\
\ldots & =\left(2^{2} 2^{n-1}+1\right)\left(2^{n-2}+1\right)\left(2^{n-3}+1\right) \ldots\left(2^{2}+1\right)(2+1)
\end{aligned}
\]
(b... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,361 |
91. Prove that if one of the numbers $2^{n}-1$ and $2^{n}+1$, where $n>2$, is prime, then the other is composite (for $n=2$, both numbers $2^{n}-1=3$ and $2^{n}+1=5$ are prime). | 91. $2^{n}$, of course, cannot be divisible by 3. If $2^{n}$ gives a remainder of 1 when divided by 3, then $2^{n}-1$ is divisible by 3; if $2^{\text {n }}$ gives a remainder of 2 when divided by 3, then $2^{n}+1$ is divisible by 3. Therefore, in all cases, one of the two numbers $2^{n}-1$ and $2^{n}+1$ is divisible by... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,362 |
92. a) Prove that if $p$ and $8 p-1$ are prime numbers, then $8 p+1$ is a composite number.
b) Prove that if $p$ and $8 p^{2}+1$ are prime numbers, then $8 p^{2}-1$ is a prime number. | 92. a) If a prime number $p>3$ gave a remainder of 2 when divided by 3, then $8 p-1$ would be divisible by 3. Therefore, the number $p$ must give a remainder of 1 when divided by 3; but in this case, $8 p+1$ is divisible by 3. If $p=3$, then $8 p+1=25$ - also a composite number.
b) If $p$ is not divisible by 3, then $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,363 |
93. Prove that the square of any prime number, except for the numbers 2 and 3, when divided by 12, leaves a remainder of 1. | 93. Prime numbers, except for 2 and 3, give a remainder of 1 or 5 when divided by 6, because if a number gave a remainder of 2 or 4 when divided by 6, it would be even, and if it gave a remainder of 3, it would be divisible by 3. Thus, any prime number greater than 3 can be written in the form $6n+1$ or $6n+5$. The squ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,364 |
94. Prove that if three prime numbers greater than the number 3 form an arithmetic progression, then the difference of the progression is divisible by 6. | 94. Out of three numbers of the form $6 n+1$ or $6 n+5$ (see the previous problem), at least two have the same form. Consequently, their difference, equal to $d$ or $2 d$, where $d$ is the common difference of the progression, is divisible by 6; therefore, $d$ is divisible by 3. Moreover, $d$, as the difference of two ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,365 |
95*. a) Ten prime numbers less than 3000 form an arithmetic progression. Find these numbers.
b) Prove that there do not exist 11 prime numbers less than 20000 that form an arithmetic progression. | 95. a) Since prime numbers (except 2) are odd, the difference of the progression is an even number. Further, if the difference of the progression did not divide by 3, then the three terms of the progression \(a_{1}, a_{1}+d, a_{1}+2d\) would all give different remainders when divided by 3 (the difference of any two of ... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,366 |
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