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48. A More Interesting Case
- Let's move on to a more interesting case, said the King.
Alice perked up her ears.
- This time, three accused individuals, A, B, and C, stood before the court, began the King. - The court knew that one of them was a knight, one a liar, and one a spy. But who was who, the court did not k... | 48. A more interesting case. The problem would have been unsolvable if the conditions did not mention that the court identified the spy after C pointed to him, since we know that the court managed to determine which of the three was the spy, and this is a very important "clue"!
Suppose that C accused A of being the sp... | C | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,527 |
49. An Even More Interesting Case
- Let's now move on to an even more interesting case. As before, three defendants $A, B$, and $C$ stood before the court. The court knew that one of them was a knight, another a liar, and the third a spy. But who was who, the court did not know.
The judge asked defendant $A$:
- Are ... | 49. An even more interesting case. We do not know what A and B answered, so we need to consider four possible cases:
1) A and B both said “yes”:
2) A said “no,” B said “yes”;
3) A said “yes,” B said “no”;
4) A and B both said “no.”
All these four cases will appear in the next two problems, so we will carefully analyze... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,528 |
50. Another Such Interesting Case
- And here's another such interesting case, the King continued. - Regarding the three accused $A, B$, and $C$, the court once again knows that one of them is a knight, another is a liar, and the third is a spy. Opening the trial, the judge stated to the accused:
- I will now ask you a... | 50. An equally interesting case. Since the judge asked defendants $A$ and $B$ the same questions as in the previous problem, we can use the familiar table.
Consider the moment in the trial when the judge asked defendant $C$ whether he was a spy. At this moment, the judge could not assert that any of the three defendan... | C | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,529 |
## 52. The First Question
- Do you know division? - asked the Black Queen.
- Of course! - Alice replied confidently.
- Excellent! Suppose you divide eleven thousand, eleven hundred, and eleven by three. What is the remainder of the division? If you like, you can use a pencil and paper.
Alice set to work and performed... | 52. The first question. Alice made a mistake by recording eleven thousand eleven hundred and eleven as 11111, which is incorrect! The number 11111 is eleven thousand one hundred and eleven! To understand how to correctly write the dividend, let's add eleven thousand, eleven hundred, and eleven in a column:
11000
1100... | 12111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,531 |
53. Another division problem
- Try to solve another division problem, the Black Queen suggested. - How much will you get if you divide a million by a quarter?
- A quarter of a million! - Alice answered quickly, or, in other words, two hundred and fifty thousand. No, what am I saying, - Alice corrected herself, - I mea... | 53. Another division problem. A million multiplied by a quarter equals a quarter of a million, while a million divided by a quarter equals a number whose quarter is one million, that is, four million.
Thus, the correct answer to the Black Queen's question is four million. | 4000000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,532 |
## 55. In a Dream or Awake?
- Let's move on to logical questions, suggested the White Queen. - When the Black King is asleep, he judges everything incorrectly. In other words, everything the Black King considers true in his dreams is actually false, and vice versa. On the other hand, when awake, the Black King judges ... | 55. In a dream or in reality? If the Black King were awake at 10 p.m. the night before, he could not falsely believe that he and the Black Queen were sleeping. Therefore, the Black King was asleep at that time. But since he judges everything incorrectly in his dreams, the Black King mistakenly believed that the Queen w... | TheBlackQueenwasawake | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,534 |
## 56. In Dreams or Awake?
- I, like the Black King, - confessed the Black Queen, - judge everything incorrectly in dreams, but correctly when awake. Last evening, shortly before eleven o'clock, the Black King thought I was asleep. At that very moment, I either thought he was asleep or thought he was awake.
What did ... | 56. In a dream or awake? At the given time, the King was either sleeping or awake. Suppose he was not sleeping. Awake, the King judges everything soundly. So, the Black Queen was sleeping. But in a dream, she judges everything incorrectly, so the Black Queen thought that the King was sleeping. Now suppose the King was ... | The\Queen\thought\that\the\King\was\sleeping | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,535 |
## 57. How Many Rattles?
- My head is aching from the last problem, - complained the White Queen. - Let's go back to arithmetic problems. Are you familiar with Tralala and Trulala?
- Of course! - rejoiced Alice.
- Excellent! So, one day Tralala and Trulala made a bet.
- What did they bet on? - asked Alice.
- The bet a... | 57. How many rattles? If Tra-la-la loses the bet, he will have half of the total number of rattles (or, the same, as many rattles as Tru-li-la), so before the bet, Tra-la-la has one rattle more than Tru-li-la. If Tra-la-la wins the bet, he will have two more rattles than half of the total number of rattles. Moreover, a... | Tralala\has\7\rattles,\\Trulala\has\5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,536 |
58. How many brothers and sisters?
- Here's another problem,- said the Black Queen. - A girl named Alice had a brother named Tony...
- But I don't have a brother named Tony,- Alice interrupted.
- I'm not talking about you,- the Black Queen sharply cut her off, - but about a completely different Alice!
- Oh, I'm sorry,... | 58. How many brothers and sisters? In Alice and Tony's family, there are four boys and three girls. Tony has three brothers and three sisters, and Alice has four brothers and two sisters. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,537 |
59. Not to the Right Address
- I want to offer you a little puzzle, - said the White Queen. - What I will tell you is a true incident that happened to me recently. Once I needed to send four letters by mail. I wrote four letters, addressed the envelopes, and absent-mindedly placed some of the letters in the wrong enve... | 59. Not to the right address. The statement about exactly three letters (out of four) being sent to the right address means the same as the statement about exactly one letter being sent to the wrong address. Therefore, one has to "choose" between two cases: when exactly three letters and exactly two letters were sent t... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,538 |
60. How much land was there?
- We'll see if you are good at practical arithmetic,- said the Black Queen. - A small farmer had no money to pay taxes. To settle the debt, the royal tax collector cut off one tenth of the land owned by the farmer, after which the farmer was left with ten acres of land.
How much land did ... | 60. How much land is there? The problem usually gives an incorrect answer: 11 acres. If the farmer originally owned 11 acres of land, the tax collector would have cut off $1 \frac{1}{10}$ acres (which is $1 / 10$ of 11 acres), and the farmer would be left with $99 / 10$ acres instead of 10, as required by the problem. ... | 11\frac{1}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,539 |
61. Another task about a piece of land
- Here's another task,- said the Black Queen. Another farmer had a piece of land. On one third of his land, he grew pumpkins, on one fourth he planted peas, on one fifth he sowed beans, and the remaining twenty-six acres he allocated for corn.
How many acres of land did the farm... | 61. Another problem about a plot of land. Let's bring all fractions to a common denominator (equal to 60): $1 / 3 + 1 / 4 + 1 / 5 = {}^{20} / 60 + {}^{15} / 60 + {}^{12} / 60 = {}^{47} / 60$. Corn occupies ${}^{13} / 60$ of the entire area. Therefore, ${}^{13} / 60$ of the plot is 26 acres, and since 13 is half of 26, ... | 120 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,540 |
62. The clock strikes twelve
- Grandfather's clock strikes six times in thirty seconds. How many seconds will it take to strike twelve times? - asked the Black Queen.
- Of course, in sixty seconds! - exclaimed Alice. - Oh no! - she corrected herself. - I was wrong. Please wait, I will give the correct answer now!
- To... | 62. The clock strikes twelve. The sixth strike is separated from the first by five time intervals (pauses). In total, these five pauses last $30 \mathrm{s}$, so the pause between two consecutive strikes is 6 s (not 5, as some mistakenly believe!). The twelfth strike is separated from the first by 11 pauses. Therefore, ... | 66\mathrm{} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,541 |
## 65. Second Round
Both brothers entered the house. Soon, one of them came out with a card in his pocket and declared:
- If I am Tralala, then the card in my pocket is not of a black suit.
Who was it?
This problem seemed much more difficult to Alice than the first one, but in the end, she solved the second problem... | 65. Second Round (Red and Black). The speaker, in essence, claims that he is not Tralala with a red card in his pocket. The statement he made must be true, because if he were Tralala with a red card in his pocket, then (since he has a red card) he could not lie and claim that he is not Tralala with a red card in his po... | Trulalawithredcard | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,544 |
## 66. Third Round
In this round, one of the brothers, coming out of the house, said:
- Either I am Tra-la-la, or I have a card of black suit in my pocket.
Who was it? | 66. The Third Round (Red and Black). "Either-or" means "at least one of the two" (and possibly both). Therefore, if the brother who came out of the cottage had a card of a black suit, it would be true that either he is Tralala or he has a card of a black suit. But this would mean that the holder of the black card made ... | Tralala | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,545 |
## 67. Fourth Round
In this round, one of the brothers, coming out of the house, said:
- Either I am Tralala with a card of black suit in my pocket, or I am Trulala with a card of red suit in my pocket.
Who was it? | 67. Fourth Round (Red and Black). This time it is impossible to determine the suit (red or black) of the card of the one who came out of the house, but in either case, it should be Tra-la-la. Suppose he has a card of the red suit. Then he is telling the truth. Therefore, in front of Alice is either Tra-la-la with a bla... | Tru-la-la | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,546 |
## 68. The Fifth Round
This time, one of the brothers, coming out of the house, said:
- Tralala has a card of black suit in his pocket.
Who was it?
- Excellent! - he said, addressing Alice. You did a wonderful job with the task of the fifth round. But the final round of our game is even more difficult. Now I will g... | 68. The Fifth Round (Red and Black). Suppose the speaker has a red card. Then the statement made by him is true. This means that in front of Alice must be Trulala. Now suppose that the speaker has a black card. Then the statement made by him is false. This means that Trula has not a black card. At the same time, the on... | Trulala | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,547 |
## 72. Third Round
In this round, the brothers declared the following:
First brother. Trulala is me.
Second brother. Tralala is me.
First brother. Our cards are of the same suit.
Who is who? | 72. The Third Round (Orange and Purple). Looking at the first two statements, it is not hard to notice that they are either both true or both false. Therefore (if we use the principle mentioned at the beginning of the solution to the previous problem), the brothers have cards of different suits. In turn, this means tha... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,551 |
## 75. The Next Round
- In this round, said one of the brothers, the rules remain the same, but the question you need to answer is different. Instead of determining who among us is Trulala and who is Tralala, you need to find out who among us is lying and who is telling the truth.
Both brothers entered the house, and... | 75. The Sixth Round (Orange and Purple). The brothers contradict each other, so one of them is lying, and the other is telling the truth. Therefore, their cards (the same principle applies!) must be of different suits.
Thus, the first brother is telling the truth (his statement is true). | The\first\brother\is\telling\the\truth | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,554 |
## 79. The Second Court Report
- Well, I've been to the most interesting court cases! - said the White Knight after Alice solved the previous problem. - What cases were heard there!
- Please tell me about another trial, - Alice asked.
The task that followed was the most interesting of all that Alice had ever heard.
... | 79. The second court report. What could the White King have learned from the White Knight that allowed him to identify the guilty party? If the White Knight had told the White King that all three defendants had given false testimony, then the White King would not have been able to find the guilty party. Indeed, $A$ cou... | C | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,557 |
80. The Next Trial
- As it is now, I remember another most curious trial, - said the White Knight. There were three defendants. Each of them accused one of the other two. The first defendant was the only one who told the truth. If each of the defendants had accused not the one they did, but the other, then the second ... | 80. The next court case. This is a very simple task. Since $A$ told the truth and accused one of the other two defendants, either $B$ or $C$ must be guilty. Therefore, $A$ is innocent. If each defendant had accused the other person instead of the one they actually accused, then $B$ would have told the truth. Since we k... | C | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,558 |
81. The Trial That Followed
- I want to tell you about a trial I myself did not attend. Barmaglot told me about it.
According to Barmaglot, there were three defendants. Each of them accused one of the other two, but Barmaglot did not tell me who accused whom. However, he informed me that the first defendant was telli... | 81. The trial, following the next. Since $A$ spoke the truth and accused either $B$ or $C$, then either $B$ or $C$ is guilty, and $A$ is innocent.
The White Knight told the White King that $C$ either lied or told the truth. If the White King had been told that $C$ lied, then the White King could not determine which of... | B | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,559 |
## 82. Another Trial
- As I remember, someone once told me about a similar trial, - the White Knight continued. - There were three defendants. Each accused one of the other two, and the first defendant was telling the truth. Unfortunately, the Bar-ma-plot did not tell me whether the second defendant was lying or telli... | 82. Another trial. As in the previous problem, since A was telling the truth and accusing one of the two defendants, A must be innocent. If the White Knight learned from the Barma that C was telling the truth, then without any additional information, the White Knight would have known that B was guilty (as we saw in the... | C | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,560 |
## 83. Another Case
"I remember it as if it were yesterday," the White Knight began, "I once attended a trial that the Baboon couldn't make it to. As is often the case with us, there were three defendants, and only one of them was guilty. When the first defendant was asked if he pleaded guilty, his answer was brief: e... | 83. Another case. There are 8 versions of the testimonies given by the defendants \(\boldsymbol{A}\), \(\boldsymbol{B}\), and \(C\) during the trial. Indeed, \(A\) could give two versions of his testimony, each of which could be combined with two versions of \(B\)'s testimony, resulting in 4 versions of the testimonies... | C | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,561 |
## 84. And One More Case
- I remember another trial I attended as if it were today, - began the White Knight. - There were three defendants, and only one of them was guilty. I remember that the first defendant accused the second, but I completely forgot what the second and third defendants said. I also remember that l... | 84. And one more case. We know that $A$ accused $B$, but we do not know what $B$ or $C$ said. Suppose we had additional information that the guilty one is the only one of the three defendants who gave false testimony. Then the guilty one could be any of the three defendants. It would be impossible to determine exactly ... | C | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,562 |
## 85. And what would you say?
- I remember one trial in great detail, - the White Knight told Alice. - I remember clearly that there were three defendants and that only one of them was guilty. I remember distinctly that the first defendant accused the second, and the second confessed to being guilty. As for the third... | 85. And what would you say? Suppose the White Knight told Humpty Dumpty that all three defendants were lying. Then Humpty Dumpty could not have preferred one of the two options: either $C$ is guilty and accused $A$, or $A$ is guilty and accused $C$ (since in both cases all three defendants were lying).
Humpty Dumpty c... | B | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,563 |
86. What happened to the Goat?
- I have seen many of the most interesting trials in my life, - continued the White Knight. - I remember one particularly curious trial! I even remember all the defendants!
"At least some variety!" - thought Alice.
- It was a very nice trial! On the bench of the accused were three: a G... | 86. What happened to the Goat? From the fact that the Goat lied, it does not follow that he is either guilty or innocent. Therefore, even if the court established that the Goat gave false testimony, the Goat could have been found guilty (based on other data of which we know nothing) or released from custody (again base... | TheGoatwasfoundguilty | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,564 |
5. Prove that among any six people, there will always be three who are pairwise acquainted or three who are pairwise unacquainted with each other.
## 7th grade | 61.5. Consider any one of these people. It is clear that among the other five, he has either three acquaintances or three strangers. Without loss of generality, we can assume that he is acquainted with at least three others: \(A\), \(B\), and \(C\). If any two of them are acquainted with each other, then we have a trip... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,569 |
7. Given a circle $O$, a square $K$, and a line $L$. Construct a segment of a given length, parallel to $L$ and such that its ends lie on $O$ and $K$ respectively. | 61.7. Let's translate the circle $O$ parallel to the line $L$ by a given distance. This operation can easily be performed using a compass and a ruler. The resulting circle $O^{\prime}$ will intersect the square $K$. Any of the intersection points $M$ will be one of the endpoints of the required segment. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,570 |
8. A three-digit number $\overline{A B C}$ is divisible by 37. Prove that the sum of the numbers $\overline{B C A}$ and $\overline{C A B}$ is also divisible by 37. | 61.8. Note. $\overline{A B C}+\overline{B C A}+\overline{C A B}=111(A+B+C), \quad$ and $111=3 \cdot 37$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,571 |
10. Given $2 n+1$ different items. Prove that from them, an odd number of items can be chosen in as many ways as an even number.
## 8th grade | 61.10. To each choice of some set of items $X$ from a given set of $2 n+1$ items, there corresponds a choice of a set $Y$ consisting precisely of those items that did not make it into $X$. Since a set $X$ with an even number of elements necessarily corresponds to a set $Y$ containing an odd number of elements, the numb... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,572 |
12. It is known that $A, B$ and $\sqrt{A}+\sqrt{B}$ are rational numbers. Prove that then $\sqrt{A}$ and $\sqrt{B}$ are rational. | 61.12. Let $\quad \sqrt{A}+\sqrt{B}=r_{1}, \quad$ then $\quad \sqrt{A B}=(r_{1}^{2}-A- B) / 2=r_{2}$. Therefore, the numbers $\sqrt{A}$ and $\sqrt{B}$ are the roots of the quadratic equation $t_{2}-r_{1} t+r_{2}=0$, i.e., equal to $\left(r_{1} \pm \sqrt{r_{1}^{2}-4 r_{2}}\right) / 2$. Since
$$
4 A=(2 \sqrt{A})^{2}=2 r... | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,573 |
15. Given $N$ numbers $x_{1}, x_{2}, \ldots, x_{n}$, each of which is either +1 or -1. It is also given that $x_{1} x_{2} + x_{2} x_{3} + \ldots + x_{n-1} x_{n} + x_{n} x_{1} = 0$. Prove that $N$ is divisible by four. | 61.15. Since the sum of $N$ addends, each of which is equal to 1 or -1, is zero, the number of addends equal to 1 and the number of addends equal to -1 must be equal, i.e., $N=2 k$. Since the product of all addends is equal to $\left(X_{1} X_{2} \ldots X_{n}\right)^{2}$ - positive, the number of (-1) must be even, i.e.... | N | Number Theory | proof | Yes | Yes | olympiads | false | 36,575 |
16. On a circle, $N$ points are marked, and it is known that for any two points, one of the arcs connecting them has a measure less than $120^{\circ}$. Prove that all points lie on an arc of measure $120^{\circ}$.
## 9th grade | 61.16. Hint. Induction on $N$. Let all points, except one (point $X$), lie on an arc $A B$ of less than $120^{\circ}$ (where $A$ and $B$ are some two of the given points). If $X$ lies on the arc $A B$, then the inductive step is complete; if not, it is clear that one of the arcs $X A, X B$, containing $A B$, has a meas... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,576 |
19. On a plane, there are $2 n$ points. Prove that they can be paired and connected in such a way that the segments do not intersect | 61.19. Let's consider all possible ways of connecting $N$ points with segments and choose the one for which the sum of the lengths of the segments is minimal. If in this set, any two segments $A B$ and $C D$ intersect, then by replacing them with the pair $A C, B D$, we obtain a set of segments with a smaller total len... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,577 |
22. Given three non-zero integers $K, M$, and $N$, where $K$ and $M$ are coprime. Prove that there exists an integer $x$ such that $M x + N$ is divisible by $K$.
23 (add.) ${ }^{1}$. Prove that all numbers of the form 1156, 111556, $11115556, \ldots$ are perfect squares.[^0]
## 10-11th grades | 61.22. Since $K$ and $M$ are coprime, there exist integers $a$ and $b$ such that $K a + M b = -1$. Multiplying this equation by $N$, we get $K N a + M N b = -N$. From this, it follows that $M(N b) + N$ is divisible by $K$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,580 |
26. Find the minimum value of the expression $\left(a^{2}+x^{2}\right) / x$, where $a>0$ is a constant, and $x>0$ is a variable. | 61.26. Answer. $2 a$. Hint. $\left(a^{2}+x^{2}\right) / x=a(a / x+x / a) \geqslant 2 a$. | 2a | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,582 |
27. Let $f(x)=x^{n}+a_{1} x^{n-1}+\ldots+a_{n}$ be a polynomial with integer coefficients, and let $P$ be its rational root. Prove that $P$ is an integer and $f(m)$ is divisible by $P-m$ for any integer $m$. | 61.27. Let $p=r / s$, where $r$ and $s$ are coprime integers. Substituting $p$ into the polynomial, we get
$$
r^{n}+a_{1} r^{n-1} s+\ldots+a_{n} s^{n}=0
$$
Since all terms starting from the second one are divisible by $s$, the first term - $r^{n}$ - must also be divisible by $s$. However, since $r$ and $s$ are coprim... | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,583 |
29. Each face of a cube is covered by two right-angled triangles, one of which is white, and the other is black. Prove that there are only two essentially different (i.e., not coinciding under rotations of the cube) ways of covering such that the sums of the white and black angles meeting at each vertex are the same. | 61.29. It is clear that if the diagonals of the faces are marked, dividing them into triangles, then from each of the eight vertices, either one or three diagonals emerge. By enumeration, it is not difficult to establish that there are exactly two vertices from which three diagonals emerge. After this, it is easy to se... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,584 |
30. A snail is crawling on a table at a constant speed. Every 15 minutes, it turns $90^{\circ}$, and between turns, it crawls in a straight line. Prove that it can return to the starting point only after an integer number of hours. | 61.30. Instruction. Prove that the time spent by the snail moving forward and backward coincides with the time spent moving left and right, and that each of these quantities is expressed as an integer number of half-hours. | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 36,585 |
5. Prove that a $201 \times 201$ chessboard can be traversed by a knight's move, visiting each square exactly once. | 62.5. Hint. Prove that a knight can traverse the border of width four cells, and then use this fact to prove by induction that such a traversal exists for any board of size $(4 n+1) \times(4 n+1)$. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,588 |
6. Can an integer, the last two digits of which are odd, be the square of another integer?
## 7th grade | 62.6. No, it cannot. One of the last two digits of a perfect square is necessarily even. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,589 |
10. In a six-digit number that is divisible by seven, the last digit was moved to the beginning. Prove that the resulting number is also divisible by seven. | 62.10. Hint. Use the fact that $10^{6}-1$ is divisible by seven. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,590 |
11*. A circle is divided into 49 regions such that no three regions touch at a single point. The resulting "map" is colored with three colors in such a way that no two adjacent regions have the same color. The boundary between two regions is considered to be colored with both colors. Prove that there will be two diamet... | 62.11. The ends of the curves - the boundaries of the regions - divide the circle into several arcs. Consider any one of them - say, the arc $A B$, colored in the first color. The diametrically opposite arc $A^{\prime} B^{\prime}$ may contain 0, 1, or more division points. If there are more than one, the picture looks ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,591 |
13. Four circles are placed on a plane such that each one touches two others externally. Prove that the points of tangency are located on one circle. | 62.13. The centers of circles $A, B, C$ and $D$ form a circumscribed quadrilateral $A B C D$. Denote the points of tangency of the circles as $K, L, M, N$ - they lie on the segments $A B, B C, C D$ and $D A$ respectively. Then $\angle N K L=180^{\circ}-\left(90^{\circ}-\angle N A K / 2\right)-\left(90^{\circ}-\right.$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,593 |
14. Let the integers $a$ and $b$ be expressible in the form $x^{2}-5 y^{2}$, where $x$ and $y$ are integers. Prove that the number $a b$ can also be expressed in this form. | 62.14. $\left(x^{2}-5 y^{2}\right)\left(a^{2}-5 b^{2}\right)=(a x+5 b y)^{2}-5(a y+b x)^{2}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,594 |
16. Let $a+b+c=1 ; m+n+p=1$. Prove that
$$
-1 \leqslant a m+b n+c p \leqslant 1
$$[^1] | 62.16. Consider the vectors $\overrightarrow{v_{1}}=(a, b, c), \overrightarrow{v_{2}}=(m, n, p)$. By condition, their lengths are equal to 1, and therefore, the absolute value of their dot product, equal to $a m+b n+c p$, does not exceed the product of their lengths, equal to 1. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,596 |
20. Given the polynomial $x^{2 n}+a_{1} x^{2 n-2}+a_{2} x^{2 n-4}+\ldots+a_{n-1} x^{2}+a_{n}$, which is divisible by $x-1$. Prove that it is divisible by $x^{2}-1$. | 62.20. Instruction. Consider the value of the polynomial at $x=$ $=-1$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Algebra | proof | Yes | Yes | olympiads | false | 36,597 | |
21. Prove that for any prime number $p$, different from two and five, there exists a natural $k$ such that the decimal representation of the number $p k$ consists only of ones.
9th grade ${ }^{1}$ | 62.21. Consider the numbers $1, 11, 111, \ldots, 111 \ldots 11$ ( $p+1$ ones). Since there are more than $p$ of them, some two of them have the same remainder when divided by $p$. Their difference has the form $111 \ldots 11000 \ldots 00$. But since $p$ is coprime with 10, it follows from the fact that $111 \ldots 1100... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,598 |
24. In a unit square grid, a random unit square is taken. Prove that one of the distances from an arbitrary lattice node to the vertices of this square is irrational. | 62.24. Let the coordinates of the given point be ( $m, n$ ). We choose from the four points - the vertices of the square - one [with coordinates $(x, y)$ ] such that the numbers $m-x, n-y$ are odd. Then the number $(m-x)^{2}+(n-y)^{2}$ cannot be the square of a rational number, since it has the form $4 k+2$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,599 |
25. The sum of ten numbers is zero. The sum of all their pairwise products is also zero. Prove that the sum of the cubes of these numbers is also zero. | 62.25. If we subtract from the square of the sum of all numbers the doubled sum of all pairwise products, we get the sum of the squares of the numbers. However, it follows from the condition that this sum is equal to zero. Therefore, all ten numbers are equal to zero. | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,600 |
29*. A cargo weighing 13.5 tons has been distributed into boxes, the weight of each of which does not exceed 350 kg after this. Prove that the entire cargo can be transported using 11 one-and-a-half-ton trucks.
## $10-$ th grade | 62.29. Let's call a box small if it weighs no more than 300 kg. Obviously, there are no more than 44 other boxes, and we can put them in four in 11 one-and-a-half-ton trucks. It remains to prove that a small box can always be placed in one of the trucks. Indeed, if this is not the case, then the load in each of them is... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 36,602 |
35. Find the sum $\sum_{k=0}^{n}(-1)^{k}\binom{n}{k} k x$. | 62.35. Answer. The sum is zero for $n>1$, $(-x)$ for $n=1$. | \begin{cases}0& | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,606 |
36. Solve the equation in natural numbers
$$
1-\frac{1}{2+\frac{1}{3+\frac{1}{4+\ldots+\frac{1}{n}}}}=\frac{1}{x_{1}+\frac{1}{x_{2}+\frac{1}{x_{3}+\ldots+\frac{1}{x_{n}}}}}
$$ | 62.36. Answer. $x_{1}=x_{2}=1, x_{k}=k$ for $k>2$. | x_{1}=x_{2}=1,x_{k}=kfork>2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,607 |
37. Given a polyhedron. The number of sides of all its faces, except one, is divisible by a given natural number $n$, greater than 1. Prove that the faces of this polyhedron cannot be colored with two colors such that adjacent faces are colored differently. | 62.37. Suppose such a coloring is possible, and the mentioned special face is painted white. Let's calculate the sum of the number of sides of the black faces - on the one hand, this will be the number of all edges of the polyhedron, and on the other hand, some natural number divisible by n. A similar calculation for t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,608 |
40. Given the polynomial $x^{k n}+a_{1} x^{k(n-1)}+a_{2} x^{k(n-2)}+\ldots+a_{n-1} x^{k}+$ $+a_{n}$, which is divisible by $x-1$. Prove that it is divisible by $x^{k}-1$. | 62.40. Hint. Prove that the polynomial is divisible by $x-a$, where $a$ is an arbitrary complex $k$-th root of 1. | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,609 |
47. Inside a $1 \times 1$ square, a convex polygon with an area greater than $1 / 2$ is placed. Prove that there exists a chord of length greater than $1 / 2$ in the polygon, parallel to any preselected side of the square.
## ELIMINATION ROUND | 62.47. Suppose this is not the case. Draw lines through all the vertices of the polygon parallel to the given side of the square. They will divide the polygon into triangles and trapezoids with bases not exceeding $1 / 2$. Since the sum of the heights in these trapezoids and triangles does not exceed 1, by adding the i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,611 |
48. Prove that seven circles of radius 1 can completely cover a circle of radius 2, but cannot cover a circle of a larger radius. | 62.48. To cover a circle of radius 2, we divide the circumference by points $A, B, C, D, E, F$ into six equal arcs and construct six circles of radius 1 on the diameters $A B, B C, C D, D E, E F$ and $F A$. Adding one circle centered at the center of the larger circle, we obtain seven circles covering the given circle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,612 |
49. In a square of area 5, nine polygons, each of area 1, are placed. Prove that some two of them have an intersection of area not less than $1 / 9$. | 62.49. This follows from the inequality
$$
5 \geqslant S_{1}+\ldots+S_{9}-S_{12}-S_{23}-\ldots-S_{89}
$$
where $S_{i}$ denotes the area of the $i$-th polygon, and $S_{ij}$ denotes the area of the intersection of the $i$-th and $j$-th polygons. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,613 |
52. Prove that from $k$ integers, one can select several such that their sum is divisible by $k$. | 62.52. Let's denote these numbers as $x_{1}, x_{2}, \ldots, x_{k}$ and consider the sums $x_{1}, x_{1}+x_{2}, x_{1}+x_{2}+x_{3}, \ldots, x_{1}+x_{2}+\ldots+x_{k}$. If none of them is divisible by $k$, then some two of them have the same remainder when divided by $k$, and their difference is divisible by $k$. Since thei... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,614 |
6. Is it possible to arrange the numbers from 1 to 1963 in a row so that any two adjacent numbers and any two numbers that are one apart are coprime?
## 7th grade | 63.6. Answer. No, it cannot. Hint. Consider even numbers. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,619 |
10. The sum of the distances between the midpoints of opposite sides of a quadrilateral is equal to its semiperimeter. Prove that this quadrilateral is a parallelogram ${ }^{1}$. | 63.10. Let's introduce the necessary notations. Let $A B C D$ be the given quadrilateral, and $M, N, P$ and $Q$ be the midpoints of sides $A B$, $B C$, $C D$, and $D A$ respectively. Since there are vector equalities $\overrightarrow{A B}+\overrightarrow{D C}=\overrightarrow{2 Q N}$, $\overrightarrow{B C}+\overrightarr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,623 |
11. In a bus without a conductor, there are 40 passengers, each having only coins of 10, 15, and 20 kopecks. In total, the passengers have 49 coins. Prove that the passengers will not be able to pay the required amount of money into the fare box and settle among themselves (the cost of a bus ticket in 1963 was 5 kopeck... | 63.11. It is clear that after the payment, each passenger should have at least one coin left. On the other hand, 2 rubles should be paid into the cash register, i.e., at least 10 coins should be spent. Thus, it is impossible to manage with fewer than 50 coins. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,624 |
15. The fraction 0.ABC... is formed according to the following rule: A and B are arbitrary digits, and each subsequent digit is the remainder when the sum of the two preceding digits is divided by 10. Prove that this fraction is purely periodic. | 63.15. Consider all pairs of the form ( $X, Y$ ), where $X$ and $Y$ are two consecutive digits from the decimal representation of the given fraction. Since there are infinitely many such pairs, there will be two identical pairs among them. Moving backward along the sequence of digits, we can assume that one of these pa... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,627 |
17. The sum of three integers, which are perfect squares, is divisible by nine. Prove that among them there are two numbers, the difference of which is divisible by nine. | 63.17. Perfect squares give only the following remainders when divided by nine: $0,1,4,7$. Obviously, the sum of three different remainders of this kind cannot be divisible by nine. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,629 |
18. Given $k+2$ integers. Prove that among them there are two integers such that either their sum or their difference is divisible by $2k$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 63.18. Consider pairs of remainders when dividing by $2 k$: $(1,2 k-1),(2,2 k-2), \ldots,(k-1, k+1)$, as well as the remainders 0 and $k$. In total, we have obtained $k+1$ sets of remainders. Since we are given $k+2$ numbers, some two of them have remainders from the same set. These are the required numbers. | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,630 | |
19. A right angle rotates around its vertex. Find the geometric locus of the midpoints of the segments connecting the points of intersection of the sides of the angle and a given circle.
## 9th grade | 63.19. Answer. Let $O$ be the center of the given circle, $R$ its radius, and $C$ the vertex of the angle. Then the geometric locus of points specified in the condition is a circle with center at the midpoint of segment $O C$ and diameter $\sqrt{2 R^{2}-|O C|^{2}}$. | \sqrt{2R^{2}-|OC|^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,631 |
20. A five-digit number $\overline{A B C D E}$ is divisible by 41. Prove that if its digits are cyclically permuted, the resulting number will also be divisible by 41 *. | 63.20. Since $\overline{b c d e a}=10 \overline{a b c d e}+a-100000 a=10 \overline{a b c d e}-$ - 99999a and 99999 is divisible by 41, then the number $\overline{b c d e a}$ is also divisible by 41. By continuing to cyclically permute the digits in a similar manner, we obtain the required property for all such numbers. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,632 |
21. There are two groups of numbers:
$$
\begin{aligned}
& A_{1} \geqslant A_{2} \geqslant \ldots \geqslant A_{n} \geqslant 0 \\
& B_{1} \geqslant B_{2} \geqslant \ldots \geqslant B_{n} \geqslant 0
\end{aligned}
$$
Prove that
$$
A_{1} B_{1}+A_{2} B_{2}+\ldots+A_{n} B_{n} \geqslant A_{1} B_{n}+A_{2} B_{n-1}+\ldots+A_{... | 63.21. Let's divide the members of the inequality into pairs as follows:
$$
A_{i} B_{i}+A_{j} B_{j} \geqslant A_{i} B_{j}+A_{j} B_{i}
$$
This inequality is true since $\left(A_{i}-A_{j}\right)\left(B_{i}-B_{j}\right) \geqslant 0$ for any $i$ and $j$. By adding all such inequalities, we obtain the required result. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,633 |
23. Given two circles with radius $R=1 / 2 \pi$. On one circle, 20 points are marked, and on the other, several arcs are marked, the total length of which is less than $1 / 20$. Prove that one circle can be superimposed on the other so that none of the marked points fall inside the marked arcs. | 63.23. Let's identify the given circles and instead of overlaps, consider rotations of the circle. We will consider all possible angles of rotation as numbers from the interval $[0 ; 1]$, i.e., the magnitude of the rotation is measured by the length of the arc that a point on the circle travels during the execution of ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,634 |
27. Given a trihedral angle and a point $A$ on one of its edges. Points $B$ and $C$ slide along the other two edges. Find the geometric locus of the centroids of triangles $A B C$. | 63.27. Answer. The part of the plane parallel to the face $B O C$ and at a distance of $d(A, B O C) / 3$ from it, lying inside the angle. | ThepartoftheplaneparalleltothefaceBOCatdistanceof(A,BOC)/3fromit,lyinginsidetheangle | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,637 |
28. Prove that the equation $A_{1} X_{1}+A_{2} X_{2}+\ldots+A_{n} X_{n}=B$ cannot be satisfied by more than half of all sets ( $X_{1}$, $\left.X_{2}, \ldots, X_{n}\right)$, where each $X_{i}$ is equal to 0 or 1, if it is known that not all of the numbers $A_{1}, A_{2}, \ldots, A_{n}, B$ are zero. | 63.28. Without loss of generality, we can assume that $A_{1}$ is not equal to zero. Then it is obvious that for each set ( $X_{1}, X_{2}, \ldots, X_{n}$ ), which is a solution to the given equation, the set ( $1-X_{1}$, $X_{2}, \ldots, X_{n}$ ) is not its solution. Therefore, the number of solutions is no more than hal... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,638 |
29. In a polyhedron, a dihedral angle $\alpha$ is called adjacent to an edge if this edge is a side of the angle $\alpha$. Prove that in any triangular pyramid, there is an edge to which only acute angles are adjacent. | 63.29. Let's consider the longest edge of the pyramid. Since in any face containing it, the largest angle is opposite to it, the angles adjacent to it can only be acute. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,639 |
32. Find the first 1963 digits after the decimal point in the decimal representation of the number $(\sqrt{26}+5)^{1963}$. | 63.32. Since the sum $(\sqrt{26}+5)^{1963}+(5-\sqrt{26})^{1963}$ is an integer (this is verified using the binomial theorem), and the inequality $-0.1 < 5 - \sqrt{26} < 0$ holds, then $-10^{-1963} < (5 - \sqrt{26})^{1963} < 0 \quad$ and, consequently, the first 1963 digits of the number $(\sqrt{26}+5)^{1963}$ after the... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,640 |
33. How should a matchbox be positioned in space so that its projection onto a plane has the largest area? | 63.33. The box-parallelepiped must be positioned so that the plane passing through the second ends of its edges, emanating from one vertex, is horizontal. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,641 |
35. A natural number $A$ is divisible by $1,2,3, \ldots, 9$. Prove that if $2 A$ is represented as the sum of several addends, each of which is equal to $1,2,3, \ldots, 9$, then among them there will be several addends whose sum is equal to $A$. | 65.35. Suppose this is not the case. Then, as is easy to verify, there are no more than seven ones in this set of numbers - otherwise, by adding other numbers, one can obtain a number from $A-8$ to $A$. Furthermore, among them, there are no more than seven twos - this is proven in the same way. Reasoning similarly, we ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,642 |
37. Solve the equation $x^{4}-2 y^{4}-4 z^{4}-8 t^{4}=$ $=0$ in integers. | 63.37. Answer. $x=y=z=t=0$. Hint. Consider an integer solution with the smallest absolute value of $x$ and try to find another solution with $x_{1}=x / 2$. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,643 |
38. Prove that a closed broken line of length 1, located on a plane, can be covered by a circle of radius $1 / 4$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 63.38. Consider two points on the broken line $A$ and $B$ such that the length of the broken line between these points is $1 / 2$. Then a circle of radius $1 / 4$ centered at the midpoint of segment $A B$ covers the entire broken line. Indeed, if the broken line intersects the circle at some point $C$, then since $A C ... | Combinatorics | MCQ | Yes | Yes | olympiads | false | 36,644 | |
40. For what values of $n$ is the expression $2^{n}+1$ a non-trivial power of a natural number? | 63.40. Only for $n=3$. Indeed, if $2^{n}+1=A^{p}$, then $2^{n}=A^{p}-1=(A-1)\left(A^{p-1}+A^{p-2}+\ldots+A+1\right)$. Then $A^{p-1}+A^{p-2}+\ldots+A+1$ is a power of two, not equal to 1, and since $A$ is odd, and the given sum is even, there is an even number of terms in it. Let $p=2 q$. Then $2^{n}=A^{2 q}-1=$ $=\left... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,645 |
41. Given a grid paper. Two parallel lines are drawn through two nodes. Prove that the closed strip formed contains infinitely many nodes of the grid paper. | 63.41. Note. If the slope coefficient $k$ of the lines is rational, then there are infinitely many nodes on the boundary of the strip. Let $k$ be irrational. Then the problem reduces to proving the existence of infinitely many pairs of integers $X$ and $Y$ such that
$$
p=m_{2}-k n_{2}<X-k Y<m_{1}-k n_{1}=q
$$
where $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,646 |
3. In the cells of a chessboard, natural numbers are placed such that each number is equal to the arithmetic mean of its neighbors. The sum of the numbers in the corners of the board is 16. Find the number standing on the field $e 2$. | 64.3. Consider the largest number standing in one of the cells. Obviously, all adjacent numbers to it are equal to it. Those adjacent to them are also equal to them, and so on. Therefore, all numbers on the board are equal. Hence, the number written in the field $e 2$ is 4. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,649 |
5. The Pioneer squad is arranged in a rectangle. In each row, the tallest is marked, and among these Pioneers, the shortest is chosen. In each column, the shortest is marked, and among them, the tallest is chosen. Which of these two Pioneers is taller ${ }^{1}$ | 64.5. Let the surname of the shortest of the giants be Ivanov, and the surname of the tallest of the dwarfs be Petrov. Consider the line in which Ivanov stands, and the row in which Petrov stands. At their intersection stands a certain pioneer who is shorter than Ivanov but taller than Petrov. Therefore, Ivanov is tall... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,651 |
7. Given a convex $n$-gon, all angles of which are obtuse. Prove that the sum of the lengths of the diagonals in it is greater than the sum of the lengths of the sides. | 64.7. Since all angles are obtuse, $n \geqslant 5$. Now consider all diagonals connecting vertices of the $n$-gon every other vertex. They form an $n$-pointed star, the length of the boundary of which is obviously greater than the perimeter of the polygon. This leads to the required inequality. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,653 |
8. Find all integer values for $X$ and $Y$ such that $X^{4}+4 Y^{4}$ is a prime number. | 64.8. Since $X^{4}+4 Y^{4}=\left(X^{2}+2 X Y+2 Y^{2}\right) \quad\left(X^{2}-2 X Y+\right.$ $+2 Y^{2}$ ), this number is prime only when $X=Y=1$ or $X=Y=-1$. | X=Y=1orX=Y=-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,654 |
12. Given triangle $ABC$; $D$ and $E$ are the midpoints of sides $AB$ and $BC$. Point $M$ lies on $AC$, and $ME > EC$. Prove that $MD < AD$.
## 8th grade | 64.12. Since the median $M E$ in triangle $M B C$ is greater than half the side $B C$, the angle $B M C$ is acute. Therefore, the angle $B M A$ is obtuse and in triangle $A B M$ the median $M D$ is less than half the side $A B$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,657 |
17. In quadrilateral $A B C D$, diagonals $A C$ and $B D$ are drawn. Prove that if the incircles of $A B C$ and $A D C$ touch each other, then the incircles of $B A D$ and $B C D$ also touch each other. | 64.17. Let the circle inscribed in triangle $ABC$ touch the diagonal at point $X$, and the second circle at point $Y$ (in this case $X=Y$). Then $AX = AB + AC - BC$; $AY = AC + AD - CD$. By equating $AX$ and $AY$, we get $AB + CD = AD + BC$, i.e., the quadrilateral is circumscribed. Conducting similar reasoning in the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,659 |
18. If numbers $A$ and $n$ are coprime, then there exist integers $X$ and $Y$ such that $|X|<\sqrt{n},|Y|<\sqrt{n}$ and $A X-Y$ is divisible by $n$. Prove this.
## 9th grade | 64.18. Hint. Consider all pairs $(X, Y)$ of non-negative integers such that $|X|<\sqrt{n},|Y|<\sqrt{n}$, and the corresponding expressions $A X-Y$. Use the Pigeonhole Principle. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,660 |
20. Inside a unit square, 51 points are placed. Prove that among them, there are three that fit into a circle with a radius of $1 / 7$. | 64.20. Let's divide the square into 25 smaller squares with side length $1 / 5$. By the pigeonhole principle, at least one of these smaller squares will contain at least three points. Since such a small square can be covered by a circle of radius $1 / 7$, the proof is complete. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,661 |
21. Prove that if for a natural number $N$
$$
\begin{gathered}
{\left[\frac{N}{1}\right]+\left[\frac{N}{2}\right]+\ldots+\left[\frac{N}{N}\right]=} \\
=2+\left[\frac{(N-1)}{1}\right]+\left[\frac{(N-1)}{2}\right]+\left[\frac{(N-1)}{3}\right]+\ldots+\left[\frac{(N-1)}{(N-1)}\right]
\end{gathered}
$$
then the number $N$... | 64.21. By reducing $[N / 1]+[N / N]$ and $2+[(N-1) / 1]$, we obtain that the equalities $[N / k]=[(N-1) / k]$ must hold for all $k$ from 2 to $N-1$. But if $N$ is a composite number and is divisible by $p \geqslant 2$, then $[N / p]>[(N-1) / p]$ - a contradiction. Therefore, $N$ is a prime number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,662 |
28. Let $\alpha_{n}$ be the exponent to which the number 2 enters the factorization of the number $n$. Let $S_{n}=\alpha_{1}+\alpha_{2}+\ldots+\alpha_{n}$. Prove that in the sequence $S_{1}, S_{2}, S_{3}, \ldots, S_{2^{n}-1}$, the number of even and odd numbers differ by one. | 64.28. Let $\alpha_{0}=0, S_{0}=0$. Then the set $\left(S_{0}, S_{1}, \ldots\right.$, $S_{2^{n}-1}$ gives the set $\left(S_{2^{n}}, \ldots, S_{2^{n+1}-1}\right)$ by adding $S_{2^{n}}$ (considering only the remainders when divided by two), since $\boldsymbol{\alpha}_{k}=\alpha_{k+2^{n}}$ (for $0<k<2^{n}$ ). Reasoning by... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,664 |
30. Let $S$ be a set of natural numbers less than a given prime number $P$ such that if the numbers $A$ and $B$ are in the set, then the remainders of $A B, A$, and $B$ when divided by $P$ are also in the set. Prove that if $S$ contains at least two numbers, then the sum of the numbers in $S$ is divisible by $P$.
## E... | 64.30. Let's denote the sum of all numbers in the set by $N$. Then, if $a$ is any number from the set $S$, $a N$ and $N$ have the same remainder when divided by $P$ because multiplying by $a$ simply permutes the remainders of the numbers in the set $S$. Choose a number $a$, not equal to 1, and we get that since $(a-1) ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,665 |
31. The vertices of two acute triangles lie on different pairs of opposite sides of a unit square. Prove that the area of their common part is no greater than four times the product of the areas of the triangles.
38 | 64.31. Let the lengths of the bases of the triangles be $a$ and $b$. Consider for each of these triangles a parallelogram, three vertices of which coincide with the vertices of the triangle, and one of its sides coincides with the base of the triangle. Then the intersection of the triangles lies within the intersection... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,666 |
32. There is a set of natural numbers $a_{1}, a_{2}, \ldots, a_{n}$, among which there may be identical ones. Denote by $f_{k}$ the number of numbers in this set that are not less than $k$. Prove that $f_{1}+f_{2}+$ $+\ldots=a_{1}+a_{2}+\ldots+a_{n}$. | 64.32. Let's represent the numbers $a_{1}, a_{2}, \ldots, a_{n}$ as columns of a certain diagram, arranged in a monotonic order (see example in Fig. 29). Then $a_{1}+a_{2}+\ldots+a_{n}$ is the total number of cells in the diagram, and $f_{1}, f_{2}, \ldots$ represent the number of cells in the rows, and their sum, of c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,667 |
33. A unit square is covered by $n$ figures such that each of its points belongs to at least $q$ figures. Prove that at least one figure has an area not less than $q / n$. | 64.33. Let's divide the square into parts that are the smallest intersections of the given figures. Then its area $S$ is equal to the sum $s_{1}+s_{2} \ldots$
Fig. 29 $\ldots+s_{n}$, where ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,668 |
34. Let all real numbers be divided into two groups. Prove that for any $k>0$ there exist three numbers $a<b<c$ from one group such that $(c-b) /(b-a)=k$.
| 64.34. First, it is necessary to prove this fact for $k=1$. This is not difficult and can be solved by enumeration. Next, let's take an arbitrary $k$. Since there will be three points $A, B$, and $C$ belonging to one group (say, group 1), such that $B$ is the midpoint of segment $A C$, we can assume that these points c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,669 |
3. Front tires of a car wear out after 25000 km of travel, while the rear tires wear out after 15000 km of travel. When should the tires be swapped to ensure they wear out simultaneously? | 65.3. Answer. After 9375 km of travel. | 9375 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,671 |
4. A rectangle of 19 cm $\times$ 65 cm is divided by lines parallel to its sides into squares with a side of 1 cm. Into how many parts will this rectangle be divided if we also draw its diagonal? | 65.4. The diagonal intersects $19+65-1$ cells, and, consequently, 83 parts are added to the initial partition. The total number of parts is $19 \cdot 65+83=1318$. | 1318 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,672 |
6. Odd numbers from 1 to 49 are written in the form of a table
| 1 | 3 | 5 | 7 | 9 |
| ---: | ---: | ---: | ---: | ---: |
| 11 | 13 | 15 | 17 | 19 |
| 21 | 23 | 25 | 27 | 29 |
| 31 | 33 | 35 | 37 | 39 |
| 41 | 43 | 45 | 47 | 49 |
Five numbers are chosen such that no two of them are in the same row or column. What is ... | 65.6. Answer. Regardless of the choice, the sum will be 125. | 125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,673 |
7. Prove that a natural number with an odd number of divisors is a perfect square. | 65.7. Let's divide all divisors of the number $N$ into pairs $(d, N / d)$. Since their number is odd, one of the divisors must form a pair with itself. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,674 |
12. Black paint was splattered onto a white plane. Prove that there will be two points of the same color, the distance between which is 1965 meters.
## 8th grade | 65.12. Consider the vertices of an equilateral triangle with a side length of 1965 m. Some two of its vertices are painted the same color. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,678 |
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