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2. In each cell of a $3 \times 3$ square, a natural number is written. Each number is distinct and different from one. It is known that the number written in each cell is a divisor of the product of all numbers in the cells adjacent to it by a side. Find the maximum possible value of the number of prime numbers among t... | Answer: Six. Solution. We will prove that it is impossible to use only two composite numbers. We will reason by contradiction. Suppose there are only two composite numbers and 7 primes. The product of several prime numbers cannot be divisible by a prime number different from them. Therefore, the composite numbers must ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,879 |
3. Can the distances from a point in a plane to the vertices of some square be equal to 1, 1, 2, and 3? | Answer: No. Solution. By the condition, point $O$ is equidistant from two vertices of a certain square $ABCD$. There are two possible cases. 1) Suppose point $O$ is equidistant from two adjacent vertices of the square (let these be vertices $A$ and $B$). Then it lies on the perpendicular bisector of side $AB$. But this... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,880 |
4. In a coffee shop, 55 Indians and Turks met, each drinking tea or coffee. All Indians tell the truth when drinking tea and lie when drinking coffee, while all Turks do the opposite. When asked "Are you drinking coffee?" 44 people answered "yes," when asked "Are you a Turk?" 33 people answered "yes," and 22 people agr... | Answer: 0. Solution. Let $h_{\psi}, h_{\kappa}$ be the number of Indians drinking tea and coffee, respectively. Similarly, define the quantities $T_{4}$ and $T_{k}$. Then from the first question, it follows that $T_{4}+T_{\kappa}=44$, and from the second question, it follows that $h_{\kappa}+T_{\kappa}=33$. Suppose tha... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,881 |
5. Each side of the triangle is divided into 2008 equal parts. Through each division point, lines parallel to the other two sides are drawn, resulting in the triangle being divided into equal triangular fields. A row will be called a series of fields enclosed between two adjacent parallel lines, or the single field sta... | Answer Vasya will win. Solution Notice that the triangle has an even number of cells, which means Vasya makes the last move. With his last move, he can always ensure that the product of all the numbers written is equal to -1. We will show that this is enough for his victory. Multiply all the numbers in the rows paralle... | Vasyawins | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,882 |
1. Alice and the White Rabbit left the Rabbit's house together at noon and went to the Duchess's reception. Having walked halfway, the Rabbit remembered that he had forgotten his gloves and fan, and ran home for them at twice the speed he had walked with Alice. Grabbing the gloves and fan, he ran to the Duchess (at the... | Answer: At 12:40. Solution: While running after his forgotten items, the Rabbit ran a distance equal to the path from his cottage to the reception area. Alice, walking at half the speed of the Rabbit, covered half of this distance in the same time. This means that when the Rabbit reached the halfway point, Alice had ju... | 12:40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,883 |
2. Inside the angle $A O B$, equal to $120^{\circ}$, rays $OC$ and $OD$ are drawn such that each of them is the bisector of one of the angles formed on the diagram. Find the measure of angle $AOC$. List all possible options.
Translate the above text into English, please keep the line breaks and format of the source te... | Answer: $30^{\circ}, 40^{\circ}, 60^{\circ}, 80^{\circ}$ or $90^{\circ}$.
Solution. If ray $O C$ is the bisector of angle $A O B$ (Fig. 1), then angle $A O C$ is equal to $60^{\circ}$ (regardless of whether ray $O D$ is the bisector of angle $A O C$ or $B O C$).
If ray $O D$ is the bisector of angle $A O B$ (Fig. 2),... | 30,40,60,80,90 | Number Theory | proof | Yes | Yes | olympiads | false | 6,884 |
3. Find all natural numbers whose decimal representation ends with two zeros and have exactly 12 natural divisors. | Answer: 200 and 500. Solution: Since the number ends with two zeros, it is divisible by 100, that is, it has the form $n \cdot 100=n \cdot 2^{2} \cdot 5^{2}$. We will prove that if a number has exactly 12 divisors, then $n$ can only be 2 or 5. The smallest number of the form $n \cdot 2^{2} \cdot 5^{2}$ is the number 10... | 200500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,885 |
4. In triangle $A B C$, an arbitrary point $D$ is marked on the median $B M$. Then a line parallel to $A B$ is drawn through $D$, and a line parallel to $B M$ is drawn through $C$. These lines intersect at point $E$. Prove that $B E = A D$. | Solution. Draw a line through point $A$ parallel to $B M$, and let $F$ be the point of its intersection with line $D E$. By Thales' theorem, from the equality $A M=M C$ it follows that $F D=D E$. Moreover, by construction, $A B D F$ is a parallelogram, from which $A B=F D$. Hence, $A B=D E$, and $A B E D$ is a parallel... | BE=AD | Geometry | proof | Yes | Yes | olympiads | false | 6,886 |
5. On an infinite checkers board, two black checkers are placed on two diagonally adjacent cells. Is it possible to add several black checkers and one white checker to the board so that the white checker can eat all the black checkers (including the two initially placed) in one move?
Recall that a checker eats a diago... | Answer. No. Solution. Let's repaint the black cells of the board in red and blue, alternating (so that blue cells border only red cells diagonally, and red cells only blue cells). A white checker, standing on a blue cell, can only eat black checkers that stand on red cells. Conversely, a white checker, standing on a re... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,887 |
1. We will call a four-digit number x amusing if each of its digits can be increased or decreased by 1 (with the digit 9 being only decreased and 0 only increased) so that the resulting number is divisible by x.
a) Find two amusing numbers. b) Find three amusing numbers.
c) Do four amusing numbers exist? (S. Berlov) | Answer. There are four funny numbers: 1111, 1091, 1109, 1089. Note. We will show that there are no other funny numbers. Notice that the number $y$ obtained after changing the digits is not less than $x$, and not equal to $x$, but its first digit can be greater than the first digit of the number $x$ by only 1. This, as ... | 1111,1091,1109,1089 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,888 |
2. Trapezoid $ABCD$ with bases $AD$ and $BC$ is such that angle $ABD$ is a right angle and $BC + CD = AD$. Find the ratio of the bases $AD: BC$. (B. Obukhov) | Answer. $A D: B C=2$. First solution. On side $A D$, we mark off a segment $A E=B C$. Then $A B C E$ is a parallelogram, and $E D=C D$. Since $A B \| C E$, the diagonal $B D$, which is perpendicular to $A B$, is also perpendicular to $C E$. Therefore, it passes through the midpoint $F$ of the base $C E$ of the isoscele... | AD:BC=2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,889 |
3. On the table, there are 100 identical-looking coins, of which 85 are counterfeit and 15 are genuine. You have a miracle tester, into which you can place two coins and get one of three results - "both coins are genuine," "both coins are counterfeit," and "the coins are different." Can you find all the counterfeit coi... | Answer. Yes. Solution. We will present one of the possible ways to define the fake coins. Divide the coins into 50 pairs and check all pairs except one. We will learn the number of fake coins in each pair. Since the total number of fake coins is known, we will also learn how many are fake in the remaining pair. We need... | 64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,890 |
4. A proper divisor of a number is any of its natural divisors, except 1 and the number itself. With a composite natural number \( a \), the following operations are allowed: divide by its smallest proper divisor or add any natural number that is divisible by its largest proper divisor. If the number becomes prime, no ... | Answer: Incorrect. Solution: We will show that the largest prime divisor of a number does not decrease under the specified operations, and therefore we can never obtain the number 2011 from a composite number that has a prime divisor greater than 2011.
Note that the largest proper divisor $b$ of a composite number $a$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,891 |
5. Do there exist 10 different rational numbers such that the product of any two of them is an integer, but the product of any three is not? Recall that a rational number is a number that can be expressed as the ratio of two integers. (O. Podlipsky) | Answer. No. Solution. Suppose such 10 numbers were found. Consider any three of them: $a, b, c$. Then the numbers $a b, b c, c a$ are integers, but the number $a b c=p / q$ is not. Then the number $(a b c)^{2}=p^{2} / q^{2}$ is not an integer. But $(a b c)^{2}=(a b)(b c)(c a)$ is an integer. Contradiction. Remark. In f... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,892 |
6. Black and white balls are arranged in a circle, with black balls being twice as many as white ones. It is known that among pairs of adjacent balls, there are three times as many monochromatic pairs as polychromatic ones. What is the smallest number of balls that could have been arranged? (B. Trushin) | Answer: 24. Solution: Since the number of black balls is twice the number of white balls, the total number of balls is divisible by three. Let's denote it by $n$. All the balls are divided into alternating groups of consecutive balls of the same color (a group can consist of just one ball). Since the colors of the grou... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,893 |
7. 1000 different positive numbers are written in a row in ascending order. Vasya divided these numbers into 500 pairs of adjacent numbers and found the sums of the numbers in all pairs. Petya divided these same numbers into 500 pairs such that there are exactly three other numbers between the numbers in each pair, and... | Solution. Lemma. Let $a0$. Solution of the problem. Divide all thousand numbers into 125 eights of consecutive numbers. Take the first eight: $a_{1}a c+b d$, is widely known as the transinequality.
Criteria. Concrete numerical examples - o points. If the statement of the lemma is used without any justification, the so... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 6,894 |
8. In a convex quadrilateral $A B C D$, angles $A B C$ and $A D C$ are right angles. Points $K, L, M, N$ are taken on sides $A B, B C, C D, DA$ respectively such that $K L M N$ is a rectangle. Prove that the midpoint of diagonal $A C$ is equidistant from the lines $K L$ and $M N$. (D. Shveцов) | Solution. Consider triangle $K N T$, where $K T \| B C$ and $N T \| C D$. It is equal to triangle $L M C$ by the side $K N=L M$ and two adjacent angles. Therefore, $K T=L C$, and $K T C L$ is a parallelogram, from which $T C \| K L$ and $T C \perp K N$.
Let $S$ and $O$ be the midpoints of segments $A T$ and $A C$ resp... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,895 |
1. On his birthday, Dadya Fyodor's parents give him an amount of money equal to the product of his father's age and his mother's age. Could it have happened that the amounts received by him in 2010 and 2011 ended with the same digit, and the amount received in 2012 was divisible by 10? | Answer: It could. Solution. For example, if in 2010 Uncle Fyodor's father was 38 years old, and his mother was 31 years old. Note. All possible examples are when the ages of Uncle Fyodor's parents in 2010 end in the digits 3 and 6, or when they end in 1 and 8. There are no other examples. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,896 |
2. Masha is practicing repainting a chessboard. In one move, she can change the color of each cell in any rectangle adjacent to a corner of the board. Will she be able to repaint the entire board in one color using such operations? | Answer: It will work. First solution. Number the cells of the board: in the top horizontal row from left to right with numbers from 1 to 8, in the second row from the top - from left to right with numbers from 9 to 16, and so on. Let the cell in the bottom right corner be white. Let $n$ be the largest number of a black... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,897 |
4. Positive numbers $x$ and $y$ are such that $x^{2} > x + y$ and $x^{4} > x^{3} + y$. Prove that $x^{3} > x^{2} + y$. | First solution. Rewrite the condition as $x^{2}-x=x(x-1)>y$, $x^{4}-x^{3}=x^{3}(x-1)>y$. We need to prove that $x^{3}-x^{2}=x^{2}(x-1)>y$. Note that $x>1$ - otherwise $x(x-1) \leq 0x(x-1)>y$. Second solution. Since $x^{2}>x+y$ and $x>1, x^{3}>x^{2}+x y>x^{2}+y$. Third solution. Multiply the inequalities $x(x-1)>y$ and ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 6,899 |
5. 40 children are standing in a circle. A child is called a tall one if they are taller than the two following them clockwise, and a small one if they are shorter than both preceding them clockwise. (A child can be both a tall one and a small one at the same time.) It is known that there are no fewer than 30 tall ones... | Solution. Let's call children who are not giants ordinary, and a group - an ordinary child and all the giants standing between him and the previous ordinary child in a clockwise direction. In each group, all children except the first two in a clockwise direction are small, because each of them is preceded by two giants... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,900 |
1. Petya thought of a natural number N, and Vasya wants to guess it. Petya tells Vasya the sum of the digits of the number $N+1$, then the sum of the digits of the number $N+2$, and so on. Is it true that sooner or later the clever Vasya will be able to determine Petya's number with certainty? (M. Diden) | Answer. Correct. Solution. Let $10^{k} \leq N<10^{k+1}$. In $10^{k+1}-N$ steps, Petya will write down the first one, and after another $10^{k+2}-10^{k+1}=9 \cdot 10^{k+1}$ steps, he will write down the second one. Knowing the number of steps between these two events, we can find $k$, and knowing $k$ and the number of s... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,901 |
2. Find the smallest natural $k$ such that for some natural number $a$, greater than 500,000, and some natural number $b$, the equality $\frac{1}{a}+\frac{1}{a+k}=\frac{1}{b}$ holds. (I. Bogdanov) | Answer. $k=1001$. Solution. Estimation. Let $a+k=c$ and $\text{GCD}(a, c)=d$. Then $a=d a_{1}, c=d c_{1}$ and $\frac{1}{a}+\frac{1}{c}=\frac{a_{1}+c_{1}}{d a_{1} c_{1}}$. Since the numbers $a_{1} c_{1}$ and $a_{1}+c_{1}$ are coprime, $d$ must divide $a_{1}+c_{1}$. Therefore, $d \geq a_{1}+c_{1}$ and $d^{2} \geq d\left(... | 1001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,902 |
3. The diagonals of a convex quadrilateral ABCD are equal and intersect at point K. Inside triangles $AKD$ and $BKC$, points $P$ and $Q$ are chosen respectively such that $\angle KAP = \angle KDP = \angle KBQ = \angle KCQ$. Prove that line $PQ$ is parallel to the bisector of angle $AKD$. (S. Berlov) | Solution. Since $\angle K A P = \angle K C Q$, $C Q \| A P$. Since $\angle K D P = \angle K B Q$, $B Q \| D P$. Let $B X$ and $C Y$ be the perpendiculars dropped from $B$ and $C$ to $D P$ and $A P$ respectively. Then the right triangles $B D X$ and $C A Y$ are equal by hypotenuse and acute angle, hence $B X = C Y$. Thi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,903 |
4. In the vertices of a regular 300-gon, numbers from 1 to 300 are placed once each in some order. It turns out that for each number a, among the 15 nearest numbers to it in the clockwise direction, there are as many numbers less than a as there are among the 15 nearest numbers to it in the counterclockwise direction. ... | Answer: 10. Solution: Estimation. To prove that there are no fewer than 10 huge numbers, it is sufficient to prove that among any 30 consecutive numbers, there will be a huge one. Indeed, suppose this is not the case. Then consider the largest of these 30 numbers. On one side of it, all 15 nearest numbers will be part ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,904 |
1. Wrote down two numbers - the first and the second. Added the second to the first - got the third, added the third to the second - got the fourth, and so on. The sum of the first six numbers is 2008. What is the fifth number? | Answer: 502. Solution: Let the first number be $a$, the second - $b$. Then the third number is $a+b$, the fourth $-a+2b$, the fifth $-2a+3b$, the sixth $-3a+5b$, and the sum of all six numbers is $8a+12b$. Thus, the fifth number is a quarter of the sum of all six numbers, that is, $2008: 4=502$.
Grading Guidelines. An... | 502 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,905 |
2. Out of four visually identical coins, three are genuine and weigh the same, while the fourth is counterfeit and its weight differs from that of a genuine coin. There are scales that can determine the exact weight of two or more coins, but the exact weight of a single coin cannot be determined. How can you find the c... | Solution. First method. Place one coin on the scales and weigh it in turn with each of the other three. If all three weighings give the same result, the first coin is counterfeit, and to find out whether it is lighter or heavier than a genuine one, a fourth weighing of two genuine coins is needed. If, however, one of t... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,906 |
3. In a right triangle, the height dropped to the hypotenuse is four times shorter than the hypotenuse. Find the acute angles of the triangle. | Answer: 15 and 75 degrees. Solution. Let $A B C$ be the given right-angled triangle with the right angle at vertex $C$. As is known, the median $C M$ is half the hypotenuse $A B$. The height $C H$ is given to be $A B / 4$. Therefore, in the right-angled triangle $C H M$, the hypotenuse $C M$ is twice the length of the ... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,907 |
4. Several ferries shuttle between points K and M across Lake B at equal constant speeds. Each ferry stays in each of the points for as long as it takes to cross the lake. A traveler noticed that ferries depart from each point at equal intervals, and his ferry departed exactly at the moment when a ferry from the opposi... | Solution. Let the ferries depart from the pier at intervals of $t$ minutes, take $T$ minutes to cross the lake, and while the ferry the traveler was on was docked, $n$ other ferries arrived at this shore. Since the last of them arrived at the moment the traveler departed, the interval $t$ fits into the time interval $T... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 6,908 |
5. On the board, 8 non-intersecting circles were drawn, and from each one, an arrow was drawn to those of the other seven that are not larger than it. In total, 33 arrows were obtained. Prove that among the circles drawn on the board, there are three equal ones. | Solution. Let's number the circles in ascending order of their radii (the order between circles of equal radii is established arbitrarily). Then, from each circle, there are arrows to all circles with smaller numbers: from the eighth - 7 arrows, from the seventh - 6, and so on. In total, there are such arrows $7+6+5+4+... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,909 |
5. Petya and Vasya simultaneously entered the same non-zero integer into their calculators. After that, every minute Petya either added 10 to his number or multiplied it by 2014; at the same time, Vasya, in the first case, subtracted 10 from his number, and in the second case, divided it by 2014. Could it happen that a... | Answer. Yes, it could have turned out that way. Solution. Suppose the last action before the numbers became equal again was that Petya multiplied by 2014, and Vasya divided. Then, before this, Petya's and Vasya's numbers were negative, and the absolute value of Vasya's number was $2014^{2}$ times greater than the absol... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,910 |
6. Let's call a natural number mountainous if its representation contains a non-edge digit (called the peak) that is greater than all others, and all other digits are non-zero and initially non-strictly increase (i.e., each subsequent digit is greater than or equal to the previous one) up to the peak, and then non-stri... | Solution. Let's call two mountainous numbers friendly if each of them is obtained from the other by writing its digits in reverse order. Note that the digits that stand in even positions in one of the friendly numbers stand in odd positions in the other. In particular, two friendly numbers cannot coincide, as their pea... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,911 |
7. The decimal representation of a natural number $N$ consists only of ones and twos. It is known that by erasing digits from this number, any of the 10000 numbers consisting of 9999 ones and one two can be obtained. Find the smallest possible number of digits in the representation of $N$. (G. Chelnokov) | Answer: 10198. Solution: Example. The number 1...121...12...21...121...1, where there are 100 twos, 99 ones at the beginning and end, and 100 ones between adjacent twos. The number consisting of 9999 ones and a two, where before the two there are $100 m+n$ ones ($0 \leq m, n \leq 99$), is obtained by deleting all twos ... | 10198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,912 |
8. We will call a diagonal of a convex 101-gon principal if on one side of it lie 50, and on the other side - 49 vertices. Several principal diagonals without common endpoints are chosen. Prove that the sum of the lengths of these diagonals is less than the sum of the lengths of the remaining principal diagonals. (I. B... | Solution. Let's call the main diagonal of a $(2n+1)$-gon $K=A_{1} A_{2} \ldots A_{2 n+1}$ any segment of the form $A_{i} A_{i+n}$ (vertex numbering is cyclic, so $A_{i+2 n+1}=A_{i}$). We will prove by induction on $n$ that the sum of the lengths of any selected set of main diagonals of the polygon $K$, which do not sha... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,913 |
5. 40 bandits crossed a river from the left bank to the right bank using a two-person boat (some trips may have been made solo). Could it have happened that each pair of bandits crossed the river together exactly once (either from the left bank to the right or from the right to the left)? (A. Shapovalov) | Answer. No, it could not. Solution. Suppose it could. Note that the boat must make an odd number of trips. Since with pairs of bandits it made $40 \cdot 39 / 2=780$ trips, there must be an odd number of trips with single bandits. Therefore, someone among the bandits must have ferried the boat alone an odd number of tim... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,914 |
6. A natural number is called perfect if it is half the sum of all its natural divisors: for example, the number 6 is perfect because $2 \cdot 6 = 1 + 2 + 3 + 6$. Can the sum of all pairwise products of the natural divisors of a perfect number $n$ be divisible by $n^{2}$? (S. Berlov) | Answer. It cannot. Solution. Note that a perfect number is equal to the sum of all its natural divisors, less than itself. Let the divisors of a perfect number $n$ be $d_{1}, \ldots, d_{k}$. The sum of all pairwise products of its divisors is $n d_{1}+\ldots+n d_{k}+d_{1} d_{2}+d_{1} d_{3}+\ldots+d_{1} d_{k}+d_{2} d_{3... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,915 |
7. In the country of Grapheinia, there are $n$ ($n \geq 2$) cities. Some cities are connected by non-stop flights (flights are operated in both directions on each route) in such a way that from any city, it is possible to reach any other city by plane (possibly with layovers), but closing any flight route would violate... | Solution. Consider a graph where vertices are cities and edges are air routes. By the condition, it is a tree, the vertices of which, as is known, can be numbered with two digits so that any two adjacent vertices are marked with different digits. Let's do this and paint in black the vertices marked with the digit that ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,916 |
8. SK is the bisector of triangle ABC. Points $L$ and $T$ are chosen on sides $BC$ and $AC$ respectively such that $CT = BL$ and $TL = BK$. Prove that the triangle with vertices at points $C, L$, and $T$ is similar to the original triangle. (S. Berlov) | Solution. Mark such a point $S$ that $B L S K$ is a parallelogram. If $S$ coincides with $T$, then the similarity is obvious. If $S$ does not coincide with $T$, then since $C T=B L=K S$ and $\cdot S K C=\cdot K C L=\cdot K C A$, the points $K, C, T, S$ are the vertices of an isosceles trapezoid. Since $T L=K B=L S$, th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,917 |
1. A $70 \times 70$ table is filled with numbers from 1 to 4900: in the first row, the numbers from 1 to 70 are written in ascending order from left to right; in the second row, the numbers from 71 to 140 are written in the same manner, and so on; in the last row, the numbers from 4831 to 4900 are written from left to ... | Answer: No. Solution: From the construction, it is clear that if the number $x$ is written in a cell, then above it is the number $x-70$, below - the number $x+70$, to the left - the number $x-1$, and to the right - the number $x+1$. The sum of these five numbers is $5x$, which is divisible by 5, while the number 2018 ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,918 |
2. At a round table, 100 people are sitting. Each of them is either a knight, a liar, or a fool. A knight always tells the truth, a liar always lies. A fool tells the truth if a liar is sitting to their left; lies if a knight is sitting to their left; and can say anything if a fool is sitting to their left. Each of the... | Answer: 0 or 50. Solution: Evaluation. Suppose there is a liar among those sitting. Then to the right of him is a knight or a fool. Any of them in this situation will tell the truth, meaning to the right of him is again a liar, and so on, which means there are exactly 50 liars. Examples. For 0 liars: at the table are o... | 0or50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,919 |
3. In trapezoid $ABCD$, the lateral side $AB$ is equal to the diagonal $BD$. Point $M$ is the midpoint of diagonal $AC$. Line $BM$ intersects segment $CD$ at point $E$. Prove that $BE = CE$. (A. Kuznetsov) | Solution. Extend triangle $A B C$ to parallelogram $A B C F$. Its diagonal $B F$ passes through point $M$, and therefore, through point $E$. Since $C F=A B=B D$, and line $C F$, being parallel to line $A B$, is not parallel to line $B D$, $B C F D$ is an isosceles trapezoid. The diagonals $B F$ and $C D$ of the trapezo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,920 |
4. On a parking lot, there are cars. Among them, there are cars of the brands "Toyota", "Honda", "Skoda", as well as cars of other brands. It is known that the number of non-"Honda" cars is one and a half times the number of non-red cars; the number of non-"Skoda" cars is one and a half times the number of non-yellow c... | The first solution. Let there be $C$ cars in total on the parking lot, among which $T$ are "Toyotas", $H$ are "Hondas", and $S$ are "Skodas", as well as $X$ red and $Y$ yellow cars. According to the conditions, $C-H=3(C-X) / 2, C-S=3(C-Y) / 2, C-T=(X+Y) / 2$. By adding the first two equations, after combining like term... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 6,921 |
5. A grasshopper starts moving in the top-left cell of a 10×10 square. It can jump one cell down or to the right. Additionally, the grasshopper can fly from the bottom cell of any column to the top cell of the same column, and from the rightmost cell of any row to the leftmost cell of the same row. Prove that the grass... | Solution. Without making a jump, the grasshopper can only get from cell $K$, where it is located, to the cells of the rectangle where $K$ is the top-left cell. Let's paint 10 cells lying on the diagonal of the square, running from the top-right cell to the bottom-left cell, red. The rectangle, whose top-left corner is ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,922 |
1. Vasya arranged all natural numbers from 1 to 100 in a circle in some order. We say that a number is well-placed if the neighboring number in the clockwise direction is greater than the neighboring number in the counterclockwise direction. Could it be that at least 99 numbers are well-placed? (A. Gолованов, I. Рубано... | Answer. It could not. First solution. Let's color the numbers in black and white so that the colors alternate. Then, a black number with the smallest white number as its neighbor clockwise, and a white number with the smallest black number as its neighbor clockwise, will not be well placed, so the total number of well-... | 98 | Combinatorics | proof | Yes | Yes | olympiads | false | 6,923 |
2. Given three positive numbers: $a, b, c$. Petya wrote on the board the numbers $\frac{1}{a}+b c, \frac{1}{b}+a c, \frac{1}{c}+a b, a$ and Vasya wrote the numbers $2 a^{2}, 2 b^{2}, 2 c^{2}$. It turned out that both wrote the same three numbers (possibly in a different order). What is the product $a b c?$ (N. Agakhano... | Answer. 1. First solution. Multiplying the expressions $\frac{1}{a}+b c, \frac{1}{b}+a c, \frac{1}{c}+a b$, we obtain the product $\frac{(a b c+1)^{3}}{a b c}$, while multiplying $2 a^{2}, 2 b^{2}, 2 c^{2}-8(a b c)^{2}$. According to the condition, the obtained products are equal. Equating them and multiplying both sid... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,924 |
3. The remainder of the division of a natural number n by 2021 is 800 more than the remainder of the division of the number n by 2020. Find the smallest such n. (A. Gолованов) | Answer. 2020.1221 = 2466 420. Solution. Let for a natural number $m$ the condition of the problem is satisfied. Then for some $d_{1}, d_{2}, d_{3}$ and $r$ the equality $m=2020 d_{1}+r=2021 d_{2}+r+800$ holds. But in this case, the equality $m-r=2020 d_{1}=2021 d_{2}+800$ also holds, so the number $m-r$ also satisfies ... | 2466420 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,925 |
4. On a $100 \times 100$ chessboard, 1975 rooks were placed (each rook occupies one cell, different rooks stand on different cells). What is the maximum number of pairs of rooks that could be attacking each other? Recall that a rook can attack any number of cells along a row or column, but does not attack a rook that i... | Answer: 3861. Solution: Sequentially remove from the $100 \times 100$ board the verticals and horizontals that do not contain rooks, each time gluing the edges of the removed strip. We will get a rectangle $\pi$, in each vertical and each horizontal of which there is at least one rook (obviously, the number of pairs of... | 3861 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,926 |
5. VM is the median of an acute-angled triangle ABC. The bisector of angle C intersects the line passing through $A$ parallel to $BC$, at point $X$. It turns out that $BM = MX$. Prove that $BC > AC$. (S. Berlov) | Solution. In triangle $A B C$, draw the altitude $A H$. The statement of the problem immediately follows from the equality $A C=B H$, which we will prove. Draw the median $M D$ of the isosceles triangle $B M X$. It is the midline of the trapezoid (or parallelogram) $A X B C$, and therefore parallel to the lines $B C$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,927 |
1. Is it possible to assign a price to each digit from 0 to 9 such that all 10 prices are distinct and there exist 20 consecutive natural numbers, each of which, except the first, is more expensive than the previous one? Here, the price of a natural number is the sum of the prices of the digits in its representation. (... | Answer. Yes. Solution. Let the digits $0,1, \ldots, 9$ cost 20, 21, $, 2,29$ rubles respectively. Then the numbers 90, $91, \ldots, 99,100, \ldots, 109$ cost, respectively, $49,50, \ldots, 58,61,62, \ldots, 70$ rubles.
Note. If we replace 20 with 21 in the problem statement, the answer becomes negative. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,928 |
2. The graph of $y=x+b \sqrt{x}+c$, where $c>0$, intersects the y-axis at point $C$, and the x-axis at points $X_{1}$ and $X_{2}$. Denote the origin by O. Prove that $\angle C X_{1} O+\angle C X_{2} O=90^{\circ}$. (A. Shklover) | Solution. It is sufficient to check that $\angle C X_{1} O=90^{\circ}-\angle C X_{2} O=\angle O C X_{2}$, which means that the right triangles $\mathrm{COX}_{1}$ and $X_{2} O C$ are similar.
Let the points $X_{1}$ and $X_{2}$ have coordinates $\left(x_{1}, 0\right)$ and $\left(x_{2}, 0\right)$ respectively. Then $x_{1... | proof | Algebra | proof | Yes | Yes | olympiads | false | 6,929 |
3. The diagonals of a convex quadrilateral ABCD intersect at point E. It is known that $A B=B C=C D=D E=1$. Prove that $A D<2$. (A. Kuznetsov) | The first solution. Note that $\angle C E D = 90^{\circ}$, so $B C > C E$. Let $B^{\prime}$ be the point symmetric to point $B$ with respect to the line $A C$. Since $B C = C D = D E$, $\angle B^{\prime} C D = \angle D C E - \angle B^{\prime} C E = \angle C E D - \angle B C E = \angle C B E = \angle C D E$. Also, $A B^... | AD<2 | Geometry | proof | Yes | Yes | olympiads | false | 6,930 |
4. Zeus has scales that allow him to find out the weight of the load placed on them, and a bag with 100 coins, among which there are 10-gram and 9-gram coins. Zeus knows the total number $N$ of 10-gram coins in the bag, but it is unknown which ones weigh how much. He would like to make four weighings on the scales and ... | Answer. For $N=15$. Solution. First, let's outline Zeus's algorithm for $N=15$. By weighing a certain number of coins, he immediately determines the number of heavy coins among the weighed ones. Since he only needs to identify one light coin, he can weigh just 8 coins in the first weighing. If there are light coins amo... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,931 |
5. 100 consecutive natural numbers were sorted in ascending order of the sum of their digits, and numbers with the same sum of digits were simply sorted in ascending order. Could the numbers 2010 and 2011 end up next to each other? (S. Volchonkov) | Answer. Could not. Solution. Let the numbers 2010 and 2011 be next to each other. This means that among the 100 taken numbers, 2010 is the largest number with the sum of digits 3, and 2011 is the smallest number with the sum of digits 4. But this means that among the 100 consecutive natural numbers taken, there are num... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,932 |
6. A convex pentagon $A B C D E$ is such that $A B\|C D, B C\| A D, A C \| D E, C E \perp B C$. Prove that E is the bisector of angle $B E D$. (P. Kozhevnikov) | Solution. Extend the segment $D E$ to intersect the line $B C$ at point $K$. From the condition, it follows that $A B C D$ and $A D K C$ are parallelograms, from which $B C=A D=C K$. Thus, $E C$ is the median and altitude, and therefore also the bisector in triangle $B E K$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,933 |
7. Five irreducible fractions with odd denominators greater than $10^{10}$ were written in red around a circle. Between each pair of adjacent red fractions, their irreducible sum was written in blue. Could it happen that all the denominators of the blue fractions are less than 100? (I. Bogdanov) | Answer. It could not. First solution. Assume the opposite. Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be the original fractions in the order of their arrangement in a circle, and the denominator of the number $a_{1}$ is greater than $10^{10}$. Note that $2 a_{1}=\left(a_{1}+a_{2}\right)+\left(a_{3}+a_{4}\right)+\left(a_{5... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,934 |
8. What is the maximum number of white and black pawns that can be placed on a 9x9 grid (a pawn, regardless of its color, can be placed on any cell of the board) so that no pawn attacks any other (including those of the same color)? A white pawn attacks two diagonally adjacent cells on the next higher horizontal row, w... | Answer: 56. Solution: An example with 56 pawns is shown in the figure. Evaluation: Note that in each rectangle of three rows and two columns, there are no more than 4 pawns. Indeed, if there are at least 5, then on one of the colors, all three cells are occupied, and the pawn in the middle row attacks one of the two re... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,935 |
1. A snail is crawling around the clock face counterclockwise at a constant speed. It started at 12:00 from the 12 o'clock mark and completed a full circle exactly at 14:00. What time did the clock show when the snail met the minute hand during its movement? | Answer: 12:40 and 13:20. Solution. From the condition, it follows that the snail moves along the clock face at half the speed of the minute hand. Therefore, by the first meeting, it has crawled a third of the way around the clock, while the hand has moved two-thirds of the way. This means that the first meeting occurs ... | 12:4013:20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,936 |
2. In each cell of a $2 \times 2$ table, one number is written. All numbers are distinct, the sum of the numbers in the first row is equal to the sum of the numbers in the second row, and the product of the numbers in the first column is equal to the product of the numbers in the second column. Find the sum of all four... | Answer: 0. Solution: Let the numbers in the top row of the table be (from left to right) $a$ and $b$, and the numbers in the bottom row (from left to right) be $c$ and $d$. By the condition, $a+b=c+d$ and $a c=b d$. Expressing $c$ from the first equation and substituting into the second, we get $a(a+b-d)=b d \Leftright... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,937 |
3. The angle bisectors of angles A and C divide the non-isosceles triangle ABC into a quadrilateral and three triangles, and among these three triangles, there are two isosceles triangles. Find the angles of triangle $A B C$. | Answer. $180^{\circ} / 7,2 \cdot 180^{\circ} / 7,4 \cdot 180^{\circ} / 7$. Solution. Let $AK$ and $CM$ be the angle bisectors, and $I$ their point of intersection. Let's see which angles in which triangles can be equal. Triangle $AIC$ cannot be isosceles: in it, $\angle AIC = 90^{\circ} + \angle ABC / 2$ is obtuse, whi... | \frac{180}{7},2\cdot\frac{180}{7},4\cdot\frac{180}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,938 |
4. All divisors of a natural number $N$, except $N$ and one, were written in a row in descending order: $d_{1}>d_{2}>\ldots>d_{k}$. It turned out that in each pair of divisors equally distant from the ends of this row, the larger divisor is divisible by the smaller one (i.e., $d_{1}$ is divisible by $d_{k}, d_{2}$ - by... | Solution. It is easy to see that the equalities $N=d_{1} d_{k}=d_{2} d_{k-1}=\ldots$ hold. This allows us to rephrase the condition of the problem as follows: if the product of two divisors of the number $N$ equals this number, then the larger of these divisors is divisible by the smaller. Suppose the number $N$ has ot... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,939 |
5. Two players are playing such a game. In one move, a player can place one pebble in one of the cells of a $1001 \times 1001$ square board (initially the board is empty; any number of pebbles can lie in one cell). They take turns. As soon as there are more than 5 pebbles in any row (vertical or horizontal), the player... | Answer. First. First solution. Before the 5005th pebble is placed on the board, the first player finds a row and a column with fewer than five pebbles (such must exist) and places a pebble in the cell at their intersection. If the second player has not made a mistake before, after the 5005th pebble is placed on the boa... | First | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,940 |
1. 200 people are standing in a circle. Each of them is either a liar or a conformist. Liars always lie. A conformist who stands next to two conformists always tells the truth. A conformist who stands next to at least one liar can either tell the truth or lie. 100 of those standing said: "I am a liar," and the other 10... | Answer: 150. Solution: A liar cannot say, "I am a liar." Therefore, 100 people who said, "I am a liar," are conformists. All of them lied, so next to each of them stands a liar. Since next to a liar there can be a maximum of two conformists, there are no fewer than 50 liars. Thus, there are no more than 150 conformists... | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,941 |
2. From a chessboard of size $13 \times 13$, two opposite corner cells have been cut out. On the remaining part of the board, several cells have been marked. Prove that it is possible to place chess kings on the marked cells such that the total number of kings does not exceed 47, and they attack all the empty marked ce... | Solution. In the diagram on the right, a $13 \times 13$ board is divided into 47 sections such that
a king standing on one of the cells of a section can attack all the other cells in that sec... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,942 |
3. The diagonals of a convex quadrilateral ABCD are equal and intersect at point O. Point P inside triangle $AOD$ is such that $CD \| BP$ and $AB \| CP$. Prove that point $P$ lies on the bisector of angle AOD. (S. Berlov) | Solution. Since $A B \| C P$, the areas of triangles $A P C$ and $B P C$ are equal. Since $C D \| B P$, the areas of triangles $B P C$ and $B P D$ are equal. Therefore, the areas of triangles $A P C$ and $B P D$ are equal. Since $A C = B D$, the heights of these triangles, dropped to sides $A C$ and $B D$ respectively,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,943 |
4. Each of the numbers $x, y$ and $z$ is not less than 0 and not greater than 1. Prove the inequality $\frac{x^{2}}{1+x+x y z}+\frac{y^{2}}{1+y+x y z}+\frac{z^{2}}{1+z+x y z} \leq 1$. (A. Khryabrov) | Solution. Note that $1+y z \geq y+z$, since (1-y)(1-z) $\geq 0$. Therefore, $1+x+x y z=1+x(1+y z) \geq 1+x(y+z) \geq x^{2}+x y+x z$. Hence, $\frac{x^{2}}{1+x+x y z} \leq \frac{x^{2}}{x^{2}+x y+x z}=\frac{x}{x+y+z}$. Applying this estimate to each of the three fractions, we obtain the required result. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 6,944 |
1. Citizen Sidorov has exactly as much money as needed to buy a ton of roundels and a ton of smuggles. If he buys 20% more roundels, he will get a 40% discount on smuggles, and the remaining money will be enough to buy a ton of smuggles. If he buys 40% more smuggles, he will get a 20% discount on roundels, and the rema... | Answer. A ton of kruglik costs twice as much as a ton of shmuglik. Solution. Let a ton of kruglik cost $x$ money, and a ton of shmuglik $y$ money. Then citizen Sidorov has ( $x+y$ ) money. If he buys 20% more kruglik, and a ton of shmuglik with a 40% discount, he will spend $1.2 x+0.6 y$ and this is no more than ( $x+y... | 2y | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,945 |
3. Three natural numbers $a, b$, and $c$ are chosen such that НОД( $a b, c)=$ НОД( $a, b c)$. Prove that after reducing the fraction $a/c$, the resulting fraction will be irreducible, with the numerator and denominator being coprime with $b$.
Note: НОД stands for GCD (Greatest Common Divisor) in English. | Solution. It is sufficient to prove that any prime number $p$ that enters the prime factorization of the number $b$, enters the factorization of the number $a$ and the factorization of the number $c$ in the same (possibly zero) powers. Let's prove this. Indeed, suppose, for example, that the number $p$ enters the facto... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,947 |
4. In pentagon $A B C D E A B=B C=C D=D E, \angle B=96^\circ$ and $\angle C=\angle D=108^\circ$. Find angle $E$. | Answer: $102^{\circ}$. Solution. Draw segments $B D$ and $C E$. Let them intersect at point $O$. Note that triangles $B C D$ and $C D E$ are isosceles with an angle of $108^{\circ}$ at the vertex, so the base angles are $36^{\circ}$ (they are marked on the diagram with one arc). Then $\angle B C E = \angle B D E = 72^{... | 102 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,948 |
5. Petya arranges cards with numbers 1, 2, ..., 9 in the cells of a $3 \times 3$ table. Then he turns away, and Vitya swaps two cards in cells that share a side, and turns all the cards face down. After this, Petya points to one or several cards, and Vitya tells him the sum of the numbers on them. Can Petya act in such... | Answer. He can. Solution. Petya arranges the cards as shown in the figure. Then Vasya swaps two cards, turns them face down, and Petya points to 4 cards, which are marked in gray in the figure (Petya initially placed $1,3,7,9$ in them).
| 8 | 1 | 2 |
| :--- | :--- | :--- |
| 3 | 5 | 9 |
| 6 | 7 | 4 |
Initially, the s... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,949 |
1. Four boys peeked into a box where colored balls were lying. In response to the question about what colors the balls were, they answered as follows. Petya: "Red, blue, and green". Vasya: "Red, blue, and yellow". Kolya: "Red, yellow, and green". Misha: "Yellow, green, and blue". Could it have happened that each boy na... | Answer: It could not. Solution: Note that each color was named by exactly three boys. Therefore, the number of correctly named colors, if each color is counted as many times as it was named, should be divisible by 3, and thus cannot equal 4. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,950 |
2. Prove that if $a+b+c+d=0$ and $a b+c d+a c+b c+a d+b d=0$, then $a=b=c=d=0$. (Folklore) | Solution. Squaring the equality $a+b+c+d=0$, we obtain
$$
a^{2}+b^{2}+c^{2}+d^{2}+2(ab+cd+ac+bc+ad+bd)=0
$$
from which $a^{2}+b^{2}+c^{2}+d^{2}=0$, which is only possible when $a=b=c=d=0$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 6,951 |
3. Borya drew nine segments, three of which are equal to the three heights of triangle ABC, three are equal to the three angle bisectors, and three are equal to the three medians. It turned out that for any of the drawn segments, among the other eight there is one equal to it. Prove that triangle $ABC$ is isosceles. (I... | Solution. Let $A A_{1}$ be the shortest of the altitudes of triangle $A B C$. If it equals the median $A A_{2}$ or the angle bisector $A A_{3}$, then the triangle is obviously isosceles. If it equals the median $B B_{2}$ or the angle bisector $B B_{3}$, then $A A_{1}$ is not shorter than the altitude $B B_{1}$. Therefo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,952 |
4. Around a circle, 49 different natural numbers less than 100 were written in red pencil. Between each pair of adjacent red numbers, their greatest common divisor was written in blue. Could it happen that all the blue numbers are different? (I. Rubanov) | Answer: It could not. Solution: Note that the GCD of two different numbers less than 100 must be less than 50, since at least one of these two numbers is more than the GCD by at least a factor of two. Therefore, if all blue numbers are distinct, then among them are all numbers from 1 to 49. In particular, among them is... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,953 |
1. Is it possible to place a chip in half of the cells of a $12 \times 12$ board such that in one $2 \times 2$ square, composed of cells of the board, there is an odd number of chips, and in all other $2 \times 2$ squares there is an even number of chips? | Answer: It is possible. First solution. Place a chip in each cell of the second, fourth, ..., twelfth row. Then, in each 2x2 square, there will be two chips. Now, move the chip from the top-left corner one cell down. In the corner square, there are still two chips, in the square below it, there are now three chips, and... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,954 |
2. In triangle $ABC$, angle $C$ is three times larger than angle $A$, and side $AB$ is twice as long as side $BC$. Prove that angle $ABC$ is 60 degrees. | Solution. Let $D$ be the midpoint of side $A B$. Since $B D=B C$, triangle $B C D$ is isosceles. Let $\angle C A D=x, \angle A C D=y$. Then $\angle D C B=3 x-y$, and $\angle C D B=x+y$. Since $\angle D C B=\angle C D B$, we have $3 x-y=x+y$, from which $y=x$. Therefore, $D C=D A=D B=B C$, which means triangle $B C D$ i... | 60 | Geometry | proof | Yes | Yes | olympiads | false | 6,955 |
3. Given 5 different natural numbers. The product of the two smallest of them is greater than 25, and the product of the two largest is less than 75. Find all these numbers (list all possible options and prove that there are no other options). | Solution. Let these numbers in ascending order be $a, b, c, d, e$. If $b$ is not greater than 5, then $a$ is not greater than 4, so $a b$ is not greater than 20. Therefore, $b$ is not less than 6. Similarly, if $d$ is not less than 9, then $e$ is not less than 10, and $d e$ is not less than 90. Therefore, $d$ is not gr... | 5,6,7,8,9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,956 |
4. Ali-Baba has 40 bags of coins. The genie can, at Ali-Baba's request, determine the number of coins in each of two specified bags, but in doing so, will take one coin from one of these bags (and Ali-Baba will not see from which one). Can Ali-Baba act in such a way that after no more than 100 such procedures, he can a... | Answer: Yes. Solution. Number the bags: $1,2, \ldots, 40$. Sequentially determine the number of coins in the following bags: $(1,2),(2,3),(3,4), \ldots(39,40)$. After the second operation, we will definitely know the number of coins in bag 1 (since we will understand whether the number of coins in bag 2 changed after t... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,957 |
5. Given nine natural numbers, where the record of the first consists only of ones, the second - only of twos, ..., the ninth - only of nines. Can the product of any two of these numbers be divisible by the product of the others? | Answer: No. Solution. Assume the opposite. Let's call the two numbers in question the chosen ones. One of the two chosen must be written with fives, because no other is divisible by 5. The other must be even. But then the highest power of two by which the product of the chosen numbers is divisible is 3, while the small... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,958 |
5. Masha took four different positive numbers and wrote down their six pairwise products in a row in ascending order. Could all five differences between consecutive numbers in this row be the same? (I. Rubanov) | Answer: No. Solution: Let's order the Machine's numbers: $0<a<b<c<d$. Then the first and second numbers in the sequence of products are $a b$ and $a c$, and the fifth and sixth are $-b d$ and $c d$. Suppose all five differences between consecutive products are the same. Then, in particular, the equation $a c-a b=c d-b ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 6,959 |
6. In the Thirtieth Kingdom, there are 100 cities, and no more than one road connects any two cities. One day, the tsar ordered that one-way traffic be introduced on each road, and at the same time, each road should be painted either white or black. The Minister of Transport proudly reported that after the order was ca... | Answer: 150. Solution: Example. Arrange the cities on a circle so that they divide it into equal arcs, and declare these arcs to be roads directed clockwise. Paint these 100 arcs in white and black colors so that the colors alternate on the circle. Direct another 50 white roads along the chords from the cities where bl... | 150 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,960 |
7. Given a convex quadrilateral $A B C D$, in which $A B=B C=C D=4$. Points $K$ and $L$ are chosen on sides $A B$ and $C D$ respectively such that $A K=D L=1$. On side $A D$, an external triangle $A M D$ is constructed, in which $A M=M D=2$. It turns out that $K L=2$. Prove that $B M=C M$. (C. Frantsuzov) | The first solution. Note that triangle $M D L$ is similar to triangle $C D M$ with a similarity coefficient of $2$ (since $C D=2 M D$, $D M=2 D L$, and the angle at vertex $D$ is common), therefore $M C=2 M L$. Similarly, $M B=2 M K$. Therefore, triangle $M L K$ is similar to triangle $M C B$. Consequently, $\angle L M... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,961 |
8. Given a natural number $k$, greater than 1. A natural number $n$, greater than 1 and coprime with $k$, is called correct if for any natural divisor $d (d < n)$ of the number $n$, the number $d + k$ is not coprime with $n$. Prove that the number of correct numbers is finite. (S. Berlov) | Solution. If the correct number $n$ is prime, then $1+k$ must be divisible by $n$, and there are only a finite number of such correct numbers. If the correct number $n$ is a power of a prime number, $p^{s}$, where $s \geq 2$, then $p+k$ and $1+k$ are not coprime with $n$, from which it follows that $k$ is divisible by ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,962 |
1. Each of the 10 gnomes is either a knight, who always tells the truth, or a liar, who always lies, and at least one of the gnomes is a knight. All the gnomes lined up, after which nine of them said: "Among those standing to my left, there is a knight," while the remaining one, Gloin, said: "Among those standing to my... | Answer: Told the truth. Solution: According to the condition, there are knights among the gnomes. Let's take the leftmost one. He could not have said, "Among those standing to the left of me, there is a knight." Therefore, this is Gloin, and consequently, he is telling the truth. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,963 |
2. On the way to the museum, a group of kindergarteners lined up in pairs, with the number of pairs of two boys being three times the number of pairs of two girls. On the way back, the same group lined up such that the number of pairs of two boys was four times the number of pairs of two girls. Prove that the same grou... | Solution. Let the number of pairs of girls on the way to the museum be $a$, and on the way back be $-b$. This means the number of pairs of boys on the way there and back were $3a$ and $4b$ respectively. Since each of the other pairs consisted of a boy and a girl, the difference between the number of boys and girls is $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,964 |
3. On the segment $A B$, a point $M$ is marked. Points $P$ and $Q$ are the midpoints of segments $A M$ and $B M$ respectively, and point $O$ is the midpoint of segment $P Q$. Choose point $C$ such that angle $A C B$ is a right angle. Let $MD$ and $ME$ be the perpendiculars dropped from point $M$ to lines $CA$ and $CB$,... | Solution. Note that CDME is a rectangle. Its diagonals are bisected by the point of intersection, so point $F$ is the midpoint of segment $C M$. Next, segments $P F$ and $F Q$ are the midlines of triangles $A C M$ and $B C M$ respectively. Therefore, they are parallel to the mutually perpendicular segments $A C$ and $C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,965 |
4. Do there exist six distinct natural numbers a, b, c, d, e, f such that the equality $(a+b+c+d+e+f):(1 / a+1 / b+1 / c+1 / d+1 / e+1 / f)=2012$ holds? (B. Trushin) | Answer. Yes, they do exist. Solution. For example, let $a=1, b=2012, c=2, d=1006, e=4$, $f=503$; then $a b=c d=e f=2012$. Therefore,
$$
1 / a+1 / b+1 / c+1 / d+1 / e+1 / f=(a+b) / a b+(c+d) / c d+(e+f) / e f=(a+b+c+d+e+f) / 2012
$$
from which the required equality follows. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,966 |
5. Sindbad has 11 visually identical dinars in his purse, among which one might be counterfeit, differing from the genuine ones in weight, but it is unknown whether it is heavier or lighter. How can he pay the merchant with eight genuine dinars, if the merchant allows him to use his balance scales twice, but without we... | Solution. Let's divide the coins into three piles I, II, III, each containing three coins. We will compare piles I and II, and then piles II and III. If all three piles weigh the same, then all the coins in them are genuine, and we have found even 9 genuine coins. Otherwise, one of the piles differs in weight from the ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,967 |
6. Given an acute-angled triangle $A B C$. The height $A A_{1}$ is extended beyond vertex $A$ to segment $A A_{2}=B C$. The height $C C_{1}$ is extended beyond vertex $C$ to segment $C C_{2}=A B$. Find the angles of triangle $A_{2} B C_{2}$. ( ( . Zhenedarov) | Answer. $\angle A_{2} B C_{2}=\angle 90^{\circ}, \angle B A_{2} C_{2}=\angle B C_{2} A_{2}=45^{\circ}$.
Solution. Triangles $A B A_{2}$ and $C C_{2} B$ are equal by two sides and the angle between them $\left(A B=C C_{2}, A A_{2}=B C\right.$, and angles $B A A_{2}$ and
 | Answer: It could not. Solution: Suppose the opposite: the products $ab, bc$, and $ca$ end with the two-digit numbers $n, n+1$, and $n+2$, respectively. Among these three consecutive numbers, there must be an odd number, meaning the product of some two of the numbers $a, b$, and $c$ is odd. This implies that at least tw... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,969 |
1. Vasya, Petya, and Kolya study in the same class. Vasya lies in response to any question, Petya alternates between lying and telling the truth, and Kolya lies in response to every third question, and tells the truth in all other cases. One day, each of them was asked six times in a row how many people are in their cl... | Answer: 27. Solution: Vasya never told the truth, Petya told the truth three times, and Kolya - four times. Thus, the truth was told exactly seven times, which is how we get the answer. | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,971 |
2. In trapezoid $ABCD$, the base $AD$ is greater than the lateral side $CD$. The bisector of angle $D$ intersects side $AB$ at point K. Prove that $AK > KB$. (S. Berlov, I. Rubanov) | Solution. Let $M$ and $N$ be the midpoints of sides $A B$ and $C D$ respectively. Mark a point $E$ on the base $A D$ such that $D E=C D$. In the isosceles triangle $C D E$, the bisector of angle $D$ is also a median. Therefore, the point $F$ of its intersection with segment $C E$ lies on the midline $M N$ of trapezoid ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,972 |
3. Petya thought of 8 different numbers, and then he started choosing two of them and dividing the larger by the smaller. He found 22 out of 28 possible quotients, and they turned out to be natural powers of two. Prove that the 6 remaining quotients are also natural powers of two. (A natural power of two is 2 raised to... | Solution. Let's write the numbers $a_{1}, \ldots, a_{8}$ that Petya thought of on a plane and connect each pair of numbers with a line: red if Petya found the quotient of the numbers it connects, and blue if he did not. In total, there will be 22 red and 6 blue lines.
Suppose two numbers, say, $a_{1}$ and $a_{2}\left(... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,973 |
4. On a circle, 48 points are marked, dividing it into equal arcs. Two players take turns. In one move, it is allowed to erase either three marked points lying at the vertices of an equilateral triangle, or four marked points lying at the vertices of a square. No point can be erased twice. The player who cannot make a ... | Answer. The one who goes second. Solution. We will paint the points in four colors, moving clockwise: kszzks...zz. The first four painted points will be assigned number 1, the second four - number 2, and so on, up to number 12 inclusive. Note that each move erases three points of the same color. We will pair the colors... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,974 |
5. There are 40 weights. The weights of any two differ by no more than 45 kg. Any ten of these weights can be divided into two groups of five weights, the sums of which differ by no more than 11 kg. Prove that there will be two weights that differ by no more than 1 kg. (S. Berlov, D. Shiryayev) | Solution. Suppose there exist weights $a_{1}0$. Moreover, by the condition $a_{40}-a_{1}=39+b_{1}+\ldots+b_{39} \leq 45$, from which $B=b_{1}+\ldots+b_{39} \leq 6$.
Let's take weights $a_{1}-a_{5}$ (call them light) and weights $a_{36}-a_{40}$ (call them heavy). By the condition, they can be divided into two groups of... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,975 |
5. The graphs of linear functions $y=a x+c, y=a x+d, y=b x+e, y=b x+f$ intersect at the vertices of square P. Can the points $K(a, c), L(a, d), M(b, e), N(b, f)$ be located at the vertices of a square equal to square $P$? (I. Rubanov) | Answer. They cannot. Solution. Suppose such a scenario is possible. Note that the lines $K L$ and $M N$ are parallel to the y-axis. Therefore, they are parallel to each other, and thus segments $K L$ and $M N$ are opposite sides of the square $Q$ with vertices $K(a, c), L(a, d), M(b, e), N(b, f)$. Consequently, the sid... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,976 |
6. Points $M$ and $N$ are the midpoints of sides $AB$ and $BC$ respectively of triangle $ABC$. On the extension of segment $CM$ beyond point $M$, point $D$ is marked. It turns out that $BC = BD = 2$ and $AN = 3$. Prove that $\angle ADC = 90^{\circ}$. (A. Kuznetsov) | Solution. Let $K$ be the point of intersection of the medians $AN$ and $CM$. By the property of medians, $KC = 2KM$ and $AK = 2KN$. Since $AN = 3$, then $KN = 1$. Therefore, in triangle $BKC$, the median to side $BC$ is $1 = BC / 2$, so $\angle BKC = 90^{\circ}$. This means that $BK$ is the altitude of triangle $BCD$, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,977 |
7. The numbers $1,2, \ldots, 1000$ are written on a board. It is allowed to erase any two numbers $a$ and $b$ and write the numbers $ab$ and $a^{2}+b^{2}$ in their place. Is it possible to achieve, using such operations, that at least 700 of the numbers written on the board are the same? (M. Antipov) | Answer: No. Solution: Let's track the number of numbers on the board that are divisible by three. Notice that if both numbers $a, b$ were divisible by 3, then both new numbers are also divisible by 3. If exactly one of the numbers $a, b$ was divisible by three, then $ab$ is divisible by three, but $a^2 + b^2$ is not. F... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,978 |
8. Given a natural number $k$. In the city, there are several children who attend several clubs. It is known that no more than $3 k$ children attend each club, each child attends exactly three clubs, and for any two children, there is a club that both of them attend. What is the maximum number of children that can be i... | Answer. $7 k$. Solution. Example. Let's divide $7 k$ children into 7 groups (numbered from 1 to 7) with $k$ children in each, and form 7 clubs from children of groups $(1,2,3) ;(1,4,5) ;(1,6,7) ;(2,4,6) ;(2,5,7) ;(3,4,7) ;(3,5,6)$. It is not hard to verify that all conditions are met.
Estimate. If all clubs consist of... | 7k | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,979 |
6. The sum of the remainders of three consecutive natural numbers when divided by 2022 is a prime number. Prove that one of the numbers is divisible by 2022. (N. Agakhanoв) | Solution. Let our numbers be $k, k+1$ and $k+2$, and none of them is divisible by 2022. Then if the remainder of dividing the number $k$ by 2022 is $r>0$, the remainders of dividing $k+1$ and $k+2$ by 2022 are, respectively, $r+1$ and $r+2$, and the sum of the three remainders is the composite number $3r+3=3(r+1)$. Con... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,980 |
7. Does there exist a triangle in which the lengths of the non-coinciding median and altitude drawn from one of its vertices are respectively equal to the lengths of two sides of this triangle? (N. Agakhanov) | Answer. There exists. Consider triangle $A B C$, where $A B=B C$ and the height $B H$, drawn from vertex $B$, equals $2 A C$ (obviously, such a triangle exists). On the extension of side
$A C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,981 |
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