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3. Let's call a four-digit number accompanying the year $\overline{20 a b}$ if it also ends in $\overline{a b}$ and, in addition, is divisible by $\overline{a b}$ (a two-digit number), for example, the number 4623 accompanies the year 2023. How many numbers accompany the year $2022?$ | Solution. $\overline{m n 22}=\overline{m n} \cdot 100+22 \Rightarrow \overline{m n} \cdot 100: 22 \Rightarrow \overline{m n}: 11: 11,22,33,44,55,66,77$, 88,99 - 9 numbers.
Answer 9.
## Grading Criteria.
Correct answer with valid reasoning - 7 points.
Valid reasoning for divisibility by 11 and description of the set... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. How many solutions does the rebus $\frac{B+O+C+b+M+O+\breve{U}}{K+J+A+C+C}=\frac{22}{29}$ have, where different letters represent different digits, and the same letters represent the same digits? It is known that the digit 0 is not used. | Solution. In the rebus, 9 different letters are used, i.e., all non-zero digits are used. $1+2+3+0+0 \leq K+Л+A+C+C \leq 6+7+8+9+9$, i.e.
$$
6 \leq K+Л+A+C+C \leq 39 . \quad \text { Since the fraction } \frac{22}{29} \text { is irreducible, the number }
$$
$K+Л+A+C+C$ must be divisible by 29, the only case that satis... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.1. Different positive numbers $x, y, z$ satisfy the equations
$$
x y z=1 \quad \text{and} \quad x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}
$$
Find the median of them. Justify your answer. | # Solution:
Method 1. Consider the polynomial $P(t)=(t-x)(t-y)(t-z)$. The numbers $x, y, z$ are its roots. Expanding the brackets or using Vieta's theorem for cubic polynomials, we get $P(t)=t^{3}+a t^{2}+b t+c$, where $a=-(x+y+z)$, $b=xy+xz+yz$, and $c=-xyz$. The condition of the problem shows that $a=-b$ and $c=-1$.... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A company is called public if it has at least 15 shareholders. A shareholder of a company is called a minority shareholder if they own no more than $25 \%$ of the shares of that company. On the stock exchange where the shares are traded, one sixth of the companies are public. Prove that among all shareholders partic... | Solution. Let the number of firms on the exchange be N. We will call a shareholder who owns more than $25 \%$ of a firm's shares a real shareholder. The number of real shareholders in one firm cannot exceed three. (If there are four, then one of them cannot own more than $25 \%$ of the shares). Therefore, the total num... | 20 | Combinatorics | proof | Yes | Yes | olympiads | false |
1. Kolya, Seryozha, and Vanya regularly go to the cinema: Kolya goes there every 4th day, Seryozha - every 5th day, and Vanya - every 6th day. Today all the boys were at the cinema. When will all three meet at the cinema again? | 1. Answer: 60.
Let's start numbering the days beginning with tomorrow (today's day is "zero"). Kolya will go to the cinema on the days whose numbers are divisible by 4, Seryozha - on the days whose numbers are divisible by 5, Vanya - on the days whose numbers are divisible by 6. For them to all be at the cinema togeth... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Four friends came back from fishing. Every two of them counted the sum of their catches. Six numbers were obtained: $7,9,14,14,19,21$. How many fish were caught in total? | 4. 28 fish.
Note that the catch of each person is counted in exactly three sums. Therefore, if we add up all six numbers, we get the tripled total catch. Thus, the total number of fish caught is $(7+9+14+14+19+21): 3=28$ fish. | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.1. From a $7 \times 7$ square grid, an equal number of $2 \times 2$ squares and $1 \times 4$ rectangles were cut out along the grid lines. What is the maximum number of these figures that could have been cut out? | Answer: 12.
Solution: Both the square and the rectangle consist of 4 cells. Therefore, the number of cut-out figures is no more than 49/4, that is, no more than 12. There are an equal number of figures of both types, so there are no more than 6 squares $2 \times 2$ and rectangles $1 \times 4$. The diagram shows how to... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.2. Find the value of the expression $a^{3}+12 a b+b^{3}$, given that $a+b=4$. | Answer: 64.
Solution. $a^{3}+12 a b+b^{3}=a^{3}+b^{3}+12 a b=(a+b)\left(a^{2}-a b+b^{2}\right)+12 a b=4 a^{2}-4 a b+4 b^{2}+12 a b=4(a+b)^{2}=64$. | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5. A natural number is called interesting if all its digits are different, and the sum of any two adjacent digits is a square of a natural number. Find the largest interesting number. | Answer: 6310972.
Solution. Let's mark 10 points on the plane, representing the digits from 0 to 9, and connect those points whose sum is a square of a natural number.
On this diagram, we ne... | 6310972 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.3. In an intergalactic hotel, there are 100 rooms with capacities of
$101, 102, \ldots, 200$ people. In these rooms, a total of $n$ people live. A VIP guest has arrived at the hotel, and a whole room needs to be vacated for them. To do this, the hotel manager selects one room and relocates all its occupants to anot... | Answer: 8824.
Solution: Suppose that with 8824 guests, the director cannot carry out the relocation. Let's divide the rooms into pairs by capacity: $101-200, 102-199, \ldots, 150-151$. Note that for each pair of rooms, the total number of people living in the two rooms is greater than the capacity of the larger room i... | 8824 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. The beetle is 3 times heavier than the cat, the mouse is 10 times lighter than the cat, the turnip is 60 times heavier than the mouse. How many times heavier is the turnip than the beetle? Justify your answer. | # Answer. 2 times.
Solution. Cat $=10$ mice, turnip $=60$ mice. Therefore, the turnip is 6 times heavier than the cat. That is, the turnip $=6$ cats. According to the condition, Zhuchka $=3$ cats. Therefore, the turnip is 2 times heavier than Zhuchka.
## Grading Criteria.
- Correct solution - 7 points.
- Correct ans... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.6. In some cells of a $200 \times 200$ square, there is one chip - either red or blue; the other cells are empty. A chip can see another if they are in the same row or column. It is known that each chip sees exactly five chips of the other color (and possibly some chips of its own color). Find the maximum possible n... | Answer: 3800 chips.
Solution: An example containing 3800 chips can be constructed as follows. Highlight the "border" of width 5 in a $200 \times 200$ square. This border consists of four corner squares $5 \times 5$ and four rectangles $5 \times 190$. Place the chips in these four rectangles: red chips in the left and ... | 3800 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.7. Initially, a natural number $N$ is written on the board. At any moment, Misha can choose a number $a>1$ on the board, erase it, and write down all natural divisors of $a$, except for $a$ itself (the same numbers can appear on the board). After some time, it turned out that there are $N^{2}$ numbers on the board. ... | Answer. Only for $N=1$.
Solution. Lemma. For any natural $n>1$, the inequality holds
$$
\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{n^{2}}<1
$$
Proof. Consider the function $f(t) = \frac{1}{t^{2}}$ for $t > 1$. The function $f(t)$ is decreasing for $t > 1$. Let $1 = d_{1} < d_{2} < \ldots < d_{k} < d_{k+1} = N$ ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. A square $10 \times 10$ is divided into unit squares. How many triangles are formed after drawing one diagonal? | Answer: 110.
Solution. The figure shows one of the obtainable triangles. All such triangles are right-angled, with the right angle vertex being any lattice node except those lying on the diagonal. There are a total of $11 \times 11$ nodes, and 11 of them are on the diagonal, so the number of triangles is $11 \times 11... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.1. Find the largest six-digit number, all digits of which are different, and each of the digits, except for the extreme ones, is either the sum or the difference of the adjacent digits. | Answer: 972538.
Solution. Let $A$ be the desired number. Let's try to find the number $A$ with the first digit being 9. We will try different options for the second digit. If the second digit is 8, then we get: $A=98176-$ it does not form a six-digit number. If the second digit is 7, then we get: $A=972538-$ a six-dig... | 972538 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.5. In the castle, there are 16 identical square rooms forming a $4 \times 4$ square. Sixteen people, who are either liars or knights (liars always lie, knights always tell the truth), moved into these rooms, one person per room. Each of these 16 people said: "At least one of the rooms adjacent to mine is occupied by ... | Answer: 8 liars.
Solution: Note that liars cannot live in adjacent rooms (otherwise, they would be telling the truth). Let's divide the rooms into 8 pairs of adjacent rooms. Then, in each pair, there can be no more than one liar. Therefore, there can be no more than 8 liars in total. Consider a chessboard coloring of ... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.1. Initially, there were 10 piles of candies on the table, containing $1, 2, \ldots, 10$ candies respectively. The Child decided to redistribute the candies. On each odd minute, he chooses one pile and divides it into two piles, each containing at least one candy. On each even minute, he chooses two piles and merges ... | Answer: Yes.
Solution. We will provide an example of how Little One can achieve such a distribution. On the first minute, he divides the pile of 10 candies into two piles of 5 candies each. Then, on the 2nd, 4th, 6th, and 8th minutes, he combines the piles of 1+9, 2+8, 3+7, 4+6 respectively, and on the 3rd, 5th, 7th, ... | 11 | Combinatorics | proof | Yes | Yes | olympiads | false |
9.2. On the board, there are $n$ different integers, any two of which differ by at least 10. The sum of the squares of the three largest of them is less than three million. The sum of the squares of the three smallest of them is also less than three million. For what largest $n$ is this possible? | Answer. For $n=202$.
Solution. Notice immediately that $990^{2}+1000^{2}+1010^{2}=(1000-10)^{2}+1000^{2}+(1000+10)^{2}=3 \cdot 1000^{2}+2 \cdot 10^{2}$, which is greater than three million. On the other hand, $989^{2}+999^{2}+1009^{2}=(1000-11)^{2}+(1000-1)^{2}+(1000+9)^{2}=3 \cdot 1000^{2}-6 \cdot 1000+(9^{2}+1^{2}+1... | 202 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Scheherazade tells the Shah mathematical tales. A mathematical tale is an equation of the form $a x+b y=c$, where $a, b, c$ are natural numbers ranging from 1 to 17 (inclusive). A tale is sad if the corresponding equation has no solutions in integers (for example, the tale $2 x+4 y=7$ is sad, while the tale $3 x+5 y... | Answer: will be able to.
Solution. The following fairy tales will be sad:
a) a and b are even, c is odd. There are 8$\cdot$8$\cdot$9=576 such fairy tales.
b) a and b are divisible by 3, c is not divisible by 3. There are 5$\cdot$5$\cdot$12 such fairy tales. However, we need to account for the overlap with a): a and ... | 1009 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In triangle $ABC$ (angle B is obtuse), the altitude $BH$ and the bisector $AK$ are drawn. Find the angle $AKB$, if the angle $KHC$ is $45^{\circ}$. | Answer: $45^{\circ}$.
Solution. Point K is equidistant from lines AB and AC (since AK is the bisector), and from lines HB and HC (since HK is the bisector of the right angle). Therefore, point K is equidistant from lines AB and BH. This means that BK is the bisector of the angle external to angle ABH. Therefore, angle... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.3 A weirdo chose 677 different natural numbers from the list $1,2,3, \ldots, 2022$. He claims that the sum of no two of the chosen numbers is divisible by 6. Did he go too far with his claim? | Solution: Suppose the eccentric is right. From the list specified in the condition, there are exactly 377 remainders of division by 6 of each type from 0 to 5. Numbers with remainders 0 and 3 can be taken no more than one each. It is impossible to take numbers with remainders 1 and 5 simultaneously, so in the eccentric... | 676 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Dwarves Glóin, Óin, and Thráin found 70 identical precious stones and want to divide them among themselves so that each of them gets no less than 10 stones. In how many ways can the dwarves do this? | Answer: 861.
Solution: Give each dwarf 9 stones, and lay out the remaining 43 stones in a row. To distribute the remaining stones among the dwarves, it is sufficient to place two dividers in the 42 spaces between the stones. Gloin will receive the stones to the left of the first divider, Oin will receive the stones be... | 861 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Given a $101 \times 101$ grid, all cells of which are painted white. It is allowed to choose several rows and repaint all cells in these rows to black. Then, choose exactly the same number of columns and repaint all cells in these columns to the opposite color (i.e., white to black, and black to white). What is the ... | Answer: 5100.
Solution. Let $k$ rows be repainted first, then $k$ columns. After the first stage of repainting, each column will contain $k$ black and $101-k$ white cells. Since $101-k$ columns will remain untouched, the total number of black cells in these columns will be $k(101-k)$. In each of the repainted columns,... | 5100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) In an $8 \times 8$ frame with a width of 2 cells (see figure), there are a total of 48 cells. How many cells are in a $254 \times 254$ frame with a width of 2 cells?
 | Answer: 2016.
Solution. First method. Cut the frame into four identical rectangles as shown in the figure. The width of the rectangles is equal to the width of the frame, i.e., 2 cells. The length of each rectangle is 2 less than the side of the frame: $254-2=252$ cells. Then the area of one rectangle is $2 \cdot 252=... | 2016 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) Anya multiplied 20 twos, and Vanya multiplied 17 fives. Now they are going to multiply their huge numbers. What will be the sum of the digits of the product? | Answer: 8.
Solution. In total, 20 twos and 17 fives are multiplied. Let's rearrange the factors, alternating twos and fives. This results in 17 pairs of $2 \cdot 5$ and three additional twos, which multiply to 8. Thus, the number 8 needs to be multiplied by 10, 17 times. This results in a number consisting of the digi... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) The graph of a reduced quadratic trinomial is shown in the figure (the y-axis is erased, the distance between adjacent marked points is 1). What is the discriminant of this trinomial?
. From the condition, it follows that $x_{2}-x_{1}=2$. Since $x_{2}=\frac{-b+\sqrt{D}}{2}, x_{1}=\frac{-b-\sqrt{D}}{2}$, we get that $x_{2}-x_{1}=\sqrt{D}$, hence $D=4$.
## Answer. 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) Replace each letter with a digit so that the operations performed horizontally and vertically are valid.
| $a b$ | + | $c d$ | $=$ | $e f$ |
| :---: | :---: | :---: | :---: | :---: |
| $\times$ | | $:$ | | + |
| $k$ | $:$ | $m$ | $=$ | $m$ |
| $m n d$ | $:$ | $e$ | $=$ | $e m$ |
Identical letters corr... | # Solution.
| 42 | + | 18 | $=$ | 60 |
| :---: | :---: | :---: | :---: | :---: |
| $\times$ | | $:$ | | + |
| 9 | $:$ | 3 | $=$ | 3 |
| 378 | $:$ | 6 | $=$ | 63 | | 63 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) The graphs of the functions $y=k x+b$ and $y=b x+k$ intersect. Find the x-coordinate of the point of intersection.
# | # Solution.
Method 1. The desired abscissa is the solution to the equation $k x + b = b x + k$. This equation can be reduced to $(k - b) x = k - b$. Since the given graphs intersect (do not coincide), $k \neq b$, so $x = 1$.
Method 2. Notice that $x = 1$ is a solution to the problem, because when $x = 1$, both given ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A three-digit number, all digits of which are different, will be called balanced if it is equal to the sum of all possible two-digit numbers formed from the different digits of this number. Provide an example of any balanced number. Justify your answer. | Solution: For example, the number $132=13+12+32+21+31+23$ is balanced (there are other options).
Grading criteria: Any suitable number with verification - 7 points, incorrect solution or only answer - 0 points. | 132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In the tournament, each participant was supposed to play exactly one game with each of the remaining participants, but two participants dropped out during the tournament, having played only 4 games each. In the end, the total number of games played turned out to be 62. How many participants were there in total? | Answer: 13.
Solution: Let the total number of participants be $n$. Then, excluding the two who dropped out, the remaining participants played $\frac{(n-2)(n-3)}{2}$ matches. If these two managed to play against each other, then 7 matches were played with their participation, and if they did not, then 8 matches. Thus, ... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In triangle $\mathrm{ABC}$ with sides $\mathrm{AB}=5, \mathrm{BC}=\sqrt{17}$, and $\mathrm{AC}=4$, a point $\mathrm{M}$ is taken on side $\mathrm{AC}$ such that $\mathrm{CM}=1$. Find the distance between the centers of the circumcircles of triangles $\mathrm{ABM}$ and $\mathrm{BCM}$. | Answer: 2.
Solution: Draw the height BH to side AC. Let $\mathrm{CH}=x$, then $\mathrm{BH}=$ 4 - $x$. By the Pythagorean theorem from two triangles, we have $B H^{2}=B C^{2}-C H^{2}=17-x^{2}$ and $B H^{2}=A B^{2}-A H^{2}=25-(4-x)^{2}$. Equating the right sides of both equations, we get $x=1$, which means points M and ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. In a school quiz, 100 students participated. After the results were tallied, it turned out that any 66 of them together earned no less than $50 \%$ of the total prize points. What is the highest percentage of points that one participant could have earned? | Answer: $25 \%$.
Solution: Suppose participant $\mathrm{X}$ scored the highest percentage of points $-x \%$. Divide the remaining participants into three groups A, B, and C, each with 33 people. Let the total percentages of points scored by these groups be $a, b$, and $c$ respectively. Then,
$$
2(100-x)=2(a+b+c)=(a+b... | 25 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
# 3. CONDITION
There is an unlimited number of chips in six colors. What is the smallest number of chips that need to be arranged in a row so that for any two different colors, there are two adjacent chips of these colors in the row? | Solution. From the condition, it follows that for each fixed color A, a chip of this color must be paired with a chip of each of the other 5 colors. In a row, a chip has no more than two neighbors, so a chip of color A must appear at least 3 times. Similarly for each other color. Thus, there should be no less than $3 \... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. Gleb placed the numbers $1,2,7,8,9,13,14$ at the vertices and the center of a regular hexagon such that in any of the 6 equilateral triangles, the sum of the numbers at the vertices is divisible by 3. What number could Gleb have written in the center? It is sufficient to provide one suitable example.
, and the centr... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. Misha suggested that Yulia move a chip from cell $A$ to cell $B$. In one step, you can move the chip to an adjacent cell by side or by corner. To make it more interesting, Misha put 30 candies in the prize fund, but said that he would take 2 candies for each horizontal or vertical move and 3 candies for ea... | Answer: 14.
Solution. From $A$ to $B$, one can get through the top or the bottom. If going through the top, the first 2 moves are diagonal (a diagonal move is more advantageous than 2 horizontal moves), and the next 5 moves are horizontal. Misha will take $2 \cdot 3 + 5 \cdot 2 = 16$ candies, and Yulia will win 14. If... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4. The diagonals of quadrilateral $A B C D$ intersect at point $K$. It turns out that $A B=B K=K D$. On segment $K C$, a point $L$ is marked such that $A K=L C$. Find $\angle B L A$, given that $\angle A B D=52^{\circ}$ and $\angle C D B=74^{\circ}$.
. We have
$$
\begin{aligned}
\angle BLA & =180^{\circ}-\angle BAL-\angle ABL=180^{\circ}-\frac{180^{\circ}-\angle ABD}{2}-\angle CDB= \\
& =90^{\circ}+\frac{1}{2} \angle ABD-\angle CDB=42^{\circ... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. A field was partially planted with corn, oats, and millet. If the remaining part is completely planted with millet, then millet will occupy half of the entire field. If the remaining part is equally divided between oats and corn, then oats will occupy half of the entire field. By what factor will the amoun... | Answer: 3.
Solution. Let the area of the entire field be 1, and the empty part be $x$. Then, from the first condition, millet occupies $\frac{1}{2}-x$, and from the second condition, oats occupy $\frac{1}{2}-\frac{1}{2} x$.
Corn is left with $1-x-\left(\frac{1}{2}-x\right)-\left(\frac{1}{2}-\frac{1}{2} x\right)=\frac... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.6. Six princesses have a magic chest. Every minute, a dress of one of 10 colors and one of 9 styles can be taken out of it. However, within one hour, it is impossible to take out two dresses from the chest that match both in color and style. What is the minimum number of dresses the princesses will have to ta... | Answer: 46.
Solution. Note that 45 dresses would not be enough, as the chest can issue exactly 5 dresses of each of the 9 styles.
We will prove that if the princesses take out 46 dresses, there will definitely be 6 dresses of the same style, and thus of different colors (identical dresses could not have occurred over... | 46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. Along an alley, maples and larches were planted in one row, a total of 75 trees. It is known that there are no two maples between which there are exactly 5 trees. What is the maximum number of maples that could have been planted along the alley? | Answer: 39.
Solution. Let's divide all the trees into groups of 12 standing in a row. There will be 6 complete groups and 3 more trees at the end. In each group, we will divide the trees into 6 pairs: the first with the seventh, the second with the eighth, ..., the sixth with the twelfth. Note that there are exactly 5... | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.8. In trapezoid $A B C D(A D \| B C) \angle A B C=108^{\circ}$ and $\angle A D C=54^{\circ}$. On ray $B A$ beyond point $A$, point $K$ is marked such that $A K=B C$. Find the angle $D K C$, given that $\angle B K C=27^{\circ}$.
Fig. 1: to the solu... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Semyon has 20 numbers: $1,2,3, \ldots, 19,20$. He formed 10 fractions by writing ten of these numbers in some order as numerators, and the remaining ten in some order as denominators. What is the maximum number of integers Semyon could obtain after simplifying all the written fractions? | Answer: 8 numbers.
## Solution:
For the fractions to have an integer value, the prime numbers $11, 13, 17, 19$ can only be numerators with a denominator of 1. Therefore, to form fractions equal to an integer, no more than 17 numbers can be used, meaning no more than 8 fractions can be formed. Example: 20/10, 19/1, $1... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Before the lesson, Nestor Petrovich wrote several words on the board. When the bell rang for the lesson, he noticed a mistake in the first word. If he corrects the mistake in it, the words with mistakes will make up $24 \%$, and if he erases the first word from the board altogether, the words with mistakes will make... | Answer: $28 \%$.
## Solution:
Let there be $n$ words written on the board before the lesson, of which $x$ have errors. If the error in the first word is corrected, then there will be $x-1$ words with errors out of $n$, and by the condition $x-1=0.24 n$. If a word with an error is erased, then there will be $x-1$ word... | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3. The integer 23713 has the following two properties:
(1) any two adjacent digits form a prime two-digit number,
(2) all these prime two-digit numbers are pairwise distinct.
Find the largest of all integers with properties (1) and (2). | Answer: 617371311979.
Solution. Since even digits and the digit 5 can only be in the highest place (and only one of these digits can form a single prime number), and the remaining prime two-digit numbers are $11, 13, 17, 19, 31, 37, 71, 73, 79$, 97, the maximum number of digits in the desired number is 12. In the desi... | 617371311979 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.4 Find the smallest positive integer $k$, for which in any coloring of the numbers of the set $M=\{1,2,3, \ldots, k\}$ in two colors, there will be ten not necessarily distinct numbers of the same color from the set $M$ | Answer: 109.
Solution. Let $k \geq 100$. Suppose that for any 10 numbers of one color, their sum is either a number of the other color or does not belong to M. Let the number 1 be of the first color, and the number 2 be of the second color. Then the numbers 10 and 20 are of the second and first colors, respectively. N... | 109 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. What is the total number of emeralds in the boxes?
Fig. 1: to the solution of problem 8.4
Notice that $\angle ABK = \angle CBL$, since they both complement $\angle ABL$ to ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots marked is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of gl... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
It remains to show that after removing any $n=5099$ stones, there will still be many stones le... | 5099 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. A circle is inscribed in trapezoid $A B C D$, touching the lateral side $A D$ at point $K$. Find the area of the trapezoid if $A K=16, D K=4$ and $C D=6$. | Answer: 432.
Solution. Let $L, M, N$ be the points of tangency of the inscribed circle with the sides $BC, AB, CD$ respectively; let $I$ be the center of the inscribed circle. Denote the radius of the circle by $r$. Immediately note that $DN = DK = 4$ (the first equality follows from the equality of the segments of ta... | 432 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Let's call a natural number "remarkable" if all its digits are different, it does not start with the digit 2, and by erasing some of its digits, the number 2018 can be obtained. How many different seven-digit "remarkable" numbers exist? | Answer: 1800.
Solution: To correctly count the number of options, it is necessary to follow the rule: before the digit 2, there must be one of the six digits $-3,4,5,6,7$ or 9. Let's assume for definiteness that this is 3, then two different digits from the remaining five ( $\frac{5 \cdot 4}{2}=10$ options) can be in ... | 1800 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Lena downloaded a new game for her smartphone, where it is allowed to conduct alchemical reactions of two types. If she combines one "fire" element and one "stone" element, she gets one "metal" element. And if she combines one "metal" element and one "stone" element, she gets three "stone" elements. Lena has 50 "fir... | Answer: 14 elements.
Solution. Consider the expression $S=2 x+y+z$, where $x$ is the number of "metal" elements, $y$ is the number of "fire" elements, and $z$ is the number of "stone" elements. It is easy to see that this expression does not change with each of the two alchemical operations:
$$
\begin{aligned}
& 2(x+... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.1. A three-digit number is 56 times greater than its last digit. How many times greater is it than its first digit? Justify your answer.
# | # Solution:
Method 1. The last digit is such that when multiplied by 6, the resulting number ends with the same digit. By exhaustive search, we confirm that this can be any even digit (and only it). Therefore, this three-digit number is either 112, 224, 336, or 448 (the option with the last digit 0 is not valid, as it... | 112 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.3. On the way from city $A$ to city $B$, there are kilometer markers every kilometer. On each marker, one side shows the distance to $A$, and the other side shows the distance to $B$. In the morning, a tourist passed by a marker where one number was twice the other. After walking another 10 km, the tourist saw a mark... | Solution: Let $C_{1}$ and $C_{2}$ be the poles mentioned in the problem ($C_{1}$ - the pole near which the tourist was in the morning). Without loss of generality, assume the tourist was walking from $A$ to $B$. Then there are two possible situations: 1) $C_{1} A = 2 C_{1} B$ or 2) $C_{1} B = 2 C_{1} A$. Since $C_{2} A... | 120 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.5. 101 people bought 212 balloons of four colors, and each of them bought at least one balloon, but no one had two balloons of the same color. The number of people who bought 4 balloons is 13 more than the number of people who bought 2 balloons. How many people bought only one balloon? Provide all possible answers an... | # Solution:
Method 1. First, exclude the 13 people who bought 4 balloons. There will be 88 people left, who bought a total of 160 balloons, with each person buying between 1 and 4 balloons, and the number of people who bought 2 and 4 balloons is now equal. Suppose now that each person who bought 4 balloons gives one b... | 52 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2.1. The numbers 7, 8, 9, 10, 11 are arranged in a row in some order. It turned out that the sum of the first three of them is 26, and the sum of the last three is 30. Determine the number standing in the middle.
. In total, $N$ is not less than some 54 numbers in the table. Therefore, $N$ is at least 54.
To... | 54 | Other | math-word-problem | Yes | Yes | olympiads | false |
# Task 8.2
For a natural number $N$, all its divisors were listed, and then the sum of digits for each of these divisors was calculated. It turned out that among these sums, all numbers from 1 to 9 were found. Find the smallest value of $\mathrm{N}$.
## Number of points 7 | Answer:
288
## Solution
Note that the number 288 has divisors $1,2,3,4,32,6,16,8,9$. Therefore, this number satisfies the condition of the problem. We will prove that there is no smaller number that satisfies the condition.
Indeed, since $\mathrm{N}$ must have a divisor with the sum of digits 9, $\mathrm{N}$ is div... | 288 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task 8.5
On the Island of Liars and Knights, a circular arrangement is called correct if each person standing in the circle can say that among their two neighbors, there is a representative of their tribe. Once, 2019 natives formed a correct arrangement in a circle. A liar approached them and said: "Now we can also ... | Answer:
1346
## Solution
We will prove that a correct arrangement around a circle is possible if and only if the number of knights is at least twice the number of liars.
Indeed, from the problem's condition, it follows that in such an arrangement, each liar has two knights as neighbors, and among the neighbors of a... | 1346 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Find the largest even three-digit number $x$ that gives a remainder of 2 when divided by 5 and satisfies the condition $\operatorname{GCD}(30, \operatorname{GCD}(x, 15))=3$. | Solution. From the condition, we get that there exist such $a, b \in \mathbb{N}$ that $3a=30$, GCD $(x, 15)=3b$, and GCD $(a, b)=1$.
Consider the equality GCD $(x, 15)=3b$. This means that $\exists c, d \in \mathbb{N}$ such that $x=3bc$ and $15=3bd$, and GCD $(c, d)=1$.
From the equality $15=3bd$ it follows that $bd=... | 972 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. A hundred people are standing in a circle, each of whom is either a knight or a liar (liars always lie, and knights always tell the truth). Each of those standing said: “I have a liar as a neighbor.” Find the minimum possible number of liars among these 100 people. | Answer: 34.
Solution: Note that 3 knights cannot stand next to each other, as in this case, the middle knight would be lying. Therefore, among any 3 standing next to each other, there is a liar. Take any liar, and divide the remaining 99 people into 33 groups of three standing next to each other. Since there is at lea... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In a football tournament where each team played against each other once, teams A, B, C, D, E, and F participated. Teams received 3 points for a win, 1 point for a draw, and 0 points for a loss. In the end, it turned out that teams A, B, C, D, and E each had 7 points. What is the maximum number of points that team $\... | 4. In a match where one of the teams won, the teams together score 3 points; in a match that ended in a draw, - 2 points. Since 7 is not divisible by 3, a team that scored 7 points must have at least one draw. Since there are five such teams, there were at least three draws in the tournament. In total, as is easy to ch... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. On the Island of Misfortune, only knights, who always tell the truth, and liars, who always lie, live. The island is governed by a group of 101 people. At the last meeting, it was decided to reduce this group by 1 person. But each member of the group stated that if they were removed from the group, the majority of t... | 2. 50 knights and 51 liars. Let there be $k$ knights in the group, then there are $101-k$ liars. Each one said that if they were removed, there would be no less than 51 liars among the remaining 100 people. Since the knights told the truth, then $k-1 \leq 49$ and
. What is the value of the expression $4 a+d$, if $b+c=1$ ? | Answer: 1.
Solution: Since the graphs pass through the point with coordinates (2; 4), then $4=4a+2b+1$ and $4=4+2c+d$. Therefore, $4a+2b=3$, and $2c+d=0$, or $4a=3-2b, d=-2c$. Summing up the obtained expressions: $4a+d=3-2b-2c=3-2(b+c)=3-2=1$. Note: The condition is satisfied by the functions $y=ax^2+bx+1$ and $y=x^2+... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.3. Given a trapezoid $A B C D$, where $A D \| B C, B C=A C=5$, and $A D=6$. The angle $A C B$ is twice the angle $A D B$. Find the area of the trapezoid. | # Answer: 22.
Solution: Drop a perpendicular CH from vertex C to the base AD. Denote the intersection of BD and CH as E.
Let $\angle \mathrm{ADB}=\alpha=\angle \mathrm{CBD}$ (alternate interior angles), then $\angle A C B=2 \alpha=\angle C A D$ (alternate interior angles). Triangle ACB is isosceles by condition, so $... | 22 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.1. In a six-digit number, the first digit, which is 2, was moved to the last place, leaving the other digits in the same order. The resulting number turned out to be three times larger than the original. Find the original number. | Answer: 285714.
Solution. According to the condition, the desired number has the form $\overline{2 a b c d e}$, then we have: $\overline{a b c d e 2}=$ $3 \cdot \overline{2 a b c d e}$, or $\overline{a b c d e} \cdot 10+2=3 \cdot(200000+\overline{a b c d e})$. Let $\overline{a b c d e}=X$ - a five-digit number, then $... | 285714 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 1. Clone 1
The figure is divided into 7 equal squares and several rectangles. The perimeter of rectangle A is 112 cm. What is the perimeter of rectangle B? Express your answer in centimeters.
![](https://cdn.mathpix.com/cropped/2024_05_06_d0dc74dfba1714359e24g-01.jpg?height=485&width=483&top_left_y=1008&top_left_x=... | # Answer: 168
## Solution
1st method. Rectangles A and B are composed of identical squares: Rectangle A consists of three, and Rectangle B consists of four. Let's call the side of such a square a "stick" and count the perimeter in sticks. The perimeter of Rectangle B is four sticks more than the perimeter of Rectangl... | 168 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 3. Clone 1
A rope was divided into 19 equal parts and arranged in a snake-like pattern. After that, a cut was made along the dotted line. The rope split into 20 pieces: the longest of them is 8 meters, and the shortest is 2 meters. What was the length of the rope before it was cut? Express your answer in meters.
-4x=32
$$ | 32 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.2. Each of the natural numbers $1,2,3, \ldots, 377$ is painted either red or blue (both colors are present). It is known that the number of red numbers is equal to the smallest red number, and the number of blue numbers is equal to the largest blue number. What is the smallest red number? | Answer: 189.
Solution. Let $N$ be the largest blue number. Then only numbers from 1 to $N$ can be painted blue. Since there are a total of $N$ blue numbers, we get that all numbers from 1 to $N$ are blue. Accordingly, all numbers from $N+1$ to 377 are red. Since the number of red numbers equals the smallest red number... | 189 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.3. Krosh and Yozhik decided to check who would run faster along a straight road from Kopyatych's house to Losyash's house. When Krosh had run 20 meters, Yozhik had run only 16 meters. And when Krosh had 30 meters left, Yozhik had 60 meters left. How many meters is the length of the road from Kopyatych's house... | Answer: 180.
Solution. When Krosh ran 20 meters, Yozhik ran only 16 meters, so their speeds are in the ratio of $5: 4$.
When Krosh had 30 meters left to run, let him have already run $5 x$ meters (where $x$ is not necessarily an integer). Then by this point, Yozhik had run $4 x$ meters. Therefore, the total length of... | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. Consider seven-digit natural numbers, in the decimal representation of which each of the digits $1,2,3,4,5,6,7$ appears exactly once.
(a) (1 point) How many of them have the digits from the first to the sixth in ascending order, and from the sixth to the seventh in descending order?
(b) (3 points) How ma... | Answer: (a) 6. (b) 15.
Solution. (a) From the condition, it follows that the sixth digit is the largest, so it is 7. The last digit can be any digit from 1 to 6, and this uniquely determines the entire number (since the first five digits must be in ascending order). Therefore, there are exactly 6 such seven-digit numb... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. In the forest, there live elves and gnomes. One day, 60 inhabitants of this forest lined up facing the same direction, at which point some of them might have been wearing hats. (There could have been from 0 to 60 elves, and inhabitants wearing hats could also have been from 0 to 60 inclusive.)
Each of the... | Answer: (a) 59. (b) 30.
Solution. (a) Note that the rightmost resident cannot be telling the truth, since there is no one to the right of him. This means he cannot be a hatless elf, so the total number of hatless elves is no more than 59.
Now let's provide an example where there are exactly 59 hatless elves. Suppose ... | 59 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.8. The cells of a $50 \times 50$ table are colored in $n$ colors such that for any cell, the union of its row and column contains cells of all $n$ colors. Find the maximum possible number of blue cells if
(a) (1 point) $n=2$
(b) (3 points) $n=25$. | Answer: (a) 2450. (b) 1300.
Solution. We will prove that the number of cells of any color $A$ present in the coloring is no less than 50. Suppose this is not the case, and the number of cells of color $A$ is no more than 49. Then there exists a row without cells of color $A$, and there also exists a column without cel... | 1300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 7.8.1. The cells of a $50 \times 50$ table are colored in $n$ colors such that for any cell, the union of its row and column contains cells of all $n$ colors. Find the maximum possible number of blue cells if
(a) (1 point) $n=2$;
(b) (3 points) $n=25$. | Answer: (a) 2450. (b) 1300. | 1300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 7.8.2. The cells of a $40 \times 40$ table are colored in $n$ colors such that for any cell, the union of its row and column contains cells of all $n$ colors. Find the maximum possible number of blue cells if
(a) (1 point) $n=2$;
(b) (3 points) $n=20$. | Answer: (a) 1560. (b) 840. | 840 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 7.8.3. The cells of a $30 \times 30$ table are colored in $n$ colors such that for any cell, the union of its row and column contains cells of all $n$ colors. Find the maximum possible number of blue cells if
(a) $(1$ point) $n=2$;
(b) $(3$ points) $n=15$. | Answer: (a) 870. (b) 480. | 480 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 7.8.4. The cells of a $20 \times 20$ table are colored in $n$ colors such that for any cell, the union of its row and column contains cells of all $n$ colors. Find the maximum possible number of blue cells if
(a) (1 point) $n=2$;
(b) (3 points) $n=10$. | Answer: (a) 380. (b) 220. | 380 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Two candles of different thickness, each 24 cm long, are lit simultaneously. The thin candle burns out in 4 hours, the thick one in 6 hours. After what time will one candle be half the length of the other? It is assumed that each candle burns uniformly, meaning it decreases by the same height over equal time interva... | Solution. Let $x$ hours $(x<4)$ have passed since the candles were lit. Since the thin candle decreases by $24: 4=6$ cm every hour, it will decrease by a length of $6 x$ in this time, leaving an unburned stub of length $24-6 x$. Similarly, the stub of the second candle will have a length of $24-4 x$. For one candle to ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.2. On a sheet of paper, three intersecting circles are drawn, forming 7 regions. We will call two regions adjacent if they share a common boundary. Regions that border at exactly one point are not considered adjacent.
Two regions already have numbers written in them. Write integers in the remaining 5 region... | Answer: -8.
Solution. Let the unknown number be $x$. Then we can place the numbers in the other regions as follows:
The sum of the numbers in the regions surrounding $x$ should be equal to $... | -8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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