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Problem 7.5. A rectangular strip of length 16 was cut into two strips of lengths 9 and 7. These two strips were placed on the table as shown in the figure.
It is known that the area of the part of the table covered only by the left strip is 27, and the area of the part of the table covered only by the right strip is 1... | Answer: 13.5.
Solution. Since the width of the two resulting strips is the same, their areas are in the ratio of their lengths, i.e., $9: 7$. Let $S$ be the area covered by both strips. Then $\frac{27+S}{18+S}=\frac{9}{7}$, from which we get $7 \cdot(27+S)=9 \cdot(18+S)$. Solving this linear equation, we get $S=13.5$. | 13.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,913 |
Problem 8.4. Given a square $A B C D$. Point $L$ is on side $C D$ and point $K$ is on the extension of side $D A$ beyond point $A$ such that $\angle K B L=90^{\circ}$. Find the length of segment $L D$, if $K D=19$ and $C L=6$.
Fig. 1: to the solution of problem 8.4
Notice that $\angle ABK = \angle CBL$, since they both complement $\angle AB... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,915 |
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 is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | 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 | 16,916 |
Problem 8.7. For quadrilateral $ABCD$, it is known that $\angle BAC = \angle CAD = 60^{\circ}$, $AB + AD = AC$. It is also known that $\angle ACD = 23^{\circ}$. How many degrees does the angle $ABC$ 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 | 16,919 |
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 | 16,922 |
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 | 16,923 |
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?
. Note that $\angle P D T = \angle B D T = 90^{\circ}$. Therefore, $\angle P H T + \angle P D T = 180^{\circ}$, which means that point $T$ lies on the circle $\omega$. Hence, $\angle P Q T = \angle P H T = 90^{\circ}$, and quadrilateral $P Q T D$ is a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,926 |
11.7. On a plane, several lines are drawn, no two of which are parallel and no three pass through the same point. Prove that the numbers can be placed in the regions into which the lines divide the plane, such that the sums of the numbers on either side of any of the drawn lines are equal. | Solution. Let us denote the drawn lines as $\ell_{1}, \ell_{2}, \ldots, \ell_{n}$, ordering their directions clockwise (see Fig. 5). Formally, this means the following. Consider an arbitrary point $O$ on the plane. Draw lines through it parallel to ours, number them clockwise, and then assign the same numbers to our li... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,927 |
11.8. Initially, 100 cards are placed on the table, each with a natural number written on it; among them, exactly 28 cards have odd numbers. Then, every minute, the following procedure is carried out. For every 12 cards lying on the table, the product of the numbers written on them is calculated, all these products are... | Answer. No, it cannot.
Solution. If at some moment there are exactly $k$ odd numbers among the numbers on the cards, then among the products of 12 numbers, there are exactly $C_{k}^{12}$ odd products; therefore, the number on the next added card will be odd if and only if $C_{k}^{12}$ is odd (and in that case, $k$ wil... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,928 |
11.3. The midpoint of the edge $S A$ of the triangular pyramid $S A B C$ is equidistant from all vertices of the pyramid. Let $SH$ be the height of the pyramid. Prove that the point $H$ does not lie inside the triangle $A B C$. | Solution. See fig.
M is the midpoint of edge SA. Project edge SA onto the plane (ABC): point S is projected to point H (so SH is the height of the pyramid), and point M is projected to point... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,929 |
11.4. Workers need to tile a square floor consisting of 36 cells (identical squares). The workers have 12 tiles. 11 strips of three cells and one corner piece of three cells. Will the workers be able to complete the task? Cutting the tiles is not allowed. | Solution. Suppose that a $6 \times 6$ grid can be divided into 11 strips of size $3 \times 1$ and one corner piece consisting of three cells. We place one of the numbers $1, 2, 3$ in each cell as shown in the figure.
| 1 | 1 | 1 | 1 | 1 | 1 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 2 | 2 | 2 | 2 | 2 | 2 |
| 3 | ... | Theywillnotbeableto | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,930 |
11.5. Do there exist natural numbers $a$ and $b$, greater than 1000, such that for any natural number $c$, which is a perfect square, the three numbers $a, b$, and $c$ cannot be the lengths of the sides of a triangle? | Solution. Positive numbers a, b, c are then and only then the lengths of the sides of a triangle when the three triangle inequalities are satisfied: $\mathrm{a}+\mathrm{b}>\mathrm{c}, \mathrm{a}+\mathrm{c}>\mathrm{b}, \mathrm{b}+\mathrm{c}>\mathrm{a}$. Given the additional condition $\mathrm{a}\mathrm{c}$, $\mathrm{a}+... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,931 |
10.1. Each of the 10 people is either a knight, who always tells the truth, or a liar, who always lies. Each of them thought of some number (not necessarily an integer). Then the first said: “My number is greater than 1”, the second said: “My number is greater than $2”, \ldots$, the tenth said: “My number is greater th... | Answer: 9 knights.
Solution. Estimation. Note that none of the knights could have said the phrase "My number is greater than 10," otherwise the number they thought of would indeed be greater than 10. But then he could not have said any of the phrases "My number is less than 1," "My number is less than 2," ..., "My num... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,932 |
10.2. Given a convex quadrilateral with a perimeter of $10^{100}$, where the lengths of all sides are natural numbers, and the sum of the lengths of any three sides is divisible by the length of the remaining fourth side. Prove that this quadrilateral is a rhombus. (P. Kozhevnikov) | Solution. Let $a, b, c$ and $d$ be the lengths of the sides, and let $N=10^{100}$.
First solution. Suppose $d$ is the largest side. According to the condition, $a+b+c$ is divisible by $d$, that is, $a+b+c=k d$ for some natural number $k$. Clearly, $a+b+c>d$ (the length of a segment is less than the length of a broken ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,933 |
10.3. The cells of a $2 \times 2019$ table must be filled with numbers (exactly one number in each cell) according to the following rules. In the top row, there should be 2019 real numbers, none of which are equal, and in the bottom row, there should be the same 2019 numbers, but in a different order. In each of the 20... | Answer: 2016.
Solution. Estimation. We will prove that in the first row of the table, where numbers are arranged according to the rules, there are no fewer than three rational numbers (and, accordingly, no more than 2016 irrational numbers). Each number appearing in the table is written in exactly two cells, one of wh... | 2016 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,934 |
10.4. An infinite sequence of non-zero numbers $a_{1}, a_{2}, a_{3}, \ldots$ is such that for all natural $n \geqslant 2018$ the number $a_{n+1}$ is the smallest root of the polynomial
$$
P_{n}(x)=x^{2 n}+a_{1} x^{2 n-2}+a_{2} x^{2 n-4}+\ldots+a_{n} .
$$
Prove that there exists an $N$ such that in the infinite sequen... | Solution. Let $n \geqslant 2018$. Notice that $P_{n}(a)=P_{n}(-a)$ for all $a$. Therefore, since $P_{n}(x)$ has a non-zero root, it also has a negative root, from which it follows that $a_{n+1}<0$.
Furthermore, since $P_{n+1}(x)=x^{2} P_{n}(x)+a_{n+1}$, we have
$$
P_{n+1}\left(a_{n+1}\right)=a_{n+1}^{2} P_{n}\left(a_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,935 |
10.5. In an isosceles triangle \(ABC\), the bisector \(BL\) is drawn. The extension of the median drawn from vertex \(B\) intersects the circumcircle \(\omega\) of triangle \(ABC\) at point \(D\). Through the center of the circumcircle of triangle \(BDL\), a line \(\ell\) is drawn parallel to the line \(AC\). Prove tha... | Solution. Let $M$ be the midpoint of segment $AC$, $S$ be the second intersection point of line $BL$ with circle $\omega$, and $N$ be the midpoint of arc $ABC$ (see Fig. 4). Then $S$ is the midpoint of the smaller arc $AC$ of circle $\omega$, and points $M, S, N$ lie on the perpendicular bisector of segment $AC$. Line ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,936 |
1. Petya was exchanging stickers. He trades one sticker for 5 others. At first, he had 1 sticker. How many stickers will he have after 50 exchanges? | Answer: 201.
Solution: After each exchange, the number of Petya's stickers increases by 4 (one sticker disappears and 5 new ones appear). After 50 exchanges, the number of stickers will increase by 50*4=200. Initially, Petya had one sticker, so after 50 exchanges, he will have $1+200=201$. | 201 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,937 |
2. One day Uncle Fyodor weighed Sharik and Matroskin. It turned out that Sharik is 6 kg heavier than Matroskin, and Matroskin is three times lighter than Sharik. How much did Matroskin weigh? | Answer: 3 kg.
Solution: Since Matroskin is three times lighter than Sharik, Matroskin is lighter than Sharik by two of his own weights. According to the condition, this is equal to 6 kg, i.e., Matroskin weighs $6: 2=3$ kg. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,938 |
4. Write 7 consecutive natural numbers such that among the digits in their notation there are exactly 16 twos. (Consecutive numbers differ by 1.) | Answer. Any of the following two sequences will do:
$2215,2216,2217,2218,2219,2220,2221$
2229, 2230, 2231, 2232, 2233, 2234, 2235 | 2215,2216,2217,2218,2219,2220,2221or2229,2230,2231,2232,2233,2234,2235 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,939 |
5. Mom bought a box of lump sugar (sugar in cubes). The children first ate the top layer - 77 cubes, then the side layer - 55 cubes, and finally, the front layer. How many sugar cubes are left in the box? | Answer: 300 or 0.
Solution. A box has three dimensions: height, width, and depth. To find out how many cubes are in the top layer, you need to multiply the width by the depth, and for the side layer, multiply the height by the depth. After the top layer is eaten, the height decreases by 1, while the depth remains the ... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,940 |
6. Cut a square with a side of 4 into rectangles, the sum of the perimeters of which is 25. | For example, two rectangles $2 \times 0.5$ and one rectangle $3.5 \times 4-$ cm. The total perimeter is $2 * 2 * (2 + 0.5) + 2 * (3.5 + 4) = 25$.
 | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,941 |
11.5. Do there exist 2013 different natural numbers such that the sum of any 2012 of them is not less than the square of the remaining one?
(O. Podlipsky) | Answer. They do not exist.
Solution. Suppose such numbers were found. Since they are distinct and there are 2013 of them, the largest one is at least 2013; let's denote it by $a$. Then the sum of all the others does not exceed $2012a$, while its square is $a^{2} \geqslant 2013a$, meaning it is greater than this sum. C... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 16,942 |
1. The sum of the factorials of the first $k$ natural numbers is equal to the square of the sum of the first $n$ natural numbers.
a) Find all such pairs $(k, n)$ (3 points);
b) prove that there are no other pairs besides those found in part a) (4 points).
(For reference: The factorial of a natural number $m$ is the ... | Solution. Obviously, the pairs $(1,1)$ and $(3,2)$ work: $1!=1^{2}=1$ and $1!+2!+3!=$ $(1+2)^{2}=9$. We will prove that there are no other suitable pairs. For $k=2$, we have $1!+2!=3$ - this is not a square of an integer. For $k=4$, we have $1!+2!+3!+4!=1+2+6+$ $24=33$ - also not a square. For $k \geqslant 5$, the subs... | (1,1)(3,2) | Number Theory | proof | Yes | Yes | olympiads | false | 16,944 |
3. There is a pile of 100 matches. Petya and Vasya take turns, starting with Petya. Petya can take one, three, or four matches on his turn. Vasya can take one, two, or three matches on his turn. The player who cannot make a move loses. Which of the players, Petya or Vasya, can win regardless of the opponent's play? | Solution. Answer: only Vasya can win. Vasya, for example, can always make the remainder of the division by 3 of the number of matches left in the pile equal to 2 after his move. If there are exactly 2 matches left in the pile, then Petya takes one and loses on the next move. If there are 5 or more matches left, then Va... | Vasya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,946 |
4. On the sides $AB, BC, CD$ and $DA$ of a convex quadrilateral $ABCD$, points $P, Q, R$ and $S$ are taken such that $BP: AB = CR: CD = 1: 3$ and $AS: AD = BQ: BC = 1: 4$. Prove that the segments $PR$ and $QS$ are divided by their point of intersection in the ratios $1: 3$ and $1: 2$. | Solution. Consider the auxiliary parallelogram $A B C D_{1}$. We can assume that points $D$ and $D_{1}$ do not coincide (otherwise, the statement of the problem is obvious). Take points $S_{1}$ and $R_{1}$ on sides $A D_{1}$ and $C D_{1}$ such that $S S_{1} \| D D_{1}$ and $R R_{1} \| D D_{1}$. Let $N$ be the intersect... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,947 |
5. In a table tennis tournament, each participant met every other one time. Each match was officiated by one referee. All referees officiated a different number of matches. Player Ivanov claims that all his matches were officiated by different referees. The same claim is made by players Petrov and Sidorov. Could it be ... | Solution. Answer: no. Let none of the three be wrong and let the number of players be $n$. We will order the referees by the non-decreasing number of matches they have officiated. Then the first referee has officiated at least one match, the second at least two, and so on. Since all of Ivanov's matches were officiated ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,948 |
8.1 On the island, $\frac{2}{3}$ of all men are married and $\frac{3}{5}$ of all women are married. What fraction of the island's population is married? | 8.1 On the island, $\frac{2}{3}$ of all men are married and $\frac{3}{5}$ of all women are married. What fraction of the island's population is married?
Solution: Let M be the number of men on the island, and W be the number of women. The number of families, if counted by husbands, is $\frac{2}{3} M$; on the other han... | \frac{12}{19} | Other | math-word-problem | Yes | Yes | olympiads | false | 16,949 |
8.2 In a triangle, one side is three times smaller than the sum of the other two. Prove that the angle opposite to it is the smallest angle in the triangle. | 8.2 In a triangle, one side is three times smaller than the sum of the other two. Prove that the angle opposite to it is the smallest angle in the triangle.
Solution: It is sufficient to prove that the given side is the smallest side of the triangle (the smaller side has the smaller angle opposite to it). Let $a$ be i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,950 |
8.3 On a grid board of $10 \times 20$, several cells are painted red. Kostya cut the board into rectangles along the grid lines such that each rectangle contained 5 red cells. Vlad cut the same board into rectangles such that each contained 7 red cells. Prove that Dima will not be able to cut the same board into rectan... | 8.3 On a grid board $10 \times 20$, several cells are painted red. Kostya cut the board into rectangles along the grid lines such that each rectangle contained 5 red cells. Vlad cut the same board into rectangles such that each contained 7 red cells. Prove that Dima cannot cut the same board into rectangles such that e... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,951 |
8.4 An infinite sequence of positive numbers $a_{1}, a_{2}, a_{3}, \ldots$ is formed according to the rule: $a_{1}=1, a_{n+1}^{2}=a_{n}^{2}+\frac{1}{a_{n}}$ for $n=1,2,3, \ldots$ Prove that the sequence $b_{1}, b_{2}, b_{3}, \ldots$, where $b_{n}=a_{n+1}-a_{n}$, is decreasing, i.e., $b_{1}>b_{2}>b_{3}>\ldots$ | 8.4 An infinite sequence of positive numbers $a_{1}, a_{2}, a_{3}, \ldots$ is formed according to the rule: $a_{1}=1, a_{n+1}^{2}=a_{n}^{2}+\frac{1}{a_{n}}$ for $n=1,2,3, \ldots$. Prove that the sequence $b_{1}, b_{2}, b_{3}, \ldots$, where $b_{n}=a_{n+1}-a_{n}$, is decreasing, i.e., $\mathrm{b}_{1}>\mathrm{b}_{2}>\mat... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,952 |
8.5 Around a circle, 10 iron weights are placed. Between each pair of adjacent weights, there is a bronze ball. The mass of each ball is equal to the difference in mass between its adjacent weights. Prove that the balls can be divided into two groups such that the scales will balance. | 8.5 Around a circle, 10 iron weights are placed. Between each pair of adjacent weights, there is a bronze ball. The mass of each ball is equal to the difference in mass between the adjacent weights. Prove that the balls can be divided into two groups such that the scales will balance.
Solution: Let \(a_{1}, a_{2}, \ld... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,953 |
3. Someone wrote down two numbers $5^{2020}$ and $2^{2020}$ in a row. How many digits will the resulting number contain? | Solution. Let the number $2^{2020}$ contain $m$ digits, and the number $5^{2020}$ contain $n$ digits. Then the following inequalities hold: $10^{m-1}<2^{2020}<10^{m}, 10^{n-1}<5^{2020}<10^{n}$ (the inequalities are strict because the power of two or five is not equal to the power of ten). Multiplying these inequalities... | 2021 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,955 |
4. Can we find two such natural numbers $x$ and $y$ that the sum of these numbers plus 2021 is less than the sum of their GCD and LCM? | Solution: In other words, we need to determine whether the equation GCD(x, y) + LCM(x, y) - (x + y) = 2021 has a solution in the set of natural numbers. Let's rewrite this equation as GCD(x, y) + LCM(x, y) + x + y = 2021. Let's analyze the obtained equation in terms of parity. If both numbers x and y are even, then the... | They\do\not\exist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,956 |
# 5. Find the sum:
$$
\frac{2}{1 \cdot 2 \cdot 3}+\frac{2}{2 \cdot 3 \cdot 4}+\frac{2}{3 \cdot 4 \cdot 5}+\ldots+\frac{2}{2008 \cdot 2009 \cdot 2010}
$$ | Solution:
Notice that $\frac{2}{n \cdot(n+1) \cdot(n+2)}=\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+1}+\frac{1}{n+2}$.
From this, it follows that the required sum is:
$\frac{1}{1}-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{3}+\frac{1}{4}+\frac{1}{3}-\frac{1}{4}-\frac{1}{4}+\frac{1}{5}+\ldots+\fra... | \frac{1009522}{2019045} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,957 |
1. The older brother noticed that in 10 years, the younger brother will be as old as he is now, and his age will be twice the current age of the younger brother. How old is the younger brother now? | Answer: 20 years.
Solution. Let the current age of the younger brother be $x$ years, and the older brother be $y$ years. In 10 years, the younger brother's age will be $y$ years, and the older brother's age will be $2x$ years. Since the age of each has changed by 10 years, we have the equations: $y+10=2x, x+10=y$. By ... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,958 |
2. Can the figure shown in the diagram be cut along the grid lines into four equal parts (equal figures can be superimposed by overlaying)?
---
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Answer: possible.
Solution. The method of cutting into four equal shapes is shown in the figure. It is not difficult to verify that these shapes can be matched by overlaying.
Criteria. Any correct cutting, even without explanations: 7 points.
Only the correct answer is given: 0 points. | Number Theory | proof | Yes | Yes | olympiads | false | 16,959 | |
3. Some number is represented as the sum of a thousand different prime numbers greater than five. Prove that it can be represented as the sum of a thousand different composite numbers. | Solution. All primes greater than five are odd. We will decrease the five hundred smallest by one and increase the five hundred largest by one. All numbers will become even and greater than 4, thus composite. The sum remains unchanged, and all numbers will still be distinct.
Criteria. Any correct solution: 7 points. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,960 |
4. A rectangle $10 \times 20$ is divided into unit squares. How many triangles are formed after drawing one diagonal?
 | Answer: 220.
Solution. The figure shows one of the obtainable triangles. All such triangles are right-angled, and the vertex of the right angle can be any lattice node,
except those lying on the diagonal. There are a total of $21 \times 11$ nodes, and 11 of them are on the
diagonal, so the number of triangles is $2... | 220 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,961 |
5. A row of 99 visually identical coins lies in front of you. Ten of them are lighter (not necessarily of the same weight) and lie in a row. The other 89 weigh the same. How can you find a lighter coin using a two-pan balance in two weighings? | Solution. Let's number the coins from 1 to 99 in the order they are arranged in a row. Consider the coins with numbers that are multiples of ten. There are exactly 9 of them. Only one of them is light. It can be found among these nine with two weighings. The first: compare two of their triplets with each other. If they... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,962 |
10.1. (7 points)
Prove that $\sqrt{\frac{11 \ldots 1}{2 n \text { digits }}-\underbrace{22 \ldots .2}_{n \text { digits }}}=\underbrace{33 \ldots 3}_{n \text { digits }}$. | # Solution:
$$
\begin{aligned}
& \sqrt{\underbrace{11 \ldots 1}_{2 n \text { digits }}-\underbrace{22 \ldots 2}_{n \text { digits }}}=\sqrt{\frac{1}{9} \cdot \underbrace{99 \ldots 9}_{2 n \text { digits }}-\frac{2}{9} \cdot \underbrace{99 \ldots 9}_{n \text { digits }}}=\frac{1}{3} \sqrt{\left(10^{2 n}-1\right)-2\lef... | 42 | Number Theory | proof | Yes | Yes | olympiads | false | 16,963 |
# 10.3. (7 points)
Prove that for any non-zero numbers $a, b$ and at least one of the quadratic equations $a x^{2}+2 b x+c=0, b x^{2}+2 c x+a=0$ and $c x^{2}+2 a x+b=0$ has a root. | Solution: We will prove by contradiction. Suppose the equations do not have real roots. Then $4 b^{2}-4 a c<0, 4 c^{2}-4 a b<0, 4 a^{2}-4 b c<0$.
Therefore, $b^{2}<a c, c^{2}<a b, a^{2}<b c$.
Since the right-hand sides are positive, the left-hand sides will also be positive, and we have the right to multiply the ineq... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,964 |
# 10.4. (7 points)
Prove that if in an arbitrary quadrilateral $A B C D$ internal angle bisectors are drawn, then the four points of intersection of the bisectors of angles $A$ and $C$ with the bisectors of angles $B$ and $D$ lie on a single circle.
# | # Solution:
Let $A P, B Q, C R, D S$ be the angle bisectors of the internal angles of quadrilateral $A B C D$, and let $\alpha, \beta, \gamma, \varphi$ be the measures of these angles. Then,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,965 |
# 10.5. (7 points)
At a joint conference of the party of liars and the party of truth-tellers, 32 people were elected to the presidium and seated in four rows of eight. During the break, each member of the presidium claimed that among their neighbors there are representatives of both parties. It is known that liars al... | Answer: with eight liars.
Solution: Divide all the seats in the presidium into eight groups as shown in the figure. If there are fewer than eight liars, then in one of these groups, only truth-tellers will be sitting, which is impossible. The contradiction obtained shows that there are no fewer than eight liars. The f... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,966 |
11.5. Given a natural number $n>1$. Prove that there exist such $n$ consecutive natural numbers that their product is divisible by all prime numbers not exceeding $2 n+1$, and is not divisible by any other prime number.
(I. Bogdanov) | 11.5. Suppose that the number $n+1$ is composite; we will show that then the numbers $n+2, \ldots, 2 n+1$ work. Clearly, their product is divisible by all prime numbers in the interval $[n+2,2 n+1]$, but is not divisible by prime numbers greater than $2 n+1$ (since all factors do not exceed $2 n+1$). For any prime $p \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,967 |
11.6. Can the 4 centers of the circles inscribed in the faces of a tetrahedron lie in the same plane? (I. Bogdanov, O. Podlipsky) | 11.6. Answer. They cannot.
Let $I_{A}, I_{B}, I_{C}, I_{D}$ be the centers of the inscribed circles of triangles $B C D, A C D, A B D, A B C$ respectively. Suppose they lie in the same plane. Then either they form a convex quadrilateral, or one of these points lies inside the triangle formed by the other three.
. It is known that any two children have different sets of subjects they are interested in (in particular, exactly one child is not interested in anyth... | 11.8. We will prove the statement of the problem in a more general form, for $n \geqslant 2$ items and $2^{n}$ children, arbitrarily divided into $2^{n-1}$ pairs of neighbors. Note that there are exactly $2^{n}$ sets of $n$ items; hence, each set of items is interesting to exactly one student.
Induction on $n$. For $n... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,970 |
9.1. Do there exist such integers $x, y$ and $z$ that the equality holds: $(x-$ $y)^{3}+(y-z)^{3}+(z-x)^{3}=2021 ?$ (7 points)
# | # Solution:
By expanding the brackets and simplifying, we obtain the polynomial
$-3 x^{2} y+3 x y^{2}-3 y^{2} z+3 y z^{2}-3 z^{2} x+3 z x^{2}=2021$. The left side is divisible by 3, while the right side is not. Therefore, such numbers do not exist.
Answer: do not exist.
Criteria: | donotexist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,971 |
9.2. Ilya has a table filled with numbers from 1 to 9 as shown in the table on the left. In one move, Ilya can swap any two rows or any two columns. Can he obtain the table on the right in several moves? (7 points)
# | # Solution.
No. Let's place the numbers 1, 2, 3 on the board (see the diagram). According to the conditions, a checker from a cell numbered 1 can move only to a cell numbered 2, from a cell numbered 2 only to a cell numbered 3, and from a cell numbered 3 only to a cell numbered 1. Therefore, if the checker has circled... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,974 |
9.5. Point $D$ outside an acute-angled triangle $ABC$ is such that $\angle ABC + \angle ABD = \angle ACB + \angle ACD = 180^{\circ}$. Prove that the center of the circumcircle of triangle $ABC$ lies on the segment
## $AD$. (6 points) | # Solution.
The angles adjacent to angles $A B D$ and $A C D$ are equal to angles $A B C$ and $A C B$, respectively, so $\mathrm{CA}$ and $\mathrm{BA}$ are the bisectors of the external angles of triangle BCD (see figure). Through point A, their intersection, passes the bisector DA of angle D of triangle BCD. The bise... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,975 |
1. On a circular highway 13 km long, there are five different settlements A, B, C, D, E. Is it possible that the shortest distance along the highway from A to B is 3 km, from B to C - 6 km, from C to D - 4 km, from D to E - 5 km, and from E to A - 6 km? | Answer: it can.
Solution. For example, the settlements are in the following order if we move clockwise: A, C, E, B, D and the distances between them in this direction are $\mathrm{AC}=3$ km, $\mathrm{CE}=4$ km, $\mathrm{EB}=3$ km, $\mathrm{BD}=2$ km, $\mathrm{DA}=1$ km. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,976 |
3. In the courtyard, there are 5 houses, in which $5, 15, 25, 35, 45$ people live. It is known that each person has at least two namesakes among the residents of the courtyard. Prove that someone has a namesake in their own house. | Solution.
Assume that all people living in the house (45) have different names (that is, there are no two people with the same name living in the house (45)). Since each of them has at least two namesakes, there should be at least $45 \cdot 3=135$ people in total. But $5+15+25+35+45=125$. Contradiction.
Therefore, th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,978 |
5. Emperor Pea has a worm in one of his rejuvenating apples. He has 13 apples in total, and they are arranged in a circle in a special box for rejuvenating apples. To find the worm, Emperor Pea decided to use a balance scale. He knows that all the apples weigh the same, except for the one with the worm, which is heavie... | # Solution.
Let's number the apples clockwise. The neighboring apples will be numbered 1 and 2, 2 and 3, ..., 12 and 13, 13 and 1.
Weigh apples 1, 2, 3, 4, 5, 6 against 7, 8, 9, 10, 11, 12.
If they are equal, then the worm is in apple 13.
Suppose 1, 2, ..., 6 are heavier.
Then after the apples are put back and the... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,979 |
7.1. Masha surveyed her friends from her ensemble and received the following answers: 25 of them study mathematics, 30 have been to Moscow, 28 have traveled by train. Among those who have traveled by train, 18 study mathematics and 17 have been to Moscow. 16 friends study mathematics and have been to Moscow, and among ... | Answer: No.
Solution. Let's calculate the number of girls who have not been to Moscow, do not study mathematics, and have not traveled by train. We get $45-25-30-28+16+18+17-15=-2<0$, which is impossible. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,980 |
7.2. In the canteen, there are sixteen cups of tea. Masha needs to make sure that all the cups have the same amount of tea, and in one step, she can only equalize the amount of tea in exactly two cups. Will Masha be able to complete the task? | Answer: Yes.
Solution. 16 is a power of two. We will solve this problem first for four cups, then for eight, and then for 16.
Divide the cups into pairs: 1-2, 3-4, 5-6, 7-8, 9-10, 11-12, 13-14, 15-16 and equalize the amount of tea in each pair of cups. Now we have two completely identical eights: $1,3,5,7,9,11,13,15$... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,981 |
7.3. In 60 chandeliers (each chandelier has 4 lampshades), lampshades need to be replaced. Each electrician spends 5 minutes replacing one lampshade. A total of 48 electricians will be working. Two lampshades in a chandelier cannot be replaced simultaneously. What is the minimum time required to replace all the lampsha... | Answer: 25 minutes
Solution. Let's show how to proceed. First, 48 electricians replace one lampshade in 48 chandeliers, which takes 5 minutes, and 48 chandeliers have one lampshade replaced, while 12 have none replaced. Then, 12 electricians replace lampshades in the chandeliers that haven't been replaced yet, while t... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,982 |
7.4. Find the perimeter of a rectangle if the sum of the lengths of two of its sides is 10 dm, and the sum of the lengths of three of its sides is 14 dm. | Answer: 18 dm, 19 dm or 20 dm
Solution. If the sum of two adjacent sides is 10 dm, then the perimeter is 20 dm, which does not contradict the condition on the sum of three sides. If the sum of opposite sides is 10 dm, then each of these sides is 5 dm. In this case, the adjacent side to them is 4 dm or 4.5 dm. | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,983 |
7.5. A basketball coach wants to take the three tallest boys to his team. In total, 25 boys came to the tryouts, each of different height. In one session, the coach can watch 5 boys and assign them places from 1 to 5. How should the tryouts be organized so that the coach can choose the boys for the team in 7 sessions? | Solution. Let's divide all the boys into five groups of five boys each. We will compare the boys within each group. This will require 5 comparisons. The sixth comparison will be between the tallest boys in each group. After this, we will denote the groups by the letters A, B, V, G, D in descending order of the height o... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,984 |
10-1-1. Select the numbers that serve as counterexamples to the given statement: “If the sum of the digits of a natural number is divisible by 27, then the number itself is divisible by $27$.”
a) 81 ; b) 999 ; c) 9918 ; d) 18 . | Answer: only 9918.
Solution option 1. The statements about 81 and 18 are definitely not counterexamples, as for them, the premise "if the sum of the digits of a natural number is divisible by 27" is not satisfied.
The statement about 999 is not a counterexample, as it does not contradict our statement.
Finally, the ... | 9918 | Number Theory | MCQ | Yes | Yes | olympiads | false | 16,985 |
10-2-1. Petya writes down a sequence of numbers: if the current number is equal to $x$, then the next one is $\frac{1}{1-x}$. The first number in the sequence is 2. What is the five hundredth number? | Answer: -1.
Solution variant 1. Let's list the first few terms of the obtained sequence:
$$
2,-1,1 / 2,2, \ldots
$$
We see that the sequence has entered a cycle with a period of 3. Since the number 500 when divided by 3 gives a remainder of 2, the 500th term will be the same as the 2nd. That is, the five hundredth n... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,986 |
10-3-1. Non-negative integers $a, b, c, d$ are such that
$$
a b+b c+c d+d a=707
$$
What is the smallest value that the sum $a+b+c+d$ can take? | Answer: 108.
Solution variant 1. The given equality can be rewritten as
\[
(a+c)(b+d)=7 \cdot 101
\]
where the numbers 7 and 101 are prime. Therefore, either one of the expressions in parentheses is 1 and the other is 707, or one of the expressions in parentheses is 7 and the other is 101. In the first case, \(a+b+c... | 108 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,987 |
10-4-1. Anya and Borya are playing rock-paper-scissors. In this game, each player chooses one of the figures: rock, scissors, or paper. Rock beats scissors, scissors beat paper, and paper beats rock. If the players choose the same figure, the game ends in a tie.
Anya and Borya played 25 rounds. Anya chose rock 12 time... | Answer: 16.
Solution version 1. Note that, since there were no draws, when Anya chose rock, Borya must have chosen scissors or paper. Anya chose rock 12 times. Borya chose scissors or paper a total of $9+3=12$ times. Therefore, all these cases must have occurred in the rounds where Anya chose rock.
This means that in... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,988 |
10-5-1. Among $n$ angles of a convex $n$-gon, $n-1$ angles are equal to $150^{\circ}$, and the remaining one is less than $150^{\circ}$. For which $n$ is this possible? List all possible answers. | Answer: $8,9,10,11$.
Solution variant 1. Let the remaining angle be $x^{\circ}$. Using the fact that the sum of the angles of an $n$-sided polygon is $180^{\circ}(n-2)$, we get
$$
\begin{gathered}
150(n-1)+x=180(n-2) \\
n=\frac{x+120}{30}
\end{gathered}
$$
Since $9<x<150$, then $7<n<12$, so $n$ can only be $8,9,10,1... | 8,9,10,11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,989 |
10-6-1. In the figure, there are two circles with centers $A$ and $B$. Additionally, points $A, B$, and $C$ lie on the same line, and points $D, B$, and $E$ also lie on the same line. Find the degree measure of the angle marked with a “?”.
. Angles $ABE$ and $DBC$ are equal as vertical angles. Note that triangle $DBC$ is isosceles wi... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,990 |
10-7-1. At a large round table, 90 people are feasting, facing the center of the table: 40 barons, 30 counts, and 20 marquises. On a signal, exactly those who have both neighbors—left and right—with the same title should stand up. What is the maximum number of people who can stand up?
For example, for a count to stand... | Answer: 86.
Solution Variant 1. We ask half of the revelers, standing every other person, to step forward. Then all the people will be divided into two circles: an inner and an outer one. Note that a person in the inner circle has both neighbors in the outer circle, and they are also neighbors there. Similarly for a p... | 86 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,991 |
10-7-3. At a large round table, 55 people are feasting, facing the center of the table: 25 barons, 20 counts, and 10 marquises. On a signal, exactly those who have both neighbors—left and right—with the same title should stand up. What is the maximum number of people who can stand up?
For example, for a count to stand... | Answer: 52.
Solution variant 3. We will place people in a new circle every other person: $1,3,5,7$, $\ldots, 55,2,4,6, \ldots, 54$, back to 1. It is sufficient to find the maximum number of neighbors with the same title in the new circle.
It is clear that there are neighbors of different titles in the new circle. It ... | 52 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,992 |
10-8-1. There is a magical grid sheet of size $2000 \times 70$, initially all cells are gray. The painter stands on a certain cell and paints it red. Every second, the painter takes two steps: one cell to the left and one cell down, and paints the cell red where he ends up after the two steps. If the painter is in the ... | Answer: 14000.
Solution version 1. We need to understand how many cells will be painted by the time the painter returns to the initial cell. Note that after every 2000 moves, the painter returns to the starting column, and after every 140 moves, he returns to the starting row. Therefore, he will return to the initial ... | 14000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,993 |
10.1 Prove that if $a$ and $b$ are positive numbers, then at least one of the numbers $a, b^{2}$, and $\frac{1}{a^{2}+b}$ is greater than 0.7 | Solution: Let $a \leq 0.7$ and $b^{2} \leq 0.7$. Then
$$
a^{2}+b \leq 0.7^{2}+\sqrt{0.7}, \frac{1}{a^{2}+b} \geq \frac{1}{0.49+\sqrt{0.7}}
$$
It remains to check that $\frac{1}{0.49+\sqrt{0.7}}>0.7$, i.e., that $0.7 \sqrt{0.7}<0.657$, i.e., that $0.49 \cdot 0.7<0.657^{2}$, i.e., that $0.343<0.657^{2}$. But it is even... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 16,996 |
10.3 On the side $\mathrm{BC}$ of triangle $\mathrm{ABC}$, there is a point K such that angle $\mathrm{CAK}$ is half of angle $\mathrm{B}$, and the intersection point O of segment $\mathrm{AK}$ with the bisector $\mathrm{BL}$ of angle $\mathrm{B}$ divides this segment into two equal parts. Prove that $\mathrm{AO} \cdot... | Solution: See Fig.
In triangle ABK, the bisector BO is, by condition, also a median. Then triangle ABK turns out to be isosceles $(\mathrm{AB}=\mathrm{BK})$, and $\mathrm{BO}$ is an altitude.
From triangle AOB we get $\alpha+\beta=90^{\circ}$. But then angle $\mathrm{A}$, which equals $\alpha+\beta,$ is a right angle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,997 |
10.4 On the Unified State Exam (USE) in mathematics, 333 students made a total of 1000 mistakes. Prove that in this case, the number of students who made more than 5 mistakes is no greater than the number of students who made fewer than 4 mistakes. | Solution: Let $x$ be the number of students who made no less than 6 mistakes, and $y_{k}$ be the number of students who made exactly $k$ mistakes, where $k=5,4,3,2,1,0$. According to the problem, we have
$$
\begin{aligned}
& x+y_{5}+y_{4}+y_{3}+y_{2}+y_{1}+y_{0}=333 \\
& 6 x+5 y_{5}+4 y_{4}+3 y_{3}+2 y_{2}+y_{1} \leq ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,998 |
9.2. Five natural numbers are written on the board. It turns out that the sum of any three of them is divisible by each of the remaining ones. Is it necessarily true that among these numbers there will be four equal ones?
-(a+b+c)$ is also divisible by $e$. Since $0 \leqslant... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,001 |
9.3. Inside parallelogram $A B C D$, a point $E$ is chosen such that $A E=D E$ and $\angle A B E=90^{\circ}$. Point $M$ is the midpoint of segment $B C$. Find the angle $D M E$. | Answer: $90^{\circ}$.
First solution. Let $N$ be the midpoint of segment $A D$. Since triangle $A E D$ is isosceles, its median $E N$ is also an altitude, that is, $E N \perp A D$. Therefore, $N E \perp B C$ (see Fig. 2).
Since $A D \| B C$ and $B M = M C = A N = N D = A D / 2$, quadrilaterals $A B M N$ and $B M D N$... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,002 |
9.4. A confectionery factory produces $N$ types of candies. For New Year, the factory gave each of 1000 school students a gift containing candies of several types (the compositions of the gifts could be different). Each student noticed that for any 11 types of candies, they received a candy of at least one of these typ... | Answer. $N=5501$.
Solution. Let $A_{1}, A_{2}, \ldots, A_{N}$ be the sets of students who did not receive candies of the 1st, 2nd, ..., $N$-th types, respectively. According to the problem, all these sets are distinct; moreover, each student is contained in no more than ten of them. Therefore, the total number of elem... | 5501 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,003 |
9.5. The numbers $x, y$, and $z$ satisfy the condition $x^{2}+y^{2}+z^{2}=1$. Prove that
$$
(x-y)(y-z)(x-z) \leqslant \frac{1}{\sqrt{2}}
$$
(L. Emelyanov, method commission) | The first solution. Since the left side of the inequality does not change or changes sign under any permutation of the variables, it is sufficient to check the inequality for any permutation of the numbers $x, y$, and $z$ for which the left side is non-negative. Therefore, we can assume that $x \geqslant y \geqslant z$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 17,004 |
9.1. Let $a_{1}, a_{2}, a_{3}, \ldots$ be an infinite increasing sequence of natural numbers, and $p_{1}, p_{2}, p_{3}, \ldots$ be a sequence of prime numbers such that for each natural $n$, the number $a_{n}$ is divisible by $p_{n}$. It turns out that for all natural $n$ and $k$ the equality $a_{n}-a_{k}=p_{n}-p_{k}$ ... | First solution. Let $c=a_{1}-p_{1} \geqslant 0$. By the condition, for any $n$ we have $a_{n}-a_{1}=p_{n}-p_{1}$, or $a_{n}-p_{n}=a_{1}-p_{1}=c$.
Assume that $c>0$. There exists an index $n$ such that $a_{n}>2c$. Then $a_{n}>p_{n}=a_{n}-c>a_{n}-\frac{a_{n}}{2}=\frac{a_{n}}{2}$, which means $2p_{n}>a_{n}>p_{n}$. Theref... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,005 |
9.2. Circle $\omega$ is tangent to sides $A B$ and $A C$ of triangle $A B C$. Circle $\Omega$ is tangent to side $A C$ and the extension of side $A B$ beyond point $B$, and also tangent to $\omega$ at point $L$, which lies on side $B C$. Line $A L$ intersects $\omega$ and $\Omega$ again at points $K$ and $M$ respective... | First solution. By symmetry, the line $A L$ is the bisector of angle $\angle B A C$ and passes through the centers of circles $\omega$ and $\Omega$.
Since the corresponding sides of triangles $K B L$ and $M C L$ are parallel, there exists a homothety with center $L$ that maps the first triangle to the second. This hom... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,006 |
9.3. Let $a_{1}, \ldots, a_{25}$ be non-negative integers, and let $k$ be the smallest of them. Prove that
$$
\left[\sqrt{a_{1}}\right]+\left[\sqrt{a_{2}}\right]+\ldots+\left[\sqrt{a_{25}}\right] \geqslant\left[\sqrt{a_{1}+\ldots+a_{25}+200 k}\right]
$$
(As usual, $[x]$ denotes the integer part of $x$, that is, the g... | Solution. Let $n_{i}=\left[\sqrt{a_{i}}\right]$. Then $a_{i}<\left(n_{i}+1\right)^{2}$, and since the numbers $a_{i}$ are integers, we have $a_{i} \leqslant n_{i}^{2}+2 n_{i}$. If we now show that
$$
\sqrt{a_{1}+\ldots+a_{25}+200 k}<n_{1}+n_{2}+\ldots+n_{25}+1
$$
then the right-hand side of the inequality to be prove... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,007 |
9.4. On a $n \times n$ chessboard, several cells are marked in such a way that the bottom-left $(L)$ and top-right $(R)$ corners of the board are not marked, and any knight's path from $L$ to $R$ necessarily includes a marked cell. For which $n>3$ can we assert with certainty that there will be three cells in a row alo... | Answer. For $n=3 k+1$, where $k$ is a natural number.
Solution. Let's number the rows from $1,2, \ldots, n$ from bottom to top, and the columns from left to right. We will denote a cell by $(a, b)$, where $a$ and $b$ are the numbers of its column and row, respectively.
First, we will show that for $n=3 k$ and $n=3 k+... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,008 |
9.5. On a circle, 99 points are marked, dividing this circle into 99 equal arcs. Petya and Vasya are playing a game, taking turns. Petya goes first; his first move is to color any marked point red or blue. Then each player, on their turn, can color any uncolored marked point red or blue, provided it is adjacent to an a... | # Answer. No.
Solution. We will present a strategy that allows Vasya to win with certainty. His first moves are made arbitrarily until 33 points are colored before his next move. Let $A$ be one of the extreme colored points, and $B$ be an uncolored point adjacent to the other extreme. Then there exists a marked point ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,009 |
9.6. Given natural numbers $a$ and $b$. Prove that there are infinitely many natural numbers $n$ such that the number $a^{n}+1$ is not divisible by $n^{b}+1$.
(A. Golev) | Solution. Let's call a natural number $n$ bad if $a^{n}+1$ is not divisible by $n^{b}+1$. Our goal is to prove that there are infinitely many bad numbers.
First Solution. We will prove that for any even $n$, one of the numbers $n$ and $n^{3}$ is bad; from this, the required result obviously follows. Assume the opposit... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,010 |
9.7. In a card game, each card is assigned a numerical value from 1 to 100, and each card beats a smaller one, with one exception: 1 beats 100. The player knows that 100 cards with different values are lying face down in front of them. The dealer, who knows the order of these cards, can inform the player, for any pair ... | Solution. Let $c_{i}$ denote the card with value $i$. Choose an arbitrary number $3 \leqslant k \leqslant 98$. Suppose the dealer informs which card beats the other in the pairs $\left(c_{k}, c_{1}\right),\left(c_{100}, c_{k}\right),\left(c_{1}, c_{100}\right)$, as well as in all pairs of the form ( $c_{i+1}, c_{i}$ ) ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,011 |
1. Is it possible to arrange the numbers $0,1,-1$ in the cells of a $4 \times 4$ square so that the ten sums in the four rows, four columns, and two main diagonals are the same (each of the numbers $0,1,-1$ must be present at least once)? | 1. Solution. Yes, it is possible. For example, the following arrangement works.
Evaluation Criteria. Any correct example - 7 points. Otherwise - 0 points. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,012 |
2. The length of a rectangular parallelepiped is 16, and the width is 12. If the centers of three of its faces, which share a common vertex, are connected, a triangle with an area of 30 is formed. Find the height of the parallelepiped. | 2. Answer: 7.2. Solution. Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be a parallelepiped, $A B=12, A D=16, A A_{1}=x, X$ be the center of the face $A A_{1} D_{1} D, Y$ be the center of the face
| 0 | 1 | -1 | 0 |
| :---: | :---: | :---: | :---: |
| -1 | 0 | 0 | 1 |
| 0 | -1 | 1 | 0 |
| 1 | 0 | 0 | -1 |
$C C_{1} D_{1} D, Z... | 7.2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,013 |
3. The equation $a x^{2}-b x+c=0$ has two distinct roots $x_{1}, x_{2}$. What is the maximum number of numbers in the set $\left\{a, b, c, x_{1}, x_{2}\right\}$ that can be prime (and, accordingly, natural) numbers? If a prime number appears in the set twice, it should be counted twice. The number 1 is not considered a... | 3. Answer: four. Solution. The quadratic trinomial $3 x^{2}-7 x+2$ has roots 2 and $1 / 3$. Therefore, in the set, prime numbers appear four times: 3, 7, 2, 2. All five numbers in the set cannot be prime, since by Vieta's theorem $a x_{1} x_{2}=c$, and a prime number cannot be the product of three primes.
Grading crit... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,014 |
4. The perimeter of a triangle is less than the diameter of its circumscribed circle. Prove that one of the angles of this triangle is greater than 150 degrees. | 4. Solution. Let $a \geq b \geq c$ be the sides of a triangle, and $R$ be the radius of the circumscribed circle. Then, by the triangle inequality, $2 R \geq 2 a$ and $\sin A=\frac{a}{2 R} \leq \frac{1}{2}$. Since the sine function is increasing from 0 to $\pi$, angle A is either greater than 150 degrees or less than 3... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,015 |
5. On a coordinate plane with the origin at point $O$, a parabola $y=x^{2}$ is drawn. Points $A, B$ are marked on the parabola such that $\angle A O B$ is a right angle. Find the smallest possible value of the area of triangle $A O B$. | 5. Answer: 1. Solution. Let $A\left(a, a^{2}\right), B\left(b, b^{2}\right)$ be arbitrary points on the parabola. Then, by the Pythagorean theorem, $a^{2}+a^{4}+b^{2}+b^{4}=(a-b)^{2}+\left(a^{2}-b^{2}\right)^{2} \Leftrightarrow a b=-1$. From this, we find the expression for the area of the triangle: $2 S=\sqrt{a^{2}+a^... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,016 |
6. The numbers $1,2, \ldots, 2016$ are written on a board. It is allowed to erase any two numbers and replace them with their arithmetic mean. How should one proceed to ensure that the number 1000 remains on the board? | 6. Solution. We will first act according to the following scheme: $1,2,3, \ldots, n-3, n-2, n-1, n \rightarrow 1,2,3, \ldots, n-3, n-1, n-1 \rightarrow 1,2,3, \ldots, n-3, n-1$. Acting in this way, we will arrive at the set $1,2,3, \ldots, 1000,1002$. Then we will act according to the same scheme from the other end:
\... | 1000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,017 |
7.1. Karlson counts 200 buns baked by Fräulein Bock: «one, two, three, ..., one hundred and nine, one hundred and ten, ..., one hundred and ninety-eight, one hundred and ninety-nine, two hundred». How many words will he say in total? (Each word is counted as many times as it was said.) | Answer: 443 words.
Solution. One word will be required to pronounce 29 numbers: $1,2,3,4,5,6,7$, $8,9,10,11,12,13,14,15,16,17,18,19,20,30,40,50,60,70,80,90,100,200$.
Among the first 99 numbers, the number of those pronounced in two words: $99-27=72$, thus, the number of words required for their pronunciation is $2 \c... | 443 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,018 |
7.2. Fill in the cells of a $3 \times 3$ table with integers such that the sum of all numbers in the table is positive, while the sum of the numbers in any $2 \times 2$ square is negative. | Answer: For example,
| 1 | 1 | 1 |
| :---: | :---: | :---: |
| 1 | -6 | 1 |
| 1 | 1 | 1 |.
## Criterion.
7 points. Any correct example.
Comment. There are several different correct examples. | \begin{pmatrix}\hline1&1&1\\\hline1&-6&1\\\hline1&1&1\\\hline\end{pmatrix} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,019 |
7.3. In the notebook, all irreducible fractions with the numerator 15 are written down, which are greater than $\frac{1}{16}$ and less than $\frac{1}{15}$. How many such fractions are written down in the notebook? | Answer: 8 fractions.
Solution. We look for all suitable irreducible fractions of the form $\frac{n}{15}$. Since $\frac{1}{16}<\frac{15}{n}$, then $\frac{15}{225}>\frac{15}{n}$, and $n>225$. Therefore, $225<n<240$. The fraction $\frac{n}{15}$ is irreducible, meaning $n$ is not divisible by 3 or 5. It is not difficult t... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,020 |
7.5. Lёsha colors cells inside a $6 \times 6$ square drawn on graph paper. Then he marks the nodes (intersections of the grid lines) to which the same number of colored and uncolored squares are adjacent. What is the maximum number of nodes that can be marked? | Answer: 45.
## Solution.
Estimation. Each grid node belongs to one, two, or four squares.
The corner vertices of the original square are adjacent to only one small square, so Lёsha will not be able to mark them. Therefore, the maximum number of marked nodes does not exceed $7 \cdot 7-4=45$.
Example.
Solution. The total weight of the weights is 34 grams, i.e., each pile weighs 17 grams. Two weights can add up to 17 grams in only one way $-16+1$: if we do not take the 16-gram weight, then with two weights we will not exceed $8+4=12$ grams. | 16+1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,024 |
2.1. All natural numbers are written in a row without spaces: 12345678910111213 .... What is the position of the twelfth ninth from the beginning?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | Answer: 174
Solution. The first nine is in the ninth position, the second is in the 29th position (between the first and second nines, we start writing two-digit numbers), the third is in the 49th position, and so on. The ninth nine (from the number 89) is in the $9+2 \cdot 80=$ 169th position (numbers from 1 to 9 occ... | 174 | Number Theory | proof | Yes | Yes | olympiads | false | 17,025 |
3.1. From a square grid with a side of 40, a rectangle $36 \times 37$ was cut out, adjacent to one of the corners of the square. Grisha wants to color a five-cell cross in the remaining piece. In how many ways can he do this?
. Due to the cut-out, $36 \times 37-1$ ways are lost (the center of the cross cannot be in the cut-out rec... | 113 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,026 |
4.1. Every evening, starting from September 1st, little Anton ate one pastry. After eating another pastry, he noticed that during this entire time, he had eaten 10 delicious pastries (the rest seemed tasteless to him). But among any seven consecutive pastries he ate, no fewer than three turned out to be delicious. What... | Answer: 26
Solution. We will show that among 27 pastries, there will be no fewer than 11 delicious pastries. Let's number the pastries from 1 to 27. Note that among the first seven pastries, at least three are delicious, among the next seven as well, and among the pastries numbered from 15 to 21, there are at least th... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,027 |
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