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11.4. The base of a right prism is a quadrilateral inscribed in a circle with a radius of $25 \mathrm{~cm}$. The areas of the lateral faces are in the ratio 7:15:20:24, and the length of the diagonal of the largest lateral face is 52 cm. Calculate the surface area of the prism. (7 points) | # Solution
The sides of the base are $7 \mathrm{x}, 15 \mathrm{x}, 20 \mathrm{x}, 24 \mathrm{x}$. Since $(7 \mathrm{x})^{2}+(24 \mathrm{x})^{2}=(15 \mathrm{x})^{2}+(20 \mathrm{x})^{2}$, two angles of the base are right angles. $(15 \mathrm{x})^{2}+(20 \mathrm{x})^{2}=50^{2}, \mathrm{x}=2$. The sides of the base are $1... | 4512\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,404 |
11.5. Solve the equation $2021 x^{2021}-2021+x=\sqrt[2021]{2022-2021 x}$. (7 points)
# | # Solution
The function $f(x)=2021x^{2021}-2021+x$ is increasing, while the function $g(x)=\sqrt[2021]{2022-2021x}$ is decreasing. Therefore, the equation $f(x)=g(x)$ has no more than one root. However, it is obvious that $f(1)=g(1)$.
Answer: $x=1$.
| criteria | points |
| :--- | :---: |
| correct solution | 7 |
| P... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,405 |
9.1. Four non-zero numbers are written on the board, and the sum of any three of them is less than the fourth number. What is the smallest number of negative numbers that can be written on the board? Justify your answer. | Solution: Let the numbers on the board be $a \geqslant b \geqslant c \geqslant d$. The condition of the problem is equivalent to the inequality $a+b+c < d$ for optimality | 7 points |
| There is a proof that there are no fewer than three negative numbers (in the absence of an example with three numbers) | 3 points |
| ... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 15,406 |
9.2. The quadratic trinomial $y=x^{2}+a x+b$ has two roots. Prove that if you add $k$ to the coefficient $a$ for any of these roots, and subtract the square of the same root from the coefficient $b$, then the resulting trinomial will also have at least one root.
# | # Solution:
Method 1. Let the roots of the original quadratic trinomial be $p$ and $s$. By Vieta's theorem, $a=-p-s, b=p s$. Suppose we add the number $p$ to the coefficient $a$, and subtract the number $p^{2}$ from the coefficient $b$. Then we get the trinomial
$$
x^{2}+(a+p) x+b-p^{2}=x^{2}-s x+p s-p^{2}
$$
Its di... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,407 |
9.3. A square is inscribed in a circle with center $O$. Let $M$ be the midpoint of its side, $X$ be an arbitrary point on the circle, and $T$ be the midpoint of segment $OX$. Prove that $\frac{M X}{M T}=\sqrt{2}$.
. Now $O(0 ; 0)$, if the radius of the circle is $R$, then point $M$ has coordinates $M\left(\frac{R}{2} ; \frac{R}{2... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,408 |
9.4. Vasya, a truant, skipped one lesson every Monday in September, two lessons every Tuesday, three on Wednesday, four on Thursday, and five on Friday. The exception was September 1: on this day, Vasya did not skip any lessons in honor of the start of the school year. It turned out that Vasya missed exactly 64 lessons... | Solution: Note that in any consecutive seven days, each day of the week occurs exactly once, so in any consecutive 7 days (starting from September 2), Vasya skips $1+2+3+4+5=15$ lessons. Therefore, from September 2 to September 29, he skips exactly $15 \cdot 4=60$ lessons, so on September 30, he skips $64-60=4$ lessons... | Wednesday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,409 |
9.5. At the exchange office, only the following operations can be performed:
1) exchange 2 gold coins for three silver coins and one copper coin;
2) exchange 5 silver coins for three gold coins and one copper coin.
Nikolai had only silver coins. After several visits to the exchange office, he had fewer silver coins, ... | Solution: As a result of each operation, Nikolai acquires exactly 1 copper coin, which means there were exactly 50 operations in total. Of these, some (let's say \(a\)) were of the first type, and the rest \(50-a\) were of the second type. On operations of the first type, Nikolai spent \(2a\) gold coins, and on operati... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,410 |
9.6. Certainly or infinitely many triples $(a, b, c)$ of integers for which the equation $a^{2}+b^{2}=2\left(c^{2}+1\right)$ holds? Justify the answer.
# | # Solution:
Method 1. $2\left(c^{2}+1\right)=(c-1)^{2}+(c+1)^{2}$, so we can set $a$ to be any integer, take $b$ as $a+2$, and $c$ as the number $a+1$. Thus, the set of such triples is infinite.
Method 2. We will prove that the set of natural numbers that can be represented as the sum of squares of two integers is cl... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,411 |
1. On an $8 \times 8$ chessboard, tokens are placed according to the following rule. Initially, the board is empty. A move consists of placing a token on any free square. With this move, exactly one of the tokens that ends up adjacent to it is removed from the board (if there is such an adjacent token). What is the max... | Answer: No more than 61 chips can be placed. | 61 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,412 |
2. $f(x)$ and $g(x)$ are quadratic trinomials, each with a leading coefficient of 1. It is known that the trinomial $h(x)=f(x)+g(x)$ has two distinct roots, and each of these roots is also a root of the equation $f(x)=g^{3}(x)+g^{2}(x)$. Prove that the trinomials $f(x)$ and $g(x)$ are equal. | Solution.
Let $x_{1}$ and $x_{2}$ be the roots of the quadratic polynomial $h(x)$. This means that $f\left(x_{1}\right)=-g\left(x_{1}\right), f\left(x_{2}\right)=$ $-g\left(x_{2}\right)$. From this, we obtain that
$$
\begin{gathered}
g^{3}\left(x_{1}\right)+g^{2}\left(x_{1}\right)+g\left(x_{1}\right)=0 \\
g\left(x_{1... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,413 |
3. The angle bisectors of angles $A$ and $C$ of triangle $ABC$ intersect the circumcircle of the triangle at points $E$ and $D$ respectively. Segment $DE$ intersects sides $AB$ and $BC$ at points $F$ and $G$ respectively. Let $I$ be the point of intersection of the angle bisectors of triangle $ABC$. Prove that quadrila... | Solution.
Arcs $B E$ and $E C$ are equal because they subtend equal angles $\angle B A E$ and $\angle E A C$. Therefore, angles $\angle B D E$ and $\angle E D C$ are equal. This means that $D... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,414 |
4. In a certain country, there are 47 cities. Each city has a bus station from which buses run to other cities in the country and possibly abroad. A traveler studied the schedule and determined the number of internal bus routes departing from each city. It turned out that if the city of Lake is not considered, then for... | Solution.
Note that external lines are not considered in this problem.
There are a total of 47 variants of the number of internal lines - from 0 to 46. Note that the existence of a city with 46 lines excludes the existence of a city with 0 lines and vice versa.
Suppose there is a city with 46 lines. Then the smalles... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,415 |
5. Find all natural numbers $n \geq 2$ such that $20^{n}+19^{n}$ is divisible by $20^{n-2}+19^{n-2}$. | Solution.
Consider the expression
$$
20^{n}+19^{n}-19^{2} \cdot\left(20^{n-2}+19^{n-2}\right)
$$
By the condition, it is divisible by $20^{n-2}+19^{n-2}$. On the other hand,
$$
20^{n}+19^{n}-19^{2} \cdot\left(20^{n-2}+19^{n-2}\right)=20^{n-2}\left(20^{2}-19^{2}\right)=20^{n-2} \cdot 39
$$
Note that $20^{n-2}$ and ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,416 |
8.1. The teacher suggested that his students - Kolya and Seryozha - solve the same number of problems during the lesson. After some time from the start of the lesson, it turned out that Kolya had solved a third of what Seryozha had left to solve, and Seryozha had left to solve half of what he had already completed. Ser... | # 8.1. Answer: 16.
Solution. Let Tanya have solved x problems, then she has $\frac{x}{2}$ problems left to solve.
Let $\mathrm{t}_{1}$ be the time interval after which Tanya and Kolya evaluated the shares of solved and remaining problems, and $\mathrm{t}_{2}$ be the remaining time. Since Tanya's problem-solving speed... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,421 |
8.2. At a new courtyard playground, an arm wrestling table was installed. Five boys organized a competition and wrestled each other. There are no ties in arm wrestling. After all the matches, it turned out that one of them had as many victories as defeats; another had 4 more victories than defeats, and each of the two ... | # 8.2. Answer. Could not.
Solution.
Suppose the last fifth participant won all the matches. In each match, one of the wrestlers wins, the other loses. Therefore, the number of wins should equal the number of losses. From the statements of the first four participants, it follows that they have 6 fewer wins than losses... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,422 |
8.3. The height AH of isosceles triangle $\mathrm{ABC}$ with base $\mathrm{AC}$ is equal to the length of the perpendicular MK dropped from the midpoint M of side AB to the base of the triangle. Find the perimeter of triangle $\mathrm{ABC}$, if $\mathrm{AK}=\mathrm{a}$. | # 8.3. Answer: 20a.
Solution.
Triangle $\mathrm{ABC}$ is isosceles, so $\angle \mathrm{MAK} = \angle \mathrm{ACH}$. Therefore, triangles $\mathrm{MAK}$ and $\mathrm{ACH}$ are congruent (rig... | 20a | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,423 |
8.5. A Lame Rook (this is a rook that can move only horizontally or only vertically exactly one square) has traversed a $10 \times 10$ board, visiting each square exactly once. In the first square the rook visited, we write the number 1, in the second square the number 2, in the third square the number 3, and so on up ... | # 8.5. Answer. Could not.
Solution.
Consider a $10 \times 10$ chessboard coloring. Notice that from a white cell, the lame rook moves to a black cell, and from a black cell, it moves to a white cell. Suppose the rook starts its tour from a white cell. Then 1 will be in a white cell, 2 in a black cell, 3 in a white ce... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,425 |
9.1. At a round table, 10 people are sitting, some of them are knights, and the rest are liars (knights always tell the truth, while liars always lie). It is known that among them, there is at least one knight and at least one liar. What is the maximum number of people sitting at the table who can say: "Both of my neig... | # Answer. 9.
Solution. Note that all 10 could not have said such a phrase. Since at the table there is both a knight and a liar, there will be a liar and a knight sitting next to each other. But then this knight does not have both neighbors as knights. If, however, at the table there are 9 liars and 1 knight, then eac... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,426 |
9.2. Let $a$ and $b$ be arbitrary different numbers. Prove that the equation $(x+a)(x+b)=2x+a+b$ has two distinct roots. | First solution. Move everything to the left side: $(x+a)(x+b)-(2x+a+b)=0$. Expanding the brackets and combining like terms, we get $x^{2}+(a+b-2)x+ab-a-b=0$. Let's calculate the discriminant of the resulting quadratic equation. It is equal to $D=(a+b-2)^{2}-4(ab-a-b)=a^{2}+b^{2}-2ab+4=(a-b)^{2}+4>0$. Therefore, the equ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,427 |
9.3. Let $A L$ be the bisector of an acute-angled triangle $A B C$, and
$\omega$ be the circumcircle of this triangle. Denote by $P$ the point of intersection of the extension of the altitude $B H$ of triangle $A B C$ with the circle $\omega$. Prove that if $\angle B L A=\angle B A C$, then $B P=C P$. | Solution. Let $\alpha=\angle B A L$. Then $\angle C A L=\alpha$, and, by the condition, $\angle B L A=2 \alpha$. Since $\angle B L A$ is the external angle in triangle $A L C$, we get $\angle A C L=\angle B L A-\angle C A L=\alpha$.
From the right triangle $B H C$ we now get $\angle C B H=$ $=90^{\circ}-\alpha$. Since... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,428 |
9.5. We will call a palindrome a natural number whose decimal representation reads the same from left to right as from right to left (the decimal representation cannot start with zero; for example, the number 1221 is a palindrome, while the numbers 1231, 1212, and 1010 are not). Which palindromes among the numbers from... | Answer. There are three times more palindromes with an even sum of digits.
Solution. Note that a 5-digit palindrome looks like $\overline{a b c b a}$, and a 6-digit palindrome looks like $\overline{a b c c b a}$. Each of these palindromes is uniquely determined by the first three digits $\overline{a b c}$. This means ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,430 |
10.1. Find all roots of the equation $(x-a)(x-b)=(x-c)(x-d)$, given that $a+d=b+c=2015$ and $a \neq c$ (the numbers $a, b, c, d$ are not given). | Answer: 1007.5.
First solution. Expanding the brackets, we get $a b-(a+b) x=$ $=c d-(c+d) x$. Substituting $d=2015-a$ and $b=2015-c$, we obtain $a(2015-c)-(a+2015-c) x=c(2015-a)-(c+2015-a) x$. Simplifying and combining like terms, we get $(a-c) 2015=2(a-c) x$. Considering that $a \neq c$, we get $2 x=2015$, which mean... | 1007.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,431 |
10.2. Vasya chose some number $x$ and wrote down the sequence $a_{1}=1+x^{2}+x^{3}, a_{2}=1+x^{3}+x^{4}, a_{3}=1+x^{4}+x^{5}, \ldots, a_{n}=$ $=1+x^{n+1}+x^{n+2}, \ldots$. It turned out that $a_{2}^{2}=a_{1} a_{3}$. Prove that for all $n \geqslant 3$ the equality $a_{n}^{2}=a_{n-1} a_{n+1}$ holds. | Solution. Write the equality $a_{2}^{2}=a_{1} a_{3}$. We have: $\left(1+x^{3}+\right.$ $\left.+x^{4}\right)^{2}=\left(1+x^{2}+x^{3}\right)\left(1+x^{4}+x^{5}\right)$. Expanding the brackets and combining like terms, we get $x^{4}+x^{3}=x^{5}+x^{2}$. Transforming: $0=x^{5}+x^{2}-x^{4}-x^{3}=x^{2}\left(x^{3}+1-x^{2}-x\ri... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,432 |
10.3. What is the minimum number of L-shaped corners consisting of 3 cells that need to be painted in a $5 \times 5$ square so that no more L-shaped corners can be painted? (Painted L-shaped corners should not overlap.) | Answer: 4.
Solution: Let the cells of a $5 \times 5$ square be painted in such a way that no more corners can be painted. Consider the 4 corners marked in Fig. 7. Since none of these corners can be painted, at least one cell in each of these corners must be painted. Note that one corner cannot paint cells of two marke... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,433 |
10.4. We will call a number greater than 25 semi-prime if it is the sum of some two different prime numbers. What is the maximum number of consecutive natural numbers that can be semi-prime | Answer: 5.
Solution: Note that an odd semiprime number can only be the sum of two and an odd prime number.
We will show that three consecutive odd numbers $2n+1$, $2n+3$, and $2n+5$, greater than 25, cannot all be semiprimes simultaneously. Assuming the contrary, we get that the numbers $2n-1$, $2n+1$, and $2n+3$ are... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,434 |
10.5. Let $A A_{1}$ and $C C_{1}$ be the altitudes of an acute, non-isosceles triangle $A B C$, and let $K, L$, and $M$ be the midpoints of sides $A B, B C$, and $C A$ respectively. Prove that if $\angle C_{1} M A_{1}=\angle A B C$, then $C_{1} K=A_{1} L$ | Solution. Segment $C_{1} M$ is the median of the right triangle $C C_{1} A$, therefore $C_{1} M = \frac{A C}{2} = M A$. Then $\angle C_{1} M A = \pi - 2 \angle B A C$. Similarly, $\angle A_{1} M C = \pi - 2 \angle B C A$. Therefore, $\angle C_{1} M A + \angle A_{1} M C = 2(\pi - \angle B A C - \angle B C A) = 2 \angle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,435 |
11.6. The sequence of numbers $a_{1}, a_{2}, \ldots, a_{2022}$ is such that $a_{n}-a_{k} \geqslant$ $\geqslant n^{3}-k^{3}$ for any $n$ and $k$ such that $1 \leqslant n \leqslant 2022$ and $1 \leqslant k \leqslant$ $\leqslant 2022$. Moreover, $a_{1011}=0$. What values can $a_{2022}$ take?
(N. Agakhanov) | Answer. $a_{2022}=2022^{3}-1011^{3}=7 \cdot 1011^{3}$.
Solution. Writing the condition for $n=2022, k=1011$ and for $n=1011, k=2022$, we get
$$
a_{2022}=a_{2022}-a_{1011} \geqslant 2022^{3}-1011^{3}
$$
and
$$
-a_{2022}=a_{1011}-a_{2022} \geqslant 1011^{3}-2022^{3}
$$
that is, $a_{2022} \geqslant 2022^{3}-1011^{3} ... | 2022^{3}-1011^{3} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 15,437 |
11.7. The product of the digits of a natural number $n$ is $x$, and the product of the digits of the number $n+1$ is $y$. Can it happen that the product of the digits of some natural number $m$ is $y-1$, and the product of the digits of the number $m+1$ is $x-1$? (A. Kuznetsov) | Answer: It cannot.
Solution: From the condition, it follows that $x, y \geqslant 1$, since the product of the digits of a natural number cannot be negative. Therefore, the numbers $n$ and $n+1$ do not contain zeros in their decimal representation. Then these numbers differ only by the last digit, and the last digit of... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,438 |
11.8. In the vertices of a regular 100-gon, 100 chips numbered $1, 2, \ldots, 100$ were placed, in exactly that order clockwise. In one move, it is allowed to swap two adjacent chips if their numbers differ by no more than $k$. For what smallest $k$ can a series of such moves result in a configuration where each chip i... | # Answer. 50.
Solution. Example. The chip 50 is sequentially exchanged 99 times with the next one counterclockwise. We get the required arrangement.
There are several ways to prove the estimate, below we provide two of them.
The first way. Suppose that for some $k<50$ the required arrangement is obtained.
At any mo... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,439 |
11.9. Trapezoid $ABCD$ with bases $AD$ and $BC$ is inscribed in circle $\omega$. Diagonals $AC$ and $BD$ intersect at point $P$. Point $M$ is the midpoint of segment $AB$. The perpendicular bisector of segment $AD$ intersects circle $\omega$ at points $K$ and $L$. Point $N$ is the midpoint of the arc $CD$ of the circum... | Solution. Let $O$ be the center of the circle circumscribed around trapezoid $ABCD$. Then $\angle COD = 2 \angle CAD = \angle PAD + \angle ADP = \angle CPD$. Here, we used the fact that the central angle is twice the inscribed angle, and that the exterior angle of a triangle is equal to the sum of the two non-adjacent ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,440 |
11.10. Given non-negative numbers $a, b, c, d$ such that $a+b+c+d=8$.
Prove that
$$
\frac{a^{3}}{a^{2}+b+c}+\frac{b^{3}}{b^{2}+c+d}+\frac{c^{3}}{c^{2}+d+a}+\frac{d^{3}}{d^{2}+a+b} \geqslant 4
$$
(A. Kuznetsov) | The first solution. Note that
$$
\frac{a^{3}}{a^{2}+b+c}=a-\frac{a(b+c)}{a^{2}+b+c} \geqslant a-\frac{a(b+c)}{2 a \sqrt{b+c}}=a-\frac{\sqrt{b+c}}{2}.
$$
Here we estimated the denominator using the inequality of means:
$$
a^{2}+b+c \geqslant 2 a \sqrt{b+c}
$$
Adding the obtained inequality with three similar ones. N... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,441 |
8.1. Petya made a mistake when writing a decimal fraction: he wrote the digits correctly, but moved the decimal point one position. As a result, he got a number that was 19.71 less than the required one. What number should Petya have written? | Answer: 21.9.
Solution. Since the number decreased as a result of the error, the decimal point must have been moved to the left. In doing so, the number was reduced by a factor of 10. Let the resulting number be \(x\), then the required number is 10x. According to the problem: \(10x - x = 19.71\), hence, \(x = 2.19, 1... | 21.9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,442 |
8.2. On the farmyard, there live six animals. A horse eats a bale of hay in 1.5 days, a bull - in 2 days, a cow - in 3 days, a calf - in 4 days, a sheep - in 6 days, and a goat - in 12 days. Explain how to divide these animals into two groups so that each group would have enough one bale of hay for the same amount of t... | Answer: There are two ways to divide the animals: 1) in one group - the horse and the cow, in the other - the bull, calf, sheep, and goat; 2) in one group - the horse, calf, and goat; in the other - the bull, cow, and sheep.
Solution. First method. Let's find what fraction of a haycock each animal eats per day: horse ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,443 |
8.3. It is known that $\frac{1}{3 a}+\frac{2}{3 b}=\frac{3}{a+2 b}$. Prove that $a=b$. | Solution. Transform the given equality by multiplying both sides by $3 a b(a+2 b)$. We get: $b(a+2 b)+2 a(a+2 b)=9 a b$. After expanding the brackets and combining like terms, the equation will take the form: $2 b^{2}+2 a^{2}-4 a b=0$. Consequently, $(a-b)^{2}=0$, from which $a=b$.
Evaluation Criteria.
«+» A complete... | b | Algebra | proof | Yes | Yes | olympiads | false | 15,444 |
8.4. Point $E$ is the midpoint of side $A B$ of parallelogram $A B C D$. On segment $D E$, there is a point $F$ such that $A D = B F$. Find the measure of angle $C F D$. | Answer: $90^{\circ}$.
Solution. Extend $D E$ to intersect line $B C$ at point $K$ (see Fig. 8.4). Since $B K \| A D$, then $\angle K B E = \angle D A E$. Moreover, $\angle K E B = \angle D E A$ and $A E = B E$, therefore, triangles $B K E$ and $A D E$ are equal. Thus, $B K = A D = B C$.
Therefore, in triangle $CFK$, ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,445 |
8.5. Kuzya cut a convex paper 67-gon along a straight line into two polygons, then similarly cut one of the two resulting polygons, then one of the three resulting ones, and so on. In the end, he got eight $n$-gons. Find all possible values of $n$. | Answer: $n=11$.
Solution. A straight-line cut can be of three types: from side to side, from a vertex to a side, and from a vertex to a vertex. Therefore, after one cut, the total number of sides of the polygons increases by 4, 3, or 2, respectively. Kuzya made 7 cuts, so the number of sides added is at least 14 but n... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,446 |
8.6. A triangle is divided into triangular cells as shown in the figure. A natural number is written in each cell. For each side of the triangle, there are four layers parallel to this side, containing seven, five, three, and one cell, respectively. It turns out that the sum of the numbers in each of these twelve layer... | Answer: 22.
Solution. Example. In each of the three corner cells, we write the number 3, and in each of the others, we write the number 1. Then the sum of the written numbers is $3 \cdot 3 + 13 \cdot 1 = 22$, and the sums of the numbers in the layers are: 11, 5, 3, and 3, respectively.
Estimation. Any corner cell is ... | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,447 |
Problem 4.1. On nine cards, the numbers from 1 to 9 are written (each one only once). These cards were laid out in a row such that there are no three consecutive cards with numbers in ascending order, and no three consecutive cards with numbers in descending order. Then, three cards were flipped over, as shown in the f... | Answer: The number 5 is written on card $A$, the number -2 on card $B$, and the number -9 on card $C$.
Solution. The missing numbers are $-2, 5$, and 9.
If the number 5 is on card $B$, then we get the consecutive numbers 3, 4, 5. If the number 5 is on card $C$, then we get the consecutive numbers 8, 7, 5. Therefore, ... | 5,2,9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,449 |
Task 4.2. Find the smallest number in which all digits are different, and the sum of all digits is 32. | Answer: 26789.
Solution. For a four-digit number with different digits, the maximum possible sum of digits is $9+8+7+6=30<32$, so the number we need must be at least five digits.
We will try to make the first digit as small as possible. It is clear that it is no less than 2. We will place 2 in the first position. The... | 26789 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,450 |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,451 |
Problem 4.4. Lev Alex decided to count the stripes on Marty the zebra (black and white stripes alternate). It turned out that there is one more black stripe than white ones. Alex also noticed that all white stripes are of the same width, while black stripes can be wide or narrow, and there are 7 more white stripes than... | # Answer: 8.
Solution. First, let's look only at the wide black stripes. There are 7 fewer of them than white ones. If we add the narrow black stripes to the wide black ones, we get all the black stripes, which are 1 more than the white ones. Therefore, to find the number of narrow black stripes, we need to first "com... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,452 |
Problem 4.5. Hooligan Dima laid out a structure in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the diagram.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire struct... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
![](https://cd... | 65 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,453 |
# 011011010011.
Information has emerged that this encoding has been deciphered, as a result of which the agent will have to use a backup encoding. In it, the letter A is replaced by 21, the letter B by 122, and the letter V by 1. What code will this message receive in the new encoding? | Answer: 211221121.
Solution. Let's look at the right end of the code. Two ones at the end can only come from the letter A: 011011010011.
Next, we look at the rightmost digit of those we haven't figured out yet, from which letter it came. The digit 0 at the end is only present in B: 011011010011.
Similarly, the previ... | 211221121 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,454 |
Problem 4.7. Schoolchildren Alexander, Boris, Sergey, Daria, and Elena attended choir classes from Monday to Friday. It is known that:
- in each of the five days, exactly three of the schoolchildren were present, and exactly two were absent
- no one was absent for two consecutive days and no one was present for three ... | Answer: Alexander, Boris, Daria.
Solution. Since no one was present for three consecutive days, the maximum number of days one can attend is 4. Four classes a week are possible only if Wednesday is skipped. Since it is not allowed to skip two days in a row, the minimum number of days one can attend is 2. Two classes a... | Alexer,Boris,Daria | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,455 |
Problem 4.8. The figure shows a road map between the houses of five children. The shortest distance by road from Asya to Galia is 12 km, from Galia to Borya - 10 km, from Asya to Borya 8 km, from Dasha to Galia - 15 km, from Vasya to Galia - 17 km. How many kilometers is the shortest distance by road from Dasha to Vasy... | Answer: 18.
Solution. Let's add the distances from Dasha to Gala and from Vasya to Gala: $15+17=32$. This will include the "main" road (from Dasha to Vasya) and twice the "branch" from it to Gala.
To the obtained sum, add the distance from Asey to Bory: $32+8=40$. Now all three branches (to Gala - twice) and the main... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,456 |
Task 5.1. In a certain month of a certain year, there are exactly 5 Fridays. At the same time, the first and last day of this month are not Fridays. What day of the week is the 12th day of the month? | Answer: Monday.
Solution. Let's consider all possible options for what day of the week the 1st of the month could be.
Suppose the first day of this month is Monday. Then the Fridays are the 5th, 12th, 19th, and 26th. There are only four - this does not fit us.
Suppose the first day of this month is Tuesday. Then the... | Monday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,457 |
Problem 5.2. Several people were seated around a round table such that the distances between adjacent people are the same. One of them was given a sign with the number 1, and then everyone was given signs with numbers 2, 3, and so on, in a clockwise direction.
The person with the sign numbered 31 noticed that the dist... | Answer: 41.
Solution. For such a situation to be possible, people from the 31st to the 14th need to be counted in a circle in the direction of decreasing numbers, and from the 31st to the 7th - in a circle in the direction of increasing numbers.
There are 16 people sitting between the 31st and the 14th. Therefore, th... | 41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,458 |
Problem 5.3. The figure shows a plan of the road system of a certain city. In this city, there are 8 straight streets, and 11 intersections are named with Latin letters $A, B, C, \ldots, J, K$.
Three police officers need to be placed at some intersections so that at least one police officer is on each of the 8 streets... | Answer: $B, G, H$.
Solution. The option will work if police officers are placed at intersections $B, G, H$. It can be shown that this is the only possible option.
Since there are only three vertical streets, and each must have one police officer, there are definitely no police officers at intersections $C$ and $K$. T... | B,G,H | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,459 |
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,460 |
Problem 5.5. Several schoolchildren have enrolled in math circles. They want to distribute them into groups evenly - so that the number of students in any two groups differs by no more than 1.
As a result of such even distribution, 6 groups were formed, among which exactly 4 groups have 13 students each. How many scho... | Answer: 76 or 80.
Solution. Since the number of students in the groups differs by no more than 1, the remaining two groups can have either 12 or 14 students each (it is clear that groups of 12 and 14 students cannot exist simultaneously). Therefore, the total number of students can be $13 \cdot 4 + 12 \cdot 2 = 76$ or... | 76or80 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,461 |
Problem 5.6. Several stones are arranged in 5 piles. It is known that
- in the fifth pile, there are six times more stones than in the third;
- in the second pile, there are twice as many stones as in the third and fifth combined;
- in the first pile, there are three times fewer stones than in the fifth, and ten fewer... | Answer: 60.
Solution. Let's denote the number of stones in the third pile as a rectangle. Then, since there are 6 times more stones in the fifth pile, this can be represented as 6 rectangles. In the second pile, the number of stones is twice the sum of the third and fifth piles, which is $(1+6) \cdot 2=14$ rectangles.... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,462 |
Problem 5.7. After the World Hockey Championship, three journalists wrote an article about the German national team - each for their own newspaper.
- The first wrote: “The German national team scored more than 10 but fewer than 17 goals throughout the championship.”
- The second: “The German national team scored more ... | Answer: $11,12,14,16$ or 17.
Solution. Let's consider all possible cases.
If no more than 10 goals were scored, then for an odd number of goals, one statement would be true, and for an even number, none would be true. This does not fit.
If 11 goals were scored, then exactly two statements are true. This fits.
If th... | 11,12,14,16,17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,463 |
Problem 5.8. Inside a large triangle with a perimeter of 120, several segments were drawn, dividing it into nine smaller triangles, as shown in the figure. It turned out that the perimeters of all nine small triangles are equal to each other. What can they be equal to? List all possible options.
The perimeter of a fig... | Answer: 40.
Solution. Let's add the perimeters of the six small triangles marked in gray in the following figure:
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,464 |
Problem 6.1. Five consecutive natural numbers are written in a row. The sum of the three smallest of them is 60. What is the sum of the three largest? | Answer: 66.
Solution. The fifth number is 4 more than the first, and the fourth is 2 more than the second. Then the sum of the three largest numbers is $2+4=6$ more than the sum of the three smallest, and it is equal to $60+6=66$. | 66 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,465 |
Problem 6.2. Arkady, Boris, Vera, Galia, Dan, and Egor formed a circle.
- Dan stood next to Vera, to her right,
- Galia stood opposite Egor
- Egor stood next to Dan,
- Arkady and Galia did not want to stand next to each other.
Who stands next to Boris | Answer: Arkady and Galya.
Solution. From the first and third statements, it follows that Vera, Dan, and Egor stand in exactly that order counterclockwise. Since there are six people in the circle, Galya stands to the left of Vera, opposite Egor. The remaining people are Arkady and Boris. Since Arkady and Galya do not ... | ArkadyGalya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,466 |
Problem 6.3. In the cells of a $4 \times 4$ table, the numbers $1,2,3,4$ are arranged such that
- each number appears in each row and each column;
- in all four parts shown in the figure, the sums of the numbers are equal.
Determine in which cells the twos are located based on the two numbers in the figure.
![](http... | Answer: In row $A$, the two is in column 2, in row $B-1$, in row $C-4$, in row $D-3$.
Solution. First, let's find what the sum of the numbers in each of the parts into which the board is divided is. In each column, the numbers from 1 to 4 appear once. Their sum is $1+2+3+4=10$, and the sum of all numbers on the board ... | A2,B1,C4,D3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,467 |
Problem 6.5. Several classmates ate a cake together. Lёsha ate the most - $\frac{1}{11}$ of the entire cake, and Alёna ate the least - $\frac{1}{14}$ of the entire cake. How many classmates ate the cake? List all possible options. | Answer: 12 or 13.
Solution. If everyone in the class had eaten $\frac{1}{11}$ of the cake, like Lёsha, there would be 11 people in the class. But since everyone except Lёsha ate less than $\frac{1}{11}$, there are more than 11 people, that is, at least 12.
Similarly, we can show that there are no more than 13 people.... | 12or13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,469 |
Problem 6.6. On an island, there live knights who always tell the truth, and liars who always lie. One day, 65 inhabitants of the island gathered for a meeting. Each of them, in turn, made the statement: "Among the previously made statements, there are exactly 20 fewer true statements than false ones." How many knights... | Answer: 23.
Solution. The first 20 people are liars, because before them, there were no more than 19 statements made, which means there cannot be exactly 20 fewer true statements than false ones.
Next, the 21st person tells the truth (since before him, there were exactly 20 false statements), so he is a knight. Then ... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,470 |
Problem 6.7. Anya places pebbles on the sand. First, she placed one stone, then added pebbles to form a pentagon, then made a larger outer pentagon with pebbles, then another outer pentagon, and so on, as shown in the picture. The number of stones she had arranged on the first four pictures: 1, 5, 12, and 22. If she co... | Answer: 145.
Solution. On the second picture, there are 5 stones. To get the third picture from it, you need to add three segments with three stones on each. The corner stones will be counted twice, so the total number of stones in the third picture will be $5+3 \cdot 3-2=12$.
$ candies. Since the total number of candies for both boys did not change after the game, we get $7 x=3(50-x)$. Solvin... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,473 |
Problem 7.2. On Teachers' Day, grateful students gave several railway tickets to Egor Sergeyevich so that he could travel across Russia.
The tickets were for travel between the following pairs of cities:
- Saint Petersburg and Tver,
- Yaroslavl and Nizhny Novgorod,
- Moscow and Kazan,
- Nizhny Novgorod and Kazan,
- M... | Answer: Saint Petersburg or Yaroslavl.
Fig. 1: to the solution of problem 7.2
Solution. All available routes are shown in Fig. 1. One road leads out of Saint Petersburg and Yaroslavl, so ... | SaintPetersburgorYaroslavl | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,474 |
Problem 7.3. Numbers were placed in the cells of a square such that the sums of the numbers in each vertical, horizontal, and each diagonal of three cells are equal. Then some numbers were hidden. What is the sum of the numbers in the two shaded cells?
| 16 | | |
| :--- | :--- | :--- |
| | | 10 |
| 8 | | 12 | | Answer: 34.
Solution. Let $a$ be the number in the top right cell, and $b$ be the number in the central cell. The sum of the diagonal containing the number 8 and the sum of the numbers in the right column have a common term $a$, so $8+b=10+12$, from which $b=14$. The sums of the three numbers along both diagonals have... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,475 |
Problem 7.4. Masha and Olya bought many identical pens for the new school year at the store. It is known that one pen costs an integer number of rubles, more than 10. Masha bought pens for exactly 357 rubles, and Olya - for exactly 441 rubles. How many pens did they buy in total? | Answer: 38.
Solution. Let the pen cost $r$ rubles. Then the numbers 357 and 441 are divisible by $d$. Since the greatest common divisor of the numbers $357=3 \cdot 7 \cdot 17$ and $441=3^{2} \cdot 7^{2}$ is $3 \cdot 7$, then 21 is also divisible by $r$. Since $r>10$, then $r=21$. Therefore, the total number of pens bo... | 38 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,476 |
Problem 7.5. In each room of the hotel, no more than 3 people can be accommodated. The hotel manager knows that a group of 100 football fans, who support three different teams, will soon arrive. In one room, only men or only women can be accommodated; also, fans of different teams cannot be accommodated together. How m... | Answer: 37.
Solution. All fans are divided into six groups: men/women, supporting the first team; men/women, supporting the second team; men/women, supporting the third team. Only people from the same group can be housed in one room. We divide the number of people in each of the six groups by 3 with a remainder. If th... | 37 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,477 |
Problem 7.6. In a class, there are 25 students, each of whom is either an excellent student or a troublemaker. Excellent students always tell the truth, while troublemakers always lie.
One day, 5 students of this class said: "If I transfer to another class, then among the remaining students, more than half will be tro... | Answer: 5 or 7.
Solution. Let's call the first group all the people who said the first phrase, and the second group all the people who said the second phrase. Consider two cases.
- Suppose there is at least one excellent student in the second group, choose any of them. Among the remaining 24 people, the number of tro... | 5or7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,478 |
Problem 7.7. The numbers from 1 to 200 were arranged in a random order on a circle such that the distances between adjacent numbers on the circle are the same.
For any number, the following is true: if you consider 99 numbers standing clockwise from it and 99 numbers standing counterclockwise from it, there will be an... | Answer: 114.
Solution. Consider the number 2. Less than it is only the number 1. Since it is unique, it cannot be in any of the groups relative to the number 2. Therefore, 1 must be opposite 2.
Consider the number 4. The numbers less than it are three in number - an odd count. This means that one of them should not b... | 114 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,479 |
Problem 7.8. On a rectangular sheet of paper, a picture in the shape of a "cross" was drawn from two rectangles $A B C D$ and $E F G H$, the sides of which are parallel to the edges of the sheet. It is known that $A B=9, B C=5, E F=3, F G=10$. Find the area of the quadrilateral $A F C H$.
![](https://cdn.mathpix.com/c... | Answer: $52.5$.
Solution. The intersection of the two original rectangles forms a "small" rectangle with sides 5 and 3. Its area is 15.
Extend the segments $D A, G H, B C, E F$ to form lines. They form a "large" rectangle with sides 9 and 10, containing the "cross" (Fig. 2). Its area is 90.
![](https://cdn.mathpix.c... | 52.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,480 |
Problem 8.2. In the chat of students from one of the schools, a vote was held: "On which day to hold the disco: October 22 or October 29?"
The graph shows how the votes were distributed an hour after the start of the voting.
Then, 80 more people participated in the voting, voting only for October 22. After that, the ... | Answer: 260.
Solution. Let $x$ be the number of people who voted an hour after the start. From the left chart, it is clear that $0.35 x$ people voted for October 22, and $-0.65 x$ people voted for October 29.
In total, $x+80$ people voted, of which $45\%$ voted for October 29. Since there are still $0.65 x$ of them, ... | 260 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,482 |
Problem 8.3. In the class, there are 29 students: some are excellent students and some are troublemakers. Excellent students always tell the truth, while troublemakers always lie.
All students in this class sat around a round table.
- Several students said: “There is exactly one troublemaker next to me.”
- All other ... | Answer: 10.
Solution. If there were three straight-A students in a row, the middle one would definitely lie. Therefore, among any three consecutive people, there must be at least one troublemaker.
Choose an arbitrary troublemaker. Assign this person the number 29, and number all the people following them clockwise fr... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,483 |
Problem 8.4. Points $D$ and $E$ are marked on sides $A C$ and $B C$ of triangle $A B C$ respectively, such that $A D=E C$. It turns out that $B D=E D, \angle B D C=\angle D E B$. Find the length of segment $A C$, given that $A B=7$ and $B E=2$. | Answer: 12.
Solution. Note that triangles $D E C$ and $B D A$ are equal. Indeed, $D E=B D, E C=$ $D A$ and $\angle D E C=180^{\circ}-\angle B E D=180^{\circ}-\angle B D C=\angle B D A$. From this, it follows that $D C=A B=7$ and $\angle D C E=\angle B A D$ (Fig. 3). From the equality of these angles, it follows that t... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,484 |
Problem 8.5. The numbers $1, 2, 3, \ldots, 235$ were written on a board. Petya erased some of them. It turned out that among the remaining numbers, no number is divisible by the difference of any two others. What is the maximum number of numbers that could remain on the board? | Answer: 118.
Solution. 118 odd numbers could remain on the board: none of them is divisible by the difference of any two others, because this difference is even.
Suppose at least 119 numbers could remain. Consider 118 sets: 117 pairs $(1,2)$, $(3,4),(5,6), \ldots,(233,234)$ and one number 235. By the Pigeonhole Princ... | 118 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,485 |
Problem 8.6. In a $3 \times 3$ table, real numbers are arranged. It turns out that the product of the numbers in any row and any column is 10, and the product of the numbers in any $2 \times 2$ square is 3. Find the number in the central cell. | Answer: $0.00081$.
## Solution.
Let's denote the numbers in the square from left to right: let $a, b, c$ be the numbers in the first row, $d, e, f$ in the second row, and $g, h, i$ in the third row. Notice that
$$
e=\frac{a b d e \cdot b c e f \cdot d e g h \cdot e f h i}{a b c \cdot \operatorname{def} \cdot g h i \... | 0.00081 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,486 |
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,487 |
Problem 8.8. In how many ways can all natural numbers from 1 to 200 be painted in red and blue so that the sum of any two different numbers of the same color is never equal to a power of two? | Answer: 256.
## Solution.
Lemma. If the sum of two natural numbers is a power of two, then the powers of 2 in the prime factorizations of these two numbers are the same.
Proof of the lemma. Let the sum of the numbers $a=2^{\alpha} \cdot a_{1}$ and $b=2^{\beta} \cdot b_{1}$, where $a_{1}$ and $b_{1}$ are odd, be $2^{... | 256 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,488 |
Problem 9.1. On an island, there live red, yellow, green, and blue chameleons.
- On a cloudy day, either one red chameleon changes its color to yellow, or one green chameleon changes its color to blue.
- On a sunny day, either one red chameleon changes its color to green, or one yellow chameleon changes its color to b... | Answer: 11.
Solution. Let $A$ be the number of green chameleons on the island, and $B$ be the number of yellow chameleons. Consider the quantity $A-B$. Note that each cloudy day it decreases by 1, and each sunny day it increases by 1. Since there were $18-12=6$ more sunny days than cloudy days in September, the quanti... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,489 |
Problem 9.2. Denis has cards with numbers from 1 to 50. How many ways are there to choose two cards such that the difference between the numbers on the cards is 11, and the product is divisible by 5?
The order of the selected cards does not matter: for example, the way to choose cards with numbers 5 and 16, and the wa... | Answer: 15.
Solution. For the product to be divisible by 5, it is necessary and sufficient that one of the factors is divisible by 5. Let $n$ be the chosen number divisible by 5, then its pair should be $n-11$ or $n+11$, and both chosen numbers must be natural numbers. Clearly, for $n=5,10,40,45,50$ there is only one ... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,490 |
Problem 9.3. Traders Andrey and Boris bought 60 bags of potatoes each from the same farmer. All the bags cost the same.
Andrey sold all his bags, increasing their price by $100 \%$. Boris first increased the price by $60 \%$, and after selling 15 bags, increased the price by another $40 \%$ and sold the remaining 45 b... | Answer: 250.
Solution. Let the bag cost the farmer $x$ rubles. Andrei and Boris spent the same amount on buying 60 bags.
From the condition, it follows that Andrei sold 60 bags for $2 x$ rubles each, i.e., he received $60 \cdot 2 x$ rubles. Boris, on the other hand, sold 15 bags at a price of $1.6 x$ rubles and then ... | 250 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,491 |
Problem 9.4. A line $\ell$ is drawn through vertex $A$ of rectangle $ABCD$, as shown in the figure. Perpendiculars $BX$ and $DY$ are dropped from points $B$ and $D$ to line $\ell$. Find the length of segment $XY$, given that $BX=4$, $DY=10$, and $BC=2AB$.
Fig. 5: to the solution of problem 9.4
Solution. Note that since $\angle Y A D=90^{\circ}-\angle X A B$ (Fig. 5), right triangles $X A B$ and $Y D A$ are similar by the acute an... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,492 |
Problem 9.5. Leonid has a white checkered rectangle. First, he painted every other column gray, starting with the leftmost one, and then every other row, starting with the topmost one. All cells adjacent to the border of the rectangle ended up being painted.
How many painted cells could there be in the rectangle if 74... | Answer: 301 or 373.
Solution. From the condition, it follows that the rectangle has an odd number of both rows and columns. Let's number the rows from top to bottom with the numbers $1,2, \ldots, 2 k+1$, and the columns from left to right with the numbers $1,2, \ldots, 2 l+1$ (for non-negative integers $k$ and $l$). W... | 301or373 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,493 |
Problem 9.6. In triangle $ABC$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $AC$, point $K$ is marked, and on side $BC$, points $L$ and $M$ are marked such that $KL=KM$ (point $L$ lies on segment $BM$).
Find the length of segment $LM$, if it is known that $AK=4$, $BL=31$, and $MC=3$.
... | Answer: 14.
Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). Since this height is also a median, then $M H=H L=... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,494 |
Problem 9.7. In a school chess tournament, 4 people participated: Andrey, Vanya, Dima, and Sasha. Each played twice against each of their opponents. In each game, 1 point was awarded for a win, 0.5 points for a draw, and 0 points for a loss.
It is known that at the end of the tournament:
- all the participants scored... | Answer: Andrey scored 4 points, Vanya - 2.5, Dima - 3.5, Sasha - 2.
Solution. In each game, two participants together receive 1 point. It is clear that there were a total of $6 \cdot 2=12$ games, so the total points accumulated by all participants is exactly 12. It is also clear that the final number of points for eac... | Andrey:4,Vanya:2.5,Dima:3.5,Sasha:2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,495 |
Problem 9.8. Integers $n$ and $m$ satisfy the inequalities $3 n - m \leq 26$, $3 m - 2 n < 46$. What can $2 n + m$ be? List all possible options. | Answer: 36.
Solution. Since the numbers $m$ and $n$ are integers, the values of the expressions in the condition $3 n-m$, $n+m$, $3 m-2 n$ are also integers. Then
$$
\left\{\begin{array}{l}
3 n-m \leqslant 4 \\
n+m \geqslant 27 \\
3 m-2 n \leqslant 45
\end{array}\right.
$$
By adding three times the first inequality ... | 36 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 15,496 |
Task 10.1. Find the largest 12-digit number $N$ that satisfies the following two conditions:
- In the decimal representation of the number $N$, there are six digits «4» and six digits «7»;
- In the decimal representation of the number $N$, no four consecutive digits form the number «7444». | Answer: 777744744744.
Solution. It is clear that the number 777744744744 meets the condition of the problem.
Suppose there exists a larger number, then it must start with four sevens. In this number, we select the largest number of consecutive fours, let there be $k$ of them. If $k \geqslant 3$, then by selecting amo... | 777744744744 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,497 |
Problem 10.2. Points $A, B, C, D, E, F, G$ are located clockwise on a circle, as shown in the figure. It is known that $A E$ is the diameter of the circle. Also, it is known that $\angle A B F=81^{\circ}, \angle E D G=76^{\circ}$. How many degrees does the angle $F C G$ measure?
. Since $\angle AHK=90^{\circ}$, segment $AK$ is the diameter of this circle, and $\angle AXK = \angle AYK = 90^{\circ}$. Therefore, $AXKY$ is a rectangle, and $XK = AY = 6$.
=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,503 |
Problem 10.8. For what least natural $a$ does the numerical interval ( $a, 3 a$ ) contain exactly 50 perfect squares? | Answer: 4486.
Solution. Let's choose the largest natural $n$ such that $n^{2} \leqslant a$. Then
$$
n^{2} \leqslant a\frac{49}{\sqrt{3}-1}=\frac{49(\sqrt{3}+1)}{2}>\frac{49(1.7+1)}{2}>\frac{132}{2}=66
$$
From the fact that $n+1>66$, it follows that $n \geqslant 66$ and $a \geqslant 66^{2}$.
If on the corresponding ... | 4486 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,504 |
Task 11.1. The product of nine consecutive natural numbers is divisible by 1111. What is the smallest possible value that the arithmetic mean of these nine numbers can take? | Answer: 97.
Solution. Let these nine numbers be $-n, n+1, \ldots, n+8$ for some natural number $n$. It is clear that their arithmetic mean is $n+4$.
For the product to be divisible by $1111=11 \cdot 101$, it is necessary and sufficient that at least one of the factors is divisible by 11, and at least one of the facto... | 97 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,505 |
Problem 11.2. A square was cut into five rectangles of equal area, as shown in the figure. The width of one of the rectangles is 5. Find the area of the square.
 | Answer: 400.
Solution. The central rectangle and the rectangle below it have a common horizontal side, and their areas are equal. Therefore, the vertical sides of these rectangles are equal, let's denote them as $x$ (Fig. 13). The vertical side of the lower left rectangle is $2x$, and we will denote its horizontal sid... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,506 |
Problem 11.3. In a football tournament, 15 teams participated, each playing against each other exactly once. For a win, 3 points were awarded, for a draw - 1 point, and for a loss - 0 points.
After the tournament ended, it turned out that some 6 teams scored at least $N$ points each. What is the greatest integer value... | # Answer: 34.
Solution. Let's call these 6 teams successful, and the remaining 9 teams unsuccessful. We will call a game between two successful teams an internal game, and a game between a successful and an unsuccessful team an external game.
First, note that for each game, the participating teams collectively earn n... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,507 |
Problem 11.4. Given a regular pentagon $A B C D E$. On side $A E$ there is a point $K$, and on side $C D$ there is a point $L$. It is known that $\angle L A E + \angle K C D = 108^{\circ}$, and $A K: K E = 3: 7$. Find $C L: A B$.
A regular pentagon is a pentagon in which all sides are equal and all angles are equal.
... | # Answer: 0.7.
Solution. Let's start by formulating the known facts that we will use in the solution.
- The sum of the angles of an $n$-sided polygon is $180^{\circ}(n-2)$. In particular, each angle of a regular pentagon is $\frac{540^{\circ}}{5}=108^{\circ}$.
- All five diagonals of a regular pentagon are equal. Eac... | 0.7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,508 |
Problem 11.5. A two-digit number is written on the board. Nезнayka claimed that it is divisible by $3, 4, 5, 9, 10, 15, 18, 30$. Upon hearing this, Znayka saddened Nезнayka by saying that he was wrong exactly 4 times. What number could have been written on the board? List all possible options. | Answer: 36, 45 or 72.
Solution. Let $N$ be the two-digit number written on the board. If $N$ were divisible by 30, it would also be divisible by $3, 5, 10, 15$, so Neznaika would have made no more than 3 mistakes, a contradiction. If $N$ were not divisible by 3, it would also not be divisible by $9, 15, 18, 30$, so Ne... | 36,45,72 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,509 |
Problem 11.6. The quadratic trinomial $P(x)$ is such that $P(P(x))=x^{4}-2 x^{3}+4 x^{2}-3 x+4$. What can $P(8)$ be? List all possible options. | Answer: 58.
Solution. Let $P(x)=a x^{2}+b x+c$. Then $P(P(x))=a\left(a x^{2}+b x+c\right)^{2}+b\left(a x^{2}+b x+c\right)+c=$
$a\left(a^{2} x^{4}+2 a b x^{3}+\left(b^{2}+2 a c\right) x^{2}+2 b c x+c^{2}\right)+b\left(a x^{2}+b x+c\right)+c$. Therefore,
\[
\begin{aligned}
& x^{4}-2 x^{3}+4 x^{2}-3 x+4=P(P(x))= \\
& =a... | 58 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,510 |
Problem 11.7. In a country, there are 110 cities. Between any two of them, there is either a road or there is not.
A motorist was in a certain city, from which there was exactly one road. After driving along this road, he arrived at a second city, from which there were already exactly two roads. Driving along one of t... | Answer: 107.
Solution. Let's number the cities in the order they are visited by the motorist: $1,2,3, \ldots, N$.
Suppose $N \geqslant 108$. From city 1, there is a road only to city 2, so from city 108, all roads lead to all 108 cities, except for 1 and 108. But then from city 2, there are at least three roads: to c... | 107 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,511 |
Problem 11.8. Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. A point $X$ is chosen on the edge $A_{1} D_{1}$, and a point $Y$ is chosen on the edge $B C$. It is known that $A_{1} X=5, B Y=3, B_{1} C_{1}=14$. The plane $C_{1} X Y$ intersects the ray $D A$ at point $Z$. Find $D Z$.
![](https://cdn.mathpix.com... | Answer: 20.
Solution. Lines $C_{1} Y$ and $Z X$ lie in parallel planes $B B_{1} C_{1} C$ and $A A_{1} D_{1} D$, so they do not intersect. Since these two lines also lie in the same plane $C_{1} X Z Y$, they are parallel. Similarly, lines $Y Z$ and $C_{1} X$ are parallel. Therefore, quadrilateral $C_{1} X Z Y$ is a par... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,512 |
6.2. On the plate, there were 15 doughnuts. Karlson took three times more doughnuts than Little Man, and Little Man's dog Bibbo took three times fewer than Little Man. How many doughnuts are left on the plate? Explain your answer. | Answer: 2 doughnuts are left.
From the condition of the problem, it follows that Little One took three times as many doughnuts as Bimbo, and Karlson took three times as many doughnuts as Little One. We can reason in different ways from here.
First method. If Bimbo took one doughnut, then Little One took three doughnu... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,514 |
6.3. A row of numbers and asterisks is written on the board: $5, *, *, *, *, *, *, 8$. Replace the asterisks with numbers so that the sum of any three consecutive numbers equals 20. | Answer: $5,8,7,5,8,7,5,8$.
The sum of the first, second, and third numbers should be equal to 20, and the sum of the second, third, and fourth numbers should also be equal to 20. Therefore, the fourth number must be equal to the first, which is five. Similarly, the seventh number must be equal to the fourth, so the se... | 5,8,7,5,8,7,5,8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,515 |
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