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4. In city $\mathrm{N}$, only blondes and brunettes live. Blondes always lie, while brunettes always tell the truth. Every day, the residents dye their hair the opposite color. On one Monday in October, everyone born in the fall was asked: "Were you born this month?" - and 200 residents answered "yes," while no one ans... | 4. Answer. 0.
Solution. Note that on Monday and Friday of the same week, the residents have the same hair color: Monday - Tuesday - Wednesday - Thursday - Friday. Then, if the same people answer differently on Monday and Friday, it means that the month has changed, and the brunettes will answer "no" to the same questi... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,176 |
5. A tournament of dodgeball was held at school. In each game, two teams competed. 15 points were awarded for a win, 11 for a draw, and no points for a loss. Each team played against each other once. By the end of the tournament, the total number of points scored was 1151. How many teams were there? | 5. Answer: 12 teams.
Solution. Let there be $\mathrm{N}$ teams. Then the number of games was $\mathrm{N}(\mathrm{N}-1) / 2$. For each game, a total of 15 or 22 points are scored. Therefore, the number of games was no less than 53 $(1151 / 22)$ and no more than $76(1151 / 15)$.
Note that if there were no more than 10 ... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,177 |
Problem 11.1. In a basket, there are 41 apples: 10 green, 13 yellow, and 18 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What is... | Answer: (a) 13 apples. (b) 39 apples.
Solution. (a) Alyona can take all 13 yellow apples from the basket, for example, if the first 13 apples turn out to be yellow. Since at no point do the yellow apples become fewer than the red ones, Alyona can indeed do this.
(b) We will show that Alyona can take all the apples ex... | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,178 |
Problem 11.2. At the zoo, oranges, bananas, and coconuts were brought for feeding three monkeys, and there were equal amounts of all three types of fruit. The first monkey was fed only oranges and bananas, with bananas being $40 \%$ more than oranges. The second monkey was fed only bananas and coconuts, with coconuts b... | Answer: $1 / 2$.
Solution. Let the first monkey eat $x$ oranges, the second monkey eat $y$ coconuts, and the third monkey eat $z$ coconuts. Then the first monkey ate $1.4 x$ bananas; the second monkey ate $1.25 y$ coconuts; the third monkey ate $2 z$ oranges. Since all the fruits were eaten equally, we have
$$
x+2 z=... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,179 |
Problem 11.3. A polynomial $G(x)$ with real coefficients takes the value 2022 at exactly five different points $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. It is known that the graph of the function $y=G(x)$ is symmetric with respect to the line $x=-8$.
(a) (2 points) Find $x_{1}+x_{3}+x_{5}$.
(b) (2 points) What is the smallest... | Answer: (a) -24. (b) 6.
Solution. (a) Since the graph of $G(x)$ is symmetric with respect to the line $x=-8$, the points at which it takes the same value must be divided into pairs of symmetric points, except possibly one point that lies on the axis of symmetry. Therefore, the middle of the five given points should li... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,180 |
Problem 11.4. On a horizontal floor, there are three volleyball balls with a radius of 18, each touching the other two. Above them, a tennis ball with a radius of 6 is placed, touching all three volleyball balls. Find the distance from the top point of the tennis ball to the floor. (All balls are spherical.)
Fig. 9: to the solution of problem 11.4
Solution. Let the centers of the volleyballs be denoted by $A, B, C$, and the center of the tennis ball by $D$; the top point of the tenn... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,181 |
Problem 11.6. Point $M$ is the midpoint of the base $B C$ of trapezoid $A B C D$. A point $P$ is chosen on the base $A D$. Ray $P M$ intersects ray $D C$ at point $Q$. The perpendicular to base $A D$, drawn through point $P$, intersects segment $B Q$ at point $K$.
It is known that $\angle K Q D=64^{\circ}$ and $\angle... | Answer: $39^{2}$.
Fig. 11: to the solution of problem 11.6
Solution. From the condition, it follows that $\angle B K D=\angle K Q D+\angle K D Q=64^{\circ}+38^{\circ}=102^{\circ}$.
Let the... | 39 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,183 |
Problem 11.7. Let $N$ be the least common multiple of ten different natural numbers $a_{1}<a_{2}<a_{3}<\ldots<a_{10}$.
(a) (2 points) What is the smallest value that $N / a_{3}$ can take?
(b) (2 points) Identify all possible values of $a_{3}$ in the interval $[1 ; 1000]$, for which the value of $N / a_{1}$ can take i... | Answer: (a) 8. (b) $315,630,945$.
Solution. (a) Note that $N$ is divisible by any of the numbers $a_{1}, a_{2}, \ldots, a_{10}$. This means that the numbers $a_{1}, a_{2}, \ldots, a_{10}$ can be represented as $a_{1}=N / b_{1}, a_{2}=N / b_{2}, \ldots, a_{10}=N / b_{10}$, where $b_{1}, b_{2}, \ldots, b_{10}$ are diffe... | 315,630,945 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,184 |
Problem 11.8. Real numbers $a, b, c$ are such that
$$
\left\{\begin{array}{l}
a^{2}+a b+b^{2}=11 \\
b^{2}+b c+c^{2}=11
\end{array}\right.
$$
(a) (1 point) What is the smallest value that the expression $c^{2}+c a+a^{2}$ can take?
(b) (3 points) What is the largest value that the expression $c^{2}+c a+a^{2}$ can take... | Answer: (a) 0; (b) 44.
Solution. (a) Note that if $a=c=0$, and $b=\sqrt{11}$, then the given system of equalities is satisfied, and $c^{2}+c a+a^{2}$ equals 0. On the other hand,
$$
c^{2}+c a+a^{2}=\frac{c^{2}}{2}+\frac{a^{2}}{2}+\frac{(c+a)^{2}}{2} \geqslant 0
$$
Thus, the smallest value of the expression $c^{2}+c ... | 44 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,185 |
Variant 11.1.1. In a basket, there are 41 apples: 10 green, 13 yellow, and 18 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What ... | Answer: (a) 13 apples. (b) 39 apples. | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,186 |
Variant 11.1.2. In a basket, there are 38 apples: 9 green, 12 yellow, and 17 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What i... | Answer: (a) 12 apples. (b) 36 apples. | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,187 |
Variant 11.1.3. In a basket, there are 35 apples: 8 green, 11 yellow, and 16 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What i... | Answer: (a) 11 apples. (b) 33 apples. | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,188 |
Variant 11.1.4. In a basket, there are 44 apples: 11 green, 14 yellow, and 19 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What ... | Answer: (a) 14 apples. (b) 42 apples. | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,189 |
Variant 11.3.1. The polynomial $G(x)$ with real coefficients takes the value 2022 at exactly five different points $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. It is known that the graph of the function $y=G(x)$ is symmetric with respect to the line $x=-8$.
(a) (2 points) Find $x_{1}+x_{3}+x_{5}$.
(b) (2 points) What is the smal... | Answer: (a) -24 . (b) 6. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,190 |
Variant 11.3.3. The polynomial $G(x)$ with real coefficients takes the value 2022 at exactly five different points $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. It is known that the graph of the function $y=G(x)$ is symmetric with respect to the line $x=-7$.
(a) (2 points) Find $x_{1}+x_{3}+x_{5}$.
(b) (2 points) What is the smal... | Answer: (a) -21 . (b) 6. | -21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,191 |
Variant 11.3.4. The polynomial $G(x)$ with real coefficients takes the value 2022 at exactly five different points $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. It is known that the graph of the function $y=G(x)$ is symmetric with respect to the line $x=-6$.
(a) (2 points) Find $x_{1}+x_{3}+x_{5}$.
(b) (2 points) What is the smal... | Answer: (a) -18 . (b) 6. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,192 |
Variant 11.7.2. Let $N$ be the least common multiple of ten different natural numbers $a_{1}<a_{2}<a_{3}<\ldots<a_{10}$.
(a) (2 points) What is the smallest value that $N / a_{4}$ can take?
(b) (2 points) List all possible values of $a_{4}$ in the interval $[1 ; 1300]$, for which the value of $N / a_{1}$ can take its... | Answer: (a) 7. (b) $360, 720, 1080$. | 360,720,1080 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,194 |
Variant 11.7.3. Let $N$ be the least common multiple of ten different natural numbers $a_{1}<a_{2}<a_{3}<\ldots<a_{10}$.
(a) (2 points) What is the smallest value that $N / a_{5}$ can take?
(b) (2 points) List all possible values of $a_{5}$ in the interval $[1 ; 1500]$, for which the value of $N / a_{1}$ can take its... | Answer: (a) 6. (b) 420, 840, 1260. | 6,420,840,1260 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,195 |
Variant 11.7.4. Let $N$ be the least common multiple of ten different natural numbers $a_{1}<a_{2}<a_{3}<\ldots<a_{10}$.
(a) (2 points) What is the smallest value that $N / a_{6}$ can take?
(b) (2 points) Identify all possible values of $a_{6}$ in the interval $[1 ; 2000]$, for which the value of $N / a_{1}$ can take... | Answer: (a) 5. (b) $504,1008,1512$. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,196 |
1. For which $p$ is one of the roots of the equation $x^{2}-p x+p=0$ the square of the other? (We assume that the roots of the equation are distinct) | Answer: $2 \pm \sqrt{5}$.
Solution. Let the roots be $a$ and $a^{2}$. By Vieta's theorem, $a + a^{2} = p$, and $a \cdot a^{2} = p$. Therefore, $a + a^{2} = a^{3}$. The root $a = 0$ is not valid because the roots of the equation $x^{2} = 0$ are identical. Considering $a \neq 0$, we get $a^{2} - a - 1 = 0$, the roots of... | 2\\sqrt{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,198 |
2. Petya found the sum of all odd divisors of some even natural number $n$, and Vasya - the sum of all its even divisors. Can the product of their results be equal to 2016? If so, find all such numbers $n$. | Answer: 88 and 192.
Solution. Let $n=m \cdot 2^{k}$ be the initial even number, and $m$ be its odd factor. The sum of the odd divisors of the number $n$ coincides with the sum $s$ of all divisors of the number $m$, and thus, Petya will get the number $s$. The sum of all even divisors of $n$ consists of the sum of divi... | 88192 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,199 |
3. In a rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, the sides $A B=2, A C=$ 3, $A A_{1}=$ 4. Find the area of the section $A M K$, where $M$ is the midpoint of $B B_{1}$ and $K$ is the midpoint of $D D_{1}$. | Answer: $2 \sqrt{22}$
Solution. Let's construct the desired section (see figure). Point $P$ is the intersection of $A M$ and $A_{1} B_{1}$. Due to the position of point $M$, $A_{1} P$ is twice as large as $A_{1} B_{1}$. Similarly, we construct point $Q$, where $A_{1} Q = 2 A_{1} D_{1}$. Therefore, $P Q$ is parallel to... | 2\sqrt{22} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,200 |
4. Let $x_{1}, x_{2}, \ldots, x_{100}$ be some numbers belonging to the interval [0; 1]. Is it true that there exists a number $x$ in this interval such that $\left|x-x_{1}\right|+\left|x-x_{2}\right|+\ldots+\mid x-$ $x_{100} \mid=50 ?$ | Answer. Correct.
Solution. Consider the function $f(x)=\left|x-x_{1}\right|+\left|x-x_{2}\right|+\ldots+\left|x-x_{100}\right|-$ 50, continuous on the interval [0;1]. We have $f(0)=x_{1}+x_{2}+\ldots+x_{100}-50, f(1)=$ $\left(1-x_{1}\right)+\left(1-x_{2}\right)+\ldots+\left(1-x_{100}\right)-50$. Therefore, $f(0)+f(1)=... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,201 |
5. On a $10 \times 10$ board, there are 10 non-attacking rooks. Can the remaining cells of the board be tiled with dominoes? (A domino is a rectangle of size $1 \times 2$ or $2 \times 1$). | Answer: No.
Solution: For each rook numbered $i$, let $s_{i}$ denote the sum of the numbers of the row and column in which it stands. Since the rooks do not attack each other, each row and column number appears exactly once. The sum of all numbers $s_{i}$, regardless of the arrangement, will be even and equal to
$$
\... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,202 |
1. $\left(225-4209520: \frac{1000795+(250+x) \cdot 50}{27}\right)=113$
4209520: $\frac{1000795+(250+x) \cdot 50}{27}=112$
$\frac{1000795+(250+x) \cdot 50}{27}=37585$
$1000795+(250+x) \cdot 50=1014795$
$250+x=280$
$x=30$ | Answer: $\boldsymbol{x}=30$.
Criteria. Correct solution - 7 points. $3 a$ for each computational error minus 1 point. Error in the algorithm for solving the equation - 0 points | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,203 |
3. 11088 is $110\%$ of the containers in December compared to November. This means that in November, $11088: 1.10=10080$ containers were manufactured, which is $105\%$ compared to October. This means that in October, $10080: 1.05=9600$ containers were manufactured, which is $120\%$ compared to September. Therefore, in ... | Answer: in September 8000, in October 9600, in November 10080 containers. | 8000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,205 |
1. In one lyceum, $76 \%$ of the students have at least once not done their homework, and $\frac{5}{37}$ sometimes forget their second pair of shoes. Find the number of students in the lyceum, if it is more than 1000 but less than 2000. | Solution. Since $76 \%=\frac{76}{100}=\frac{19}{25}$, and the numbers 25 and 37 are coprime, the number of students is a multiple of $25 \cdot 37$, i.e., $925 \mathrm{k}$, where k is a natural number. Since $1000<925 k<2000$, then $\mathrm{k}=2$, and the number of students $925 \cdot 2$ $=1850$.
Answer. 1850
Recommen... | 1850 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,206 |
2. Denis housed chameleons that can change color only to two colors: red and brown. Initially, the number of red chameleons was five times the number of brown chameleons. After two brown chameleons turned red, the number of red chameleons became eight times the number of brown chameleons. Find out how many chameleons D... | Solution. Let $t$ be the number of brown chameleons Denis had. Then the number of red chameleons was $5t$. From the problem statement, we get the equation $5 t+2=8(t-2)$. Solving this, we find $t=6$. Therefore, the total number of chameleons is $6 t$, which is 36.
Answer. 36
Recommendations for checking. Only the cor... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,207 |
3. An odd six-digit number is called "simply cool" if it consists of digits that are prime numbers, and no two identical digits stand next to each other. How many "simply cool" numbers exist? | Solution. There are four single-digit prime numbers in total - 2, 3, 5, 7. We will place the digits starting from the least significant digit. In the units place, three of them can stand $-3, 5, 7$. In the tens place, there are also three out of the four (all except the one placed in the units place). In the hundreds p... | 729 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,208 |
4. On the extreme horizontal rows of a $15 \times 15$ board, two rows of chips are placed: on the bottom horizontal row, there is a white chip in each cell, and on the top horizontal row, there is a black chip in each cell. Each move, the players slide one of their chips (the first player a white one, the second a blac... | Solution. The first player wins. The first move is to slide his token (for example, the leftmost one) to the maximum possible number of cells and forget about the vertical with this token. The remaining part of the board is divided into strips of two verticals. On each such strip, the following strategy is applied: if ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,209 |
5. In the $3000-$th year, the World Hockey Championship will be held under new rules: 12 points will be awarded for a win, 5 points will be deducted for a loss, and no points will be awarded for a draw. If the Brazilian national team plays 38 matches in this championship, scores 60 points, and loses at least once, how ... | Solution. Let Brazil win in x matches and lose in y matches. We form the equation $12 x-5 y=60$. We see that $12 \mathrm{x} \vdots 12$ and $60 \vdots 12$. GCD(5, 12)=1, i.e., $y \vdots 12$. Possible: a) $y=12$. Then we get the equation $12 x-60=60$. Thus, $x=10$. This is possible. b) $y=24$. We get the equation: $12 x-... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,210 |
6. Three boys and 20 girls stood in a row. Each child counted the number of girls who were to the left of them, the number of boys who were to the right of them, and added the results. What is the maximum number of different sums that the children could have obtained? (Provide an example of how such a number could be o... | Solution. Let's consider how the number changes when moving from left to right by one person. If the adjacent children are of different genders, the number does not change. If we move from a girl to a girl, the number increases by 1, and if from a boy to a boy, it decreases by 1. Thus, the smallest number in the row co... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,211 |
10.1. Given various real numbers $a, b, c$. Prove that at least two of the equations $(x-a)(x-b)=x-c$, $(x-b)(x-c)=x-a$, $(x-c)(x-a)=x-b$ have a solution.
(I. Bogdanov) | The first solution. Let $f_{1}(x)=(x-b)(x-c)-(x-a)$, $f_{2}(x)=(x-c)(x-a)-(x-b)$, and $f_{3}(x)=(x-a)(x-b)-(x-c)$. Suppose the statement of the problem is false, that is, the maximum of one of these functions is a root. Then, two of them, say $f_{1}$ and $f_{2}$, do not have roots. Since the leading coefficients of the... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,212 |
10.2. On a circle, $n$ points are marked, dividing it into $n$ arcs. The circle is rotated around its center by an angle of $2 \pi k / n$ (for some natural number $k$), as a result of which the marked points move to $n$ new points, dividing the circle into $n$ new arcs. Prove that there will be a new arc that lies enti... | Solution. We will assume that the radius of the circle is 1, and the rotation occurred clockwise. If the two new points lie on the same old arc, then the new arc between them is the required one. Suppose there are no such new points. Since there are $n$ old arcs and $n$ new points, this is only possible if there is exa... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,213 |
10.3. Find all natural $k$ such that the product of the first $k$ prime numbers, decreased by 1, is a perfect power of a natural number (greater than the first).
(V. Senderov) | Answer. $k=1$.
Solution. Let $n \geqslant 2$, and $2=p_{1}<p_{2}<\ldots<p_{k}$; then $k>1$. The number $a$ is odd, so it has an odd prime divisor $q$. Then $q>p_{k}$, otherwise the left side of the equation $(*)$ would be divisible by $q$, which is impossible. Therefore, $a>p_{k}$.
Without loss of generality, we can ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,214 |
10.4. Inside the inscribed quadrilateral $A B C D$, points $P$ and $Q$ are marked such that $\angle P D C + \angle P C B = \angle P A B + \angle P B C = \angle Q C D + \angle Q D A = \angle Q B A + \angle Q A D = 90^{\circ}$. Prove that the line $P Q$ forms equal angles with the lines $A D$ and $B C$.
(A. Pastor) | Solution. Let the circumcircles of quadrilateral $ABCD$ and triangles $ABP, CDP, ABQ, CDQ$ be denoted by $\Omega, \omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}$ respectively (see Fig. 6).
Let $X$ be the projection of $P$ onto $BC$; denote the line $PX$ as
$ are divisible by 5, four numbers $(4,8,12,20)$ are divisible by 4, and the total sum is 71.
Remark. It can be proven (but, of course, this is not required in the problem), that in any example satisfying the problem's c... | 71 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,216 |
9.2. On the board, a certain natural number $N$ was written nine times (one under the other). Petya appended a non-zero digit to the left or right of each of the 9 numbers; all the appended digits are different. What is the maximum number of prime numbers that could result among the 9 obtained numbers?
(I. Efremov) | Answer: 6.
Solution. Let $S$ be the sum of the digits of the number $N$. Then the sums of the digits of the obtained numbers will be $S+1, S+2, \ldots, S+9$. Three of these sums will be divisible by 3. By the divisibility rule for 3, the corresponding three numbers on the board will also be divisible by 3. Since these... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,217 |
9.3. Given a quadratic trinomial $P(x)$, not necessarily with integer coefficients. It is known that for some integers $a$ and $b$, the difference $P(a)-P(b)$ is a square of a natural number. Prove that there are more than a million such pairs of integers $(c, d)$ for which the difference $P(c)-P(d)$ is also a square o... | Solution. Let $P(x)=u x^{2}+v x+w$. By the condition,
$$
\begin{aligned}
(a-b)(u(a+b)+v)=\left(u a^{2}+v a+w\right)-\left(u b^{2}\right. & +v b+w)= \\
& =P(a)-P(b)=n^{2}
\end{aligned}
$$
Since $n$ is a natural number, $a-b \neq 0$. We will look for suitable pairs of numbers $(c, d)$ in the form $c=a+k$ and $d=b-k$. T... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,218 |
9.4. In the company, some pairs of people are friends (if $A$ is friends with $B$, then $B$ is also friends with $A$). It turned out that among any 100 people in the company, the number of pairs of friends is odd. Find the largest possible number of people in such a company.
(E. Bakayev) | Answer: 101.
Solution: In all solutions below, we consider the friendship graph, where vertices are people in the company, and two people are connected by an edge if they are friends.
If the graph is a cycle containing 101 vertices, then any 100 vertices will have exactly 99 edges, so such a company satisfies the con... | 101 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,219 |
9.5. Let $CE$ be the bisector in an acute-angled triangle $ABC$. On the external bisector of angle $ACB$, point $D$ is marked, and on side $BC$, point $F$ is marked, such that $\angle BAD = 90^{\circ} = \angle DEF$. Prove that the center of the circumcircle of triangle $CEF$ lies on the line $BD$.
(I. Frolov) | Solution. In all solutions below, the circumcircle of triangle $C E F$ is denoted by $\omega$, and its center by $O$.
First Solution. The external $(C D)$ and internal $(C E)$ bisectors of angle $A C B$ are perpendicular. Then $\angle D A E = \angle D C E = 90^{\circ}$, and quadrilateral $A D C E$ is cyclic. Therefore... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,220 |
1. Vasya claims that he drew a rectangle on graph paper that can be cut along the cell edges into one strip of $1 \times 37$ cells and 135 three-cell corners. Is Vasya correct? | Answer: Vasya is wrong.
Solution. If such a rectangle existed, its area would be $37+135 \cdot 3=$ 442 cells. To be able to cut out a strip $1 \times 37$ from the rectangle, one of its sides must be at least 37. Since $442=2 \times 13 \times 17$, the only suitable rectangles are $1 \times 442$ and $2 \times 221$. Howe... | Vasya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,221 |
3. The numbers $1,2,3, \ldots, 29,30$ were written in a row in a random order, and partial sums were calculated: the first sum $S_{1}$ equals the first number, the second sum $S_{2}$ equals the sum of the first and second numbers, $S_{3}$ equals the sum of the first, second, and third numbers, and so on. The last sum $... | Answer: 23.
Solution: Evaluation: adding an odd number changes the parity of the sum, there are 15 odd numbers, so the parity of the sums changes at least 14 times. Therefore, there will be at least 7 even sums, and thus no more than 23 odd sums.
Implementation: arrange the numbers as follows: 1, then all even number... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,223 |
4. A three-digit number, in which there are no zeros, is written on the board. The sum of all different numbers that can be obtained by rearranging the digits of the written number is 2775. What number could have been written on the board | Answer: $889,988,898,997,799,979$.
Solution. The number on the board could not have been written with three identical digits, as then the number on the board would be 2775, which is not a three-digit number. But it also could not have been written with three different digits, because then each digit in the units place... | 889,997 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,224 |
5. On the side $B C$ of triangle $A B C$, a point $K$ is marked. In triangles $A K B$ and $A K C$, the angle bisectors $K M$ and $K P$ are drawn, respectively. It turns out that triangles $B M K$ and $P M K$ are equal. Prove that point $M$ bisects $A B$. | Solution. Lines KM and KR are perpendicular, as the bisectors of adjacent angles. Therefore, triangle $B M K$ is also a right triangle. However, angle $B K M$ is acute, and if angle $K B M$ were right, the equality of segments $M K = M P$ would hold. But one of these segments is a leg, and the other is the hypotenuse o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,225 |
3. Planes $\alpha$ and $\beta$ intersect. Points $\mathrm{M}$ and $\mathrm{N}$ do not lie in these planes. A perpendicular is dropped from point $\mathrm{M}$ to plane $\alpha$ and a perpendicular is dropped from point $\mathrm{N}$ to plane $\beta$, and these perpendiculars turn out to be in the same plane. Then, a perp... | # Solution
Let line a be the line of intersection of planes $\alpha$ and $\beta$. O is the foot of the perpendicular dropped from point $\mathrm{M}$ to plane $\alpha$, Q is the foot of the perpendicular dropped from point $\mathrm{N}$ to plane $\beta$. A is the foot of the perpendicular dropped from point $\mathrm{M}$... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,228 |
4. In a joint-stock company, there are 2017 shareholders, and any 1500 of them hold a controlling stake (not less than $50 \%$ of the shares). What is the largest share of shares that one shareholder can have
# | # Solution
Let's order the shareholders by the ascending share of their stocks. The 1500th shareholder has no less than $\frac{50}{1500} \%=\frac{1}{30} \%$ of the shares. If the 1500th has less, then the first 1500 shareholders together have less than $50 \%$ of the shares. Therefore, the first 2016 shareholders toge... | 32.8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,229 |
113. In the forest, there grew pines, cedars, and larches, and there were cones on all the trees, and they were equal in number. A gentle breeze blew, and several cones fell to the ground. It turned out that $11\%$ of the cones fell from each pine, $54\%$ from each cedar, and $97\%$ from each larch. At the same time, e... | 11.3. Let $m$ be the number of cones on each tree, $s, k, l-$ the number of pines, cedars, and deciduous trees in the forest. Then the condition of the problem corresponds to the equation $0.3 m(s+k+l)=0.11 m s+0.54 m k+0.97 m l$. This is equivalent to $19(s+k+l)=43 k+86 l$. The right side of this expression is divisib... | 43 | Number Theory | proof | Yes | Yes | olympiads | false | 12,231 |
114. At the base of the heptagonal pyramid $S A_{1} A_{2} \ldots A_{7}$ lies a convex heptagon $A_{1} A_{2} \ldots A_{7}$. It is known that the projections of the vertices $A_{2}, A_{4}, A_{6}$ onto the line $S A_{1}$ fall into one point. Prove that the projections of the vertices $A_{3}, A_{5}, A_{7}$ also fall into o... | 114. Let $P$ be the point where the projections of vertices $A_{2}, A_{4}, A_{6}$ onto the line $S A_{1}$ fall. Draw a plane $\alpha$ through point $P$ perpendicular to the line $S A_{1}$. The points $A_{2}, A_{4}, A_{6}$ are three points not lying on the same line (the polygon $A_{1} A_{2} \ldots A_{4}$ is convex), an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,232 |
115. In how many non-empty subsets of the set $\{1,2,3, \ldots, 10\}$ will there be no two consecutive numbers? | 115. 143.
Let $A_{n}$ denote the set of non-empty subsets of the set $\{1,2,3, \ldots, n\}$ that do not contain two consecutive numbers, and let $a_{n}$ be the number of such subsets. Clearly, the set $A_{1}$ consists of the subset $\{1\}$, and the set $A_{2}$ consists of the subsets $\{1\}$ and $\{2\}$. Thus, $a_{1}... | 143 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,233 |
10.1. Fill the cells of a $5 \times 5$ table with integers such that the sum of all numbers in the table is positive, while the sum of the numbers in any $3 \times 3$ square is negative. | Answer: For example,
| 1 | 1 | 1 | 1 | 1 |
| :---: | :---: | :---: | :---: | :---: |
| 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | -10 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 |.
## Criterion.
7 points. Any correct example.
Comment. There are several different correct examples. | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,234 |
10.3. 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.
## Reasoning.
Evaluation. Each grid node belongs to one, two, or four squares.
The corner vertices of the original square are adjacent to only one small square each, so Lёsha will not be able to mark them. Therefore, the maximum number of marked nodes does not exceed $7 \cdot 7-4=45$.
 | Answer: It could not.
Solution: Suppose such a thing could be. Let the given numbers be $a_{1}, a_{2}, \ldots, a_{1111}$ and denote their sum by $S$. By the condition, for each index $k$ the numbers $a_{k}$ and $S-a_{k}$ end with the same digit. This implies that the difference between these numbers, which is $S-2 a_{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,237 |
9.2. In quadrilateral $A B C D$, sides $A D$ and $B C$ are parallel. Prove that if the angle bisectors of angles $D A C, D B C, A C B$ and $A D B$ form a rhombus, then $A B=C D$.
(L. Emelyanov) | The first solution. Let $O$ be the point of intersection of the diagonals $AC$ and $BD$ (see Fig. 1). The angle bisectors of $\angle ADB$ and $\angle DAC$ intersect at the center $O_1$ of the circle inscribed in triangle $AOD$, and the angle bisectors of $\angle ACB$ and $\angle DBC$ intersect at the center $O_2$ of th... | AB=CD | Geometry | proof | Yes | Yes | olympiads | false | 12,238 |
9.3. The teacher wrote four different natural numbers in Petya's notebook. For each pair of these numbers, Petya found their greatest common divisor. He got six numbers: 1, 2, 3, 4, 5, and \( N \), where \( N > 5 \). What is the smallest value that \( N \) can have?
(O. Dmitriev)
# | # Answer: 14.
Solution. The number $N$ can equal 14, as shown, for example, by the quartet of numbers $4, 15, 70, 84$. It remains to show that $N \geqslant 14$.
Lemma. Among the pairwise GCDs of four numbers, there cannot be exactly two numbers divisible by some natural number $k$.
Proof. If among the original four ... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,239 |
9.4. All cells of a $100 \times 100$ square table are numbered in some order with numbers from 1 to 10000. Petya colors the cells according to the following rules. Initially, he colors $k$ cells at his discretion. Then, on each move, Petya can color one more uncolored cell with number $a$ if at least one of the two con... | Answer. $k=1$.
Solution. First, let's prove the following statement.
Lemma. For any two cells $A$ and $B$, there exists a cell $C$ such that by coloring it, one can then color both $A$ and $B$ (possibly $C$ coincides with $A$ or $B$).
Proof. We can assume that the number $a$ of cell $A$ is less than the number $b$ o... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,240 |
10.1. A student received 17 grades (each of them 2, 3, 4, or 5) over one week. The arithmetic mean of these 17 grades is an integer. Prove that the student received some grade no more than twice. | Solution. Assume the opposite. Then each of the grades $2,3,4,5$ was received by the student at least three times. Take three grades of each kind; the sum of the 12 taken grades is 42. Since each of the remaining five grades is not less than 2 and not more than 5, the sum of all 17 grades is not less than $42+5 \cdot 2... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,241 |
10.2. A hundred-digit number $n$ is called unusual if the decimal representation of the number $n^{3}$ ends with $n$, and the decimal representation of the number $n^{2}$ does not end with $n$. Prove that there are at least two hundred-digit unusual numbers.
(V. Senderov) | Solution. For example, such numbers are $n_{1}=$ $=10^{100}-1=99 \ldots 9$ and $n_{2}=\frac{10^{100}}{2}-1=49 \ldots 9$. Indeed, the numbers $n_{1}^{3}-n_{1}=\left(n_{1}+1\right) n_{1}\left(n_{1}-1\right)=10^{100} \cdot n_{1}\left(n_{1}-1\right)$ and $n_{2}^{3}-n_{2}=\left(n_{2}+1\right) n_{2}\left(n_{2}-1\right)=10^{1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,242 |
10.3. In the language of the AU tribe, there are two letters - "a" and "u". Some sequences of these letters are words, and each word has no more than 13 letters. It is known that if you write down any two words in a row, the resulting sequence of letters will not be a word. Find the maximum possible number of words in ... | Answer. $2^{14}-2^{7}=16056$.
First solution. If all sequences, the number of letters in which is not less than 7 and not more than 13, are words, then, obviously, the condition of the problem is satisfied; in this case, the number of such words is $2^{7}+\ldots+2^{13}=2^{14}-2^{7}$. It remains to show that this numbe... | 2^{14}-2^{7}=16056 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,243 |
10.4. On side $AB$ of triangle $ABC$, points $C_{1}$ and $C_{2}$ are chosen. Similarly, on side $BC$, points $A_{1}$ and $A_{2}$ are chosen, and on side $AC$, points $B_{1}$ and $B_{2}$ are chosen. It turns out that segments $A_{1} B_{2}$, $B_{1} C_{2}$, and $C_{1} A_{2}$ have equal lengths, intersect at one point, and... | Solution. Note that
$$
\overrightarrow{A_{1} B_{2}}+\overrightarrow{B_{2} B_{1}}+\overrightarrow{B_{1} C_{2}}+\overrightarrow{C_{2} C_{1}}+\overrightarrow{C_{1} A_{2}}+\overrightarrow{A_{2} A_{1}}=\overrightarrow{0}
$$
By the condition, we have $A_{1} B_{2}=B_{1} C_{2}=C_{1} A_{2}$, and the angle between any two of t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,244 |
11.1. Given a convex heptagon. Four of its angles are chosen arbitrarily, and their sines are calculated; for the remaining three angles, their cosines are calculated. It turns out that the sum of these seven numbers does not depend on the initial choice of the four angles. Prove that this heptagon has four equal angle... | Solution. Consider one of the sums from the condition. Then, we will swap the arguments of one sine and one cosine (let's call these arguments $\alpha$ and $\beta$, respectively; the sum will change by
$(\sin \beta + \cos \alpha) - (\sin \alpha + \cos \beta) = \sqrt{2}(\sin (\beta - \pi / 4) - \sin (\alpha - \pi / 4))... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,245 |
11.2. On the board, the expression $\frac{a}{b} \cdot \frac{c}{d} \cdot \frac{e}{f}$ is written, where $a, b, c, d, e, f$ are natural numbers. If the number $a$ is increased by 1, then the value of this expression increases by 3. If in the original expression the number $c$ is increased by 1, then its value increases b... | # Answer: 60.
First solution. Let the value of the original expression be $A$. Then, as a result of the first operation, the product will take the value $\frac{a+1}{a} \cdot A=A+3$, from which $A=3a$. This means that $A$ is a natural number. Moreover, from this equality, it follows that it is divisible by 3. Similarly... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,246 |
11.3. All cells of a square table $n \times n$ are numbered in some order with numbers from 1 to $n^{2}$. Petya makes moves according to the following rules. On the first move, he places a rook in any cell. On each subsequent move, Petya can either place a new rook in some cell or move a rook from a cell numbered $a$ h... | Answer: $n$.
Solution. We will show that $n$ rooks are sufficient. For this, note that one rook is enough for each row: it can be placed in the cell of the row with the smallest number, and then visit all the cells of the row in increasing order of their numbers.
On the other hand, we will show that fewer than $n$ ro... | n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,247 |
11.4. The plane $\alpha$ intersects the edges $A B, B C, C D$ and $D A$ of the tetrahedron $A B C D$ at points $K, L, M$ and $N$ respectively. It turns out that the dihedral angles $\angle(K L A, K L M)$, $\angle(L M B, L M N)$, $\angle(M N C, M N K)$ and $\angle(N K D, N K L)$ are equal. (Here, $\angle(P Q R, P Q S)$ ... | Solution. Let $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ be the projections of vertices $A, B, C, D$ respectively onto the plane $\alpha$. Let $X$ be an arbitrary point on the extension of segment $K L$ beyond point $K$. Then we have $\angle(K X A, K X N)=\angle(K L A, K L M)$ and $\angle(K N A, K N X) = \angle(N... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,248 |
3. Potatoes and beets are transported on 79 trucks, not necessarily identical, with each truck loaded either with potatoes or beets. Prove that it is possible to select 40 trucks from these such that they transport no less than $50 \%$ of all the potatoes and no less than $50 \%$ of all the beets. | # Solution.
Let potatoes be transported by $n$ trucks. Arrange the trucks in descending order of the mass of potatoes they carry.
$$
a_{1} \geq a_{2} \geq a_{3} \geq \ldots \geq a_{n-1} \geq a_{n}
$$
Let beets be transported by $k$ trucks. Arrange the trucks in descending order of the mass of beets they carry.
$$
b... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,251 |
4. $\mathrm{ABCD}$ is a rhombus. Points $E$ and $F$ lie on sides $AB$ and $BC$ respectively, such that $\frac{AE}{BE}=\frac{BF}{CF}=5$. Triangle $DEF$ is equilateral. Find the angles of the rhombus. Solution.
The notation is shown in the figure. The situation where angle $D$ of the rhombus is acute is impossible (see ... | Answer. The angles of the rhombus are $60^{\circ}$ and $120^{\circ}$. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,252 |
10.1 At a competition in meaningless activity, a participant recorded 2022 non-zero numbers in a circle, and it turned out that the absolute value of each number coincides with the absolute value of the sum of its two neighbors. Can you do that? | Solution: We can arrange the repeating triplets of numbers $1,1,-2$ in a circle.
Criteria:
- Correct example provided - 7 points. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,253 |
10.2 About numbers a and b, it is known that the system of equations
$$
\left\{\begin{array}{l}
y^{2}=x^{2}+a x+b \\
x^{2}=y^{2}+a y+b
\end{array}\right.
$$
has no solutions. Find a. | Solution: Since the system has no solutions, in particular, there are no solutions with $x=y$. When $x=y$, both equations of the system are equivalent to the equation $a x+b=0$. This linear equation has no roots only when its slope coefficient $a$ is zero.
## Criteria:
- Points are not deducted for the absence of an ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,254 |
10.3 On the board, an acute-angled $\triangle A B C$ was drawn, and the feet of the altitudes $A_{1}, B_{1}$ and $C_{1}$ were marked. Then the entire drawing was erased, except for the points $A_{1}, B_{1}$ and $C_{1}$. Can the original $\triangle A B C$ be reconstructed using a compass and a straightedge? | Solution: Angles $\angle A A_{1} B$ and $\angle B B_{1} A$ are equal, so the quadrilateral $A B_{1} A_{1} B$ is cyclic, $\angle B_{1} A_{1} C=\angle A$. Similarly, $\angle C_{1} A_{1} B=\angle A$. Therefore, the sides of triangle $A B C$ are the external angle bisectors of triangle $A_{1} B_{1} C_{1}$. This constructio... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,255 |
10.4 Another participant in the competition for meaningless activity marked the centers of 13 cells in a grid rectangle of size $(N-1) \times(N+1)$ such that the distance between any two marked points is greater than 2. What is the smallest value that $N$ can take? | Solution: We will show that it is impossible to mark cells in a $6 \times 8$ rectangle (and thus in any smaller size) in such a way. Indeed, let's divide the rectangle into $2 \times 2$ squares. In each of them, the pairwise distances between the centers of the cells do not exceed $\sqrt{2}$, so no more than one cell i... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,256 |
10.5 Can three such irrational numbers be found, such that their sum is an integer and the sum of their reciprocals is also an integer? | Solution 1: Consider the polynomial $f(x)=x^{3}-30 x^{2}+31 x-1$. Since $f(0) < 0$ and $f(2) > 0$, the equation $f(x)=0$ has three real roots $x_{1}, x_{2}, x_{3}$. The numbers $\pm 1$ are not roots, so all roots of the equation are irrational. According to Vieta's theorem, their sum is 30, and the sum of their recipro... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,257 |
10.1. Given a quadratic trinomial $p(x)=a x^{2}+b x+c$, where $a, b, c$ are odd integers. Will its roots $x_{1}$ and $x_{2}$ (assuming they exist) be integers? | Answer: They will not.
Proof. Suppose there exists a quadratic trinomial $p(x) = a x^{2} + b x + c$, where $a, b, c$ are odd integers, and $x_{1}$ and $x_{2}$ are its integer roots. But then the product $x_{1} \cdot x_{2} = \frac{c}{a}$ is odd, which implies that $x_{1}$ and $x_{2}$ are odd. Their sum is even, but $x_... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,258 |
10.2. Does the inequality
$$
x^{3}(y+1)+y^{3}(x+1) \geq x^{2}\left(y+y^{2}\right)+y^{2}\left(x+x^{2}\right)
$$
hold for non-negative $x$ and $y$? | Answer. Yes.
Solution. $x^{3}(y+1)+y^{3}(x+1)-\left(x^{2}\left(y+y^{2}\right)+y^{2}\left(x+x^{2}\right)\right)=(x-y)^{2}(x y+x+y) \geq 0$, since $(x-y)^{2} \geq 0$ and $(x y+x+y) \geq 0$. | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 12,259 |
10.3. Anna, Berta, Vika, Galia, and Dasha (in some order) are waiting for the moment when a rare flower will bloom in their greenhouse. Each of the girls observed the flower once, and Anna watched the flower for twice as long as Berta, Berta waited for the flower to bloom for twice as long as Vika, and Galia, Dasha, an... | Answer: Could not.
Solution: Let's assume the scientific supervisor marked each girl once. Then Anna kept watch for less than 2 hours, Berta for less than an hour, but then Vika, Galya, and Dasha for less than half an hour. Since Vika, Galya, and Dasha kept watch for less than half an hour each, no two of them could h... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,260 |
10.4. In pentagon $M N P Q S: \quad M N=N P=P Q=Q S=S M$ and $\angle M N P=2 \angle Q N S$. Find the measure of angle $M N P$. | Answer: $60^{\circ}$
Solution. Since $\angle S N Q=\angle M N S+\angle P N Q$, we can take a point $T$ on the side $S Q$ such that $\angle S N T=\angle M N S=\angle M S N$, i.e., $N T \| M S$.
Then $\angle T N Q=\angle S N Q-\angle S N T=\angle P N Q=\angle N Q P$, i.e., $N T \| P Q$. Therefore, $M S \| P Q$, and sin... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,261 |
10.5. The teams participating in the quiz need to answer 50 questions. The cost (in integer points) of a correct answer to each question was determined by experts after the quiz, the cost of an incorrect answer - 0 points. The final score of the team was determined by the sum of points received for correct answers. Whe... | Answer: 50.
Solution: We will prove that with 50 teams, such a distribution of points can exist. The example is obvious - let the $k$-th team answer only the $k$-th question. Then, by assigning the costs of the questions as $a_{1}, a_{2}, \ldots, a_{50}$, where $\left\{a_{1}, a_{2}, \ldots, a_{50}\right\}=\{1,2, \ldot... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,262 |
9-1-1. A snail is crawling along a straight line. On the first day, it crawls 1 m forward and $1 / 2$ m backward. On the second day, it crawls $1 / 2$ m forward and $1 / 3$ m backward. On the third day, it crawls $1 / 3$ m forward and $1 / 4$ m backward, and so on. How far from the starting point will it be at the end ... | Express the answer in meters.
Answer. $74 / 75$.
Solution Variant 1. The snail will be at a distance of
$$
\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{74}-\frac{1}{75}\right)=\frac{1}{1}-\frac{1}{75}=\frac{74}{75}
$$ | \frac{74}{75} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,264 |
9-1-2. A snail is crawling along a straight line. On the first day, it crawls 1 m forward and $1 / 2$ m backward. On the second day, it crawls $1 / 2$ m forward and $1 / 3$ m backward. On the third day, it crawls $1 / 3$ m forward and $1 / 4$ m backward, and so on. How far from the starting point will it be at the end ... | Express the answer in meters.
Answer. $55 / 56$. | \frac{55}{56} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,265 |
9-1-3. A snail is crawling along a straight line. On the first day, it crawls 1 m forward and $1 / 2$ m backward. On the second day, it crawls $1 / 2$ m forward and $1 / 3$ m backward. On the third day, it crawls $1 / 3$ m forward and $1 / 4$ m backward, and so on. How far from the starting point will it be at the end ... | Express the answer in meters.
Answer. $44 / 45$. | \frac{44}{45} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,266 |
9-1-4. A snail is crawling along a straight line. On the first day, it crawls 1 m forward and $1 / 2$ m backward. On the second day, it crawls $1 / 2$ m forward and $1 / 3$ m backward. On the third day, it crawls $1 / 3$ m forward and $1 / 4$ m backward, and so on. How far from the starting point will it be at the end ... | Express the answer in meters.
Answer. $96 / 97$. | \frac{96}{97} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,267 |
9-2-1. Non-zero numbers $a, b, c, d, e, f$ are such that among the products $a c d, a c e, b d e$, $b d f$ and $b e f$ exactly one is positive. Which one? | Answer. $b d e$.
Solution 1. Notice that $(a c e) \cdot(b d f)=(a c d) \cdot(b e f)$. Therefore, exactly one of these four numbers cannot be positive. This means that the remaining number $b d e$ is positive.
Comment. There are examples where the product $b d e$ is indeed positive, while the others are negative. For ... | e | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,268 |
9-3-1. The numbers from 1 to 217 are divided into two groups: one group has 10 numbers, and the other has 207. It turns out that the arithmetic means of the numbers in the two groups are equal. Find the sum of the numbers in the group of 10 numbers. | Answer: 1090.
Solution Variant 1. The shortest solution to this problem is based on the following statement:
If the arithmetic means of the numbers in two groups are equal, then this number is also equal to the arithmetic mean of all the numbers.
A formal proof of this is not difficult: it is sufficient to denote th... | 1090 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,269 |
9-4-1. In the figure, $O$ is the center of the circle, $A B \| C D$. Find the degree measure of the angle marked with a «?».
 | Answer: $54^{\circ}$.
Solution variant 1. Quadrilateral $A D C B$ is an inscribed trapezoid in a circle. As is known, such a trapezoid is isosceles, and in an isosceles trapezoid, the angles at the base are equal: $\angle B A D=\angle C B A=63^{\circ}$. Triangle $D O A$ is isosceles ($O A$ and $O D$ are equal as radii... | 54 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,270 |
9-5-1. The numbers 1 and 7 are written on the board. In one move, the greatest common divisor of both numbers on the board is added to each of them. For example, if at some point the numbers 20 and 50 are on the board, they will be replaced by the numbers 30 and 60.
What numbers will be on the board after 100 moves? P... | Solution option 1. Let's perform the first few actions described in the condition.
| actions | first number | second number | their GCD |
| :---: | :---: | :---: | :---: |
| none | 1 | 7 | 1 |
| after the first | 2 | 8 | 2 |
| after the second | 4 | 10 | 2 |
| after the third | 6 | 12 | 6 |
| after the fourth | 12 | 1... | 588594 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,271 |
9-6-1. Six pirates - a captain and five members of his crew - are sitting around a campfire facing the center. They need to divide a treasure: 180 gold coins. The captain proposes a way to divide the treasure (i.e., how many coins each pirate should receive: each pirate will receive a non-negative integer number of coi... | Answer: 59.
Solution variant 1. Let's number the pirates clockwise, starting from the captain: First, Second, ..., Fifth. The main observation in this problem is:
two pirates sitting next to each other cannot both vote "for": one of them will not vote "for" because the captain offers no more to him than to his neighb... | 59 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,272 |
9-7-1. The equation $x^{4}-7 x-3=0$ has exactly two real roots $a$ and $b$, $a>b$. Find the value of the expression $\frac{a-b}{a^{4}-b^{4}}$. | Answer: $1 / 7$.
Solution variant 1. The fact that $a$ and $b$ are roots of the equation simply means that $a^{4}=7 a+3$ and $b^{4}=7 b+3$. Therefore,
$$
\frac{a-b}{a^{4}-b^{4}}=\frac{a-b}{(7 a+3)-(7 b+3)}=\frac{a-b}{7(a-b)}=\frac{1}{7}
$$
As a comment, note that the equation from the condition indeed has two real r... | \frac{1}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,273 |
9-8-1. In Midcity, houses stand along one side of a street, each house can have $1,2,3, \ldots, 9$ floors. According to an ancient law of Midcity, if two houses on one side of the street have the same number of floors, then no matter how far apart they are from each other, there must be a house with more floors between... | Answer: 511.
Solution Variant 1. Estimation. Let $a_{k}$ be the maximum number of houses if all of them have no more than $k$ floors. Clearly, $a_{1}=1$. Let's find $a_{k}$. Consider the tallest house with $k$ floors, which is only one by the condition. Both to the left and to the right of it stand houses with $1,2,3,... | 511 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,274 |
1. Given a finite set of cards. On each of them, either the number 1 or the number -1 is written (exactly one number on each card), and there are 100 more cards with -1 than cards with 1. If for each pair of different cards, the product of the numbers on them is found, and all these products are summed, the result is 1... | Answer: 3950.
Solution: Let the number of cards with 1 be $m$, and the number of cards with -1 be $k$. Then, among all pairs, there are $\frac{m(m-1)}{2}$ pairs of two 1s, $\frac{k(k-1)}{2}$ pairs of two -1s, and $m k$ pairs of 1 and -1. Therefore, the sum in the condition is $\frac{m(m-1)}{2}+\frac{k(k-1)}{2}-m k$, f... | 3950 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,275 |
2. Given a quadratic trinomial $f(x)$ with a non-zero coefficient of $x^{2}$. Prove the existence of a natural number $n$ such that the polynomial $g(x)=f(x)+f(x+1)+\ldots+$ $f(x+n)$ has no real roots. | First solution. Let $f(x)=a x^{2}+b x+c$, without loss of generality, we can assume that $a>0$. $f(x)>0$ outside some interval centered at $-\frac{b}{2 a}$ (or everywhere), so there exists a natural number $t$ such that $f\left(-\frac{b}{2 a}+t\right)>0$ and $f\left(-\frac{b}{2 a}-t\right)>0$. Consider the parabola $h(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,276 |
3. Prove that if in a tetrahedron two segments, extending from the ends of a certain edge to the centers of the inscribed circles of the opposite faces, intersect, then the segments extending from the ends of the intersecting edge to the centers of the inscribed circles of the opposite faces also intersect. | Solution. Let the tetrahedron be $ABCD$, the edge given in the condition is $AB$, $E$ is the center of the circle inscribed in triangle $BCD$, and $F$ is the center of the circle inscribed in triangle $ACD$. Since segments $AE$ and $BF$ intersect, they lie in the same plane $\alpha$. $CD$ and $AB$ are skew lines, so $C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,277 |
4. For natural numbers $n$ and $m$, it is known that $\sqrt{3}-\frac{m}{n}>0$.
Prove that $\sqrt{3}-\frac{m}{n}>\frac{1}{2 m n}$. | Solution. Since $\sqrt{3}-\frac{m}{n}>0$, then $\sqrt{3} n-m>0, \sqrt{3} n>m, 3 n^{2}>m^{2}$. Since $n$ and $m$ are natural numbers, then $3 n^{2} \geqslant m^{2}+1$. The square of an integer when divided by three can give remainders 0 and 1, so $m^{2}+1$ can give remainders 1 and 2 when divided by three. Therefore, th... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,278 |
5. Each point of the plane is painted either blue or red. Prove that there exists a triangle with vertices of the same color and the shortest side of length 1, with the ratio of the angles being $1: 2: 4$. | Solution. Suppose that such a triangle does not exist and we will arrive at a contradiction. Consider a regular heptagon $A_{1} A_{2} \ldots A_{7}$ with side length 1. Notice that the triangle $A_{1} A_{2} A_{4}$ is arranged as specified in the condition. Indeed, $\angle A_{1} A_{4} A_{2}=\frac{\pi}{7}, \angle A_{4} A_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,279 |
1. Move everything to the right side
$(x+c)(x+d)-(2 x+c+d)=0$,
$x^{2}+c x+d x+c d-2 x-c-d=0$,
$x^{2}+(c+d-2) x+c d-c-d=0$.
Find the discriminant of the quadratic equation
$D=(c+d-2)^{2}-4(cd-c-d)=c^{2}+d^{2}+4+2cd-4c-4d-4cd+4c+4d=$ $=c^{2}+d^{2}-2cd+4=(c-d)^{2}+4>0$ Therefore, the equation has two distinct roots. | Solution 2. Consider $\mathrm{f}(\mathrm{x})=(\mathrm{x}+\mathrm{c})(\mathrm{x}+\mathrm{d})-(2 \mathrm{x}+\mathrm{c}+\mathrm{d})$, then the equation from the condition will take the form $\mathrm{f}(\mathrm{x})=0$. Notice that $\mathrm{f}(-\mathrm{c})=\mathrm{c}-\mathrm{d}$ and $\mathrm{f}(-\mathrm{d})=\mathrm{d}-\math... | (-)^{2}+4>0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,280 |
9.1. Donkey Eeyore made two triangles from six sticks. Then he took the triangles apart and painted the six sticks in two colors: the three shortest ones - in yellow, and the other three - in green. Is it necessarily possible for Eeyore to make two triangles, one - from three yellow sticks, and the other - from three g... | Answer. No, it is not necessarily so.
Solution. For example, if Io-Io had two equal triangles with sides $1,2,2$, then in the first pile there would be sticks with lengths $1,1,2$, from which a triangle cannot be formed.
Remark. There are many other examples. It is worth noting that a triangle can always be formed fr... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,283 |
9.2. Non-zero numbers $x$ and $y$ satisfy the inequalities $x^{2}-x>y^{2}$ and $y^{2}-y>x^{2}$. What sign can the product $x y$ have?
(N. Agakhanov) | Answer. It is positive
First solution. Adding the inequalities from the condition, we get that $-x-y>0$. Multiplying the inequalities from the condition (this can be done since their right-hand sides are non-negative), we get that $x y(1-x-y)>0$. The expression in parentheses is positive, so the product $x y$ is also ... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 12,284 |
9.3. Consider natural numbers $a, b$, and $c$ such that the fraction
$$
k=\frac{a b+c^{2}}{a+b}
$$
is a natural number less than $a$ and $b$. What is the smallest number of natural divisors that the number $a+b$ can have?
(P. Kozlov) | Answer. Three divisors.
First solution. Since the number $a+b$ is greater than one, it has at least two distinct divisors. We will prove that there cannot be exactly two, i.e., that the number $a+b$ cannot be prime. Multiplying the equality from the condition by the denominator, we get $a b+c^{2}=k a+k b$ or, equivale... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,285 |
9.4. Circles $\Omega$ and $\omega$ touch each other internally at point $A$. In the larger circle $\Omega$, a chord $C D$ is drawn, touching $\omega$ at point $B$ (chord $A B$ is not a diameter of $\omega$). Point $M$ is the midpoint of segment $A B$. Prove that the circle circumscribed around triangle $C M D$ passes t... | First solution. Let $O$ be the center of the circle $\omega$. Draw a common tangent to our circles through point $A$; let it intersect the line $C D$ at point $P$. Since $P A = P B$, point $P$ lies on the perpendicular bisector of segment $A B$, which also passes through points $M$ and $O$.
Since $A M$ is the altitude... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,286 |
9.5. Petya and Vasya are playing on a $100 \times 100$ board. Initially, all cells of the board are white. On each of his turns, Petya paints one or several white cells black, which are located consecutively along a diagonal. On each of his turns, Vasya paints one or several white cells black, which are located consecu... | Answer. Petya wins.
Solution. We will present one of the possible winning strategies for Petya. He will always make moves parallel to one of the
diagonals of the board (let's call it the ma... | Petyawins | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,287 |
10.1. Natural numbers starting from 1 are written in a row. This results in a sequence of digits: 1234567891011121314... What digit is in the 2021st position? | Answer. 1.
Solution. Note that the sum of the digits of all single-digit and two-digit numbers is $1892021$. Therefore, the digit in the 2021st position belongs to the recording of some three-digit number.
Let $x$ be some three-digit number, then the sum of the digits in the sequence from 1 to $x$ is $n=189+3(x-99)$.... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,288 |
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