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3. All natural numbers are divided into "good" and "bad" according to the following rules:
a) From any bad number, you can subtract some natural number not exceeding its half so that the resulting difference becomes "good".
b) From a "good" number, you cannot subtract no more than half of it so that it remains "good"... | Answer: 2047. Solution. Note that good numbers have the form $2^{n}-1$. Indeed, let a number have the form $M=2^{n}+k$, where $k=0,1, \ldots, 2^{n}-2$. Then from such a number, you can subtract $k+1 \leqslant \frac{1}{2}\left(2^{n}+k\right)=\frac{M}{2}$. On the other hand, from a number of the form $N=2^{n}-1$, you nee... | 2047 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Calculate
$$
\frac{2 a b\left(a^{3}-b^{3}\right)}{a^{2}+a b+b^{2}}-\frac{(a-b)\left(a^{4}-b^{4}\right)}{a^{2}-b^{2}} \quad \text { for } \quad a=-1, \underbrace{5 \ldots 5}_{2010} 6, \quad b=5, \underbrace{4 \ldots 44}_{2011}
$$
Answer: 343. | Solution. Given $a$ and $b$, we get
$$
\begin{gathered}
\frac{2 a b\left(a^{3}-b^{3}\right)}{a^{2}+a b+b^{2}}-\frac{(a-b)\left(a^{4}-b^{4}\right)}{a^{2}-b^{2}}=2 a b(a-b)-(a-b)\left(a^{2}+b^{2}\right)= \\
\quad=-(a-b)\left(a^{2}-2 a b+b^{2}\right)=-(a-b)^{3}=-(-7)^{3}=343
\end{gathered}
$$ | 343 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Find the smallest natural number that is greater than the sum of its digits by 1755 (the year of the founding of Moscow University). | Answer: 1770.
Solution. From the condition, it follows that the desired number cannot consist of three or fewer digits. We will look for the smallest such number in the form $\overline{a b c d}$, where $a, b, c, d$ are digits, and $a \neq 0$. We will form the equation:
$$
\begin{gathered}
\overline{a b c d}=a+b+c+d+1... | 1770 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 9.
Let $A(n)$ denote the greatest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9, A(64)=1$. Find the sum $A(111)+A(112)+\ldots+A(218)+A(219)$. | Answer: 12045
Solution. The largest odd divisors of any two of the given numbers cannot coincide, as numbers with the same largest odd divisors are either equal or differ by at least a factor of 2. Therefore, the largest odd divisors of the numbers $111, 112, \ldots, 218, 219$ are distinct. This means that the largest... | 12045 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Two trains, each containing 15 identical cars, were moving towards each other at constant speeds. Exactly 28 seconds after the first cars of the trains met, passenger Sasha, sitting in the third car, passed passenger Valera from the oncoming train, and another 32 seconds later, the last cars of these trains had comp... | Solution. Since 60 seconds have passed from the moment the "zero" cars parted ways (which is also the moment the first cars met) to the moment the 15th cars parted ways, the next cars parted ways every $60: 15=4$ seconds. Therefore, after 28 seconds, the 7th cars of the trains had just parted ways, meaning the 7th car ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Find the area of the figure defined on the coordinate plane by the system
$$
\left\{\begin{array}{l}
\sqrt{1-x}+2 x \geqslant 0 \\
-1-x^{2} \leqslant y \leqslant 2+\sqrt{x}
\end{array}\right.
$$ | Answer: 4.
Solution. The system is defined under the condition $x \in[0,1]$, under which the first inequality of the system is satisfied. Since the functions $y=x^{2}$ and $y=\sqrt{x}$ are inverses of each other for $x \in[0,1]$, the area of the figure defined by the conditions $x \in[0,1],-1-x^{2} \leqslant y \leqsla... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. Two circles touch each other internally at point K. A chord $AB$ of the larger circle touches the smaller circle at point $L$, and $AL=10$. Find $BL$, if $AK: BK=2: 5$. | Answer: $B L=25$.
Solution. Draw the chord $P Q$ and the common tangent $M N$ (see fig.), then
$$
\angle P Q K=\frac{1}{2} \smile P K=\angle P K M=\frac{1}{2} \smile A K=\angle A B K
$$
th... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. What is the smallest (same) number of pencils that need to be placed in each of 6 boxes so that in any 4 boxes there are pencils of any of 26 predefined colors (there are enough pencils available)? | Answer: 13.
Solution. On the one hand, pencils of each color should appear in at least three out of six boxes, since if pencils of a certain color are in no more than two boxes, then in the remaining boxes, which are at least four, there are no pencils of this color. Therefore, the total number of pencils should be no... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3.
In triangle $A B C$, the bisector $B E$ and the median $A D$ are equal and perpendicular. Find the area of triangle $A B C$, if $A B=\sqrt{13}$. | Answer: 12
Solution. Let $A B=c, B E=A D=2 a$. Since triangle $A B D$ is isosceles (the bisector is perpendicular to the base, $A B=B D=c, B C=2 c$), then by the formula for the length of the bisector (where $\angle A B C=\beta$) $2 a=\frac{4 c^{2}}{3 c} \cos \frac{\beta}{2} \Leftrightarrow \frac{3 a}{2}=c \cos \frac{... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 4.
Masha tightly packed 165 identical balls into the shape of a regular triangular pyramid. How many balls are at the base? | Answer: 45
Solution. We will solve the problem in general for $\frac{1}{6} n(n+1)(n+2)$ balls. At the base of a triangular pyramid of height $n$ rows lies the $n$-th triangular number of balls $1+2+3+\ldots+n=\frac{1}{2} n(n+1)$. Then the total number of balls in the pyramid is $\sum_{k=1}^{n} \frac{k(k+1)}{2}=\frac{1... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6.
B-1
Find the minimum value of the expression $4 x+9 y+\frac{1}{x-4}+\frac{1}{y-5}$ given that $x>4$ and $y>5$. | Answer: 71
Solution. $4 x+9 y+\frac{1}{x-4}+\frac{1}{y-5}=4 x-16+\frac{4}{4 x-16}+9 y-45+\frac{9}{9 y-45}+61 \geq 2 \sqrt{4}+2 \sqrt{9}+61=$ 71 (inequality of means). Equality is achieved when $4 x-16=2,9 y-45=3$, that is, when $x=\frac{9}{2}, y=\frac{16}{3}$.
B-2
Find the minimum value of the expression $9 x+4 y+\f... | 71 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find the sum of the digits of the number $A$, if $A=2^{63} \cdot 4^{25} \cdot 5^{106}-2^{22} \cdot 4^{44} \cdot 5^{105}-1$. Answer: 959. | Solution. Since $2^{63} \cdot 4^{25} \cdot 5^{106}=2^{113} \cdot 5^{106}=2^{7} \cdot 10^{106}$ and $2^{22} \cdot 4^{44} \cdot 5^{105}=$ $2^{110} \cdot 5^{105}=2^{5} \cdot 10^{105}$, then $A+1=2^{7} \cdot 10^{106}-2^{5} \cdot 10^{105}=\left(2^{7} \cdot 10-2^{5}\right) \cdot 10^{105}=$ $1248 \cdot 10^{105}$. Therefore, t... | 959 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Each Kinder Surprise contains exactly 3 different gnomes, and there are 12 different types of gnomes in total. In the box, there are enough Kinder Surprises, and in any two of them, the triplets of gnomes are not the same. What is the minimum number of Kinder Surprises that need to be bought to ensure that after the... | Answer: 166.
Solution. From 11 different gnomes, a maximum of $C_{11}^{3}=\frac{11 \cdot 10 \cdot 9}{3 \cdot 2 \cdot 1}=165$ different Kinder can be formed, so if you take one more Kinder, there cannot be fewer than 12 different gnomes among them. | 166 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Points $A$ and $B$ lie on a circle with center $O$ and radius 6, and point $C$ is equidistant from points $A, B$, and $O$. Another circle with center $Q$ and radius 8 is circumscribed around triangle $A C O$. Find $B Q$. Answer: 10 | Solution. Since $A Q=O Q=8$ and $A C=O C$, triangles $\triangle A C Q$ and $\triangle O C Q$ are equal. Therefore, points $A$ and $O$ are symmetric with respect to the line $C Q$, so $A O \perp C Q$.
Next, $\angle B O Q=\angle B O C+\angle C O Q=\angle A O C+\angle O C Q=\pi / 2$. Therefore, $B Q=\sqrt{B O^{2}+O Q^{2}... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. A motorcycle and a quad bike are driving on a circular road, a quarter of which passes through a forest, and the remaining part - through a field. The motorcycle's speed when driving through the forest is 20 km/h, and through the field - 60 km/h. The quad bike's speed when driving through the forest is 40 km/h, and ... | Answer: The quad will overtake the motorcycle on its $10-\mathrm{th}$ lap while the quad is on its 11-th lap. | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that $m, n, k$ are distinct natural numbers greater than 1, the number $\log _{m} n$ is rational, and, moreover,
$$
k^{\sqrt{\log _{m} n}}=m^{\sqrt{\log _{n} k}}
$$
Find the minimum of the possible values of the sum $k+5 m+n$. | Answer: 278.
Solution. Transform the original equation, taking into account that the numbers $m, n, k$ are greater than 1 and the corresponding logarithms are positive:
$$
\begin{aligned}
& k \sqrt{\log _{m} n}=m \sqrt{\log _{n} k} \Leftrightarrow \log _{m} k \cdot \sqrt{\log _{m} n}=\sqrt{\log _{n} k} \Leftrightarro... | 278 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. To guard the object, the client agreed with the guards on the following: all of them will indicate the time intervals of their proposed shifts with the only condition that their union should form a predetermined time interval set by the client, and he will choose any set of these intervals that also satisfies the sa... | 6. $\frac{90000}{2 \cdot 300}=150 \text{h}$. | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2.1. Find the sum of all integer values of $a$ from the interval $[-2012 ; 2013]$, for which the equation $(a-3) x^{2}+2(3-a) x+\frac{a-7}{a+2}=0$ has at least one solution. | Answer: 2011. Solution. When $a=3$ there are no solutions. For other $a$, it should be $\frac{D}{4}=(a-3)^{2}-\frac{(a-3)(a-7)}{a+2} \geq 0$. The last inequality (plus the condition $a \neq 3$) has the solution $a \in(-\infty ;-2) \bigcup\{1\} \cup(3 ;+\infty)$. The desired sum is: $-2012-2011-\ldots-5-4-3+1+4+5+\ldots... | 2011 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. Among the numbers exceeding 2013, find the smallest even number $N$ for which the fraction $\frac{15 N-7}{22 N-5}$ is reducible. | Answer: 2144. Solution: The presence of a common factor in the numbers $15 N-7$ and $22 N-5$ implies that the same factor is present in the number $(22 N-5)-(15 N-7)=7 N+2$, and subsequently in the numbers $(15 N-7)-2 \cdot(7 N+2)=N-11,(7 N+2)-7 \cdot(N-11)=79$. Since 79 is a prime number, the fraction is reducible by ... | 2144 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.1. Find the area of the figure defined on the coordinate plane by the inequality $\sqrt{\arcsin \frac{x}{3}} \leq \sqrt{\arccos \frac{y}{3}} \cdot$ In your answer, specify the integer closest to the found value of the area. | Answer: 16. Solution. The domain of admissible values of the inequality is determined by the system $\left\{\begin{array}{l}\arcsin \frac{x}{3} \geq 0, \\ \arccos \frac{y}{3} \geq 0\end{array} \Leftrightarrow\left\{\begin{array}{c}0 \leq x \leq 3, \\ -3 \leq y \leq 3 .\end{array}\right.\right.$ If $-3 \leq y \leq 0$, t... | 16 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
7.1. Given two different geometric progressions, the first terms of which are equal to 1, and the sum of the denominators is 3. Find the sum of the fifth terms of these progressions, if the sum of the sixth terms is 573. If the answer to the question is not unique, specify the sum of all possible values of the desired ... | Answer: 161. Solution. Let the denominators of the progressions be $p$ and $q$. According to the condition, we get: $\left\{\begin{array}{c}p+q=3, \\ p^{5}+q^{5}=573\end{array}\right.$. We need to find $p^{4}+q^{4}$. Denoting $p+q=a, p q=b$, we express:
$$
\begin{gathered}
p^{4}+q^{4}=\left((p+q)^{2}-2 p q\right)^{2}-... | 161 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. Chords $A A^{\prime}, B B^{\prime}$, and $C C^{\prime}$ of a sphere intersect at a common point $S$. Find the sum $S A^{\prime}+S B^{\prime}+S C^{\prime}$, if $A S=6, B S=3, C S=2$, and the volumes of pyramids $S A B C$ and $S A^{\prime} B^{\prime} C^{\prime}$ are in the ratio $2: 9$. If the answer is not an integ... | Answer: 18. Solution. If $a, b, c$ are the given edges, $x, y, z$ are the corresponding extensions of these edges, and $k$ is the given ratio, then by the theorem of intersecting chords and the lemma on the ratio of volumes of pyramids with equal (vertical) trihedral angles at the vertex, we have: $\left\{\begin{array}... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.1. Find the sum of all such integers $a \in[0 ; 400]$, for each of which the equation $x^{4}-6 x^{2}+4=\sin \frac{\pi a}{200}-2\left[x^{2}\right]$ has exactly six roots. Here the standard notation is used: $[t]$ - the integer part of the number $t$ (the greatest integer not exceeding $t$). | Answer: 60100. Solution. Let $b=\sin \frac{\pi a}{200}, t=x^{2}$. Using the equality $[t]=t-\{t\}$ (here $\{t\}$ is the fractional part of $t$), we get that the original equation is equivalent to the equation $t^{2}-6 t+4=b-2 t+2\{t\}$. Note that any positive $t$ corresponds to two different $x$, two different positive... | 60100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In a box, there are a hundred colorful balls: 28 red, 20 green, 13 yellow, 19 blue, 11 white, and 9 black. What is the smallest number of balls that need to be pulled out without looking into the box to ensure that there are at least 15 balls of the same color among them? | Answer: 76. Solution. Worst-case scenario: 14 red, 14 green, 13 yellow, 14 blue, 11 white, and 9 black balls will be drawn - a total of 75 balls. The next ball will definitely be the 15th ball of one of the colors: either red, green, or blue.
Answer of variant 152: 109.
Answer of variant 153: 83.
Answer of variant 1... | 76 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In a convex quadrilateral $A B C D$, diagonals $A C$ and $D B$ are perpendicular to sides $D C$ and $A B$ respectively. A perpendicular is drawn from point $B$ to side $A D$, intersecting $A C$ at point $O$. Find $A O$, if $A B=4, O C=6$. | Answer: 2. Solution. Since $\angle A B D=\angle A C D=90^{\circ}$, the quadrilateral is inscribed in a circle with diameter $A D$. Therefore, $\angle A C B=\angle A D B$ - as angles subtending the same arc. Since $B H$ is the altitude in the right $\triangle A B D$, then $\angle A D B=\angle A B H$. Then triangles $A B... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Masha, bored during a math lesson, performed the following operation on a certain 2015-digit natural number: she discarded the last digit of its decimal representation and added twice the discarded digit to the number obtained by multiplying the remaining number by 3. She then performed the same operation on the res... | Answer: a) 17; b) 09 (the number $100 \ldots 0009$).
Solution. a) Let this number be $n$, ending in the digit $y$. Then $n=10 x+y$ after the next operation will become $3 x+2 y$. The equation $10 x+y=3 x+2 y$ is equivalent to $7 x=y$ and, since $y$ is a digit, then $y=7, x=1$. Therefore, $n=17$.
b) Note that if the i... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Find the largest four-digit number in which all digits are different, and moreover, no two of them can be swapped to form a smaller number. | Answer: 7089.
Solution. If the digit 0 is not used, then the digits in each such number must be in ascending order, and the largest of such numbers is 6789. Since a number cannot start with zero, using 0, one can obtain additional suitable numbers where 0 is in the hundreds place, and the other digits are in ascending... | 7089 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 3. On the board, all such natural numbers from 3 to 223 inclusive are written, which when divided by 4 leave a remainder of 3. Every minute, Borya erases any two of the written numbers and instead writes their sum, decreased by 2. In the end, only one number remains on the board. What can it be? | Answer: 6218.
Solution. Initially, the total number of written numbers is $224: 4=56$, and their sum is $(3+223) \cdot 56: 2=6328$. Every minute, the number of written numbers decreases by 1, and their sum decreases by 2. Therefore, one number will remain on the board after 55 minutes and will be equal to $6328-55 \cd... | 6218 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Masha has seven different dolls, which she arranges in six different doll houses so that each house has at least one doll. In how many ways can Masha do this? It is important which doll ends up in which house. It does not matter how the dolls are seated in the house where there are two of them. | Answer: 15120.
Solution. In each such seating arrangement, in one of the cabins there will be two dolls, and in the other cabins - one each. First, we choose which two dolls will sit in one cabin. This can be done in $7 \cdot 6: 2=21$ ways. Then we choose which cabin to seat these two dolls in. This can be done in 6 w... | 15120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. It is known that $x+y+z=2016$ and $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{1}{32}$. Find $\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}$
ANSWER: 60. | Solution: Add 1 to each fraction, we get $\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}+1=\frac{x+y+z}{x+y}+$ $\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}=2016 \cdot\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{2016}{32}=63$. | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. There are 5 identical buckets, each with a maximum capacity of some integer number of liters, and a 30-liter barrel containing an integer number of liters of water. The water from the barrel was distributed among the buckets, with the first bucket being half full, the second one-third full, the third one-quarter ful... | The solution $-\frac{x}{2}+\frac{x}{3}+\frac{x}{4}+\frac{x}{5}+\frac{x}{6}=\frac{87 x}{60}=\frac{29 x}{20}$ will be an integer only when $x$ is a multiple of 20. But $x>20$ cannot be taken, as the sum will exceed 30 liters. | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Two oligarchs, Alejandro and Maximilian, looted their country in 2012. It is known that Alejandro's wealth at the end of 2012 equals twice Maximilian's wealth at the end of 2011. And Maximilian's wealth at the end of 2012 is less than Alejandro's wealth at the end of 2011. Which is greater: Maximilian's wealth or th... | Answer: Maximilian's state is greater.
Solution: Consider the table where $z, x-$ are the states of Alejandro and Maximilian in 2011:
| | 2011 | 2012 |
| :--- | :--- | :--- |
| $\mathrm{A}$ | $z$ | $2 x$ |
| $\mathrm{M}$ | $x$ | $y$ |
Then $N=(2 x+y)-(x+z)-$ are the national riches of the country. Subtracting $x$ f... | 80 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Several boys and girls gathered around a round table. It is known that exactly for 7 girls, girls are sitting to their left, and for 12 - boys. It is also known that for $75\%$ of the boys, girls are sitting to their right. How many people are sitting at the table? | Answer: 35 people.
Solution: From the condition, it is clear that there are exactly 19 girls.
Note that the number of girls who have boys sitting to their left is equal to the number of boys who have girls sitting to their right. Thus, $75\%$ of the boys equals 12, i.e., there are 16 boys in total sitting at the tabl... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$ | Answer: 18108.
Solution: Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace{4 \ldots 4}_{2012}=\underbrace{4 \ldots 4}_{2011} 3 \underbrace{5 \ldots 5}_{2011} 6$. The sum of the digits is $4 \cdot 2011+3+5 \cdot 2011+... | 18108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. A trip in Moscow using the "Troyka" card in 2016 costs 32 rubles for one subway ride and 31 rubles for one trip on surface transport. What is the minimum total number of trips that can be made under these rates, spending exactly 5000 rubles? | Answer: 157.
Solution: If $x$ is the number of trips by surface transport and $y$ is the number of trips by subway, then we have $31x + 32y = 5000$, from which $32(x + y) = 5000 + x \geqslant 5000$. Therefore, $x + y \geqslant 5000 / 32 = 156 \frac{1}{4}$. The smallest integer value of $x + y$ is 157, which is achieve... | 157 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.1. Determine the number of natural divisors of the number $11!=1 \cdot 2 \cdot \ldots \cdot 10 \cdot 11$ that are multiples of three. Answer. 432. | Solution. The prime factorization of the given number is $11!=2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7 \cdot 11$. All multiples of three that are divisors of this number have the form $2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} \cdot 7^{\delta} \cdot 11^{\varphi}$, where $\alpha \in[0 ; 8], \beta \in[1 ; 4], \gamma \in[0... | 432 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. Solve the inequality
$$
8 \cdot \frac{|x+1|-|x-7|}{|2 x-3|-|2 x-9|}+3 \cdot \frac{|x+1|+|x-7|}{|2 x-3|+|2 x-9|} \leqslant 8
$$
In the answer, write the sum of its integer solutions that satisfy the condition $|x|<120$. | Answer: 6 (solution of the inequality: $\left.\left[\frac{3}{2} ; 3\right) \cup\left(3 ; \frac{9}{2}\right]\right)$.
Solution. Let $a$ and $b$ be the first and second terms in the left part of the inequality, respectively. Then $b>0$ and
$$
a b=24 \cdot \frac{(x+1)^{2}-(x-7)^{2}}{(2 x-3)^{2}-(2 x-9)^{2}}=24 \cdot \fr... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5.3. Solve the inequality
$$
12 \cdot \frac{|x+10|-|x-20|}{|4 x-25|-|4 x-15|}-\frac{|x+10|+|x-20|}{|4 x-25|+|4 x-15|} \geqslant-6
$$
In the answer, write the sum of its integer solutions that satisfy the condition $|x|<100$. | Answer: 10 (solution of the inequality: $\left.\left[\frac{15}{4} ; 5\right) \cup\left(5 ; \frac{25}{4}\right]\right)$.) | 10 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5.4. Solve the inequality
$$
9 \cdot \frac{|x+4|-|x-2|}{|3 x+14|-|3 x-8|}+11 \cdot \frac{|x+4|+|x-2|}{|3 x+14|+|3 x-8|} \leqslant 6
$$
In the answer, write the sum of its integer solutions that satisfy the condition $|x|<110$. | Answer: -6 (solution of the inequality: $[-4 ;-1) \cup(-1 ; 2]$ ). | -6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6.1. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $-1,4,6,9,10,11,15,16,20,22$. Determine which numbers are written on the board. In your answer, write their product. | Answer: -4914 (numbers on the board: $-3,2,7,9,13$).
Solution. The sum of the numbers in the obtained set is 112. Each of the original five numbers appears 4 times in this sum. Therefore, the sum of the required numbers is $112: 4=28$. The sum of the two smallest numbers is -1, and the sum of the two largest numbers i... | -4914 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.2. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $3,8,9,16,17,17,18,22,23,31$. Determine which numbers are written on the board. In your answer, write their product. | Answer: 3360 (numbers on the board: $1,2,7,15,16$). | 3360 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.3. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $2,6,10,10,12,14,16,18,20,24$. Determine which numbers are written on the board. In your answer, write their product | Answer: -3003 (numbers on the board: $-1,3,7,11,13$ ). | -3003 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.4. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $6,9,10,13,13,14,17,17,20,21$. Determine which numbers are written on the board. In your answer, write their product. | Answer: 4320 (numbers on the board: $1,5,8,9,12$ ). | 4320 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.5. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $-1,5,8,9,11,12,14,18,20,24$. Determine which numbers are written on the board. In your answer, write their product. | Answer: -2002 (numbers on the board: $-2,1,7,11,13$ ). | -2002 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.6. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $5,8,9,13,14,14,15,17,18,23$. Determine which numbers are written on the board. In your answer, write their product. | Answer: 4752 (numbers on the board: $2,3,6,11,12$). | 4752 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.7. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $-1,2,6,7,8,11,13,14,16,20$. Determine which numbers are written on the board. In your answer, write their product. | Answer: -2970 (numbers on the board: $-3,2,5,9,11$). | -2970 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.8. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $5,9,10,11,12,16,16,17,21,23$. Determine which numbers are written on the board. In your answer, write their product. | Answer: 5292 (numbers on the board: $2,3,7,9,14$). | 5292 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. In the expansion of the function $f(x)=\left(1+x-x^{2}\right)^{20}$ in powers of $x$, find the coefficient of $x^{3 n}$, where $n$ is equal to the sum of all coefficients of the expansion. | Answer: 760.
Solution. The expansion in powers of $x$ has the form $f(x)=\sum_{k=0}^{40} a_{k} x^{k}$. The sum of all coefficients is $\sum_{k=0}^{40} a_{k}=f(1)=1$. Therefore, we need to find the coefficient of $x^{3}$. We have:
$$
\begin{aligned}
f(x)=\left(\left(1-x^{2}\right)+x\right)^{20}=(1- & \left.x^{2}\right... | 760 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. Find the smallest four-digit number that is not a multiple of 10 and has the following property: if the digits are reversed, the resulting number is a divisor of the original number, and the quotient is different from one.
Answer. 8712. | Solution. Let's find all four-digit numbers with the property specified in the problem. Let $\overline{a b c d}$ be the desired number, then $\overline{c b a d}$ is the number obtained from it by reversing the digits, $a, d \neq 0$. According to the condition, $\overline{a b c d}=k \cdot \overline{c b a d}$, where $k$ ... | 8712 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.1. Propose a word problem that reduces to solving the inequality
$$
\frac{11}{x+1.5}+\frac{8}{x} \geqslant \frac{12}{x+2}+2
$$
Write the problem statement, its solution, and the answer.
Example of the required problem. Points A and B are connected by two roads: one is 19 km long, and the other is 12 km long. At 1... | Answer: no more than 4 km/h. | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. It is known that $x+y+z=2016$ and $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{7}{224}$. Find $\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}$ ANSWER:60. | Solution: Add 1 to each fraction, we get $\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}+1=\frac{x+y+z}{x+y}+$ $\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}=2016 \cdot\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{2016}{32}=63$. | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Construct on the plane the set of points whose coordinates satisfy the inequality $|3 x+4|+|4 y-3| \leq 12$. In your answer, specify the area of the resulting figure.
ANSWER: 24. | Solution: We will shift the coordinate axes by $-\frac{4}{3}$ along the Ox axis, and by $\frac{3}{4}$ along the Oy axis. In the new coordinates, the obtained figure is described by the inequality $|3 x|+|4 y| \leq 12$. It is not difficult to show that this will be a rhombus with diagonals of 6 and 8, therefore, its are... | 24 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. Find the largest ten-digit number of the form $\overline{a_{9} a_{8} a_{7} a_{6} a_{5} a_{4} a_{3} a_{2} a_{1} a_{0}}$, possessing the following property: the digit equal to $\mathrm{a}_{\mathrm{i}}$ appears in its representation exactly $\mathrm{a}_{9-\mathrm{i}}$ times (for example, the digit equal to $\mathrm{a}_... | Solution: We will divide all digits into pairs $a_{0}-a_{9}, \ldots, a_{4}-a_{5}$. We will prove that the digit 9 cannot be present in the given number. Suppose $a_{i}=9$, then the paired digit $a_{9-i}$ appears 9 times. If $a_{9-i} \neq 9$, then 9 appears only once, so we get a number consisting of one nine and nine o... | 8888228888 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.1. (2 points) On two bookshelves, there are mathematics books, with an equal number on each. If 5 books are moved from the first shelf to the second, then the second shelf will have twice as many books as the first. How many books are there in total on both shelves? | Answer: 30.
Solution. If $x$ is the number of books on both shelves, then $\frac{x}{3}+5=\frac{2 x}{3}-5$, from which $x=30$. | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.1. (14 points) A team consisting of juniors and masters from the "Vimpel" sports society went to a shooting tournament. The average number of points scored by the juniors turned out to be 22, by the masters - 47, and the average number of points for the entire team - 41. What is the share (in percent) of masters in t... | Answer: 76.
Solution. Let there be $x$ juniors and $y$ masters in the team. Then the total number of points scored by the team is $22x + 47y = 41(x + y)$, from which we find $19x = 6y$. Therefore, the proportion of masters is $\frac{y}{x+y} = \frac{19y}{19x + 19y} = \frac{19y}{25y} = 0.76$, i.e., $76\%$. | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. (14 points) Calculate $x^{3}+\frac{1}{x^{3}}$, given that $x+\frac{1}{x}=3$. | Answer: 18.
Solution. If $x+\frac{1}{x}=a$, then $a^{3}=\left(x+\frac{1}{x}\right)^{3}=x^{3}+3\left(x+\frac{1}{x}\right)+\frac{1}{x^{3}}$, from which $x^{3}+\frac{1}{x^{3}}=a^{3}-3 a=18$. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. (14 points) An eraser, 3 pens, and 2 markers cost 240 rubles, while 2 erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers? | Answer: 520.
Solution. From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will cost $2 \cdot 240=480$ rubles. Therefore, one pen costs $480-440=40$ rubles. Then, 3 erasers, 4 pens, and 6 markers cost $680-4 \cdot 40=520$ rubles. | 520 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. (14 points) In an acute-angled triangle $A B C$, angle $A$ is equal to $35^{\circ}$, segments $B B_{1}$ and $C C_{1}$ are altitudes, points $B_{2}$ and $C_{2}$ are the midpoints of sides $A C$ and $A B$ respectively. Lines $B_{1} C_{2}$ and $C_{1} B_{2}$ intersect at point $K$. Find the measure (in degrees) of ang... | Answer: 75.
Solution. Note that angles $B$ and $C$ of triangle $ABC$ are greater than $\angle A=35^{\circ}$ (otherwise it would be an obtuse triangle), so point $C_{1}$ lies on side $AB$ be... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.1. (14 points) In a hut, several residents of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The residents of the Ah tribe always tell the truth, while the residents of the Ukh tribe always lie. One of the residents said: "There are no more than 16 of us in the hut," and then added: "We... | Answer: 15.
Solution. A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 peop... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.1. (14 points) Find the greatest integer value of $a$ for which the equation
$$
(x-a)(x-7)+3=0
$$
has at least one integer root. | Answer: 11.
Solution. For the desired values of $a$, the numbers $x-a$ and $x-7$ are integers, and their product is -3, so there are 4 possible cases:
$$
\left\{\begin{array} { l }
{ x - a = 1 , } \\
{ x - 7 = - 3 ; }
\end{array} \quad \left\{\begin{array} { l }
{ x - a = 3 , } \\
{ x - 7 = - 1 ; }
\end{array} \qua... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The second term of a geometric progression is 5, and the third term is 1. Find the first term of this progression. | Answer: 25. Solution. Since $\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=q$, then $a_{1}=\frac{a_{2}^{2}}{a_{3}}=25$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the area of a right triangle, one of whose legs is 6, and the hypotenuse is 10. | Answer: 24. Solution. According to the Pythagorean theorem, the second leg is equal to $\sqrt{10^{2}-6^{2}}=8$. Therefore, the area of the triangle is $\frac{6 \cdot 8}{2}=24$.
## Main Task | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.1. Businessmen Ivanov, Petrov, and Sidorov decided to create an automobile enterprise. Ivanov bought 70 identical cars for the enterprise, Petrov - 40 such cars, and Sidorov contributed 44 million rubles to the enterprise. It is known that Ivanov and Petrov can divide this money between themselves so that the contrib... | Answer: 40. Solution. Method 1. Each of the businessmen should contribute as much as Sidorov, that is, 44 million rubles. If the car costs $x$, then $\frac{70 x+40 x}{3}=44, x=1.2$. It turns out that Ivanov contributed $70 \cdot 1.2=84$ million rubles, so he should get back $84-44=40$ million rubles. Petrov contributed... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.1. How many 9-digit numbers divisible by 5 can be formed by rearranging the digits of the number 377353752? | Answer: 1120. Solution. Since the number is divisible by 5, the digit in the 9th place can only be five. After this, we need to distribute the remaining 8 digits across the 8 remaining places: 3 sevens, 3 threes, a five, and a two. The total number of permutations will be 8!, but since there are repeating digits, the a... | 1120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2.2. How many 9-digit numbers divisible by 2 can be formed by rearranging the digits of the number 131152152? | Answer: 840. Instructions. The number of numbers will be $\frac{8!}{4!2!}=840$. | 840 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2.3. How many 9-digit numbers divisible by 5 can be formed by rearranging the digits of the number 137153751? | Answer: 1680. Instructions. The number of numbers will be $\frac{8!}{3!2!2!}=1680$. | 1680 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2.4. How many 9-digit numbers divisible by 2 can be formed by rearranging the digits of the number 231157152? | Answer: 3360. Instructions. The number of numbers will be $\frac{8!}{3!2!}=3360$. | 3360 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3.1. Determine the total surface area of a cube if the distance between non-intersecting diagonals of two adjacent faces of this cube is 8. If the answer is not an integer, round it to the nearest integer. | Answer: 1152. Solution. Let's take as the diagonals specified in the condition the diagonals $A_{1} C_{1}$ and $A D_{1}$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$. Construct two parallel planes $A_{1} C_{1} B$ and $A D_{1} C$, each containing one of these diagonals. The distance between these planes is equal to the... | 1152 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.1. Find the sum of all roots of the equation $x^{2}-31 x+220=2^{x}\left(31-2 x-2^{x}\right)$. | Answer: 7. Solution. The original equation is equivalent to the equation $\left(x+2^{x}\right)^{2}-31\left(x+2^{x}\right)+220=0 \Leftrightarrow\left[\begin{array}{l}x+2^{x}=11, \\ x+2^{x}=20 .\end{array}\right.$
Each of the equations in this system has no more than one root, as the function $f(x)=x+2^{x}$ is monotonic... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.2. Find the sum of all roots of the equation $x^{2}-41 x+330=3^{x}\left(41-2 x-3^{x}\right)$. | Answer: 5. Instructions. Roots of the equation: 2 and 3. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. Among all the simple fractions, where the numerator and denominator are two-digit numbers, find the smallest fraction greater than $\frac{3}{4}$. In your answer, specify its numerator. | Answer: 73. Solution. It is required to find a fraction $\frac{a}{b}$ such that $\frac{a}{b}-\frac{3}{4}=\frac{4 a-3 b}{4 b}$ reaches a minimum. Therefore, the maximum two-digit $b$ is sought, for which $4 a-3 b=1$. If in this case $b \geq 50$, then the fraction $\frac{a}{b}-\frac{3}{4}=\frac{1}{4 b}$ will always be le... | 73 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.2. Among all the irreducible fractions, where the numerator and denominator are two-digit numbers, find the smallest fraction greater than $\frac{5}{6}$. In your answer, specify its numerator. | Answer: 81. Instructions. The required fraction: $\frac{81}{97}$. | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.3. Among all the irreducible fractions, where the numerator and denominator are two-digit numbers, find the smallest fraction greater than $\frac{4}{5}$. In your answer, specify its numerator. | Answer: 77. Instructions. The required fraction: $\frac{77}{96}$. | 77 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.6. Among all the irreducible fractions, where the numerator and denominator are two-digit numbers, find the smallest fraction greater than $\frac{4}{9}$. In your answer, specify its numerator. | Answer: 41. Instructions. The required fraction: $\frac{41}{92}$. | 41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.1. In an isosceles triangle \(ABC\), one of the angles is equal to the difference of the other two, and one of the angles is twice another. The angle bisectors of angles \(A\), \(B\), and \(C\) intersect the circumcircle of the triangle at points \(L\), \(O\), and \(M\) respectively. Find the area of triangle \(LOM\)... | Answer: 11. Solution. Let in triangle $ABC$ the angle $\alpha$ be equal to the difference of angles $\beta-\gamma$. Since $\alpha+\beta+\gamma=180^{\circ}$, then $\beta-\gamma+\beta+\gamma=180^{\circ}$ and therefore $\beta=90^{\circ}$. If $\beta=90^{\circ}$ is twice one of the angles $\alpha$ or $\gamma$, then the tria... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.4. In an isosceles triangle \(ABC\), one of the angles is equal to the difference of the other two, and one of the angles is twice another. The angle bisectors of angles \(A\), \(B\), and \(C\) intersect the circumcircle of the triangle at points \(L\), \(O\), and \(M\) respectively. Find the area of triangle \(LOM\)... | Answer: 27. Instructions. Exact answer: $10 \sqrt{3}+10$. | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.2. Solve the inequality $\sqrt{x^{2}-x-56}-\sqrt{x^{2}-25 x+136}<8 \sqrt{\frac{x+7}{x-8}}$, and find the sum of its integer solutions that belong to the interval $[-25 ; 25]$. | Answer: -285. Instructions. Solution of the inequality: $x \in(-\infty ;-7] \cup(18 ; 20)$. | -285 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
8.3. Solve the inequality $\sqrt{x^{2}+3 x-54}-\sqrt{x^{2}+27 x+162}<8 \sqrt{\frac{x-6}{x+9}}$, and find the sum of its integer solutions that belong to the interval $[-25 ; 25]$. | Answer: 290 . Instructions. Solution of the inequality: $x \in(-21 ;-19) \bigcup[6 ;+\infty)$. | 290 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
8.4. Solve the inequality $\sqrt{x^{2}+x-56}-\sqrt{x^{2}+25 x+136}<8 \sqrt{\frac{x-7}{x+8}}$, and find the sum of its integer solutions that belong to the interval $[-25 ; 25]$. | Answer: 285. Instructions. Solution of the inequality: $x \in(-20 ;-18) \cup[7 ;+\infty)$. | 285 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9.1. Find the maximum value of the expression $(\sqrt{8-4 \sqrt{3}} \sin x-3 \sqrt{2(1+\cos 2 x)}-2) \cdot(3+2 \sqrt{11-\sqrt{3}} \cos y-\cos 2 y)$. If the answer is not an integer, round it to the nearest integer. | Answer: 33. Solution. Let $f(x)=\sqrt{8-4 \sqrt{3}} \sin x-3 \sqrt{2(1+\cos 2 x)}-2$, $g(y)=3+2 \sqrt{11-\sqrt{3}} \cos y-\cos 2 y$ and estimate these two functions.
1) $f(x)=\sqrt{8-4 \sqrt{3}} \sin x-6|\cos x|-2 \leq \sqrt{8-4 \sqrt{3}} \cdot 1-6 \cdot 0-2=\sqrt{8-4 \sqrt{3}}-2=\sqrt{(\sqrt{6}-\sqrt{2})^{2}}-2$ $=\s... | 33 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4. Find the minimum value of the expression $(\sqrt{2(1+\cos 2 x)}-\sqrt{3-\sqrt{2}} \sin x+1) \cdot(3+2 \sqrt{7-\sqrt{2}} \cos y-\cos 2 y)$. If the answer is not an integer, round it to the nearest integer. | Answer: -9 . Instructions. Exact answer: $2 \sqrt{2}-12$. | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.5. Find the maximum value of the expression $(\sqrt{36-4 \sqrt{5}} \sin x-\sqrt{2(1+\cos 2 x)}-2) \cdot(3+2 \sqrt{10-\sqrt{5}} \cos y-\cos 2 y) \cdot$ If the answer is not an integer, round it to the nearest integer. | Answer: 27. Instructions. Exact answer: $36-4 \sqrt{5}$. | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.6. Find the minimum value of the expression $(\sqrt{2(1+\cos 2 x)}-\sqrt{36-4 \sqrt{5}} \sin x+2) \cdot(3+2 \sqrt{10-\sqrt{5}} \cos y-\cos 2 y) \cdot$ If the answer is not an integer, round it to the nearest integer. | Answer: -27 . Instructions. Exact answer: $4 \sqrt{5}-36$. | -27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. Find the largest integer $k$ such that for at least one natural number $n>1000$, the number $n!=1 \cdot 2 \cdot \ldots \cdot n$ is divisible by $2^{n+k+2}$. | Answer: -3. Solution. The highest power of two that divides $n!$ is the finite sum $\left[\frac{n}{2^{1}}\right]+\left[\frac{n}{2^{2}}\right]+\ldots \leq \frac{n}{2^{1}}+\frac{n}{2^{2}}+\ldots<n$, so it does not exceed $n-1$. At the same time, the equality of this power to the value $n-1$ is achieved at powers of two (... | -3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Several boys and girls gathered around a round table. It is known that exactly for 7 girls, girls are sitting to their left, and for 12 - boys. It is also known that for $75\%$ of the boys, girls are sitting to their right. How many people are sitting at the table? | Answer: 35 people.
Solution: From the condition, it is clear that there are exactly 19 girls.
Note that the number of girls who have boys sitting to their left is equal to the number of boys who have girls sitting to their right. Thus, $75\%$ of the boys equals 12, i.e., there are 16 boys in total sitting at the tabl... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$ | Answer: 18108.
Solution: Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace{4 \ldots 4}_{2012}=\underbrace{4 \ldots 4}_{2011} 3 \underbrace{5 \ldots 5}_{2011} 6$. The sum of the digits is $4 \cdot 2011+3+5 \cdot 2011+... | 18108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. A set of natural numbers is called bad if it is possible to select several numbers from it such that their sum equals 2012. Find the smallest such $n$ that the numbers $503, 504, \ldots, 2011$ can be partitioned into $n$ sets such that none of these sets are bad. | Answer: $n=2$.
Solution: We will show that the given set can be divided into two subsets, neither of which are bad. Consider the sets $M_{1}=\{503,504, \ldots, 671\}$, $M_{2}=\{672,673, \ldots, 1006\}$, $M_{3}=\{1007,1008, \ldots, 1340\}$, $M_{4}=\{1341,1342, \ldots, 2011\}$, and show that $M_{1} \cup M_{3}$ and $M_{2... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1.1.1. (2 points) Find the sum of the squares of two numbers if it is known that their arithmetic mean is 8, and the geometric mean is $2 \sqrt{5}$. | Answer: 216.
Solution. If $a$ and $b$ are the numbers in question, then $a+b=16, a b=(2 \sqrt{5})^{2}=20$, therefore
$$
a^{2}+b^{2}=(a+b)^{2}-2 a b=256-40=216 .
$$ | 216 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.1.1. (2 points) Find the greatest value of $x$ that satisfies the inequality
$$
\left(6+5 x+x^{2}\right) \sqrt{2 x^{2}-x^{3}-x} \leqslant 0
$$ | Answer: 1.
Solution.
$$
\left[\begin{array} { c }
{ 2 x ^ { 2 } - x ^ { 3 } - x = 0 ; } \\
{ \{ \begin{array} { c }
{ 2 x ^ { 2 } - x ^ { 3 } - x > 0 ; } \\
{ 6 + 5 x + x ^ { 2 } \leqslant 0 ; }
\end{array} }
\end{array} \Longleftrightarrow \left[\begin{array} { l }
{ - x ( x - 1 ) ^ { 2 } = 0 ; } \\
{ \{ \begin{a... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3.2.1. (12 points) The equation $x^{2}+5 x+1=0$ has roots $x_{1}$ and $x_{2}$. Find the value of the expression
$$
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}
$$ | Answer: 220.
Solution. Since $x_{1}^{2}=-5 x_{1}-1$, then $\left(1+x_{1}\right)^{2}=1+2 x_{1}-5 x_{1}-1=-3 x_{1}$. Therefore,
$$
\begin{gathered}
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}=6\left(\frac{-5 x_{1}-1}{-3 x_{2}}+\frac{-5 x_{2}-1}{-3 x_{1}}\right)=\frac{... | 220 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.3.1. (12 points) Simplify the expression
$$
\left(\frac{2}{\sqrt[3]{3}}+3\right)-\left(\frac{\sqrt[3]{3}+1}{2}-\frac{1}{\sqrt[3]{9}+\sqrt[3]{3}+1}-\frac{2}{1-\sqrt[3]{3}}\right): \frac{3+\sqrt[3]{9}+2 \sqrt[3]{3}}{2}
$$
Answer: 3. | Solution. Let's introduce the notation $a=\sqrt[3]{3}$. Then the expression transforms to
$$
\left(\frac{2}{a}+3\right)-\left(\frac{a+1}{a^{3}-1}-\frac{1}{a^{2}+a+1}-\frac{2}{1-a}\right): \frac{a^{3}+a^{2}+2 a}{a^{3}-1}
$$
and after simplifications and performing the division, it transforms to $\left(\frac{2}{a}+3\ri... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1.1. (12 points) From point $A$ to point $B$, a bus and a cyclist departed simultaneously at 13:00. After arriving at point $B$, the bus, without stopping, headed back and met the cyclist at point $C$ at 13:10. Upon returning to point $A$, the bus again, without stopping, headed to point $B$ and caught up with the cy... | Answer: 40.
Solution. Let $v_{1}$ and $v_{2}$ be the speeds (in km/h) of the cyclist and the bus, respectively. By the time of the first meeting, they have collectively traveled $\frac{1}{6} v_{1}+\frac{1}{6} v_{2}=8$ km. Until the next meeting, the time that has passed is equal to $\frac{s_{0}}{v_{1}}=\frac{2 \cdot \... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.2.1. (12 points) Every morning, each member of the Ivanov family drinks an 180-gram cup of coffee with milk. The amount of milk and coffee in their cups varies. Masha Ivanova found out that she drank $\frac{2}{9}$ of all the milk consumed that morning and $\frac{1}{6}$ of all the coffee consumed that morning. How man... | Answer: 5.
Solution. Let there be $x$ (for example, grams) of milk and $y$ grams of coffee, and $n$ people in the family. Since each family member drank the same amount of coffee with milk, then $\left(\frac{2 x}{9}+\frac{y}{6}\right) n=$ $=x+y \Leftrightarrow(4 x+3 y) n=18 x+18 y \Leftrightarrow 2 x(2 n-9)=3 y(6-n)$.... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.3.1. (12 points) On the table, there are 13 weights arranged in a row by mass (the lightest on the left, the heaviest on the right). It is known that the mass of each weight is an integer number of grams, the masses of any two adjacent weights differ by no more than 5 grams, and the total mass of the weights does not... | Answer: 185.
Solution. If the mass of the heaviest weight is $m$, then the masses of the other weights will be no less than $m-5, m-10, \ldots, m-60$ grams, and their total mass will be no less than $13 m-390$ grams. Then $13 m-390 \leq 2019$, from which $m \leq 185$. It remains to verify that the set of weights with ... | 185 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4.4.1. (12 points) A goat eats 1 hay wagon in 6 weeks, a sheep in 8 weeks, and a cow in 3 weeks. How many weeks will it take for 5 goats, 3 sheep, and 2 cows to eat 30 such hay wagons together? | Answer: 16.
Solution. The goat eats hay at a rate of $1 / 6$ cart per week, the sheep - at a rate of $1 / 8$ cart per week, the cow - at a rate of $1 / 3$ cart per week. Then 5 goats, 3 sheep, and 2 cows together will eat hay at a rate of $\frac{5}{6}+\frac{3}{8}+\frac{2}{3}=\frac{20+9+16}{24}=\frac{45}{24}=\frac{15}{... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1.1. (12 points) In an acute-angled triangle $A B C$, angle $A$ is equal to $35^{\circ}$, segments $B B_{1}$ and $C C_{1}$ are altitudes, points $B_{2}$ and $C_{2}$ are the midpoints of sides $A C$ and $A B$ respectively. Lines $B_{1} C_{2}$ and $C_{1} B_{2}$ intersect at point $K$. Find the measure (in degrees) of a... | Answer: 75.
Solution. Note that angles $B$ and $C$ of triangle $ABC$ are greater than $\angle A=35^{\circ}$ (otherwise it would be an obtuse triangle), so point $C_{1}$ lies on side $AB$ between points $B$ and $C_{2}$, and point $B_{1}$ lies on side $AC$ between points $C$ and $B_{2}$. Therefore, the point $K$ of inte... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.3.1. (12 points) Among all possible triangles $ABC$ such that $BC=2 \sqrt[4]{3}, \angle BAC=\frac{\pi}{3}$, find the one with the maximum area. What is this area? | Answer: 3.
Solution. The locus of points from which the segment $B C$ is "seen" at an angle $\alpha$ consists of arcs of two circles, from the centers of which the segment $B C$ is "seen" at an angle $2 \pi-2 \alpha$. The points on these arcs that are farthest from the segment are their midpoints. Therefore, the desir... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.2.1. (12 points) Solve the system of equations $\left\{\begin{array}{l}9 y^{2}-4 x^{2}=144-48 x, \\ 9 y^{2}+4 x^{2}=144+18 x y .\end{array}\right.$
Given the solutions $\left(x_{1} ; y_{1}\right),\left(x_{2} ; y_{2}\right), \ldots,\left(x_{n} ; y_{n}\right)$, write the sum of the squares
$$
x_{1}^{2}+x_{2}^{2}+\ldo... | Answer: 68.
Solution. From the first equation, we get $(3 y)^{2}=(2 x-12)^{2}$ and sequentially substitute into the second equation $y=\frac{2 x-12}{3}$ and $y=\frac{12-2 x}{3}$. The solutions obtained are: $(x ; y)=(0 ; 4),(0 ;-4),(6 ; 0)$. Answer: $6^{2}+4^{2}+(-4)^{2}=68$. | 68 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.4.1. (12 points) Solve the inequality $\sqrt{x-4}+\sqrt{x+1}+\sqrt{2 x}-\sqrt{33-x}>4$. In your answer, write the sum of all its integer solutions. | Answer: 525.
Solution. Rewrite the inequality as
$$
\sqrt{x-4}+\sqrt{x+1}+\sqrt{2 x}>\sqrt{33-x}+4
$$
Find the domain of the variable $x$:
$$
\left\{\begin{array}{l}
x-4 \geqslant 0 \\
x+1 \geqslant 0 \\
2 x \geqslant 0 \\
33-x \geqslant 0
\end{array}\right.
$$
from which $x \in[4,33]$. We are interested in the in... | 525 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
7.2.1. (12 points) Calculate the value of the expression $\arccos \frac{\sqrt{6}+1}{2 \sqrt{3}}-\arccos \sqrt{\frac{2}{3}}$. Write the obtained expression in the form $\frac{a \pi}{b}$, where $a$ and $b$ are integers, and are coprime, and specify the value of $|a-b|$.
# | # Answer: 7.
Solution. Since $\alpha=\arccos \frac{\sqrt{6}+1}{2 \sqrt{3}} \in\left(0 ; \frac{\pi}{2}\right)$ and $\beta=\arccos \sqrt{\frac{2}{3}} \in\left(0 ; \frac{\pi}{2}\right)$, then $A=\alpha-\beta \in\left(-\frac{\pi}{2} ; \frac{\pi}{2}\right)$. Therefore, $\sin A=\sin \alpha \cos \beta-\cos \alpha \sin \beta$... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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