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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 |
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 |
6.4. 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 |
9.5. 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 |
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 |
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 |
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 |
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 |
4.3.1. (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 |
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 |
8.1.1. (12 points) Among the first hundred elements of the arithmetic progression $3,7,11, \ldots$ find those that are also elements of the arithmetic progression $2,9,16, \ldots$ In your answer, indicate the sum of the found numbers. | Answer: 2870.
Solution.
$$
\begin{gathered}
a_{n}=a_{1}+(n-1) d_{1}=3+(n-1) \cdot 4 \\
b_{m}=b_{1}+(m-1) d_{2}=2+(m-1) \cdot 7 \\
3+(n-1) \cdot 4=2+(m-1) \cdot 7 \\
4(n+1)=7 m, \quad m=4 k, \quad n=7 k-1
\end{gathered}
$$
Consider the sequence of coinciding terms of the progressions $A_{k}$. For $k=1$, we find $n=6$... | 2870 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The sequence $a_{1}, a_{2}, \ldots$ is defined by the equalities
$$
a_{1}=100, \quad a_{n+1}=a_{n}+\frac{1}{a_{n}}, \quad n \in \mathbb{N}
$$
Find the integer closest to $a_{2013}$. | # Answer: 118.
Solution.
\[
\begin{aligned}
a_{2013}^{2}=\left(a_{2012}+\frac{1}{a_{2012}}\right)^{2}=a_{2012}^{2}+2+\frac{1}{a_{2012}^{2}}=a_{2011}^{2} & +2 \cdot 2+\frac{1}{a_{2011}^{2}}+\frac{1}{a_{2012}^{2}}=\ldots \\
& =a_{1}^{2}+2 \cdot 2012+\frac{1}{a_{1}^{2}}+\ldots+\frac{1}{a_{2011}^{2}}+\frac{1}{a_{2012}^{2... | 118 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The quiz participants were asked four questions: 90 participants answered the first question correctly, 50 answered the second, 40 answered the third, and 20 answered the fourth, and no one was able to answer more than two questions correctly. What is the minimum number of participants in the quiz under these condit... | Answer: 100.
Solution. The total number of answers is 200. Since no one answered more than two questions, the minimum possible number of participants in the quiz is 100, and in this case, each participant in the quiz must answer exactly 2 questions correctly. We will provide an example of a quiz where the described si... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. The numbers $a, b$, and $c$ satisfy the equation $\sqrt{a}=\sqrt{b}+\sqrt{c}$. Find $a$, if $b=52-30 \sqrt{3}$ and $c=a-2$. | Answer: $a=27$.
Solution. We have
$$
\sqrt{b}=\sqrt{52-30 \sqrt{3}}=\sqrt{27-2 \cdot 5 \cdot 3 \sqrt{3}+25}=3 \sqrt{3}-5
$$
Therefore, $\sqrt{a}-\sqrt{a-2}=3 \sqrt{3}-5, \frac{2}{\sqrt{a}+\sqrt{a-2}}=\frac{2}{3 \sqrt{3}+5}, \sqrt{a}+\sqrt{a-2}=\sqrt{27}+\sqrt{25}$. Since the function $f(a)=\sqrt{a}+\sqrt{a-2}$ is in... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Lёsha's cottage plot has the shape of a nonagon, which has three pairs of equal and parallel sides (see figure). Lёsha knows that the area of the triangle with vertices at the midpoints of the remaining sides of the nonagon is 12 acres. Help him find the area of the entire cottage plot.
 sides. From the condition, it follows that the quadrilateral $D F G E$ is a parallelogram, since its opposite sides $D E$ and $F G$ are equal and parallel. In... | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. In a right triangle $ABC$ with a right angle at $C$, points $P$ and $Q$ are the midpoints of the angle bisectors drawn from vertices $A$ and $B$. The inscribed circle of the triangle touches the hypotenuse at point $H$. Find the angle $PHQ$. | Answer: $90^{\circ}$.
Solution. First, let's prove the following lemma.
Lemma. Let $ABC (\angle C=90^{\circ})$ be a right triangle, $I$ the intersection point of the angle bisectors $AM$ and $BK$, $S$ the midpoint of $KM$, and the inscribed circle of the triangle touches the hypotenuse at point $H$. Then the points $... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 2. Find the number of natural numbers not exceeding 2022 and not belonging to either the arithmetic progression $1,3,5, \ldots$ or the arithmetic progression $1,4,7, \ldots$ | Answer: 674.
Solution. These two progressions define numbers of the form $1+2n$ and $1+3n$. This indicates that the desired numbers are of the form $6n$ and $6n-4, n \in \mathbb{N}$. Since 2022 is divisible by 6, the numbers of the form $6n$ will be $\frac{2022}{6}=337$, and there will be as many numbers of the form $... | 674 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 3. Find the three last digits of the number $10^{2022}-9^{2022}$. | Answer: 119.
Solution. Since $A=10^{2022}-(10-1)^{2022}=10^{2022}-10^{2022}+2022 \cdot 10^{2021}-C_{2022}^{2} \cdot 10^{2022}+\ldots+$ $C_{2022}^{3} \cdot 10^{3}-C_{2022}^{2} \cdot 10^{2}+C_{2022}^{1} \cdot 10-1$, then $A(\bmod 1000) \equiv-C_{2022}^{2} \cdot 100+C_{2022}^{1} \cdot 10-1(\bmod 1000) \equiv$ $-\frac{202... | 119 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. There is a certain number of identical plastic bags that can be placed inside each other. If all the other bags end up inside one of the bags, we will call this situation a "bag of bags." Calculate the number of ways to form a "bag of bags" from 10 bags.
Explanation. Denote the bag with parentheses.
If we ... | Answer: 719.
Solution. If $\Pi_{n}$ denotes the number of ways for $n$ packages, then:
$$
\begin{gathered}
\Pi_{1}=1, \Pi_{2}=1, \Pi_{3}=2, \Pi_{4}=4, \Pi_{5}=9, \Pi_{6}=20, \Pi_{7}=48, \Pi_{8}=115, \Pi_{9}=286 \\
\Pi_{10}=719
\end{gathered}
$$
The problem is solved by enumerating the cases. For example, if we take ... | 719 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5.1. (12 points) A guard has detained a stranger and wants to drive him away. But the person caught said that he had made a bet with his friends for 100 coins that the guard would not drive him away (if the guard does, he has to pay his friends 100 coins, otherwise they pay him), and, deciding to bribe the guard, offer... | Answer: 199.
Solution. If the guard asks for 199 coins, then the outsider, agreeing, will give him this amount, but will win the dispute and receive 100 coins. In total, he will lose 99 coins. If the outsider refuses, he will lose the dispute and lose 100 coins, which is less favorable (by 1 coin) for the one caught. ... | 199 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.1. (12 points) Find the smallest natural number such that after multiplying it by 9, the result is a number written with the same digits but in some different order. | Answer: 1089.
Solution. Note that the number must start with one, otherwise multiplying by 9 would increase the number of digits. After multiplying 1 by 9, we get 9, so the original number must contain the digit 9. The number 19 does not work, so two-digit numbers do not work. Let's consider three-digit numbers. The s... | 1089 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.1. (12 points) The surface of a round table is divided into $n$ identical sectors, in which numbers from 1 to $n (n \geqslant 4)$ are written sequentially clockwise. Around the table sit $n$ players with numbers $1,2, \ldots, n$, going clockwise. The table can rotate around its axis in both directions, while the pla... | Answer: 69.
Solution. Let player No. 3 get one coin exactly $k$ times, then we get the equation $(m-k)-(n-1) k=50$. Exactly in $7 k$ cases, one coin was received by someone from players $2,3,4$, then we get the equation $3(m-7 k)-7 k(n-3)=74$. We expand the brackets, combine like terms, and get a system of two equatio... | 69 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Find the area of a right triangle if the height drawn from the right angle divides it into two triangles, the radii of the inscribed circles of which are 3 and 4. | Answer: 150.
Solution. The ratio of the radii of the inscribed circles is equal to the similarity coefficient of the right triangles into which the height divides the original triangle. This coefficient is equal to the ratio of the legs of the original triangle. Let these legs be denoted as $3x$ and $4x$. By the Pytha... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. All natural numbers, the sum of the digits of each of which is equal to 5, were arranged in ascending order. What number is in the 125th place
# | # Answer: 41000.
Solution. Let's calculate the number of $n$-digit numbers, the sum of the digits of each of which is equal to 5, for each natural $n$. Subtract 1 from the leading digit, we get a number (which can now start with zero), the sum of the digits of which is equal to 4. Represent the digits of this number a... | 41000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. To what power must the root $x_{0}$ of the equation $x^{11} + x^{7} + x^{3} = 1$ be raised to obtain the number $x_{0}^{4} + x_{0}^{3} - 1 ?$ | Answer: 15.
Solution. If $x_{0}=1$, then $x_{0}^{4}+x_{0}^{3}-1=1$, so in this case the degree can be any. But the number $x_{0}=1$ does not satisfy the equation $x^{11}+x^{7}+x^{3}=1$, therefore $x_{0} \neq 1$.
Since $1=x_{0}^{11}+x_{0}^{7}+x_{0}^{3}$, we get
$$
x_{0}^{4}+x_{0}^{3}-1=x_{0}^{4}+x_{0}^{3}-x_{0}^{11}-... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. At the disco, 42 people arrived: boys and girls. Each girl danced with all the boys, except for four, and each boy danced with all the girls, except for three. How many boys were at the dance? (Folklore) | Answer: 24. Solution: Let $m$ be the number of boys and $d$ be the number of girls who came to the disco. Denote the boys by blue dots, the girls by red dots, and connect the boys and girls who did not dance with each other by segments. Let there be a total of $k$ segments. Since 4 segments come out of each red dot and... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. The numbers from 1 to 2150 are written on a board. Every minute, each number undergoes the following operation: if the number is divisible by 100, it is divided by 100; if it is not divisible by 100, 1 is subtracted from it. Find the largest number on the board after 87 minutes. | Answer: 2012. Solution. All numbers, the last two digits of which are 86 or less, will transform into numbers ending in 00 within 87 minutes, and in the next step, they will decrease by a factor of 100. In the end, all such numbers will be no more than $2100 / 100=21$ after 87 minutes. Those numbers that end in 87 or m... | 2012 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. 99 wise men sat at a round table. They know that fifty of them are wearing hats of one of two colors, and the other forty-nine are wearing hats of the other color (but it is not known in advance which of the two colors 50 hats are, and which 49 are). Each wise man can see the colors of all the hats except his own. A... | Solution. Let there be 50 white and 49 black hats among the hats. It is clear that the 49 sages who see 50 white and 48 black hats know that they are wearing black hats. Now let each of those who see 49 white and black hats name the color that predominates among the 49 people following them clockwise. If A is one of th... | 74 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Let's call two positive integers almost adjacent if each of them is divisible (without a remainder) by their difference. During a math lesson, Vova was asked to write down in his notebook all numbers that are almost adjacent to $2^{10}$. How many numbers will he have to write down? | Answer: 21. Solution. The number $2^{10}$ is divisible only by powers of two: from $2^{0}$ to $2^{10}$. Therefore, the numbers that are almost adjacent to it can only be $2^{10}-2^{9}, 2^{10}-2^{8}, \ldots, 2^{10}-2^{0}, 2^{10}+2^{0}, \ldots, 2^{10}+2^{10}$ (the number $0=2^{10}-2^{10}$ does not count, as it is not pos... | 21 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Among the natural numbers $a_{1}, \ldots, a_{k}$, there are no identical ones, and the difference between the largest and the smallest of them is less than 1000. For what largest $k$ can it happen that all quadratic equations $a_{i} x^{2}+2 a_{i+1} x+a_{i+2}=0$, where $1 \leq i \leq k-2$, have no roots? (I. Bogdanov... | Answer. For $k=88$. Solution. Let the sequence $a_{1}, \ldots, a_{k}$ satisfy the condition of the problem. The absence of roots in the equations specified in the condition is equivalent to the inequality $\left(a_{i+1}\right)^{2}0$, from which $e-d>0$.
Let $a_{m}$ be the smallest number in the sequence. Obviously, on... | 88 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. There are 2009 piles, each containing 2 stones. It is allowed to take the largest pile from those in which the number of stones is even (if there are several, then any of them), and move exactly half of the stones from it to any other pile. What is the maximum number of stones that can be obtained in one pile using ... | Answer: 2010. Solution. The operations described in the condition do not reduce the numbers of stones in the heaps, so at any moment, there is at least one stone in each of them. Therefore, it is impossible to accumulate more than $2009 \cdot 2$ $2008=2010$ stones in one heap.
We will show how to obtain a heap with 20... | 2010 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. On each side of the square, 100 points are chosen, and from each chosen point, a segment perpendicular to the corresponding side of the square is drawn inside the square. It turns out that no two of the drawn segments lie on the same line. Mark all the intersection points of these segments. For what largest $k<200$ ... | Answer. For $k=150$. Solution. Estimation. Suppose there is an example with $k>150$. We will associate it with a $200 \times 200$ table, the rows of which correspond to horizontal segments (ordered from bottom to top), and the columns to vertical segments (ordered from left to right). In the cell of the table, there is... | 150 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. On the side AB of triangle ABC with an angle of $100^{\circ}$ at vertex C, points $P$ and $Q$ are taken such that $A P=B C$ and $B Q=A C$. Let $M, N, K-$ be the midpoints of segments $A B, C P, C Q$ respectively. Find the angle NМК. (M. Kungozhin + jury) | Answer: $40^{\circ}$. Solution: Extend the triangle to form a parallelogram $A C B D$. Then $M$ is the midpoint of segment $C D$.
Since $A P = B C = A D$ and $B Q = A C = B D$, triangles $A P... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. On the hundredth year of his reign, the Immortal Treasurer decided to start issuing new coins. In this year, he put into circulation an unlimited supply of coins with a value of $2^{100}-1$, the following year - with a value of $2^{101}-1$, and so on. As soon as the value of the next new coin can be exactly matched ... | Answer. On the two hundredth. Solution. Let on the $k$-th year of reign $2^{k}-1$ can be made up of previously issued coins: $2^{k}-1=a_{1}+\ldots+a_{n}=N-n$, where $N$ is the sum of powers of two, each of which is divisible by $2^{100}$. Since $2^{k}$ is also divisible by $2^{100}$, the number $n-1$ must be divisible ... | 200 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. In a convex quadrilateral $ABCD$, angles $A$ and $C$ are both 100°. Points $X$ and $Y$ are chosen on sides $AB$ and $BC$ respectively such that $AX = CY$. It turns out that line $YD$ is parallel to the bisector of angle $ABC$. Find angle $AXY$. (A. Kuznetsov, S. Berlov) | Solution. Draw a line through point $Y$ parallel to $AB$. Let it intersect $AD$ at point $K$. Then $\angle DYC = \angle DYK$ and $\angle C = 100^{\circ} = \angle BAD = \angle YKD$, so triangles $DYC$ and $DYK$ are congruent by two angles and a side. Therefore, $YK = YC = AX$ and $AXYK$ is a parallelogram. Hence, $\angl... | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. It is allowed to cut out any 18 cells from a $20 \times 20$ chessboard, and then place several rooks on the remaining cells so that they do not attack each other. What is the maximum number of rooks that can be placed in this way? Rooks attack each other if they stand on the same row or column of the board and there... | Answer: 38 rooks. Solution: Let's call the cut-out cells holes. In addition to them, add a hole at the bottom of each vertical line of the board, and a hole to the right of each horizontal line; a total of $2 \cdot 20=40$ holes are added. Suppose several rooks are placed on the board, not attacking each other. We will ... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In quadrilateral $A B C D$, side $A B$ is equal to diagonal $A C$ and is perpendicular to side $A D$, and diagonal $A C$ is perpendicular to side $C D$. A point $K$ is taken on side $A D$ such that $A C=A K$. The bisector of angle $A D C$ intersects $B K$ at point $M$. Find the angle $A C M$. (R. Zhenodarov) | Answer: $\angle A C M=45^{\circ}$. Solution. Since triangle $B A K$ is a right isosceles triangle, $\angle A K B=45^{\circ}$. Let the bisector of angle $C A D$ intersect segment $B K$ at point $N$. Triangles $A N K$ and $A N C$ are equal: $A N$ is common, $A C=A K, \angle C A N=\angle K A N$. Therefore, $\angle N C A=\... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. At the vertices of a cube, the numbers $1^{2}, 2^{2}, \ldots, 8^{2}$ (one number per vertex) are placed. For each edge, the product of the numbers at its ends is calculated. Find the maximum possible sum of all these products. | Answer: 9420. Solution. Let's color the vertices of the cube in two colors so that the ends of each edge are of different colors. Let the numbers $a_{1}, a_{2}, a_{3}, a_{4}$ be placed in the vertices of one color, and the numbers $b_{1}, b_{2}, b_{3}, b_{4}$ in the vertices of the other color, with numbers having the ... | 9420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. (I. Bogdanov) On an infinite tape, numbers are written in a row. The first is one, and each subsequent number is obtained from the previous one by adding to it the smallest non-zero digit of its decimal representation. How many digits are in the decimal representation of the number standing in this sequence at the $... | Answer: 3001. Solution. Since each number in the sequence, starting from the second, is at least one greater than the previous one, the $9 \cdot 1000^{1000}$-th number in the sequence is greater than $9 \cdot 1000^{1000}$, meaning it has at least 3001 digits. Let's denote the $n$-th number in the sequence by $a_{n}$, a... | 3001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Squares with sides of 11, 9, 7, and 5 are arranged approximately as shown in the figure. It turned out that the area of the gray parts is twice the area of the black parts. Find the area of the white parts. | Answer: 42. Solution: Let the area of the white parts be $x$, and the area of the black parts be $y$. The total area of the white and black parts is $9^{2}+5^{2}=106=x+y$, and the total area of the white and gray parts is $11^{2}+7^{2}=170=x+2 y$. By subtracting the first equation from the second, we find that $y=64$, ... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. 10 runners start simultaneously: five in blue T-shirts from one end of the running track, and five in red T-shirts from the other. Their speeds are constant and different, with each runner's speed being greater than 9 km/h but less than 12 km/h. Upon reaching the end of the track, each runner immediately turns aroun... | Answer: 50. Solution: We will show that by the time the fastest runner finishes, any two runners of different colors have met exactly twice, from which the answer 2$\cdot$5$\cdot$5 = 50 will follow.
Let $s$ (km) be the length of the track. Set $T=2 s / 12$ (hours). Since the speed of the fastest runner is less than 12... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Twenty-two people are standing in a circle, each of them is either a knight (who always tells the truth) or a liar (who always lies). Each of them said: "The next 10 people clockwise after me are liars." How many of these 22 people are liars? | Answer: 20 liars
Solution: If more than 10 liars stand in a row, then one of them is telling the truth, which is impossible. There are 22 people in total, so there must be a knight among them. Consider the knight, who tells the truth, meaning that the 10 people following him are liars. Since 11 liars cannot stand in a... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. We took four natural numbers. For each pair of these numbers, we wrote down their greatest common divisor. Six numbers were obtained: 1, 2, 3, 4, 5, N, where $N>5$. What is the smallest value that the number $N$ can take? (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 $N \geq 14$.
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 numbers t... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. On a grid board of size $2014 \times 2014$, several (no less than one) cells are colored such that in each $3 \times 3$ square, an even number of cells are colored. What is the smallest possible number of colored cells? (M. Antipov) | Answer: 1342. Solution: Example. We will color the second, third, fifth, sixth, ..., 2012th, and 2013th cells in the first vertical column of the board. Then, in all 3x3 squares adjacent to the left edge of the board, exactly two cells are colored, and in all other colored cells, there are none. In this case, a total o... | 1342 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In triangle $ABC$, the angle bisectors $BK$ and $CL$ are drawn. A point $N$ is marked on segment $BK$ such that $LN \parallel AC$. It turns out that $NK = LN$. Find the measure of angle $ABC$. (A. Kuznetsov) | Answer: $120^{\circ}$. Solution. In the isosceles triangle $L N K$, $\angle K L N = \angle L K N$. Moreover, the angles $\angle K L N$ and $\angle L K A$ are equal as alternate interior angles when the lines $L N$ and $A C$ are parallel. Thus, $\angle K L N = \angle L K A$, which means that the ray $K L$ is the bisecto... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In the city of liars and knights, there are 366 residents, all born on different days of a leap year. All residents of the city answered two questions. To the question “Were you born in February?” 100 people answered affirmatively, and to the question “Were you born on the 30th?” 60 people answered affirmatively. Ho... | Answer: 29. Solution: To the first question, knights born in February and liars born in other months answered affirmatively. Let $x$ be the number of knights born in February, where $x$ does not exceed 29. Then, $29-x$ liars were born in February, and $100-x$ liars were born in other months. The total number of liars i... | 29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Triangles are inscribed in some cells of an $8 \times 8$ board, with one side coinciding with the side of the cell, and the third vertex lying on the opposite side of the cell. The triangles have no common points. What is the least possible number of empty cells? | Answer: 24. Solution: Estimation. On each side of a triangle, there are no fewer than two vertices of cells, with a total of $9 * 9=81$ vertices. Then, the total number of triangles is no more than 40, and the number of free cells is no less than 24. Example. Filled and unfilled concentric rings alternate (see figure).... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. A mad constructor created a clock with 150 hands. The first hand rotates at a speed of one revolution per hour, the second hand makes 2 revolutions per hour, ..., the 150th hand makes 150 revolutions per hour. The clock was started from a position where all hands were pointing straight up. When two or more hands mee... | Answer: In 20 minutes. Solution. The first meeting of the hands will occur when the fastest 150th hand catches up with the slowest first hand. After this, they will fall off, and we can forget about them. The second meeting will happen when the fastest of the remaining, the 149th, catches up with the slowest of the rem... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In the election in Sunny City, one could vote for Vintik, Shpuntik, or Knopochka. After the results were announced, it turned out that all candidates together received $146\%$ of the votes. The vote counter, Neznaika, explained that by mistake he calculated the percentage of votes for Vintik not from the total numbe... | Solution. Let Shpuntik receive $a$ votes, Vintik - $k a$ votes, and Knopochka - $b$ votes. According to the condition, $k a /(a+k a)+(a+b) /(a+k a+b)=1.46 \Rightarrow k a /(a+k a)>0.46 \Rightarrow k>0.46(1+k) \Rightarrow k>46 / 54>0.85$. Since Shpuntik received more than 1000 votes, Vintik received $k a>1000 k>1000 \cd... | 850 | Logic and Puzzles | proof | Yes | Yes | olympiads | false |
6. In a convex quadrilateral $A B C D$, the bisector of angle $B$ passes through the midpoint of side $A D$, and $\angle C=\angle A+\angle D$. Find the angle $A C D$. (S. Berlov) | Answer. $\angle A C D=90^{\circ}$. Solution. Let $E$ be the midpoint of side $A D$, and $F$ be the
intersection point of $B E$ and $A C$. From the given condition, we have: $\angle B=360^{\ci... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. At 9:00, a pedestrian set off on a journey. An hour later, a cyclist set off after him from the same starting point. At 10:30, the cyclist caught up with the pedestrian and continued on, but after some time, the bicycle broke down. After $3a$ minutes of repair, the cyclist resumed his journey, following the pedestri... | Answer: 100 minutes. Solution: The cyclist caught up with the pedestrian half an hour after his start and one and a half hours after the pedestrian's start. This means he is moving three times faster than the pedestrian. By the time of the second meeting with the cyclist, the pedestrian had been walking for 4 hours = 2... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In triangle $ABC$, the bisector $BD$ is drawn, and in triangles $ABD$ and $CBD$ - the bisectors $DE$ and $DF$ respectively. It turned out that $EF \parallel AC$. Find the angle $DEF$. (I. Rubanov) | Answer: 45 degrees. Solution: Let segments $B D$ and $E F$ intersect at point $G$. From the condition, we have $\angle E D G = \angle E D A = \angle D E G$, hence $G E = G D$. Similarly, $G F = G D$. Therefore, $G E = G F$, which means $B G$ is the bisector and median, and thus the altitude in triangle $B E F$. Therefo... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. A square with a side of 100 was cut into squares (not necessarily the same) with sides parallel to the sides of the original square and less than 10. Prove that the sum of the perimeters of the resulting squares is not less than 4400. (I. Rubanov) | Solution. Draw 11 parallel segments, two of which are sides of the $100 \times 100$ square, and the other nine divide this square into rectangles $10 \times 100$. Then each square of our dissection intersects exactly one of these segments with a segment equal to its side. Therefore, the sum of the sides of the squares ... | 4400 | Geometry | proof | Yes | Yes | olympiads | false |
3. 30 people are lined up in six rows of five people each. Each of them is either a knight, who always tells the truth, or a liar, who always lies, and they all know who among them is a knight and who is a liar. A journalist asked each of them: “Is it true that there will be at least 4 rows in which liars are in the ma... | Answer: 21. Solution: Let's call a row blue if it contains more than half (that is, no less than three) liars and red if there are no more than two liars in it.
Suppose the knights said "yes." Then we have no more than two red and no less than four blue rows. In the red rows, there are no more than 10 knights, and in ... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Misha and Masha were traveling by train to Kirov. Misha was lying on the bunk, while Masha was looking out the window. "How far is it to Kirov?" Misha asked Masha at 12:00. "73 kilometers," Masha replied. To the same question asked at 12:15 and 12:45, Masha answered: "62 kilometers" and "37 kilometers." It is known ... | Answer: 48 km/h. Solution: Since Masha rounded the distance to the nearest whole number each time, at 12:00 the distance to Kirov was no less than 72.5 and no more than 73.5 km, at 12:15 - no less than 61.5 and no more than 62.5 km, and at 12:45 - no less than 36.5 and no more than 37.5 km. Therefore, in 15 minutes fro... | 48 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. On an island, there are 1000 villages, each with 99 residents. Each resident of the island is either a knight, who always tells the truth, or a liar, who always lies. It is known that there are exactly 54054 knights on the island. One fine day, every resident of the island was asked the question: "Who is more in you... | Answer: 638. Solution. There cannot be an equal number of knights and liars in any village, because then all its inhabitants would have lied. If there are more knights in a village, then obviously 66 people told the truth and 33 people lied, and if there are more liars - then the opposite. Let there be $n$ villages whe... | 638 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. A square $15 \times 15$ is divided into $1 \times 1$ squares. Some of these squares were chosen, and one or two diagonals were drawn in each of the chosen squares. It turned out that no two drawn diagonals share a common endpoint. What is the maximum number of diagonals that can be drawn? (In your solution, provide ... | Answer: 128. Solution. Let's number the rows and columns of the square in order with numbers from 1 to 15 and draw two diagonals in the cells located at the intersection of odd-numbered rows with odd-numbered columns. There are 64 such cells, meaning 128 diagonals will be drawn. On the other hand, the diagonals should ... | 128 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In a row, 2011 consecutive five-digit numbers are written. It turned out that the sum of the digits of the 21st number is 37, and the sum of the digits of the 54th number is 7. Find the sum of the digits of the 2011th number. (Provide all possible answers and prove that there are no other answers). | Answer: 29. Solution. When transitioning from the 21st number to the 54th, the sum of the digits decreases by 30. It is easy to verify that this is possible only if there is a transition through a number divisible by 10000 along the way. This means that the thousands and hundreds places of the 21st number are nines, an... | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The numbers 1, 2, 3, ..., 10 are written in a circle in some order. Petya calculated 10 sums of all triples of adjacent numbers and wrote the smallest of the calculated numbers on the board. What is the largest number that could have been written on the board? | Answer: 15. Solution: First, we prove that the written number is not greater than 15. We single out the number 10, and divide the remaining 9 numbers into three groups of three consecutive numbers. The sum of the numbers in these three groups is $1+2+3+\ldots+9=45$, so in at least one of the considered groups, the sum ... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. In a football tournament, 8 teams participated, and each played against each other exactly once. It is known that any two teams that drew with each other ended up with a different number of points. Find the maximum possible total number of draws in this tournament. (A team earns 3 points for a win, 1 point for a dra... | Answer: 22. Solution: We will prove that no more than two teams can have exactly 6 draws. Indeed, any such team has either 6 or $6+$ 3 points (depending on whether they won or lost their decisive match). If there are three such teams, then two of them have the same number of points, meaning they did not play to a draw ... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A car is driving at a constant speed in one direction along a straight road, near which there are two houses. At noon, when the car had not yet reached the houses, the sum of the distances from it to these houses was 10 km. After 10 minutes, when the car had already passed both houses, it turned out that the sum of ... | Answer: 60 km/h. Solution: Let the houses be at points $A$ and $B$, the car was at point $C$ at noon, and at point $D$ 10 minutes later (see figure). According to the problem, $C A + C B = D A + D B = 10$. Note that $C A + C B = 2 C A + A B$, and $D A + D B = 2 D B + A B$, from which we get $C A = D B$.
, $KLMN$ be the cut-out square with side $x$, where $KL$ lies on $AB$ (point $K$ is closer to $A$ than $L$). Then the perimeter of the octagon $AKNMLBCD$ exceeds the perimeter of the square $ABCD$ by $KN + LM = 2x = 0.4 \cdot 4AB = ... | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. There are three consecutive even numbers. For the first one, we found the largest even proper divisor, for the second one - the largest odd proper divisor, and for the third one - again the largest even proper divisor. Can the sum of the three obtained divisors be equal to 2013? (A divisor of a natural number is cal... | Answer: Yes, it can. Solution. Here is an example: 1340, 1342, and 1344. For the first number, the largest even divisor is 670, for the third number, it is 672, and for the second number, the largest odd divisor is 671. $670+671+672=2013$. Remark. There are two natural ways to come up with this example. One can try to ... | 2013 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Sharik and Matroskin are skiing on a circular track, half of which is an uphill climb and the other half is a downhill descent. Their speeds on the climb are the same and four times slower than their speeds on the descent. The minimum distance by which Sharik lags behind Matroskin is 4 km, and the maximum distance i... | Answer: 24 km. Solution. The minimum lag of Sharik behind Matroskin occurs when Sharik is at the lowest point of the track, and Matroskin is climbing the hill (if Matroskin were descending at this time, it would mean that half of the track is less than 4 km, which is obviously impossible). And it remains until Matroski... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the rebus UHA = LCM(UX, UA, HA). Here U, X, A are three different digits. Two-digit and three-digit numbers cannot start with zero. Recall that the LCM of several natural numbers is the smallest natural number that is divisible by each of them. | Answer. $150=\operatorname{HOK}(15,10,50)$. Solution. Since UHA is divisible by UX, A $=0$, that is, the number has the form UX0. Since UX0 is divisible by U0 and X0, UX is divisible by U and X, from which it follows that X is divisible by U, and 10 U is divisible by X. Let $\mathrm{X}=\mathrm{M} \cdot \mathrm{U}$. The... | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. The robbers filled a chest to the brim with gold and silver sand, with twice as much gold sand by volume as silver sand. Ali-Baba calculated that if half of the silver sand were removed and the chest topped up with gold sand, the value of the chest would increase by 20 percent. By what percentage and how would the v... | Answer. It will decrease by $40 \%$. Solution. Let the initial volume of silver sand in the chest be $A$, the cost of a unit volume of gold sand be $x$, and the cost of a unit volume of silver sand be $y$. Then the initial cost of the chest is $2 A x+A y$. The cost of the chest according to Ali-Baba's calculations is $... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The sides of 100 identical equilateral triangles are painted in 150 colors such that exactly two sides are painted in each color. If two triangles are placed with their same-colored sides together, the resulting rhombus will be called good. Petya wants to form as many good rhombi as possible from these triangles, wi... | Answer: 25. Solution: Why is it always possible to get 25 good rhombuses? Let's construct the maximum possible number of good rhombuses. Suppose there are fewer than 25. Then there are at least 52 triangles that have not entered any good rhombus. Let's call such triangles single. For each triangle, choose a color, exac... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Foma and Yerema were walking in the same direction along a road with kilometer markers. In one hour, Foma passed by five markers, while Yerema passed by six. Could Foma's speed have been greater than Yerema's? | Answer. Could. Solution. Let Erema be 50 m from the first of the poles he passed at the beginning of the hour, and 50 m beyond the sixth pole at the end of the hour. Then he walked 5100 m in an hour. Let Foma be 600 m from the first of the poles he passed at the beginning of the hour, and 600 m beyond the fifth pole he... | 5200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Let $p_{1}, p_{2}, \ldots, p_{100}$ be one hundred distinct prime numbers. Natural numbers $a_{1}, \ldots, a_{k}$, greater than 1, are such that each of the numbers $p_{1} p_{2}^{3}, p_{2} p_{3}^{3}, \ldots, p_{99} p_{100}^{3}, p_{100} p_{1}^{3}$ is equal to the product of some two of the numbers $a_{1}, \ldots, a_{... | Solution. We will call a natural number white if it is divisible by the square of some prime number, and black otherwise. Let's say that the number $a_{i}$ serves a number from the list $p_{1} p_{2}^{3}, p_{2} p_{3}^{3}, \ldots, p_{99} p_{100}^{3}, p_{100} p_{1}^{3}(*)$, if it gives it in the product with some $a_{j}$.... | 150 | Number Theory | proof | Yes | Yes | olympiads | false |
3. In the cells of a $10 \times 10$ table, the natural numbers 1, 2, ..., 99, 100 are arranged. Let's call a corner the figure that results from removing one cell from a 2x2 square. We will call a corner good if the number in its cell, which borders by sides with two other cells, is greater than the numbers in these tw... | Answer: 162. Solution: Let's call the center of a corner the cell that borders by sides with two other cells of the corner. In any $2 \times 2$ square, no more than two of the four corners contained in it can be good: those with the two largest numbers in the center of this square. There are 81 $2 \times 2$ squares in ... | 162 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Ninety-nine positive numbers are arranged in a circle. It turns out that for any four consecutive numbers, the sum of the first two of them in a clockwise direction is equal to the product of the last two of them in a clockwise direction. What can the sum of all 99 arranged numbers be? (S. Berlov) | Answer: 198. Solution: Let the consecutive numbers be $a, b, c, d, e$. Notice that if $a > c$, then $a + b > b + c$, which implies $c < a$. Similarly, if $c d > d e$, then $c > e$. Continuing this reasoning, we get that each number is greater than the number that is two positions clockwise from it. However, writing a c... | 198 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. A row of 100 coins is laid out. All the coins look the same, but somewhere among them are 50 counterfeit ones (and the rest are genuine). All genuine coins weigh the same, and the counterfeit ones may weigh differently, but each counterfeit is lighter than a genuine one. Can at least 34 genuine coins be found with a... | Answer. It is possible. Solution. Number the coins from left to right with numbers from 1 to 100. Compare coins 17 and 84. At least one of them is genuine. Therefore, if the scales are in balance, both coins are genuine; in this case, 34 coins with numbers 1-17 and 84-100 will be genuine, since 50 counterfeit coins do ... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Inside parallelogram $A B C D$, a point $E$ is chosen such that $A E=D E$ and $\angle A B E=90^{\circ}$. Point $M$ is the midpoint of segment $BC$. Find the angle $DME$. (A. Kuznetsov) | Answer: $90^{\circ}$. Solution: Let $N$ be the midpoint of segment $A D$. Since triangle $A E D$ is isosceles, $E N \perp A D$. Since $A B \| M N$ and $\angle A B E=90^{\circ}$, then $B E \perp M N$. Therefore, $E$ is the orthocenter of triangle $B M N$. This means $M E \perp B N$. Since $B M D N$ is a parallelogram, $... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. The school table tennis championship was held according to the Olympic system. The winner won 6 matches. How many participants in the championship won more matches than they lost? (In the first round of the championship held according to the Olympic system, participants are paired. Those who lost the first game are ... | Answer: 16. Solution. Since in each round every player found a partner and in each pair one of the players was eliminated, the total number of players decreased by half after each round. The winner participated in every round and won, so there were a total of six rounds. Since the winner was determined unequivocally af... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In triangle $ABC$, the median $BM$ is twice as short as side $AB$ and forms an angle of 40 degrees with it. Find angle $ABC$.
| Answer: $110^{\circ}$. Solution. Extend the median $B M$ beyond point $M$ by its length and obtain point $D$. Since $A B=2 B M$, then $A B=B D$, which means triangle $A B D$ is isosceles. Therefore, angles $B A D$ and $B D A$ are each equal to $\left(180^{\circ}-40^{\circ}\right): 2=70^{\circ}$. $A B C D$ is a parallel... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. The length of a rectangle was reduced by $10 \%$, and the width was reduced by $20 \%$. As a result, the perimeter of the rectangle decreased by $12 \%$. By what percentage will the perimeter of the rectangle decrease if its length is reduced by $20 \%$ and its width is reduced by $10 \%$? | Answer: By 18%. Solution. Let the length be $a$, and the width be $b$. According to the condition, $2(0.1 a + 0.2 b) = 0.12(2 a + 2 b)$, from which we get $a = 4 b$. If the length is reduced by $20\%$, and the width by $10\%$, then the perimeter will decrease by $2(0.2 a + 0.1 b) = 1.8 b$, and it was $10 b$. Therefore,... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. On the table, there are 100 identical-looking coins, of which 85 are counterfeit and 15 are genuine. You have a miracle tester, into which you can place two coins and get one of three results - "both coins are genuine," "both coins are counterfeit," and "the coins are different." Can you find all the counterfeit coi... | Answer. Yes. Solution. We will present one of the possible ways to define the fake coins. Divide the coins into 50 pairs and check all pairs except one. We will learn the number of fake coins in each pair. Since the total number of fake coins is known, we will also learn how many are fake in the remaining pair. We need... | 64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Black and white balls are arranged in a circle, with black balls being twice as many as white ones. It is known that among pairs of adjacent balls, there are three times as many monochromatic pairs as polychromatic ones. What is the smallest number of balls that could have been arranged? (B. Trushin) | Answer: 24. Solution: Since the number of black balls is twice the number of white balls, the total number of balls is divisible by three. Let's denote it by $n$. All the balls are divided into alternating groups of consecutive balls of the same color (a group can consist of just one ball). Since the colors of the grou... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Find the smallest natural $k$ such that for some natural number $a$, greater than 500,000, and some natural number $b$, the equality $\frac{1}{a}+\frac{1}{a+k}=\frac{1}{b}$ holds. (I. Bogdanov) | Answer. $k=1001$. Solution. Estimation. Let $a+k=c$ and $\text{GCD}(a, c)=d$. Then $a=d a_{1}, c=d c_{1}$ and $\frac{1}{a}+\frac{1}{c}=\frac{a_{1}+c_{1}}{d a_{1} c_{1}}$. Since the numbers $a_{1} c_{1}$ and $a_{1}+c_{1}$ are coprime, $d$ must divide $a_{1}+c_{1}$. Therefore, $d \geq a_{1}+c_{1}$ and $d^{2} \geq d\left(... | 1001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Wrote down two numbers - the first and the second. Added the second to the first - got the third, added the third to the second - got the fourth, and so on. The sum of the first six numbers is 2008. What is the fifth number? | Answer: 502. Solution: Let the first number be $a$, the second - $b$. Then the third number is $a+b$, the fourth $-a+2b$, the fifth $-2a+3b$, the sixth $-3a+5b$, and the sum of all six numbers is $8a+12b$. Thus, the fifth number is a quarter of the sum of all six numbers, that is, $2008: 4=502$.
Grading Guidelines. An... | 502 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In a right triangle, the height dropped to the hypotenuse is four times shorter than the hypotenuse. Find the acute angles of the triangle. | Answer: 15 and 75 degrees. Solution. Let $A B C$ be the given right-angled triangle with the right angle at vertex $C$. As is known, the median $C M$ is half the hypotenuse $A B$. The height $C H$ is given to be $A B / 4$. Therefore, in the right-angled triangle $C H M$, the hypotenuse $C M$ is twice the length of the ... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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