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9. For what values of the parameter $\boldsymbol{a}$ does the equation $|f(x)-5|=p(x)$, where $f(x)=\left|\frac{x^{2}-10 x+25}{x-5}-\frac{x^{2}-3 x}{3-x}\right|$, $p(x)=a \quad$ have three solutions? If there is more than one value of the parameter, indicate their product in the answer. | # Solution:
Simplify $f(x)=\left|\frac{x^{2}-10 x+25}{x-5}-\frac{x^{2}-3 x}{3-x}\right|$, we get $f(x)=$ $|2 x-5|$, where $x \neq 5, x \neq 3$.
Solve the equation || $2 x-5|-5|=a$, where $x \neq 5, x \neq 3$ graphically in the system $x O a$.
The equation has three solutions when $a=4, a=5$.
The product is 20.
Ans... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. If a two-digit number is decreased by 54, the result is a number written with the same digits but in reverse order. In the answer, specify the arithmetic mean of the resulting numbers.
# | # Solution:
$\overline{x y}=10 x+y-$ the original two-digit number, then $\overline{y x}=10 y+x$ - the number written with the same digits but in reverse order. We get the equation $10 x+y=10 y+x+54$.
From the equation, it is clear that the two-digit number is greater than 54. Let's start the investigation with the t... | 82 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. Given triangle $A B C . \angle A=\alpha, \angle B=\beta$. Lines $O_{1} O_{2}, O_{2} O_{3}, O_{1} O_{3}$ are the bisectors of the external angles of triangle $A B C$, as shown in the figure. Point O is the center of the inscribed circle of triangle $A B C$. Find the angle between the lines $\mathrm{CO}_{1}$ and $\mat... | # Solution:
We will prove that the bisectors of two external angles and one internal angle intersect at one point. Let \( O_{1} G, O_{1} H, O_{1} F \) be the perpendiculars to \( B C, A C \)... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In how many ways can two knights - a white one and a black one - be placed on a chessboard consisting of $16 \times 16$ cells so that they threaten each other? (A knight moves in an "L" shape, i.e., it can move to one of
$ rectangular coordinate system to the board. Denote the coordinates of the two knights as $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$, where $x_{k}, y_{k} \in\{1,2, \ldots, n\}, k=1,2$. The knights threaten each other if 1) $\left|x_{1}-x_{2}\right|=1, \qua... | 1680 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the smallest natural number that has exactly 70 natural divisors (including 1 and the number itself).
$(16$ points $)$ | Solution: Let $n$ be the required natural number, $n=p_{1}^{k_{1}} \cdot p_{2}^{k_{2}} \cdot \ldots \cdot p_{m}^{k_{m}}$ - the prime factorization of the number $n$. Any natural divisor of this number has the form $d=p_{1}^{h_{1}} \cdot p_{2}^{l_{2}} \cdot \ldots \cdot p_{m}^{l_{m}}$, where $l_{i} \in\left\{0,1, \ldots... | 25920 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. A student wrote a program for recoloring a pixel into one of 128 different colors. These colors he numbered with natural numbers from 1 to 128, and the primary colors received the following numbers: white color - number 1, red - 5, orange - 13, yellow - 21, green - 45, blue - 75, indigo - 87, violet - 91, black - 12... | Solution. The final pixel color number is equal to $f^{[2019]}(5)$ where $f^{[k]}(n)=\underbrace{f(f(f(\ldots(f}_{k \text { times }}(n) \ldots)-k$-fold composition of the function $f(n)$, which is equal to $n+4$ when $n \leq 19$, and equal to $|129-2 n|$ when $n \geq 20$. Let's compute and write down the first few valu... | 75 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points) On a tourist base, the number of two-room cottages is twice the number of one-room cottages. The number of three-room cottages is a multiple of the number of one-room cottages. If the number of three-room cottages is tripled, it will be 25 more than the number of two-room cottages. How many cottages are ... | Solution. Let the number of one-bedroom houses be $x$, two-bedroom houses $2 \mathrm{x}$, and three-bedroom houses nx.
$3 n x-25=2 x ; x(3 n-2)=25 ; \Rightarrow 25:(3 n-2) \Rightarrow 3 n-2=1 ; 5 ; 25$. If
$3 n-2=1 ; n=1 ; x=25 ; 25+50+25=100$ total houses.
If $3 n-2=5 ; 3 n=7$, which is impossible. If $3 n-2=25 ; 3... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. According to the theorem inverse to Vieta's theorem, we form a quadratic equation. We get $x^{2}-\sqrt{2019} x+248.75=0$.
Next, solving it, we find the roots $a$ and $b$: $a=\frac{\sqrt{2019}}{2}+\frac{32}{2}$ and $b=\frac{\sqrt{2019}}{2}-\frac{32}{2}$, and consequently, the distance between the points $a$ and $b$:... | Answer: 32
| 15 points | The correct answer is obtained justifiably |
| :---: | :---: |
| 10 points | The quadratic equation is solved, but an arithmetic error is made or the distance between the points is not found |
| 5 points | The quadratic equation is correctly formulated according to the problem statement. |
| 0... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Initially, the farmer placed his produce in boxes with a capacity of 8 kg each, but one box was not fully loaded. Then the farmer repackaged all the produce into boxes with a capacity of 6 kg each, which required 8 more boxes, but in this case, one box was also not fully loaded. When the produce was finally placed i... | Solution. Let $x$ kg be the weight of the farmer's produce. Then $\quad 8(n-1)<x<8 n$, $6(n+7)<x<6(n+8), \quad 5(n+13)=x, \Rightarrow 8(n-1)<5(n+13)<8 n, \quad 6(n+7)<5(n+13)<6(n+8)$, $\Rightarrow 21 \frac{2}{3}<n<23, \quad n=22, \quad x=35 \cdot 5=175$.
Answer: 175. | 175 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In the country of Landia, which breeds an elite breed of horses, an annual festival is held to test their speed, in which only one-year-old, two-year-old, three-year-old, and four-year-old horses can participate. For each horse that meets the speed standard, the festival organizers pay a fixed amount of money to the... | Solution. A four-year-old horse can earn a maximum of 4 landricks over its entire participation in festivals. If the horse starts participating in festivals at 1 year old, it can participate for another 3 years after that. In the case of annual victories, it will earn $1+2+3+4=10$ landricks over 4 years. If the horse s... | 200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. The number $N$ is written as the product of consecutive natural numbers from 2019 to 4036: $N=2019 \cdot 2020 \cdot 2021 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036$. Determine the power of two in the prime factorization of the number $N$.
(points) | Solution. The number $N$ can be represented as
$$
\begin{aligned}
& N=\frac{(2 \cdot 2018)!}{2018!}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036}{2018!}=\frac{(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot(2 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4036)}{2018!}= \\
& =\frac{(1 \cdot 3 \cdot \... | 2018 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Determine the smallest natural number $N$, among the divisors of which are all numbers of the form $x+y$, where $x$ and $y$ are natural solutions to the equation $6 x y-y^{2}-5 x^{2}=7$. points) | Solution. Transform the equation by factoring the right-hand side
$6 x y-y^{2}-5 x^{2}-x^{2}+x^{2}=7 \Rightarrow 6 x(y-x)-(y+x)(y-x)=7 \Rightarrow(y-x)(6 x-y-x)=7 \Rightarrow$ $(y-x)(5 x-y)=7$.
Considering that the variables are natural numbers, and 7 is a prime number, we get
$$
\left\{\begin{array} { l }
{ y - x ... | 55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. From point $A$ of a circular track, a car and a motorcycle started simultaneously and in the same direction. The car drove two laps without stopping in one direction. At the moment when the car caught up with the motorcyclist, the motorcyclist turned around and increased his speed by 16 km/h, and after $3 / 8$ hours... | Solution. Let $x$ (km/h) be the speed of the motorcyclist, $x$ (km/h) be the speed of the car, $S$ (km) be the distance the motorcyclist travels before turning around, then the total length of the track is $2 S+5.25$. We have $\frac{S}{x}=\frac{3 S+5.25}{y}, \frac{3 x}{8}+6=S, \frac{3 y}{8}=S+5.25$. This leads to the q... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the sum of all numbers of the form $x+y$, where $x$ and $y$ are natural number solutions to the equation $5 x+17 y=307$.
points) | Solution. We solve the auxiliary equation $5 x+17 y=1$. For example, its solutions can be 7 and 2. Multiplying them by 307 and considering linear combinations for integer $t$, we get values in natural numbers
$\left\{\begin{array}{l}x=7 \cdot 307-17 t, \\ y=-2 \cdot 307+5 t,\end{array} t \in Z, x>0, y>0 \Rightarrow t ... | 164 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In 100 containers of three types with capacities of $1 \pi$, 10 liters, and 50 liters, 500 liters of oil have been distributed. How many containers of each type were used, if the amount of oil in each container corresponds to its capacity?
(12 points) | Solution. Let $n, m, k$ be the number of containers with capacities of 1 liter, 10 liters, and 50 liters, respectively. Then $\left\{\begin{array}{c}n+10 m+50 k=500, \\ n+m+k=100 .\end{array}\right.$ Since $n=500-10 m-50 k=10(50-m-5 k)$, $n$ is divisible by 10, i.e., $n=10 l, l \in N, n \leq 100, l \leq 10 .\left\{\beg... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Find all natural numbers $n \geq 2$, for which the equality $4 x_{n}+2 y_{n}=55 n^{2}+61 n-116$ holds, where $x_{n}=1 \cdot 2+2 \cdot 3+\cdots+(n-1) \cdot n, y_{n}=1^{2}+2^{2}+3^{2}+\cdots+(n-1)^{2}$.
(20 points) | Solution: Let $z_{n}=1+2+\cdots+n=\frac{(n+1) n}{2}=\frac{n^{2}+n}{2}$. We have
$$
\begin{aligned}
& x_{n}=1 \cdot 2+2 \cdot 3+\cdots+(n-1) \cdot n=(2+3+\cdots n)+(3+4+\cdots n)+\cdots((n-1)+n)+n= \\
& =\left(z_{n}-z_{1}\right)+\left(z_{n}-z_{2}\right)+\cdots+\left(z_{n}-z_{n-1}\right)=(n-1) z_{n}-\left(z_{1}+z_{2}+\c... | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. In triangle $ABC$, lines parallel to the sides of the triangle are drawn through an arbitrary point $O$. As a result, triangle $ABC$ is divided into three parallelograms and three triangles. The areas of the resulting triangles are $6 \text{ cm}^2$, $24 \text{ cm}^2$, and $54 \text{ cm}^2$. Find the area of triangle... | # Solution:
The triangles are similar. Let the first and third be similar with a similarity coefficient $k_{1}$, and the second and third with a similarity coefficient $k_{2}$. Then for the areas of these triangles, we have the ratios $\frac{S_{1}}{S_{3}}=k_{1}^{2}, \frac{S_{2}}{S_{3}}=k_{2}^{2}$.
 Two medians of a triangle, equal to 18 and 24, are perpendicular to each other. Find the length of the third median of this triangle. | Solution. Let $\mathrm{AP}=18$ and $\mathrm{BH}=24$. We will find the length of the median $\mathrm{CM}$. By the property of medians in $\triangle A B C$ we have:
$A O: O P=B O: O H=2: 1 \Rightarrow A O=$
$=\frac{2}{3} A P=\frac{2}{3} \cdot 18=12 ; B O=\frac{2}{3} B H=\frac{2}{3} \cdot 24=16$.
Then, by the Pythagore... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (20 points) Calculate the value of the expression:
$$
1 \cdot 2 \cdot(1+2)-2 \cdot 3 \cdot(2+3)+3 \cdot 4 \cdot(3+4)-\cdots+2019 \cdot 2020 \cdot(2019+2020)
$$ | Solution. We will prove by mathematical induction for natural n that:
$$
\begin{gathered}
-0 \cdot 1 \cdot(0+1)+1 \cdot 2 \cdot(1+2)-2 \cdot 3 \cdot(2+3)+3 \cdot 4 \cdot(3+4)-\cdots- \\
-(2 n-2) \cdot(2 n-1) \cdot(4 n-3)+(2 n-1) \cdot 2 n \cdot(4 n-1)= \\
=(2 n-1) \cdot 2 n \cdot(2 n+1) .
\end{gathered}
$$
Base: \( n... | 8242405980 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Variant №1
№1. The number of male students in the university is $35 \%$ more than the number of female students. All students are distributed between two buildings, with $\frac{2}{5}$ of all male students in the first building and $\frac{4}{7}$ of all female students in the second building. How many students are the... | # Solution:
Let $n$ be the number of girls studying at the university, then the number of boys is $1.35 n = \frac{27}{20} n$, from which we can conclude that $n$ is a multiple of 20, i.e., $n = 20 m$.
Thus, the number of girls is $20 m$, and the number of boys is $27 m$.
In the first building, $\frac{2}{5}$ of all b... | 4935 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Given the vertices of a regular 100-gon $A_{1}, A_{2}, A_{3}, \ldots, A_{100}$. In how many ways can three vertices be chosen from them to form an obtuse triangle? (10 points) | Solution. Let the vertices be numbered clockwise.
Denote the selected vertices clockwise as $K, L, M$, where angle $K L M$ is obtuse. If $K=A_{k}, L=A_{l}, M=A_{m}$, then $\alpha=\angle K L M=\frac{180^{\circ}}{100}(100-(m-k))>90^{\circ}, 0<m-k<50$.
The difference $m-k$ is taken modulo 100
(for example, $15-\left.70... | 117600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given the vertices of a regular 120-gon $A_{1}, A_{2}, A_{3}, \ldots, A_{120}$. In how many ways can three vertices be chosen from them to form an obtuse triangle? (10 points) | Solution. Let the vertices be numbered clockwise.
Denote the selected vertices clockwise as $K, L, M$, where angle $K L M$ is obtuse. If $K=A_{k}, L=A_{l}, M=A_{m}$, then $\alpha=\angle K L M=\frac{180^{\circ}}{120}(120-(m-k))>90^{\circ}, 0<m-k<60$.
The difference $m-k$ is taken modulo 120
(for example, $15-\left.90... | 205320 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. There are two types of containers: 27 kg and 65 kg. How many containers of the first and second type were there in total, if the cargo in the first type of containers exceeds the cargo in the second type of containers by 34 kg, and the number of 65 kg containers does not exceed 44 units? | Solution. Let $x$ be the number of containers weighing 27 kg, and $y$ be the number of containers weighing 65 kg. We obtain the equation $27 x - 65 y = 34$.
$27(x - 2y) - 11y = 34$, let $x - 2y = k$.
$27k - 11y = 34$,
$11(2k - y) + 5k = 34$, let $2k - y = t$.
$11t + 5k = 34$,
$5(2t + k) + t = 34$, let $2t + k = n$... | 66 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. In triangle $A B C$ with angle $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Find the degree measure of angle $C_{1} B_{1} A_{1}$. | # Solution.
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$. Considerin... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Given a rectangular trapezoid $\mathrm{ABCE}$, the bases of which $\mathrm{BC}$ and $\mathrm{AE}$ are equal to 5 and 7, respectively. The smaller lateral side $\mathrm{AB}$ is equal to $\mathrm{BC}$. On $\mathrm{AB}$, a point $\mathrm{F}$ is marked such that $\mathrm{BF}: \mathrm{FA}=2: 3$, on $\mathrm{AC}$, a point... | Solution. Construct perpendiculars GI and GH.
1) $\Delta G I D = \Delta G F H$ - by two legs, since $G I = G H = 4; F H = I D = 1$, therefore $F G = G D$, $\angle F G H = \angle D G I = \alp... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. A chemistry student conducted an experiment: from a tank filled with syrup solution, he poured out several liters of liquid, refilled the tank with water, then poured out twice as much liquid and refilled the tank with water again. As a result, the amount of syrup in the tank decreased by $\frac{8}{3}$ times. Determ... | # Solution.
1) Let the syrup content in the original solution be $p \%$ and let $\mathcal{X}$ liters of the solution be poured out the first time.
2) Then after pouring out the liquid, there are $(1000-x)$ liters of solution left, and in it $(1000-x) \cdot \frac{p}{100}$ liters of syrup and $(1000-x) \cdot \frac{100-p... | 250 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Elvira takes a triplet of numbers and transforms it according to the rule: at each step, each number is replaced by the sum of the other two. What is the difference between the largest and the smallest numbers in the triplet after 2019 applications of this rule, if the initial triplet was $\{100 ; 89 ; 60\}$? If the... | Solution. Let's denote the 3 numbers as $\{x ; x+a ; x+b\}$, where $0<a<b$. Then the difference between the largest and the smallest number at any step, starting from the zeroth step, will be an invariant, that is, unchanged and equal to $b$.
$b=100-60=40$.
Answer: 40. | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. On the sides $\mathrm{AB}$ and $\mathrm{AC}$ of the right triangle $\mathrm{ABC}\left(\angle B A C=90^{\circ}\right)$, right triangles АВТ and АСК are constructed externally such that $\angle A T B=\angle A K C=90^{\circ}$, $\angle A B T=\angle A C K=30^{\circ}$. On the side $\mathrm{BC}$, a point М is chosen such t... | Solution. Mark points P and O at the midpoints of sides AB and AC, respectively. Connect point P with points M and T, and point O with points K and M.
Then: 1) $\triangle T P M = \Delta K O... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Inside the square $A B C D$ with a side length of 5, there is a point $X$. The areas of triangles $A X B, B X C$, and $C X D$ are in the ratio $1: 5: 9$. Find the sum of the squares of the distances from point $X$ to the sides of the square. | Solution:
Let the side of the square be $a=5$, and the distances from point $X$ to sides $AB$, $BC$, $CD$, and $DA$ be $h_{1}$, $h_{2}$, $h_{3}$, and $h_{4}$, respectively. Since the area of a triangle is $S=\frac{1}{2} a h$, we conclude that $h_{1}: h_{2}: h_{3}=1: 5: 9$ or $h_{1}=x$, $h_{2}=5 x, h_{3}=9 x$. However,... | 33 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. The boat traveled 165 km upstream against the current and then returned. On the return trip, it took 4 hours less than the trip there. Find the boat's own speed if the river's current speed is 2 km/h. | # Solution:
$$
\begin{aligned}
& v_{\text {down}}=v_{\mathrm{c}}+v_{\text {current}} \\
& v_{\text {up}}=v_{\mathrm{c}}-v_{\text {current}}
\end{aligned}=>v_{\text {down}}-v_{\text {up}}=2 v_{\text {current}}=2 \cdot 2=4
$$
Then $\quad v_{\text {up}}=x$ km/h, $v_{\text {down}}=x+4$ km/h
Preliminary (online) stage of... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.3. How many solutions in integers $x, y$ does the inequality $|y-x|+|3 x-2 y| \leq a$ have a) for $a=2$; b) for $a=20$? | Answer: a) 13; b) 841. Solution. Let $m=y-x, n=3 x-2 y$, then (expressing $x$, y from these equations) we get $x=2 m+n$ and $y=3 m+n$ and thus, for any integers m, n there correspond integers $x, y$ (and vice versa), i.e., there is a one-to-one correspondence between ordered pairs (m.n) and ( $x, y$ ). Therefore, we ne... | 13 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
7.2. In 7a grade, there are 33 students. At the beginning of the school year, two clubs were organized in the class. According to school rules, a club can be organized if at least $70 \%$ of all students in the class sign up for it. What is the smallest number of students who could have signed up for both clubs simulta... | Answer: 15 students. Solution. In each club, there should be no less than $33 \cdot 0.7=23.1$ people, which means no less than 24 people. Let $n_{1}, n_{2}$ be the number of students who signed up for the first and second club, respectively, and $n$ be the number of students who signed up for at least one club. Obvious... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.3. In a three-digit number, the first digit was crossed out to obtain a two-digit number. If the original number is divided by the obtained number, the quotient is 8 and the remainder is 6. Find the original number. | Answer: 342. Solution. Let $a$ be the first digit of the original number, and $b$ be the two-digit number formed by the last two digits. According to the condition, $100 a+b=8 b+6 \Leftrightarrow 7 b=2(50 a-3)=2 \cdot 49 a+2(a-3)$. Thus, the number $a-3$ must be divisible by 7. Considering that $0<a \leq 9$, we get tha... | 342 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. There are 200 matches. How many ways are there to form, using all the matches, a square and (separately) an equilateral triangle? (Different ways differ in the sizes of the square and the triangle). | Answer: 16. Solution. Let $x$ (matches) be the side length of the square, and $y$ (matches) be the side length of the triangle. Then $4 x+3 y=200 \Leftrightarrow 3 y=4(50-x)$. Thus, for divisibility by 3, we need to consider all natural numbers up to 50 as $x$ that give a remainder of 2 when divided by 3 (i.e., the sam... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.3. How many points on the hyperbola $y=\frac{2013}{x}$ have integer coordinates $(x ; y)$? | Answer: 16. Solution. Integer points in the first quadrant correspond to the natural divisors of the number $2013=3 \cdot 11 \cdot 61$. The number of such divisors is 8 (they can be listed directly or the formula $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \ldots\left(\alpha_{k}+1\right)$ for the number of natu... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.2. a) Given the quadratic equation $x^{2}-9 x-10=0$. Let $a$ be its smallest root. Find $a^{4}-909 a$. b) For the quadratic equation $x^{2}-9 x+10=0$, where $b$ is the smallest root, find $b^{4}-549 b$. | Answer: a) 910; b) -710. Solution. a) Let's solve the problem in general. Let $x^{2}-c x+d=0-$ be a quadratic equation, and $a-$ be its root. Then $a^{4}=(c a-d)^{2}=c^{2} a^{2}-2 a c d+d^{2}=$ $c^{2}(c a-d)-2 a c d+d^{2}=a\left(c^{3}-2 c d\right)+d^{2}-c^{2} d$. Therefore, $a^{4}-\left(c^{3}-2 c d\right) a=d^{2}-c^{2}... | 910 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.3. There are 12 matches, each 2 cm long. Can a polygon with an area of 16 cm $^{2}$ be formed from them? (The matches cannot be broken, and all matches must be used.) | Answer: It is possible. Solution. The result follows from the Pythagorean theorem (since $10^{2}=6^{2}+8^{2}$) and construction (see figure). The area of the polygon is $\frac{6 \cdot 8}{2}-8=16$.
. Thus, we get $\sqrt[3]{a^{3}}=a=2017$ | 2017 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. How many five-digit natural numbers exist, each of which has adjacent digits with different parity | Answer: 5625. Solution. If the first digit is even, then it can be chosen in four ways $(2,4,6,8)$. And all subsequent ones can be chosen in five ways $(1,3,5,7,9$ - for the second and fourth digits and $0,2,4,6,8$ - for the third and fifth). In the end (by the rule of product), we have a total of $4 \cdot 5^{4}=2500$ ... | 5625 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. Find the value of the expression $\frac{\left(a^{2}+b^{2}\right)^{2}-c^{2}-4 a^{2} b^{2}}{a^{2}+c-b^{2}}$ for $a=2017, b=2016, c=2015$. Justify your result. | Answer: 2018. Solution. The numerator equals $a^{4}+2 a^{2} b^{2}+b^{4}-4 a^{2} b^{2}-c^{2}=\left(a^{2}-b^{2}\right)^{2}-c^{2}=$ $\left(a^{2}-b^{2}+c\right)\left(a^{2}-b^{2}-c\right)$, and after dividing by the denominator we get $a^{2}-b^{2}-c=$ $(a-b)(a+b)-c=2017+2016-2015=2018$. | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.4. Find the natural number $x$ that satisfies the equation
$$
x^{3}=2011^{2}+2011 \cdot 2012+2012^{2}+2011^{3}
$$ | Answer: 2012. Solution: Let $a=2011, \quad b=2012$. Then $a^{2}+a b+b^{2}=\frac{b^{3}-a^{3}}{b-a}=b^{3}-a^{3}$ (since $b-a=1$). Therefore, $x^{3}=b^{3}-a^{3}+a^{3}=b^{3}$. Hence, $x=b=2012$. | 2012 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. In a five-digit number, one of the digits was crossed out, and this four-digit number was subtracted from the original number. As a result, the number 54321 was obtained. Find the original number. | Answer: 60356. Solution. Let $x$ be the four-digit number obtained after crossing out a digit. Note that the crossed-out digit was the last digit of the five-digit number, because otherwise the last digit after subtraction would be zero. Let this digit be $y$. We have the equation $10 x+y-x=54321 \Leftrightarrow 9 x+y=... | 60356 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2. Is it possible to place the 12 numbers $1,2, \ldots, 12$ on the edges of a cube such that the product of the four numbers on the top face equals the product of the four numbers on the bottom face? | Answer: It is possible. Solution. An example of such an arrangement can be as follows: on the top face, place the numbers $2,4,9,10$; on the bottom face, place the numbers $3,5,6,8$; the remaining numbers $1,7,11,12$ are placed on the side edges. The products on the top and bottom faces are the same and equal to 720. (... | 720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.4. There are ten coins of different weights and a balance scale without weights. It is required to select the heaviest and the lightest coin. Can this be achieved in 13 weighings? | Answer: It is possible. Solution. First, divide all the coins into 5 pairs and in 5 weighings compare the weight of each pair. By selecting the heavier coin in each pair, we form a "heavy" group of 5 coins. The remaining 5 coins will form the "light" group. Now, in the heavy group, over 4 sequential weighings, select t... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.2. In a five-digit number, one digit was crossed out, and the resulting four-digit number was added to the original. The sum turned out to be 54321. Find the original number. | Answer: 49383. Solution. Note that the crossed-out digit must be the last digit of the number $N$, because otherwise the sum of the two numbers would have an even digit in the last place. Let's denote this crossed-out digit by $x$ and let $y$ be the four-digit number obtained after the crossing out. Then the condition ... | 49383 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.2. The number $a$ is a root of the quadratic equation $x^{2}-x-50=0$. Find the value of $a^{4}-101 a$. | Answer: 2550. Solution. We have $a^{2}=a+50$, therefore $a^{4}=(a+50)^{2}=a^{2}+100 a+2500=$ $a+50+100 a+2500=101 a+2550$. Hence $a^{4}-101 a=2550$. | 2550 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. In triangle $A B C$, side $B C$ is equal to segment $A M$, where $M$ is the point of intersection of the medians. Find the angle $\angle B M C$. | Answer: $90^{\circ}$. Solution. Let $A N-$ be the median. By the property of the intersection point of medians, $M N=\frac{1}{2} A M=\frac{1}{2} B C$. Thus, in triangle $B M C$, the median $M N$ is equal to half of side $B C$. Therefore, $\triangle B M N$ is a right triangle with a right angle at $B M N$ (this is a kno... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1. In 7a class, $52\%$ are girls. All students in the class can line up in such a way that boys and girls alternate. How many students are in the class? | Answer: 25 students. Solution: Considering that there are more girls than boys in the class, from the condition of alternation, we get that there are exactly one more girl than boys. Therefore, one person constitutes $52-48=4 \%$ of the class size. Thus, the number of students in the class (i.e., $100 \%$) is $100 / 4=... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.3. Given a rectangular grid of $7 \times 14$ (cells). What is the maximum number of three-cell corners that can be cut out from this rectangle? | Answer: 32 corners. Solution: It is obvious that no more than 32 corners can be cut out, as otherwise, the rectangle must contain no fewer than $33 \cdot 3=99>98$ cells. The image below shows an example of cutting one
 Prove that there exists a pair of two-digit numbers such that if 15 is added to the first number and 20 is subtracted from the second, the resulting numbers will remain two-digit, and their product will be equal to the product of the original numbers. b) How many such pairs are there? | Answer: b) 16 pairs. Solution. The sought numbers $x$ and $y$ must satisfy the condition $(x+15)(y-20)=xy$. Expanding the brackets and transforming, we get the equation $3y-4x=60$. From this equation, it follows that the number $x$ must be divisible by 3, and $y$ by 4, i.e., $x=3x_{1}, y=4y_{1}$ for some natural $x_{1}... | 16 | Algebra | proof | Yes | Yes | olympiads | false |
8.3. In triangle $A B C$, the bisector $A M$ and the median $B N$ intersect at point $O$. It turns out that the areas of triangles $A B M$ and $M N C$ are equal. Find $\angle M O N$. | Answer: $90^{\circ}$. Solution. Since $S_{A M N}=S_{M N C}$ (because $A N=N C$), from the condition of the problem we have $S_{A B M}=S_{A N M}$. Therefore, in triangles $A B M$ and $A M N$, the heights drawn from vertices $B$ and $N$ are equal. Let $B_{1}$ and $N_{1}$ be the bases of these heights. Then the right tria... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.5. 10 girls and 10 boys stood in a row such that girls and boys alternate, specifically from left to right: girl-boy-girl-boy and so on. Every minute, in one (any) pair of neighbors "girl-boy," the children can swap places, provided that the girl is to the left of the boy. Can such an "exchange process" continue for ... | Answer: It cannot. Solution. Consider the numbers in order (from left to right) of all ten boys. Initially, these were all even numbers $2,4,6, \ldots, 20$. Every minute, the number of one of the boys decreases by one, and the process will continue until the numbers of the boys become $1,2,3, \ldots, 10$ (i.e., until a... | 55 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.2. The number $a$ is a root of the quadratic equation $x^{2}-x-50=0$. Find the value of $a^{3}-51 a$. | Answer: 50. Solution. We have $a^{2}=a+50$, therefore $a^{3}=a^{2}+50 a=a+50+50 a=51 a+50$. Hence $a^{3}-51 a=50$. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.5. How many solutions in natural numbers $x, y$ does the system of equations have
$$
\left\{\begin{array}{l}
\text { GCD }(x, y)=20! \\
\text { LCM }(x, y)=30!
\end{array} \quad(\text { where } n!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot n) ?\right.
$$ | Answer: $2^{8}$. Solution. If for the given two numbers $x, y$ the set of their prime divisors is denoted as $p_{1}, p_{2}, \cdots, p_{k}$ and we write $x=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}, y=p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdots p_{k}^{\beta_{k}}$ (where $\alpha_{i}, \beta_{i}$ are n... | 256 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.2. In triangle $A B C$, angle $A$ is the largest. Points $M$ and $N$ are symmetric to vertex $A$ with respect to the angle bisectors of angles $B$ and $C$ respectively. Find $\angle A$, if $\angle M A N=50^{\circ}$. | Answer: $80^{\circ}$. Solution. See problem 8.2 | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.4. In a class of 30 people, for New Year's, each person sent greeting cards to no fewer than 16 classmates. Prove that there were no fewer than 45 pairs of mutual greetings. | Solution. A total of no less than $30 \cdot 16=480$ letters were sent, while the number of pairs of classmates is $(30 \cdot 29) / 2=435$. For each pair of classmates, there can be one of three situations: a) neither of them wrote to the other; b) only one wrote to the other; c) they exchanged letters. Let the number o... | 45 | Combinatorics | proof | Yes | Yes | olympiads | false |
11.5. On the coordinate plane, the graph of $y=\frac{2020}{x}$ is constructed. How many points on the graph have a tangent that intersects both coordinate axes at points with integer coordinates? | Answer: 40 points. Solution. The equation of the tangent at the point $\left(x_{0}, y_{0}\right)$ to the hyperbola $y=k / x$ is $y-y_{0}=-\left(k / x_{0}^{2}\right)\left(x-x_{0}\right)$, where $y_{0}=k / x_{0}$. From this equation, the coordinates $x_{1}$ and $y_{1}$ of the points of intersection with the axes O $x$ an... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. Given two coprime natural numbers $p$ and $q$, differing by more than one. a) Prove that there exists a natural number $n$ such that the numbers $p+n$ and $q+n$ will not be coprime. b) Find the smallest such $n$ for $p=2, q=2023$. | Answer: b) 41.
Solution. See the solution to problem 8.2 (including the notation and comments). If both numbers $p+n$ and $q+n$ are divisible by some $k>1$, then $m=q-p$ is also divisible by $k$, and therefore $k$ is not less than the smallest prime divisor of the number $m$. For $p=2, q=2023$, we have $m=2021=43 \cdo... | 41 | Number Theory | proof | Yes | Yes | olympiads | false |
9.1. Append a digit to the left and right of the eight-digit number 20222023 so that the resulting 10-digit number is divisible by 72. (List all possible solutions.) | Answer: 3202220232.
Solution. Since $72=8 \cdot 9$, it is required to append digits so that the resulting number is divisible by both 8 and 9. Divisibility by 8 is determined by the last three digits: thus, to the two-digit number 23, we need to append a digit on the right to form a three-digit number that is a multip... | 3202220232 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.2. The number $a$ is a root of the quadratic equation $x^{2}-x-100=0$. Find the value of $a^{4}-201 a$ | Answer: 10100.
Solution. Squaring the expression $a^{2}=a+100$, we get $a^{4}=a^{2}+200a+10000=a+100+200a+10000=201a+10100$. From this, we obtain the answer. | 10100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.3. How many points are there on the hyperbola $y=\frac{2013}{x}$ such that the tangent at these points intersects both coordinate axes at points with integer coordinates | Answer: 48 points. Solution. Let $k=2013$. The equation of the tangent to the hyperbola $y=\frac{k}{x}$ at the point $\left(x_{0}, y_{0}\right)$ is $y-y_{0}=-\frac{k}{x_{0}^{2}}\left(x-x_{0}\right)$, where $y_{0}=\frac{k}{x_{0}}$. From this, we find the coordinates of the points of intersection of the tangent with the ... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.4. How many solutions in integers $x, y$ does the equation $|3 x+2 y|+|2 x+y|=100$ have? | Answer: 400. Solution. Note that for any integers $a, b$ the system of equations $\left\lvert\,\left\{\begin{array}{l}3 x+2 y=a \\ 2 x+y=b\end{array}\right.$ has an integer solution $\left\lvert\,\left\{\begin{array}{l}x=2 b-a \\ y=2 a-3 b\end{array}\right.$, and different ordered pairs $(a, b)$ correspond to different... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4. How many three-digit natural numbers $n$ exist for which the number $n^{3}-n^{2}$ is a perfect square | Answer: 22. Solution. Let $n^{3}-n^{2}=m^{2}$ for some natural number $m$. Then $n^{2}(n-1)=m^{2}$, and therefore $n-1$ must also be a perfect square: $n-1=a^{2}$. Thus, $n=a^{2}+1$ and $m=\left(a^{2}+1\right) a$. Therefore, we need to find all three-digit numbers $n$ that are one more than perfect squares. Such number... | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.3. Given a right triangle, the height dropped to the hypotenuse is 4 times smaller than the hypotenuse. Find the acute angles of this triangle. | Answer: $15^{\circ}$ and $75^{\circ}$. Solution. See problem 8.4. | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.2. In triangle $A B C$, angle $A$ is the largest. Points $M$ and $N$ are symmetric to vertex $A$ with respect to the angle bisectors of angles $B$ and $C$ respectively. Find $\angle A$, if $\angle M A N=50^{\circ}$. | Answer: $80^{\circ}$. Solution. Let $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$. Then $\angle A M B=90^{\circ}-\frac{\beta}{2}, \angle A N C=90^{\circ}-\frac{\gamma}{2}$ (since triangles $A M B$ and $\quad$ are isosceles). Therefore, $\angle M A N=180^{\circ}-\left(90^{\circ}-\frac{\beta}{2}\right)-\left(90^{\ci... | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.2. The bisector of angle $ABC$ forms an angle with its sides that is three times smaller than the adjacent angle to $ABC$. Find the measure of angle $ABC$. | Answer. $\quad 72^{\circ}$. Solution. Let $x$ be the degree measure of angle $ABC$. From the condition of the problem, we get the equation $\frac{x}{2}=\frac{180-x}{3} \Leftrightarrow 5 x=360 \Leftrightarrow x=72$ (degrees). | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.3. A natural number is called curious if, after subtracting the sum of its digits from it, the result is a number consisting of identical digits. How many three-digit curious numbers exist? | Answer: 30 numbers. Solution: Let $\overline{x y z}$ be a curious three-digit number. Then the number
$$
A=\overline{x y z}-(x+y+z)=100 x+10 y+z-(x+y+z)=9(11 x+y)
$$
is divisible by 9 and consists of identical digits, and $100-27 \leq A \leq 999-1$. Thus, $A$ can be either 99, 333, or 666.
In the case of $A=99$, we ... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. In the 7th grade, there are 25 people, and each attends a dance club or a drama club (some attend both clubs). Everyone took a math test, and the teacher calculated the percentage of students who received a failing grade (2) separately among the "dancers" and among the "actors." It turned out that the percentage i... | Answer. It can. Solution. Consider the following example. Suppose there are 10 "dancers" and 20 "actors" in the class, then the number of "dancing actors" will be $10+20-25=5$ people, and the number of "pure dancers" will be $10-5=5$ and the number of "pure actors" will be $20-5=15$. Suppose that three "dancers" and si... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.3. Find a six-digit number that, when multiplied by 9, is written with the same digits as the original number but in reverse order? How many such six-digit numbers are there? | Answer. 109989 is the only number. Solution See problem 7.3. | 109989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.3. Given a triangle $A B C$. On the side $A C$, the largest in the triangle, points $M$ and $N$ are marked such that $A M=A B$ and $C N=C B$. It turns out that angle $N B M$ is three times smaller than angle $A B C$. Find $\angle A B C$. | Answer: $108^{\circ}$.
Solution. Point $N$ lies between $A$ and $M$, since $A M+C V=A B+B C > AC$. Let $\angle N B M = x$, then $\angle A B M + \angle N B C = \angle A B C + \angle N B M = 3... | 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1. Find the smallest natural number divisible by 5, with the sum of its digits being 100. Justify your answer. | Answer: 599999999995 (between two fives there are 10 nines). Solution. Due to divisibility by 5, the last digit of the number $N$ can be either 5 or 0. If the last digit is 0, without changing the sum of the digits, we can replace 0 with 5 and subtract one from each of the five non-zero digits in other positions. Then ... | 599999999995 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.3. On the edges of a cube, numbers $1,2, \ldots, 12$ were placed in some order, and for each face, the sum of the four numbers on its edges was calculated. Prove that there is a face for which this sum is greater than 25. | Solution. Let's calculate the corresponding sum on each face and then add these sums for all six faces. We will get $(1+2+\ldots+12) \cdot 2$ as a result, since in this calculation any edge will be counted twice. Thus, the total sum is 156, and then the sum for at least one face is not less than $\frac{156}{6}=26$. (In... | 26 | Combinatorics | proof | Yes | Yes | olympiads | false |
8.4. Given a right triangle, the height dropped to the hypotenuse is 4 times smaller than the hypotenuse. Find the acute angles of this triangle. | Answer: $15^{\circ}$ and $75^{\circ}$. Solution. Let $A B C$ be the given triangle, $C M$ the height from vertex $C$ of the right angle, and $C O$ the median. Consider the right triangle $C M O$. By the property of the median from the vertex of the right angle, $C O=\frac{A B}{2}$, and therefore (by the condition) $C O... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.2. Petya says to Kolya: «I placed some numbers at the vertices of a cube, and then on each face I wrote the sum of the four numbers at its vertices. Then I added all six numbers on the faces and got 2019. Can you figure out what the sum of the eight numbers at the vertices of the cube is?» How would you answer this q... | Answer: 673. Solution. Each vertex of the cube belongs to three faces (which meet at this vertex) and therefore the number placed at this vertex participates three times in the calculation of the sums on the faces. Thus, the sum of the numbers at the vertices is $2019: 3=673$. Comment: another way to solve (direct) is ... | 673 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.2 In a company of $n$ people, 100000 ruble coins need to be distributed equally. How many different values of $n$ exist for which such a distribution is possible? | Answer: 36. Solution: Let's calculate the number of natural divisors of the number $100000=2^{5} \cdot 5^{5}$. Any such divisor has the form $2^{i} \cdot 5^{j}$, where the integers $i, j$ can take six values: from 0 to 5. Then the number of different ordered pairs ( $i ; j$ ) will be $6 \cdot 6=36$. | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.4. How many Pythagorean triangles exist with one of the legs equal to 2013? (A Pythagorean triangle is a right triangle with integer sides. Equal triangles are counted as one.). | Answer: 13. Solution. From the Pythagorean theorem, we obtain the equation in integers $2013^{2}+x^{2}=y^{2} \Leftrightarrow(y-x)(y+x)=2013^{2}=3^{2} \cdot 11^{2} \cdot 61^{2}$. This equation is equivalent to the system $\left\{\begin{array}{l}y-x=d_{1} \\ y+x=d_{2}\end{array}\right.$, where $d_{1}, d_{2}=\frac{2013^{2... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1 The length of a rectangle is $25 \%$ greater than its width. By a straight cut, parallel to the shorter side, this rectangle is cut into a square and a rectangular strip. By what percentage is the perimeter of the square greater than the perimeter of the strip? | Answer: By $60 \%$. Solution: Let $a$ be the width of the rectangle, then its length is $1.25 a$. The perimeter of the resulting square is $4 a$, and the perimeter of the strip is $2 a+0.5 a=2.5 a$. The ratio of the perimeters is $4 a: 2.5 a=1.6$. Therefore, the perimeter of the square is greater than the perimeter of ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.2 Can 8 numbers be chosen from the first hundred natural numbers so that their sum is divisible by each of them | Answer. Yes. Solution. An example of the required numbers can be $1 ; 2 ; 3 ; 6 ; 12 ; 24 ; 48 ; 96.3$ here the sum 192 is divisible by each of the numbers. This example is not unique; others can be provided, such as $1 ; 2 ; 3 ; 4 ; 5 ; 15 ; 30 ; 60$ with a sum of 120. | 120 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2 In the glass, there was a solution in which water made up $99 \%$. The glass with the solution was weighed, and the weight turned out to be 500 gr. After that, part of the water evaporated, so that in the end, the proportion of water was $98 \%$. What will be the weight of the glass with the resulting solution, if ... | Answer: 400 g. Indication. Initially, the weight of the solution was $500-300=200$ (g)., and the amount of water was $0.99 \cdot 200=198$ (g.), and thus, the substance was $200-198=2$ (g.). After the evaporation of water, 2 g of the substance make up $100 \% -98 \% =2 \%$ of the weight of the solution, so the entire so... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3. Given a triangle $A B C$, point $O$ is the center of the inscribed circle. Find the angle $A$ if it is known that the radii of the circumscribed circles of triangles $A B C$ and $B O C$ are the same. | Answer. $\quad 60^{\circ}$.
Solution. Let $\angle A=\alpha$. Note that $\angle B O C=90^{\circ}+\frac{\alpha}{2}$; indeed, $\angle B O C=180^{\circ}-(\angle O B C+\angle B C O)=180^{\circ}-\frac{1}{2}(\angle B+\angle C)=180^{\circ}-\frac{1}{2}\left(180^{\circ}-\alpha\right)=90^{\circ}+\frac{\alpha}{2}$. Then $B C=2 R_... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1. A natural number $n$ was multiplied by the sum of the digits of the number $3 n$, and the resulting number was then multiplied by 2. As a result, 2022 was obtained. Find $n$. | Answer: 337. Solution. Let $s(N)$ denote the sum of the digits of the number $N$. Then the condition of the problem can be written as $2 n \cdot s(3 n)=2 \cdot 3 \cdot 337$, i.e., $n \cdot s(3 n)=3 \cdot 337=1011$. Therefore, $n \leq 1011$, so $3 n$ is at most a four-digit number, and $s(3 n)<4 \cdot 9=36$. Thus, $s(3 ... | 337 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2. Let $a$ be the number of six-digit numbers divisible by 13 but not divisible by 17, and $b$ be the number of six-digit numbers divisible by 17 but not divisible by 13.
Find $a-b$. | Answer: 16290. Solution. Let $c$ be the number of six-digit numbers divisible by both 13 and 17. Then $a+c$ is the number of all six-digit numbers divisible by 13, and therefore $a+c=\left[\frac{999999}{13}\right]-\left[\frac{99999}{13}\right]=76923-7692=69231$.
denotes the integer part of the number $x$ ). Similarly,... | 16290 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.4. Find the largest natural number, all digits of which are different, and the product of these digits is a cube of some natural number. | Answer: 984321. Solution. Obviously, among the digits of the desired number $x$, there is no zero. The product of all digits from 1 to 9 is $2^{7} \cdot 3^{4} \cdot 5^{1} \cdot 7^{1}$. Therefore, $x$ cannot contain the digit 5 or the digit 7 (otherwise, the product of the digits of the number $x$ would have to be divis... | 984321 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.3. Find the largest natural number, all digits of which are different, and the product of these digits is a cube of some natural number. | Answer: 984321. Solution. See problem 7.4 | 984321 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.3. The inscribed circle of triangle $A B C$ with center $O$ touches the sides $A B, B C$ and $A C$ at points $M, N$ and $K$ respectively. It turns out that angle $A O C$ is four times the angle $M K N$. Find angle $B$. | Answer: $108^{\circ}$. Solution. Let $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$. Then, in the isosceles triangle $A K M$, the angle at the base is $\angle A K M=\left(180^{\circ}-\alpha\right) / 2$. Similarly, in triangle $C K N$, we find $\angle C K N=\left(180^{\circ}-\gamma\right) / 2$. Therefore, $\angle M ... | 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.1. Append a digit to the left and right of the eight-digit number 20222023 so that the resulting 10-digit number is divisible by 72. (List all possible solutions.) | Answer: 3202220232.
Solution. Since $72=8 \cdot 9$, it is required to append digits so that the resulting number is divisible by both 8 and 9. Divisibility by 8 is determined by the last three digits: thus, to the two-digit number 23, we need to append a digit on the right to form a three-digit number that is a multip... | 3202220232 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. Given an isosceles triangle $A B C$ with base $A C$. Points $K$ and $N$ are marked on the side $B C$ ( $K$ lies between $B$ and $N$ ). It turns out that $K N=A N$ and $\angle B A K=\angle N A C$. Find $\angle B A N$. | Answer: $60^{\circ}$. Solution. Let $\angle B A K=\angle N A C=x, \angle K A N=y$. In the isosceles triangle $A K N$, the angle $A K N$ (at the base) is also equal to $y$. The angles at the base of triangle $A B C$ are equal to $2 x+y$, and the angle at the vertex $B$ is therefore equal to $180^{\circ}-2(2 x+y)$. The e... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.2. Find the sum of all three-digit natural numbers that do not contain the digits 0 or 9 in their representation. | Answer: 255 744. Solution. We will add the numbers in a column. Each last digit (for example, two) appears in the units place as many times as there are three-digit numbers of the specified form with this last digit (in our case -- these are numbers of the form $\overline{x y 2}$, where $x, y$ are any digits from 1 to ... | 255744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2. Find the sum of all four-digit natural numbers composed of the digits 3, 6, and 9. | Answer: 539946. Solution. We will add the numbers in a column. Each last digit (for example, six) appears in the units place as many times as there are three-digit numbers of the specified type with this last digit (in our case -- these are numbers of the form $\overline{x y z 6}$, where $x, y, z$ are arbitrary sets of... | 539946 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. Find the smallest natural number divisible by 5, with the sum of its digits being 100. Justify your answer. | Answer: 599999999995 (between two fives there are 10 nines).
Solution. Due to divisibility by 5, the last digit of the number $N$ can be either 5 or 0. If the last digit is 0, without changing the sum of the digits, we can replace 0 with 5 and subtract one from each of the five non-zero digits in other positions. Then... | 599999999995 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2. What is the smallest number of digits that can be appended to the right of the number 2013 so that the resulting number is divisible by all natural numbers less than 10? | Answer: three digits. Solution. The least common multiple of the numbers $(1,2, \ldots, 9)$ is 2520. If two digits are appended to 2013, the resulting number will not exceed 201399. Dividing 201399 by 2520 with a remainder, we get a remainder of 2319. Since 2319 > 99, there is no integer multiple of 2520 between the nu... | 2013480 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2. Find the sum of all three-digit natural numbers that do not contain the digit 0 or the digit 5. | Answer: 284160. Solution. We will add the numbers in a column. Each last digit appears in the units place as many times as there are three-digit numbers ending with this digit. Therefore, it will appear $8 \cdot 8=64$ times (since a total of 8 digits are used for the hundreds and tens places). Thus, the sum of the digi... | 284160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1 The average age of the teaching staff of a school, consisting of 20 teachers, was 49 years. When another teacher joined the school, the average age became 48 years. How old is the new teacher? | Answer: 28 years old. Solution: Before the new teacher arrived, the total age of the teachers was 49*20=980. Then the total age became 48*21=1008. Therefore, the new teacher is $1008-$ $980=28$ years old. | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.4 On a grid sheet of paper sized $60 \times 70$ cells (horizontally and vertically respectively), Lena drew a coordinate system (the origin is at the center of the sheet, the $x$-axis is horizontal, the $y$-axis is vertical, and the axes are drawn to the edges of the sheet) and plotted the graph $y=0.83 x$. Then Lena... | Answer: 108. Solution: The graph passes through the first and third quadrants. Let's count the number of shaded cells in the first quadrant (in the third quadrant, there will be the same number, since the graph is a straight line passing through the origin, and therefore centrally symmetric). The graph intersects 29 ve... | 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.2 Let $s(n)$ denote the sum of the digits of a natural number $n$. Solve the equation $n+s(n)=2018$. | Answer: $n=2008$. Solution. Since $n2018-29=1989$, i.e., $n$ is written in the form $\overline{199 x}$ or $\overline{200 x}$ or $\overline{201 x}$ where $x$ is some digit. In the first case, we have the equation $1990+x+19+x=2018$, which gives a non-integer value of $x$. Similarly, in the third case, the equation $2010... | 2008 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. Kolya and Petya exchanged stamps. Before the exchange, Kolya had 5 more stamps than Petya. After Kolya exchanged $24\%$ of his stamps for $20\%$ of Petya's stamps, Kolya had one stamp less than Petya. How many stamps did the boys have before the exchange? | Answer. Petya had 45 stamps, Kolya had 50 stamps. Solution. Let Petya have $x$ stamps before the exchange, then Kolya had $(x+5)$ stamps. After the exchange, Petya had $x-\frac{x}{5}+(x+5) \cdot \frac{6}{25}$, and Kolya had $x+5-(x+5) \cdot \frac{6}{25}+\frac{x}{5}$. Solving the equation $x-\frac{x}{5}+(x+5) \cdot \fra... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.2. For a 92-digit natural number n, the first 90 digits are known: from the 1st to the 10th - ones, from the 11th to the 20th - twos, and so on, from the 81st to the 90th - nines. Find the last two digits of the number n, given that n is divisible by 72. | Answer: 36. Solution: Let the last digits be $x$ and $y$. The number $n$ must be divisible by 9 and 8. The number consisting of the first 90 digits is divisible by 9, since the sum of its digits is divisible by 9. Therefore, the number $\overline{x y}$ is also divisible by 9. In addition, by the divisibility rule for 4... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. $\quad$ In the 7a class, $60 \%$ of the students are girls. When two boys and one girl were absent due to illness, the percentage of girls present was $62.5 \%$. How many girls and boys are there in the class according to the list? | Answer: 21 girls and 14 boys. Solution. Let there be $d$ girls and $m$ boys in the class. From the conditions of the problem, we have two equations: $\frac{d}{d+m}=0.6$ and $\frac{d-1}{d+m-3}=0.625$. From the first equation, $2 d=3 m$. Substituting $m=\frac{2}{3} d$ into the second equation and solving it, we get $d=21... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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