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1. Masha has an integer number of times more toys than Lena, and Lena has the same number of times more toys than Katya. Masha gave Lena 3 toys, and Katya gave Lena 2 toys. After that, the number of toys the girls had formed an arithmetic progression. How many toys did each girl originally have? Indicate the total numb... | 1. Let initially Katya has $a$ toys, then Lena has $k a$, and Masha has $\kappa^{2} a$, where $\kappa-$ is a natural number $\geq 2$. After the changes: Katya has $a-2$, Lena has $a \kappa+5$, and Masha has $a \kappa^{2}-3$. Then these numbers form an arithmetic progression in some order. Let's consider the possible ca... | 105 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,762 |
3. Given a triangle $\mathrm{PEF}$ with sides $\mathrm{PE}=3, \mathrm{PF}=5, \mathrm{EF}=7$. On the extension of side $\mathrm{FP}$ beyond point $\mathrm{P}$, a segment $\mathrm{PA}=1.5$ is laid out. Find the distance $d$ between the centers of the circumcircles of triangles $\mathrm{EPA}$ and $\mathrm{EAF}$. In the an... | 3. By the cosine theorem for angle EPF, we find that $\operatorname{Cos} E P F=-\frac{1}{2}$. Therefore, the angle $\mathrm{EPF}=120^{\circ}$; hence the adjacent angle will be $60^{\circ}$. Draw a perpendicular from point E to PF, meeting at point D. Since angle EPF is obtuse, point D will lie outside triangle EPF. In ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,764 |
4. Find the minimum value of the sum
$$
\left|x-1^{2}\right|+\left|x-2^{2}\right|+\left|x-3^{2}\right|+\ldots+\left|x-10^{2}\right|
$$ | 4. Grouping the terms, we get $\left(\left|x-1^{2}\right|+\left|x-10^{2}\right|\right)+\left(\left|x-2^{2}\right|+\left|x-9^{2}\right|\right)+\left(\left|x-3^{2}\right|+\left|x-8^{2}\right|\right)+\left(\left|x-4^{2}\right|+\left|x-7^{2}\right|\right)+\left(\left|x-5^{2}\right|+\left|x-6^{2}\right|\right)$. It is known... | 275 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,765 |
5. Find all integers $n$ for which the equality $\frac{n^{2}+3 n+5}{n+2}=1+\sqrt{6-2 n}$ holds. | 5. We have: $\frac{n^{2}+3 n+5}{n+2}-1=\sqrt{6-2 n} ; \frac{n^{2}+2 n+3}{n+2}=\sqrt{6-2 n} ; n+\frac{3}{n+2}=\sqrt{6-2 n}$. Since for integer $n$, the right-hand side can be either an integer or irrational, while the left-hand side is either an integer or rational, the given equality is possible only if $\frac{3}{n+2}$... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,766 |
6. Solve the system
$$
\left\{\begin{array}{l}
\operatorname{tg}^{3} x+\operatorname{tg}^{3} y+\operatorname{tg}^{3} z=36 \\
\operatorname{tg}^{2} x+\operatorname{tg}^{2} y+\operatorname{tg}^{2} z=14 \\
\left(\operatorname{tg}^{2} x+\operatorname{tg} y\right)(\operatorname{tg} x+\operatorname{tg} z)(\operatorname{tg} ... | 6. Let $\operatorname{tg} x=x_{1}, \operatorname{tg} y=x_{2}, \operatorname{tg} z=x_{3}$.
Then $\left\{\begin{array}{l}x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=36 \\ x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=14 \\ \left(x_{1}+x_{2}\right)\left(x_{1}+x_{3}\right)\left(x_{2}+x_{3}\right)=60\end{array}\right.$. Further, let $\left\{\begin{arra... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,767 |
7. Solve the system
$\left\{\begin{array}{l}a+c=4 \\ a d+b c=5 \\ a c+b+d=8 \\ b d=1\end{array}\right.$
In the answer, write the sum of all solutions of the given system. | 7. We have
$\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=x^{4}+(a+c) x^{3}+(a c+b+d) x^{2}+(a d+b c) x+b d=$
$=x^{4}+4 x^{3}+6 x^{2}+5 x+2=(x+1)(x+2)\left(x^{2}+x+1\right)$. Since the polynomial $x^{2}+x+1$ does not have
real roots, the identity $\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=\left(x^{2}+3 x+2... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,768 |
9. Find the integer part of the number $\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}}$ | 9. Let $a=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}}$. It is easy to see that $a>\sqrt{6}>2 \Rightarrow$ $a>2$. Let's find an upper bound for this number. At the end of the expression for $a$, replace $\sqrt{6}$ with $\sqrt{9}$. Then we get that $a<\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{9}}}}}=\sqrt{6+\sqrt{6+\sqrt{6+... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,770 |
# Variant 4.
1) Let's draw an arbitrary line $y=k-b$, intersecting the parabola at points $A$ and $B$. Then the abscissas of the endpoints of the chord $A$ and $B$ must satisfy the conditions: $k x - b = x^2$, i.e., $x^2 - k x + b = 0$, $x_A = \frac{k - \sqrt{k^2 - 4b}}{2}$, $x_B = \frac{k + \sqrt{k^2 - 4b}}{2}$. Then... | Answer: $f(x)=\frac{5}{8 x^{2}}-\frac{x^{3}}{8}$.
3) Let
$$
2 x+y-3 z-3=m, x-2 y-4 z-1=n, y+7 z-3 x+7=k
$$
Then we find that $m+n+k=3$. Since by condition $x, y, z$ are integers, then $m, n, k$ are also integers and $m, n, k>0$. Therefore, we get that $m=n=k=1$ and then we have the inequality: $0>z^{2}-9 z+18$. Solv... | f(x)=\frac{5}{8x^{2}}-\frac{x^{3}}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,774 |
# Task 1.
Since the pedestrian covers 1 km in 10 minutes, his speed is 6 km/h. There are more oncoming trams than overtaking ones because, relative to the pedestrian, the speed of the former is greater than that of the latter. If we assume the pedestrian is standing still, the speed of the oncoming trams is the sum of... | Answer: 15 km $/$ hour.
# | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,776 |
# Task 3.
The domain of definition is $x \neq \frac{\pi K}{3}$. We have:
$$
\begin{gathered}
\cos ^{2} x=\sin ^{2} 2 x+\frac{\cos (2 x+x)}{\sin 3 x} \\
\cos ^{2} x=4 \sin ^{2} x \cos ^{2} x+\frac{\cos 2 x \cos x-\sin 2 x \sin x}{\sin 3 x} \\
\cos ^{2} x\left(1-4 \sin ^{2} x\right)=\frac{\cos x\left(\cos 2 x-2 \sin ^{... | Answer: $\left\{\frac{\pi}{2}+\pi n ; \pm \frac{\pi}{6}+\pi k ; k, n \in \mathrm{Z}\right\}$. | {\frac{\pi}{2}+\pin;\\frac{\pi}{6}+\pik;k,n\in\mathrm{Z}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,778 |
# Task 7.
If interest is compounded annually, then according to the compound interest formula, after 10 years, the depositor will receive an amount equal to $(1+0.05)^{10} \cdot 1000$. Similarly, if interest is compounded monthly, then after 10 years (i.e., 120 months), the depositor will receive the amount: $1000 \cd... | Answer: the second case is greater
# | proof | Algebra | proof | Yes | Yes | olympiads | false | 4,779 |
# Task 8.
Constructing a section of a cube.
The line $\mathrm{KL}$, where $\mathrm{K}$ is the midpoint of $D_{1} C_{1}$ and $\mathrm{L}$ is the midpoint of $C_{1} B_{1}$, lies in the plane... | Answer: $\frac{7 \sqrt{17}}{24}$.
U
Chairman of the Methodological Commission for the "Mathematics" profile
FGAOU VO RUT (MIIT) INTERREGIONAL INDUSTRY OLYMPIAD FOR SCHOOL STUDENTS "SAILOF... | \frac{7\sqrt{17}}{24} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,780 |
Task 6.
$$
x^{2}=(y+1)^{2}+1, \quad x^{2}-(y+1)^{2}=1, \quad(x-y-1) \cdot(x+y+1)=1
$$
Since $x, y$ are integers, the equality is possible in two cases:
1) $\left\{\begin{array}{l}x-y-1=1 \\ x+y+1=1\end{array}\right.$
2) $\left\{\begin{array}{l}x-y-1=-1 \\ x+y+1=-1\end{array}\right.$
From the first system, we find t... | Answer: $(1,-1),(-1 ;-1)$. | (1,-1),(-1,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,781 |
# Task 7.
If interest is compounded annually, then according to the compound interest formula, after 8 years, the depositor will receive an amount equal to $(1+0.04)^{8} \cdot 1000$. Similarly, if interest is compounded monthly, then after 8 years (i.e., 96 months), the depositor will receive the amount: $1000 \cdot\l... | Answer: the second case is greater
# | proof | Algebra | proof | Yes | Yes | olympiads | false | 4,782 |
# Task 1.
The pedestrian's speed is 5 km/h. There are more oncoming trams than overtaking ones, because the speed of the former relative to the pedestrian is greater than that of the latter. If we assume that the pedestrian is standing still, the speed of the oncoming trams is the sum of the tram's own speed $V+$ the ... | Answer: $11 \mathrm{km} /$ hour. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,784 |
# Task 3.
Domain of definition: $\sin 3 x \neq 0, x \neq \frac{\pi k}{3}$.
\[
\begin{gathered}
\sin ^{2} 2 x=\cos ^{2} x+\frac{\cos 3 x}{\sin 3 x} \\
4 \sin ^{2} x \cos ^{2} x-\cos ^{2} x=\frac{\cos (2 x+x)}{\sin 3 x} \\
\cos ^{2} x\left(4 \sin ^{2} x-1\right)=\frac{\cos 2 x \cos x-\sin 2 x \sin x}{\sin 3 x}=\frac{\c... | Answer: $\left\{\frac{\pi}{2}+\pi n, \pm \frac{\pi}{6}+\pi k, k, n \in Z\right\}$. | {\frac{\pi}{2}+\pin,\\frac{\pi}{6}+\pik,k,n\inZ} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,785 |
# Task 6.
$$
\begin{gathered}
4 x^{2}=y^{2}+2 y+1+3 \\
(2 x)^{2}-(y+1)^{2}=3 \\
(2 x-y-1) \cdot(2 x+y+1)=3
\end{gathered}
$$
Since $x, y$ are integers, we get:
1) $\left\{\begin{array}{l}2 x-y-1=1 \\ 2 x+y+1=3\end{array}\right.$
2) $\left\{\begin{array}{l}2 x-y-1=3 \\ 2 x+y+1=1\end{array}\right.$
3) $\left\{\begin{a... | Answer: $\{(1,0),(1,-2),(-1,0),(-1,-2)\}$. | {(1,0),(1,-2),(-1,0),(-1,-2)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,786 |
# Task 7.
If interest is compounded annually, then according to the compound interest formula, after 5 years, the depositor will receive an amount equal to $(1+0.03)^{5} \cdot 1000$. Similarly, if interest is compounded monthly, then after 5 years (i.e., 60 months), the depositor will receive the amount: $1000 \cdot\l... | Answer: The second case is greater.
# | proof | Algebra | proof | Yes | Yes | olympiads | false | 4,787 |
1. Solve the equation $\arccos \left(3 x^{2}\right)+2 \arcsin x=0$.
- Compute the cosine of both sides of the equation $\arccos \left(3 x^{2}\right)=-2 \arcsin x$
$3 x^{2}=\cos (-2 \arcsin x)=1-2 \sin ^{2}(\arcsin x)=1-2 x^{2}$
$$
\begin{gathered}
3 x^{2}=1-2 x^{2} \\
x^{2}=\frac{1}{5} ; \quad x= \pm \frac{1}{\sqrt{... | Answer: $-\frac{1}{\sqrt{5}}$. | -\frac{1}{\sqrt{5}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,789 |
2. Find the solution of the system
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=a^{2}, a>0 \\
z^{2}+t^{2}=b^{2}, \quad b>0 \\
\quad x t+y z=a b
\end{array}\right.
$$
for which the value of $x+z$ is maximized.
- The left side of the last equation resembles the dot product in coordinate form. Let's use this. Consider it as ... | # Answer:
$\left(\frac{a^{2}}{\sqrt{a^{2}+b^{2}}} ; \frac{a b}{\sqrt{a^{2}+b^{2}}} ; \frac{b^{2}}{\sqrt{a^{2}+b^{2}}}, \frac{a b}{\sqrt{a^{2}+b^{2}}}\right),\left(\frac{a^{2}}{\sqrt{a^{2}+b^{2}}} ;-\frac{a b}{\sqrt{a^{2}+b^{2}}} ;-\frac{b^{2}}{\sqrt{a^{2}+b^{2}}}, \frac{a b}{\sqrt{a^{2}+b^{2}}}\right)$, $\left(-\frac{... | (\frac{^{2}}{\sqrt{^{2}+b^{2}}};\frac{}{\sqrt{^{2}+b^{2}}};\frac{b^{2}}{\sqrt{^{2}+b^{2}}},\frac{}{\sqrt{^{2}+b^{2}}}),(\frac{^{2}}{\sqrt{^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,790 |
5. Point $E$ is the midpoint of edge $B B_{1}$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$. Find the tangent of the angle between the lines $A E$ and $C A_{1}$.
- Let the edge of the cube be 1. Then $C A_{1}=\sqrt{3}$. Draw a line through point $A_{1}$ parallel to $A E$. It intersects the extension of edge $B B_{1}$... | Answer: $\sqrt{14}$.
Grading criteria:
Maximum points for one problem 7 points - a complete justified solution is provided,
6 points - the problem is almost solved,
5 points - paths to a complete solution have been identified,
4 points - reasoning or formulas that show significant progress have been provided,
3 p... | \sqrt{14} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,791 |
1. A professor from the Department of Mathematical Modeling at FEFU last academic year gave 6480 twos, thereby overachieving the commitments taken at the beginning of the year. In the next academic year, the number of professors increased by 3, and each of them started giving more twos. As a result, a new record was se... | 1. The professors of the Department of Mathematical Modeling at DVFU in the previous academic year gave 6480 twos, thereby over-fulfilling the obligations they had taken on at the beginning of the year. In the next academic year, the number of professors increased by 3, and each of them started giving more twos. As a r... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,792 |
2. Find natural numbers $n$ such that for all positive numbers $a, b, c$, satisfying the inequality
$$
n(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there exists a triangle with sides $a, b, c$. | 2. Find natural numbers $n$ such that for all positive numbers $a, b, c$ satisfying the inequality
$$
n(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there exists a triangle with sides $a, b, c$.
$\square$ Since $(a b+b c+c a) \leq\left(a^{2}+b^{2}+c^{2}\right)$, then $n>5$. If $n \geq 7$, then the numbers $a=1, b... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,793 |
3. The sequence of polynomials is defined by the conditions:
$$
P_{0}(x)=1, P_{1}(x)=x, P_{n+1}(x)=x P_{n}(x)-P_{n-1}(x), n=1,2, \ldots
$$
How many distinct real roots does the polynomial $P_{2018}(x)$ have? | 3. The sequence of polynomials is defined by the conditions:
$$
P_{0}(x)=1, P_{1}(x)=x, P_{n+1}(x)=x P_{n}(x)-P_{n-1}(x), n=1,2, \ldots
$$
How many distinct real roots does the polynomial $P_{2018}(x)$ have?
$\square$ We will prove by induction that the polynomial $P_{n}(x)=x^{n}+\ldots$ has exactly $n$ distinct rea... | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,794 |
4. On one of the sides of the right triangle $A B C$, whose legs are equal to 1, a point $P$ is chosen. Find the maximum value of $P A \cdot P B \cdot P C$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 4. On one of the sides of the right-angled triangle $ABC$, with legs equal to 1, a point $P$ is chosen. Find the maximum value of $PA \cdot PB \cdot PC$.
$\square \mathrm{B}$ in the Cartesian coordinate system $A(1,0) ; B(0,1) ; C(0,0)$. If $P(x, 0)$ is on side $AC$, then
$$
PA \cdot PB \cdot PC = x(1-x) \sqrt{1+x^{2... | \frac{\sqrt{2}}{4} | Logic and Puzzles | other | Yes | Yes | olympiads | false | 4,795 |
5. Stepanka, celebrating his victory in the "Ocean of Knowledge" Olympiad, poured champagne into 2018 glasses. Fair Hryusha is trying to ensure that all glasses have the same amount of champagne. He takes two glasses and equalizes the amount of champagne in them. Can Stepanka pour the champagne in such a way that Hryus... | 5. Stepan, celebrating his victory in the "Ocean of Knowledge" Olympiad, poured champagne into 2018 glasses. Fair-minded Hryusha tries to ensure that all glasses have the same amount. Can Stepan pour the champagne in such a way that Hryusha's attempts are in vain?
$\square$ Suppose Stepan pours 1 unit of champagne int... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,796 |
1. A rectangle is drawn on the board. It is known that if its width is increased by $30 \%$, and its length is decreased by $20 \%$, then its perimeter remains unchanged. How would the perimeter of the original rectangle change if its width were decreased by $20 \%$, and its length were increased by $30 \%$? | Answer. It will increase by $10 \%$.
Solution. Let's denote the width of the original rectangle as $a$, and the height as $b$. In the first case, the modified width and length will be $1.3a$ and $0.8b$ respectively. According to the condition, $2(a+b) = 2(1.3a + 0.8b)$, from which we get $b = 1.5a$. This means the ori... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,797 |
2. The quadratic trinomial $x^{2}+u x-v$ has distinct non-zero roots $p$ and $q$, and the quadratic trinomial $x^{2}+p x-q-$ has distinct non-zero roots $u$ and $v$. Find all possible values of $p, q, u$ and $v$. | Answer. $p=-1, q=2, u=-1, v=2$.
Solution. Applying Vieta's theorem to both equations, we form the system of equations:
$$
\left\{\begin{array}{l}
p+q=-u \\
u+v=-p \\
p q=-v \\
u v=-q
\end{array}\right.
$$
Subtracting the second equation from the first, after simplification, we get: $q=v$. Substituting $q$ for $v$ in... | p=-1,q=2,u=-1,v=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,798 |
3. In trapezoid $K L M N$ the base $L M$ is twice as short as $K N$. Inside the trapezoid, there is a point $O$ such that $K L=O L$. Prove that the line connecting point $M$ with the midpoint of segment $O N$ is perpendicular to $O K$. | Solution. Let $P$ be the midpoint of $O N$, and $Q$ be the midpoint of $K O$. Segment $Q P$ is the midline of triangle $K O N$, so $Q P\|K N\| L M$ and $Q P=\frac{1}{2} K N=L M$. Therefore, $Q L M P$ is a parallelogram and $L Q \| M P$. On the other hand, segment $L Q$ is the median of isosceles triangle $K L O$, so $L... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,799 |
4. Misha, over the course of a week, picked an apple each day and weighed it. Each apple weighed a different amount, but the weight of each apple was a whole number of grams and ranged from 221 grams to 230 grams (inclusive). Misha also calculated the average weight of all the apples he picked, and it was always a whol... | Answer: 230 grams.
Solution: Each apple weighed 220 grams plus an integer from 1 to 10. From the numbers 1 to 10, we need to choose 7 numbers such that their sum is divisible by 7. One of these numbers is 5, so we need to choose 6 from the numbers $1,2,3,4,6,7,8,9,10$. The smallest sum of six of these numbers is 23, a... | 230 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,800 |
5. Several teams held a hockey tournament - each team played against each other once. 2 points were awarded for a win, 1 point for a draw, and no points for a loss. The team "Squirrels" won the most games and scored the fewest points. What is the minimum number of teams that could have participated in the tournament | Answer: 6 teams.
Solution. In a tournament with $n$ teams, the "Squirrels" scored less than the average number, i.e., no more than $n-2$ points. Therefore, some team must have scored more than the average, i.e., no less than $n$ points. For this, they had to win at least one match. Consequently, the "Squirrels" must h... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,801 |
2. Mom, dad, and nine children stood in a row on a straight path. Mom and dad are standing next to each other, 1 meter apart. Can the children stand in such a way that the total distance from all the children to mom equals the total distance from all the children to dad? If yes, provide an example; if no, explain why. | Answer. No.
Solution. Let the mother stand to the left of the father. Then the child standing to the left of the mother is 1 meter closer to the mother than to the father, and the child standing to the right of the father is 1 meter closer to the father than to the mother. For the sums of the distances to be equal, th... | No | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,804 |
4. One of the four squirrels broke the honey jar. Grey claimed that the jar was broken by Blackie. But Blackie claimed that it was Firetail who was to blame. Reddy said that he did not break the jar, and Firetail said that Blackie was lying. Only one of the squirrels told the truth. Who told the truth, and who broke th... | Answer. Firetail spoke the truth, Redpaw broke the jar.
Solution. Blackie and Firetail contradict each other. Therefore, one of them is telling the truth. Since only one of them is telling the truth, Gray and Redpaw lied. Therefore, Redpaw broke the jar. Blackie said that Firetail was to blame, so he also lied. That l... | Firetailspokethetruth,Redpawbrokethejar | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,806 |
5. On the glade, there are 6 fewer bushes than trees. Birds flew in and sat both on the bushes and on the trees. They sat so that there were an equal number of birds on all the trees and an equal number of birds on all the bushes, but there were at least 10 more birds on a tree than on a bush. There were a total of 128... | Answer: 2.
Solution: There are no fewer than 7 trees, as there are 6 more trees than shrubs. There are no fewer than 11 birds on one tree, as there are at least 10 more birds on one tree than on one shrub. There cannot be 12 or more trees, as then there would be $12 \cdot 11=$ 132 or more birds on the trees. Therefore... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,807 |
1. Senya thought of two numbers, then subtracted the smaller from the larger, added both numbers and the difference, and got 68. What was the larger of the numbers Senya thought of? | Answer: 34.
Solution. The subtrahend plus the difference equals the minuend. Therefore, the doubled minuend equals 68.
Comment. Correct solution - 20 points. The result is obtained based on examples and the pattern is noticed but not explained - 15 points. The result is obtained based on one example - 10 points. The ... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,808 |
2. There are two hollows in the cedar. The baby squirrel sitting in front of the second hollow said:
1) there are no nuts in the other hollow,
2) there are nuts in at least one hollow.
Red baby squirrels always tell the truth, while gray ones always lie. What color is this baby squirrel? | Answer. Red.
Solution. Suppose he is gray, then both statements are false. If the second statement is false, there are no nuts in any hollow. And if the first statement is false, there are nuts in another hollow, which is a contradiction. Let the squirrel be red, then both statements are true. Then from the second sta... | Red | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,809 |
4. Each cell of a $6 \times 6$ board is occupied by ants sitting motionless. The number of ants in cells adjacent by side differs by 1. There are 4 ants on one cell and 14 ants on another cell. Fill the entire board with numbers indicating how many ants are in each cell. | Answer. Up to rotation:
| 9 | 10 | 11 | 12 | 13 | 14 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 8 | 9 | 10 | 11 | 12 | 13 |
| 7 | 8 | 9 | 10 | 11 | 12 |
| 6 | 7 | 8 | 9 | 10 | 11 |
| 5 | 6 | 7 | 8 | 9 | 10 |
| 4 | 5 | 6 | 7 | 8 | 9 |
Solution. From one cell to another, a path can be laid consisting of no m... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,811 |
5. At a children's party, there were 8 children. The adults prepared 16 bags of candies. The first one had 1 candy, the second had 2 candies, and so on, with the 16th bag having 16 candies. Each of the eight children received one bag at the beginning of the party, and one bag at the end of the party. Could it have been... | Answer: Yes.
Solution. First, divide the packages into pairs with equal sums:
| Start | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| End | 16 | 15 | 14 | 13 | 12 | 11 | 10 | 9 |
Each column in the table corresponds to one child, meaning each will receive... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,812 |
1. Seven children are standing in a circle: Anya, Borya, Vasya, Gena, Dasha, Eva, Zhena. Starting with one of the children, every third person clockwise leaves, and the counting continues until only one person remains. For example, if the counting starts with Anya (Anya - first), then Vasya leaves, in the next trio Gen... | Answer. From Vasya.
Solution. Let's start counting from Anya. Gena will be left. Therefore, there needs to be a shift of 2 people, and we need to start from Vasya. Then the children leave in the following order: Dasha, Anya, Gena, Borya, Zhenya, Vasya.
Comment. Correct solution - 20 points. The solution is started, b... | FromVasya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,813 |
3. In the first quarter, Masha and Lena together received 23 fives, Svetlana and Masha together received 18 fives, and Svetlana and Lena together received 15 fives. How many fives did each of the girls receive? | Answer. Masha -13, Lena -10, Svetlana -5.
Solution. All three girls together received half of $(23+18+15)=28$ fives. Therefore, Svetlana received $28-23=5$ fives, Masha received $18-5=13$ fives, and Lena received $23-13=10$ fives.
Comment. Correct solution - 20 points. There is a gap in the justification - 15 points.... | Masha-13,Lena-10,Svetlana-5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,815 |
4. Place the numbers $1,2,3$ in the cells of a $6 \times 6$ square so that all 12 sums of the rows and columns are different. | Answer. For example, like this
| 3 | 3 | 3 | 3 | 3 | 3 | $\mathbf{1 8}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 3 | 3 | 3 | 3 | 3 | 1 | $\mathbf{1 6}$ |
| 3 | 3 | 3 | 3 | 1 | 1 | $\mathbf{1 4}$ |
| 3 | 3 | 2 | 1 | 1 | 1 | $\mathbf{1 1}$ |
| 3 | 2 | 1 | 1 | 1 | 1 | $\mathbf{9}$ |
| 2 | 1 | 1 | 1 |... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,816 |
5. Mom, Dad, Vasya, and Nina were making dumplings. Mom and Nina made 4 more dumplings than Dad and Vasya, and Mom and Dad made 2 more dumplings than Vasya and Nina. Then Vasya and Mom left, and Nina and Dad continued making dumplings at the same speed, and each made twice as many dumplings as they did initially. Who m... | Answer: Equally
Solution: Two moms, Nina and dad made 6 more dumplings than two Vasya, Nina and dad, which means mom made 3 more dumplings than Vasya. Then, Nina made 1 more dumpling than dad. After the break, Nina made 2 more dumplings than dad, and together they made 3 more dumplings than dad. Mom had 3 more dumplin... | Equally | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,817 |
1. There are 28 students in the class. 17 have a cat at home, and 10 have a dog. 5 students have neither a cat nor a dog. How many students have both a cat and a dog? | Answer: 4.
Solution: The number of students with cats or dogs is $28-5=23$. If we add $17+10=27$, it exceeds 23 due to those who have both a cat and a dog, who were counted twice. The number of such students is $27-23=4$.
Comment: Correct solution - 20 points. Solution started, with some progress - 5 points. Solution... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,818 |
2. The little squirrel has several bags of nuts. In two bags, there are 2 nuts each, in three bags, there are 3 nuts each, in four bags, there are 4 nuts each, and in five bags, there are 5 nuts each. Help the little squirrel arrange the bags on two shelves so that there are an equal number of bags and nuts on each she... | Solution. For example, $5+5+5+4+4+2+2=27$ nuts in 7 bags - the first shelf, $5+5+4+4+3+3+3=27$ nuts in 7 bags - the second shelf.
Comment. Correct solution - 20 points. Solution started, some progress made - 5 points. Solution started, but progress insignificant - 1 point. Solution incorrect or absent - 0 points. | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,819 |
3. Vasya, Dima, Kolya, and Sergey, who study in grades 5, 6, 7, and 8, decided to form a rock band. Among them, there is a saxophonist, a keyboardist, a drummer, and a guitarist. Vasya plays the saxophone and is not in grade 8. The keyboardist is in grade 6. The drummer is not Dima, Sergey is not the keyboardist and no... | Answer. Dima studies in the 8th grade and plays the guitar.
Solution. According to the condition, Vasya plays the saxophone, and Dima is not a drummer. But Dima is also not a keyboardist since he does not study in the 6th grade, so he plays the guitar. Since Sergey is not a keyboardist, he is a drummer, and the keyboa... | Dimastudiesinthe8thgradeplaystheguitar | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,820 |
4. On graph paper, large and small triangles are drawn (all cells are square and of the same size). How many small triangles can be cut out from the large triangle? The triangles cannot be rotated or flipped (the large triangle has a right angle at the bottom left, the small triangle has a right angle at the top right)... | Answer: 12.
Solution. See the figure.
Comment. Correct drawing with 12 triangles - 20 points. Incorrect answer with drawing - 1 point. Any answer without drawing - 0 points. | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,821 |
5. In a row, all natural numbers from 1 to 100 inclusive are written in ascending order. Under each number in this row, the product of its digits is written. The same procedure is applied to the resulting row, and so on. How many odd numbers will be in the fifth row? | Answer: 19.
Solution: Note that if a number contains an even digit in its notation, then the product of the digits will be even. However, then in all subsequent rows, the product will also be even. Let's find all the products of two digits in the multiplication table that are written using only odd digits:
$$
\begin{... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,822 |
1. On a certain day, there were several nuts in the bag. The next day, as many nuts were added to the bag as there already were, but eight nuts were taken away. On the third day, again as many nuts were added as there already were, but eight were taken away. The same thing happened on the fourth day, and after that, th... | Answer: 7 nuts.
Solution. At the beginning of the fourth day, there were $(0+8): 2=4$ nuts in the bag. Therefore, at the beginning of the third day, there were $(4+8): 2=6$, and at the very beginning $-(6+8): 2=7$. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,823 |
3. All 10 dumbbells weighing $4, 5, 6, 9, 10, 11, 14, 19, 23, 24$ kilograms need to be placed on three racks so that the weight of the dumbbells on the first rack is half the weight of the dumbbells on the second rack. And the weight of the dumbbells on the second rack is half the weight of the dumbbells on the third r... | Answer: No.
Solution: Suppose we managed to solve the problem. Then, if the total weight of the dumbbells on the first stand is $N$ kg, the total weight of the dumbbells on the second stand will be $2 N$ kg, and on the third stand, it will be twice as much, that is, $4 N$ kg. Then the total weight of all the dumbbells... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,825 |
5. Sixth-graders were discussing how old their principal is. Anya said: "He is older than 38 years." Borya said: "He is younger than 35 years." Vova: "He is younger than 40 years." Galya: "He is older than 40 years." Dima: "Borya and Vova are right." Sasha: "You are all wrong." It turned out that the boys and girls mad... | Solution. Note that Anya and Vova cannot be wrong at the same time, so Sasha is wrong. Also, at least one of the pair "Anya-Borya" and at least one of the pair "Vova-Galya" is wrong. Thus, there are no fewer than three wrong answers, and due to the evenness, there are exactly four: two boys and two girls. This means th... | 39 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,827 |
1. Find all integers $a$ for which the modulus of the number $\left|a^{2}-3 a-6\right|$ is a prime number. Answer. $-1 ; 4$. | Solution. The number $a^{2}-3 a$ is always even (as the difference of two even or two odd numbers), so the prime number $\left|a^{2}-3 a-6\right|$ can only be 2. The equation $a^{2}-3 a-6=2$, or $a^{2}-3 a-8=0$, has no integer solutions. Solving the equation $a^{2}-3 a-6=-2$, we find the roots $a_{1}=-1, a_{2}=4$.
Com... | -1,4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,828 |
3. Alice wrote several positive integers. Sasha rewrote these numbers and added one integer, smaller than all of Alice's numbers. Each of them found the sum and the product of their numbers and divided the sum by the product. Sasha got a number 5 times smaller than Alice. What number could he have added? | Answer: 6 or -20.
Solution: Let Alice's sum be $a$, Alice's product be $b$, and the added number be $c$. Then the condition can be written as: $\frac{a}{b}=\frac{a+c}{b c} \cdot 5$, or $a c=5 a+5 c$. Factoring, we get: $(a-5)(c-5)=25$. The factor $a-5$ is a divisor of 25 and can take the values $\pm 1, \pm 5, \pm 25$.... | 6or-20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,830 |
4. In trapezoid $ABCD$, the bases $AD$ and $BC$ are in the ratio $AD: BC = 3: 2$, the lateral side $AB$ is perpendicular to the bases. A point $K$ is chosen on side $AB$ such that $KA: AB = 3: 5$. From point $K$, a perpendicular is drawn to $CD$, intersecting segment $CD$ at point $P$. Prove that $\angle KPA = \angle K... | Solution. From the condition, it follows that $K A: K B=3: 2$. Triangles $K D A$ and $K B C$ are similar as right triangles with proportional legs, so $\angle K C B=\angle K D A$. Quadrilateral $A D P K$ is inscribed, since $\angle K P D=\angle K A D=90^{\circ}$, therefore, $\angle K D A=$ $\angle K P A$ (these angles ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,831 |
5. Natural numbers divisible by 3 were painted in two colors: red and blue, such that the sum of a blue and a red number is red, and the product of a blue and a red number is blue. In how many ways can the numbers be colored so that the number 546 is blue? | Answer: 7.
Solution: We will prove that all blue numbers are divisible by the same number. First, we will prove that the product of two blue numbers is blue. Assume the statement is false. Let $a$ and $b$ be two such blue numbers that their product is a red number. Let $d$ be some red number (we can take $d = ab$). Th... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,832 |
1. The sum of two numbers is 2022. If the last digit 5 of the first number is erased, and the digit 1 is appended to the right of the second number, the sum of the modified numbers will become 2252. Find the original numbers. | Answer: 1815 and 207.
Solution. Let the original numbers be $x$ and $y$. We form the system
$$
\left\{\begin{array}{l}
x+y=2022 \\
\frac{x-5}{10}+10 y+1=2252
\end{array}\right.
$$
Solving the system, we find $x=1815, y=207$.
Comment. Correct solution - 20 points. Solution is not complete, but there is progress - 5 ... | 1815207 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,833 |
2. The perimeter of a triangle is 4. Prove that the sum of the squares of the lengths of the sides is greater than 5. | Solution. If the triangle is equilateral with side $a=\frac{4}{3}$, then the inequality holds, since $3 a^{2}=3 \cdot\left(\frac{4}{3}\right)^{2}=\frac{16}{3}>5$. Let an arbitrary triangle have sides $a+x, a+y, a+z$, where $x+y+z=0$. Then
$(a+x)^{2}+(a+y)^{2}+(a+z)^{2}=3 a^{2}+2 a(x+y+z)+x^{2}+y^{2}+z^{2} \geq 3 a^{2}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 4,834 |
3. Two baby squirrels are located at points $A$ and $B$, and start jumping simultaneously along lines $A O$ and $B O$ towards point $O$ (after passing point $O$, each continues moving along its own line). The distance $A O=120$ meters, $B O=80$ meters, and the angle $A O B=60^{\circ}$. The baby squirrels have a constan... | Answer: $20 \sqrt{3}$ meters.
Solution: Let the squirrels travel a distance $d = A E = B F$. Draw a line through point $E$ parallel to $O B$, and mark a segment $E K$ on it, equal to $A E$. Draw the line $A K$, and let it intersect the line $O B$ at point $C$.
Triangle $A E K$ is isosceles with an angle of $60^{\circ... | 20\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,835 |
4. Find the smallest natural $n>1$, for which the sum of no two natural powers is a perfect square of a natural number.
# | # Answer.4.
Solution. For $n=2: 2^{1}+2^{1}=2^{2}$. For $n=3: 3^{3}+3^{2}=6^{2}$. Let $n=4$. Suppose there exist such $m, k$ that $4^{m}+4^{k}=a^{2}$.
Any power of 4 gives a remainder of 1 when divided by 3 (since $\left.4^{m}=(3+1)^{m}\right)$. Therefore, $4^{m}+4^{k}$ will give a remainder of 2 when divided by 3. B... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,836 |
5. Can 5 intersection points of grid lines be marked on graph paper so that these points are vertices of a regular pentagon | Answer. No.
Solution. Suppose such integer-sided pentagons exist; let's take a pentagon with the smallest side. Choose the origin of coordinates, draw the axes. Each side will correspond to a vector $\bar{a}_{i}=\left(x_{i}, y_{i}\right), x_{i}^{2}+y_{i}^{2}=a^{2}$. We choose the direction of the vectors clockwise, so... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,837 |
1. How many irreducible fractions exist where the sum of the numerator and denominator is 100? | Answer: 20.
Solution: Consider the equation $a+b=100$. If $a$ and $b$ have a common divisor, then this divisor also divides 100, meaning that the only possible prime common divisors of $a$ and $b$ satisfying the equation can be 2 and 5. We will pair all numbers as ( $a, 100-a$ ). Since the fraction $\frac{a}{b}$ is ir... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,838 |
2. Two squirrels had the same number of pine cones and the same number of cedar cones. In total, each squirrel had fewer than 25 cones. The first squirrel gathered as many pine cones as it already had and 26 cedar cones. It ended up with more pine cones than cedar cones. The second squirrel gathered as many cedar cones... | Answer: 17 pine cones, 7 cedar cones.
Solution. Let the number of pine cones be $x$ and the number of cedar cones be $y$. We can write the conditions as: $\left\{\begin{array}{l}x+y=26 \\ 2 y>x-4\end{array}\right.$
Add the second and third inequalities: $2(x+y)>x+y+22$, or $x+y>22$. But $x+y=26$, so $x+y>22$ is alway... | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,839 |
3. The sum of $m$ consecutive natural numbers is equal to a prime number $p$. What can $m$ be? | Answer: 1 and 2.
Solution. The sum of $m$ consecutive numbers from $k$ to $k+m-1$ inclusive is $\frac{(2 k+m-1) m}{2}$. This sum must equal $p$, that is, the factors are 1 and $p$. If $m>2$ and $m$ is odd, then $m$ equals $p$. For $m=p: \frac{2 k+p-1}{2}=1, 2 k+p=3$, which is impossible. If $m>2$ and even, then $m / 2... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,840 |
5. A $7 \times 7$ table is filled with non-zero integers. First, the border of the table is filled with negative numbers. Then, the cells are filled in any order, and the next number is equal to the product of the previously placed numbers that are closest to it in the same row or column. What is the maximum number of ... | # Answer. 24.
Solution. Evaluation. We will prove that there must be at least one negative number in the shaded area. Suppose this is not the case. Consider the corner cell of the shaded area where the number was placed last among the four corners (it is shaded black). By assumption, if some other numbers are placed i... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,841 |
1. In a convex hexagon $A B C D E F$, point $M$ is the midpoint of segment $A C$, and point $N$ is the midpoint of segment $C E$. Prove that the area of the shaded figure is half the area of the hexagon $A B C D E F$. | Solution. Segments $B M, F M, F N, D N$ are medians of triangles $A B C, A C F, F C E, E C D$. Each median divides the corresponding triangle into two triangles of equal area, due to the equality of heights and bases.
. Let the first group consist of squirrels numbered $1,5,9$, the second group - squirrels numbered $2,6,7$, and the third group - squirrels numbered $3,4,8$. When the first ... | Yes | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 4,843 |
3. Two inequalities were written on the board, each of which holds for some numbers $a \geq b \geq c \geq 0$.
$$
\text { (1) } a^{2}+b^{2}+c^{2} \leq 2(a b+b c+a c)
$$
(2) $a^{4}+b^{4}+c^{4} \leq 2\left(a^{2} b^{2}+b^{2} c^{2}+a^{2} c^{2}\right)$.
Anya believes that inequalities (1) and (2) are equivalent, Petya - t... | Answer. Dani.
Solution. When $a=4, b=1, c=1$, inequality (1) is satisfied, but inequality (2) is not, since $16+1+1=18 \leq 2(4+4+1)=18$, but $4^{4}+1^{4}+1^{4}=258>2\left(4^{2} 1^{2}+4^{2} 1^{2}+\right.$ $\left.1^{2} 1^{2}\right)=66$. Therefore, (1) does not imply (2).
When $a=4, b=3, c=-2$, inequality (2) is satisf... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,844 |
4. A circle passing through the vertices $L$ and $M$ of trapezoid $K L M N$ intersects the lateral sides $K L$ and $M N$ at points $P$ and $Q$ respectively and touches the base $K N$ at point $S$. It turns out that $\angle L S M=50^{\circ}$, and $\angle K L S=\angle S N M$. Find $\angle P S Q$. | Answer: $65^{\circ}$.
Solution. Since $K S$ is a tangent to the circle, then
$$
\angle K S P=\angle S Q P=\angle S L P=\angle K L S=\angle S N M .
$$
But $\angle K S P$ and $\angle S N M$ are corresponding angles for the lines $P S$ and $M N$, so $P S \| M N$ and the inscribed quadrilateral $S Q M P$ is a trapezoid.... | 65 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,845 |
5. A little squirrel collected 15 nuts in the forest weighing $50, 51, \ldots, 64$ grams. He knows the weight of each nut. Using a balance scale, the little squirrel tries to prove to his friends that the first nut weighs 50 g, the second 51 g, the third 52 g, and so on (initially, the friends know nothing about the we... | Answer. One weight.
Solution. Note that it is impossible to do without weights altogether, since if the weight of each nut is halved, the results of all weighings will not change. We will show that the baby squirrel can manage with the help of one 1 g weight. First, he will convince his friends that the nuts weigh $n,... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,846 |
1. It is known that $3^{3}+4^{3}+5^{3}=6^{3}$. Prove that for any odd natural $n$ there exist $n$ distinct numbers, the sum of whose cubes is the cube of a natural number. | Solution. Multiply both sides of the equality $3^{3}+4^{3}+5^{3}=6^{3}$ by $2^{3}$, we get $6^{3}+8^{3}+$ $10^{3}=12^{3}$. Replace $6^{3}$ with $3^{3}+4^{3}+5^{3}$, then we get the equality $3^{3}+4^{3}+5^{3}+8^{3}+10^{3}=$ $12^{3}$. By sequentially performing this operation, we can obtain the required representation f... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 4,847 |
2. On 8 balls, numbers are written: $2,3,4,5,6,7,8,9$. In how many ways can the balls be placed into three boxes so that no box contains a number and its divisor? | Answer: 432.
Solution. The numbers 5 and 7 can be placed in any box, the number of ways is $3 \cdot 3=9$. The numbers $2, 4, 8$ must be in different boxes, the number of ways is $3 \cdot 2 \cdot 1=6$. Thus, the numbers $2, 4, 5, 7, 8$ can be arranged in 54 ways. Suppose the numbers 2 and 3 are placed in the same box (... | 432 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,848 |
3. A three-digit number consists of different digits. Between the first and second digits, as well as between the second and third digits, $n$ zeros are inserted. Prove that there is more than one original three-digit number such that the resulting $(2 n+3)$-digit number is a square of an integer for any natural $n$. | Answer: 169 and 961.
Solution. Let's find two such numbers. Let the original number $A=\overline{a b c}$. Then $A=a \cdot 10^{2 n+2}+$ $b \cdot 10^{n+1}+c$. This expression is a perfect square $\left(\sqrt{a} \cdot 10^{n+1}+\sqrt{c}\right)^{2}$ under the condition that $b=2 \sqrt{a c}$. Since the square root must be a... | 169961 | Number Theory | proof | Yes | Yes | olympiads | false | 4,849 |
4. Point $O$ is the center of the circumcircle of an acute-angled triangle $K L M$ with angle $\angle L=30^{\circ}$. Ray $L O$ intersects segment $K M$ at point $Q$. Point $P$ is the midpoint of the arc $O M$ of the circumcircle of triangle $Q O M$, not containing point $Q$. Prove that points $K, L, P$, $Q$ lie on the ... | Solution. With respect to the circumcircle of triangle $K L M$, the angle $\angle K L M$, equal to $30^{\circ}$, is an inscribed angle, therefore, the central angle $\angle K O M=60^{\circ}$. Thus, triangle $K O M$ is equilateral. Then quadrilateral $K O R M$ is composed of an equilateral and an isosceles triangle, lin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,850 |
5. What is the maximum number of cells in an $8 \times 8$ square that can be colored so that the centers of any four colored cells do not form the vertices of a rectangle with sides parallel to the edges of the square? | Answer: 24 cells.
Solution: Suppose that no less than 25 cells are marked. We will say that a pair of marked cells in the same row covers a pair of columns in which these cells are located. A prohibited quartet arises when a pair of columns is covered twice. If in one row there are $m$ marked cells, and in another $n>... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,851 |
1. Two identical triangles were placed on top of each other such that a hexagon was formed at their intersection (see figure). Prove that the areas of the gray and white parts are equal. | Solution. The area of the first triangle is equal to the sum of the area of the black hexagon and the white parts; the area of the second triangle is equal to the sum of the area of the black hexagon and the gray parts. But the areas of the triangles are equal.
Comment. Correct example - 20 points. | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,854 |
4. The sum of several natural numbers is 2022. Arina divided each even number by two, and multiplied each odd number by three. Could the sum of the numbers remain unchanged? | Answer. No.
Solution. Let the sum of even numbers be $A$, and the sum of odd numbers be $B$. Then $A+B=\frac{A}{2}+3 B$, from which $A=4 B$ and the total sum is $5 B$. But 2022 is not divisible by 5.
Comment. Correct solution - 20 points. There are gaps in the justification - 15 points. The equation is correctly set ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,855 |
5. Vasya added four numbers in pairs. The four largest of the six sums obtained were $20, 16, 13, 9$. Find the two remaining sums and determine which numbers Vasya could have added. | Answer. The sums are 2 and 6. The numbers: $-0.5 ; 2.5 ; 6.5 ; 13.5$ or $-2.5; 4.5; 8.5; 11.5$.
Solution. Let the numbers be $a, b, c, d$; Assume $a \leq b \leq c \leq d$. The largest sum is obtained by adding $d$ and $c$, the next largest by adding $d$ and $b$. That is, $d+c=20, d+b=16$. There are further possibiliti... | 26 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,856 |
1. In a $5 \times 5$ square, color as many cells black as possible, so that the following condition is met: any segment connecting two black cells must necessarily pass through a white cell. | Answer. We will prove that it is impossible to color more than 9 cells. Divide the square into 9 parts:
In each part, no more than one cell can be colored.
9 cells can be colored as follows:... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,857 |
3. In a rectangle of size $7 \times 9$ cells, some cells contain one baby squirrel each, such that in every rectangle of size $2 \times 3$ (or $3 \times 2$) there are exactly 2 baby squirrels. Draw how they can be seated. | Answer. For example, like this (21 squirrels sit in the shaded cells).
Comment. A correct example - 20 points. | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,858 |
4. Select five more natural numbers to the number 34 so that two conditions are met:
(1) among the six numbers, there are no three numbers that have a common divisor (greater than one);
(2) among any three numbers from these six, there will be two numbers that have a common divisor (greater than one). | Answer. For example, $6,34,35,51,55,77$.
Solution. Since $34=2 \cdot 17$, let's take the prime number 3 and add the numbers $2 \cdot 3=6$ and $17 \cdot 3=51$. The numbers $6,34,51$ satisfy the conditions. We will construct another set of three numbers using the prime numbers $5,7,11$, obtaining the numbers $35,55,77$.... | 6,34,35,51,55,77 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,859 |
1. After pulling out the turnip, a lot of earth was left. The Grandpa removed half of the pile, Grandma removed a third of the remaining earth, the Granddaughter removed a quarter of the remainder, the Dog removed a fifth of the remainder, the Cat removed a sixth, and the Mouse removed a seventh. What part of the pile ... | Answer: $\frac{1}{7}$.
Solution: First, half of the pile remained, then $\frac{2}{3}$ of the remainder, then $\frac{3}{4}$ of the new remainder, and so on. In the end, $1 \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdot \frac{6}{7}=\frac{1}{7}$ of the original pile remain... | \frac{1}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,861 |
2. Diana wrote a two-digit number, and appended to it a two-digit number that was a permutation of the digits of the first number. It turned out that the difference between the first and second numbers is equal to the sum of the digits of the first number. What four-digit number was written? | # Answer: 5445.
Solution. The difference between such numbers is always divisible by 9, since $10a + b - (10b + a) = 9(a - b)$. This difference equals the sum of two digits, so it is no more than 18. However, it cannot be 18, because then both the first and second numbers would be 99. Therefore, the difference between... | 5445 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,862 |
3. For a children's party, pastries were prepared: 10 eclairs, 20 mini tarts, 30 chocolate brownies, 40 cream puffs. What is the maximum number of children who can each take three different pastries? | Answer: 30.
Solution: From eclairs, baskets, and brownies, at least 2 pastries must be taken, and there are 60 of them in total, meaning no more than 30 children can take three different pastries. They can do this as follows: 10 children will take an eclair, a brownie, and a roll, and 20 children will take a basket, a... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,863 |
5. On 900 cards, all natural numbers from 1 to 900 are written. Cards with squares of integers are removed, and the remaining cards are renumbered, starting from 1.
Then the operation of removing squares is repeated. How many times will this operation have to be repeated to remove all the cards | Answer: 59.
Solution: During the first operation, 30 cards will be removed, leaving $900-30=30 \cdot 29$ cards. Since $30 \cdot 29 > 29^2$, all squares except $30^2$ remain. During the second operation, 29 cards will be removed. There will be $30 \cdot 29 - 29 = 29^2$ cards left. Thus, in two operations, we transition... | 59 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,865 |
1. Andrey and Kolya are not the same age, but last December each of them turned as many years old as the sum of the digits of their birth year. How old are they now? | Answer: 7 and 25.
## Solution.
Let the birth year be $\overline{a b c d}$, then $2021-\overline{a b c d}=a+b+c+d$, or $1001 a+101 b+$ $11 c+2 d=2021$. When $a=2$, we get $101 b+11 c+2 d=19$, that is, $b=0,11 c+2 d=$ 19. The digit $c$ must be odd and no more than 1, so $c=1$, then $d=4, \overline{a b c d}=$ 2014
When... | 725 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,866 |
2. A notebook costs 10 rubles. Eight children bought notebooks, and each had a different non-zero amount of rubles left, but none had enough for another notebook. The children pooled their remaining rubles, and it was exactly enough to buy several more notebooks. How much money did each child have left before pooling? | Answer: $1,2,3,4,6,7,8,9$.
Solution. Let's take all possible non-zero remainders when dividing by $10: 1,2, \ldots, 9$. The sum of all 9 remainders is 45. We need to exclude one remainder so that the sum of the remaining ones is divisible by 10. Clearly, there is only one possibility: to exclude the remainder 5.
Comm... | 1,2,3,4,6,7,8,9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,867 |
3. Find all possible values of the perimeter of a rectangle, given that it can be cut into three rectangles, the perimeter of each of which is 10, and the side lengths are integers. | Answer: $14,16,18,22$ or 26.
Solution. There are two possible ways of cutting.
1) The cut lines are parallel. Let $x$ be the length of one side of one of the rectangles in the partition, then it is not difficult to express the other sides. In this case, $1 \leq x \leq 4$, so the perimeter of the original rectangle is... | 14,16,18,22,26 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,868 |
4. Squirrel Pusistik and Lohmatik ate a basket of berries and a bag of seeds, which contained more than 50 but less than 65 seeds, starting and finishing at the same time. At first, Pusistik ate berries, and Lohmatik ate seeds, then (at some point) they switched. Lohmatik ate berries six times faster than Pusistik, and... | # Answer: 54.
Solution. Divide the berries into 3 equal parts. Each part, Lomhatic ate 6 times faster than Pushistik, but there are two parts, so he spent only 3 times less time on the berries than Pushistik. Therefore, Pushistik ate the seeds in one-third the time of Lomhatic. Since Pushistik eats three times slower,... | 54 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,869 |
5. Out of six indistinguishable nuts, two are artificial (real nuts weigh the same, artificial nuts also weigh the same but are lighter than real ones). There are balance scales, for each weighing on which you have to give up one nut. If the nut given is real, the scales show the correct result, but if it is artificial... | Solution. Let's number the nuts from 1 to 6. Place the 1st and 2nd nuts on one side of the balance, and the 3rd and 4th nuts on the other side, giving away the 6th nut. There are two possible outcomes:
1) the balance is in equilibrium;
2) one of the pans, for example, with the 1st and 2nd nuts, outweighs the other.
I... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,870 |
1. Boris distributes 8 white and 8 black balls into two boxes. Nastya randomly chooses a box and then takes a ball from it without looking. Can Boris distribute the balls in such a way that the probability of drawing a white ball is greater than $\frac{2}{3}$? | Answer. Yes.
Solution. Boris will put 1 white ball in the first box, and all the others in the second. Then the probability of drawing a white ball is $\frac{1}{2}+\frac{1}{2} \cdot \frac{7}{15}=\frac{22}{30}>\frac{2}{3}$.
Comment. Correct solution - 20 points. Correct solution with arithmetic errors - 15 points. Con... | \frac{22}{30}>\frac{2}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,871 |
2. On the extension of side $B C$ of an equilateral triangle $A B C$, a point $M$ is chosen. A line parallel to $A C$ is drawn through it. This line intersects the extension of side $A B$ at point $N$. The medians of triangle $B N M$ intersect at point $O$. Point $D$ is the midpoint of $A M$. Find the angles of triangl... | Answer: $30^{\circ}, 60^{\circ}, 90^{\circ}$.
Solution. Let's take a point $K$ on $NM$ such that $AK \| CM$, then $ACMK$ is a parallelogram. We will prove that triangles $KON$ and $COM$ are equal by two sides and the angle between them. In triangle $AKN$, angle $N$ is $60^{\circ}$, and $AK \| BC$, so triangle $AKN$ is... | 30,60,90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,872 |
4. Find all natural numbers $a$ for which the number
$$
\frac{a+1+\sqrt{a^{5}+2 a^{2}+1}}{a^{2}+1}
$$
is also a natural number. | Answer. $a=1$.
Solution. Let $a+1=b, \sqrt{a^{5}+2 a^{2}+1}=c$. Write the polynomial $c^{2}-b^{2}=$ $a^{5}+2 a^{2}+1-(a+1)^{2}=a^{5}+a^{2}-2 a$, and divide it by $a^{2}+1$ using long division, we get a remainder of $-(a+1)$. For $a>1$, the absolute value of the remainder is less than $a^{2}+1$, so the remainder cannot... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,873 |
1. To qualify for the competition, wrestler Vladimir had to conduct three bouts and win at least two in a row. His opponents were Andrey (A) and Boris (B). Vladimir could choose the sequence of matches: ABA or BAB. The probability of Vladimir losing a single bout to Boris is 0.3, and to Andrey is 0.4; the probabilities... | Answer. ABA; $p=0.588$.
Solution. Let Vladimir meet a weaker opponent twice, that is, the scheme is
Vladimir chooses the ABA scheme. Then $p_{2}=0.6 \cdot 0.7 \cdot 0.4 + 0.6 \cdot 0.7 \cdot... | 0.588 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,875 |
2. Find for all natural $n>1$ the positive solutions of the system
$$
\left\{\begin{array}{l}
x_{1}+2 x_{2}+\cdots+n x_{n}=3 \\
\frac{1}{x_{1}}+\frac{1}{2 x_{2}}+\cdots+\frac{1}{n x_{n}}=3
\end{array}\right.
$$ | Answer. For $n=3: x_{1}=1, x_{2}=1 / 2, x_{3}=1 / 3$; for $\quad n=2: x_{1}=\frac{3 \pm \sqrt{5}}{2}, x_{2}=\frac{3 \mp \sqrt{5}}{4} . \quad$ For other values of $n$, no solutions exist.
Solution. Let $y_{k}=k x_{k}$ and add the equations of the system:
$$
\left(y_{1}+\frac{1}{y_{1}}\right)+\left(y_{2}+\frac{1}{y_{2}... | x_{1}=\frac{3\\sqrt{5}}{2},x_{2}=\frac{3\\sqrt{5}}{4}forn=2;\quadx_{1}=1,x_{2}=1/2,x_{3}=1/3forn=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,876 |
3. Prove that $\cos \frac{\pi}{7}+\cos \frac{3 \pi}{7}+\cos \frac{5 \pi}{7}=\frac{1}{2}$. | Solution. Let $S=\cos \frac{\pi}{7}+\cos \frac{3 \pi}{7}+\cos \frac{5 \pi}{7}$. Consider the sum $A$ of seven cosines:
$$
A=\cos \frac{\pi}{7}+\cos \frac{3 \pi}{7}+\cos \frac{5 \pi}{7}+\cos \frac{7 \pi}{7}+\cos \frac{9 \pi}{7}+\cos \frac{11 \pi}{7}+\cos \frac{13 \pi}{7}
$$
Since $\cos \alpha=\cos (2 \pi-\alpha), A=2 ... | \frac{1}{2} | Algebra | proof | Yes | Yes | olympiads | false | 4,877 |
4. Kolya and Vlad drew identical convex quadrilaterals $A B C D$. On side $A B$, each of them chose a point $E$, and on side $C D$, each chose a point $F$. Kolya chose the points at the midpoints of the sides, while Vlad chose the points at a distance of $\frac{1}{3}$ of the length of side $A B$ from $A$ and at a dista... | Answer: No. The areas are equal.
Solution. First, we will prove that if $KLMN$ is a quadrilateral (not a segment or a triangle), then it is convex. Since $MK$ is the midline of triangle $AFB$, segment $MK$ intersects segment $EF$ at its midpoint $O$. Similarly, $LN$ is the midline of triangle $CED$, and segment $LN$ i... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,878 |
5. Three squirrels usually eat porridge for breakfast: semolina (M), buckwheat (B), oatmeal (O), and millet (R). No porridge is liked by all three squirrels, but for each pair of squirrels, there is at least one porridge that they both like. How many different tables can be made where each cell contains a plus (if it i... | Answer: 132.
Solution: If two different pairs like the same porridge, then the porridge is liked by three squirrels, violating the first condition. From three squirrels, three different pairs can be formed, and these three pairs like different porridges. There are 4 ways to choose these 3 porridges, and $3!=6$ ways to... | 132 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,879 |
1. A snowdrift 468 cm high decreased in height by 6 cm in the first hour, by 12 cm in the second hour, ..., by $6 k$ cm in the -th hour. After some time $T$, the snowdrift melted completely. What fraction of the snowdrift's height melted in the time $\frac{T}{2}$? | Answer: $\frac{7}{26}$.
Solution: From the equation $6+12+\cdots+6n=468$, we find $6 \cdot \frac{n(n+1)}{2}=468$ or $n^{2}+n-$ $156=0$. From this, $n=12$, meaning the snowdrift melted in 12 hours. In 6 hours, it melted by $6 \cdot \frac{6 \cdot 7}{2}=126$ cm in height, which is $\frac{126}{468}=\frac{7}{26}$.
Comment... | \frac{7}{26} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,880 |
2. On the border of a circular glade, points $A, B, C, D$ are marked clockwise. At point $A$ is a squirrel named An, at point $B$ is a squirrel named Bim, at point $C$ stands a pine tree, and at point $D$ stands an oak tree. The squirrels started running simultaneously, An towards the pine tree, and Bim towards the oak... | Answer: No.
Solution: Let the speed of Ana be $-v$, and the speed of Bim be $-u$. They both reached point $M$ at the same time, so $\frac{A M}{v}=\frac{B M}{u}$. Triangles $D A M$ and $C B M$ are similar (by three angles).
Therefore, $\frac{A D}{B C}=\frac{A M}{B M}$. But $\frac{A M}{B M}=\frac{v}{u}$, so $\frac{A D}... | 61 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,881 |
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