problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
4. In a convex heptagon $A B C D E F G$, seven quadrilaterals are considered, the vertices of which are four consecutive vertices of the heptagon in a clockwise direction. Can four of the seven quadrilaterals be cyclic? | Answer. No.
Solution. Two quadrilaterals adjacent in a clockwise direction share two sides. Let these quadrilaterals be $A B C D$ and $B C D E$. Suppose both are circumscribed. Then $A B + C D = B C + A D, C D + B E = B C + D E$. Subtracting, we get: $A B - B E = A D - D E$ or $A D + B E = A B + D E$. Since the heptag... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,882 |
5. Find all pairs $(x ; y)$ of natural numbers for which both numbers $x^{2}+8 y ; y^{2}-8 x$ are perfect squares. | Answer. $(x ; y)=(n ; n+2)$, where $n$ is a natural number, as well as (7; 15), (33; 17), $(45 ; 23)$.
Solution. It is easy to verify that pairs of the form $(n ; n+2)$, where $n$ is a natural number, satisfy the condition of the problem. Let $(x ; y)$ be any other pair satisfying the condition of the problem. Conside... | (n;n+2),(7;15),(33;17),(45;23) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,883 |
1. What is the probability that in a random sequence of 8 ones and two zeros, there are exactly three ones between the two zeros? | Answer: $\frac{2}{15}$.
Solution: Number the positions in the sequence from the first to the tenth. If one of the zeros is in positions $1-4$ or $7-10$, then it can have only one zero that is three digits away. If one of the zeros is in position 5 or 6, then it has two such zeros. The number of all outcomes is $10 \cd... | \frac{2}{15} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,884 |
2. Each of the 8 squirrels threw a pine cone at some other squirrel, independently of the others. Prove that there will always be a group of three squirrels who did not throw a pine cone at a squirrel from this group. | Solution. By the Pigeonhole Principle, there will be a baby squirrel A, at which no more than 1 acorn was thrown (otherwise, they would have had to throw 16 acorns, but there are only 8). We will write down baby squirrel A in a separate group. We will cross out from the list the baby squirrel at which A threw an acorn ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,885 |
3. Is the equation $y^{2}+y=x^{3}-x$ solvable in coprime natural numbers? Answer. No. | Solution. Rewrite the equation as $y(y+1)=x\left(x^{2}-1\right)$. By the condition $\operatorname{GCD}(x ; y)=1$, it follows that $y+1=k x$ for some natural number $k$. Then $x^{2}-1=k y$. Eliminating $y$, we get $x^{2}-k^{2} x+k-1=0$. The discriminant of this quadratic equation $D=k^{4}-4 k+4$ must be a perfect square... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,886 |
4. Points $P$ and $Q$ are the midpoints of arcs $K L$ and $L M$ of the circumcircle of triangle $K L M$, and $L S$ is the angle bisector of this triangle. It turns out that $\angle K L M = 2 \angle K M L$ and $\angle P S Q = 90^{\circ}$. Find the angles of triangle $K L M$ | Answer. $\angle L=90^{\circ}, \angle K=\angle M=45^{\circ}$.
Solution. By the condition $\angle M=\frac{\angle L}{2}=\angle S L M$, i.e., triangle $L S M$ is isosceles: $L S = M S$. In addition, $L Q = Q M$, so triangles $L S Q$ and $M S Q$ are equal by three sides, and $S Q$ is the bisector of angle $L S M$. By the c... | \angleL=90,\angleK=\angleM=45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,887 |
5. Prove that for all natural $n$ the inequality
$$
\frac{1}{n^{2}+1}+\frac{2}{n^{2}+2}+\cdots+\frac{n}{n^{2}+n}<\frac{1}{2}+\frac{1}{6 n}
$$ | Solution. Let's represent each term $\frac{k}{n^{2}+k}$ on the left side of the inequality as a difference $\frac{k}{n^{2}}-\frac{k^{2}}{n^{2}\left(n^{2}+k\right)}$ and add the resulting equalities term by term:
$$
\begin{aligned}
& \frac{1}{n^{2}+1}+\frac{2}{n^{2}+2}+\cdots+\frac{n}{n^{2}+n} \\
& \quad=\frac{1+2+\cdo... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 4,888 |
2. (7-8 grades) Each inhabitant of the island is either a knight, who always tells the truth, or a liar, who always lies. One day, 50 islanders sat around a round table, and each one said whether their right-hand neighbor was a knight or a liar. In this case, the residents sitting in the $1-\mathrm{m}, 3-\mathrm{m}, \l... | Answer: "Knight".
Solution. It is enough to consider two cases, who the first one was - a knight or a liar.
1) If the first one is a knight, then from the given statements we get a chain of those sitting at the table: RRLLRRLLRR...: pairs of knights and liars alternate. Since the total number of inhabitants is 50 - n... | Knight | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,889 |
3. (7-8 grade) Vasya thought of 5 natural numbers and told Petya all their pairwise sums (in some order): $122,124,126,127,128,129,130,131,132,135$. Help Petya determine the smallest of the numbers Vasya thought of. Answer: 60. | Solution. By adding all 10 pairwise sums, we get the sum of all 5 numbers, multiplied by 4, since each number participates in it exactly 4 times. $122+124+126+127+128+129+130+131+132+135=1284=4 S$, therefore, $S=321$. The sum of the two smallest numbers is 122, and the sum of the two largest is 135, so the third larges... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,890 |
4. In a right-angled triangle, the lengths of all sides are natural numbers, with one of the legs equal to 2012.
a) (7th grade) Find at least one such triangle.
b) (8th-9th grades) Find all such triangles. Answer: b) $(2012,253005,253013); (2012,506016, 506020); (2012,1012035,1012037)$ and $(2012,1509,2515)$. | # Solution.
Let's solve the equation $x^{2}+2012^{2}=y^{2}$ in natural numbers. $2012^{2}=y^{2}-x^{2}=(y-x) \cdot(y+x)$. Factorize $2012^{2}=2^{4} \cdot 503^{2}$. Note that $y-x<y+x$ and these two numbers $(x+y$ and $y-x)$ must both be even. That is, the twos must be present in the factorization of both numbers. In ad... | (2012,253005,253013);(2012,506016,506020);(2012,1012035,1012037);(2012,1509,2515) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,891 |
5. (7-8 grade) Maria Ivanovna is a strict algebra teacher. She only puts twos, threes, and fours in the grade book, and she never gives the same student two twos in a row. It is known that she gave Vovochka 6 grades for the quarter. In how many different ways could she have done this? Answer: 448 ways. | Solution. Let $a_{n}$ be the number of ways to assign $n$ grades. It is easy to notice that $a_{1}=3$, $a_{2}=8$. Note that 3 or 4 can be placed after any grade, while 2 can only be placed if a 3 or 4 preceded it. Thus, a sequence of length $n$ can be obtained by appending 3 or 4 to a sequence of length $n-1$ or by app... | 448 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,892 |
7. 7-8 grade Excellent student Kolya found the sum of the digits of all numbers from 0 to 2012 and added them all together. What number did he get? Answer: 28077. | Solution. Note that the numbers 0 and 999, 1 and 998, ..., 499 and 500 complement each other to 999, i.e., the sum of their digit sums is 27. Adding their digit sums, we get $27 \times 500 = 13500$. The sum of the digits of the numbers from 1000 to 1999, by similar reasoning, is 14500 (here we account for the fact that... | 28077 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,894 |
9. (8-9 grade) Compare the numbers $\frac{1}{2!}+\frac{2}{3!}+\ldots+\frac{2012}{2013!}$ and 1. ( $n$ ! denotes the product $1 \cdot 2 \cdot \ldots \cdot n$ ) | Answer: The first number is smaller.
Solution. Note that $\frac{n}{(n+1)!}=\frac{(n+1)}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$. When adding, the inner terms will cancel out, leaving $1-\frac{1}{2013!}$. | 1-\frac{1}{2013!} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,896 |
10. (9th grade) On the side $D E$ of a regular hexagon $A B C D E F$, a point $K$ is chosen such that the line $A K$ divides the area of the hexagon in the ratio $3: 1$. In what ratio does the point $K$ divide the side $D E$? Answer: $3: 1$. | Solution. Without loss of generality, we can assume that $S(A B C D E F)=6$. Note that then the areas of triangles $A B C$ and $A E F$ are 1, and the areas of triangles $A C D$ and $A D E$ are 2. Denoting $x=D K: D E$, we get that $S(\triangle A D K)=2 x, S(\triangle A E K)=2(1-x)$, from which, $3+2 x=3(1+2(1-x))$. Sol... | 3:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,897 |
11. (9th grade) In how many ways can the numbers $1,2,3,4,5,6$ be written in a row so that for any three consecutive numbers $a, b, c$, the quantity $a c-b^{2}$ is divisible by 7? Answer: 12. | Solution. Each subsequent member will be obtained from the previous one by multiplying by a certain fixed value (modulo 7). This value can only be 3 or 5. And the first term can be any, so there are 12 options in total. | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,898 |
12. (9th grade) Find all increasing arithmetic progressions of natural numbers, in which for any natural $n$ the product of the first $n$ terms divides the product of the next $n$ terms (from $n+1$ to $2n$). | Answer: $k, 2 k, 3 k, \ldots, k \in \mathbb{N}$.
Solution. 1. Divide the given progression by the GCD of all its terms and obtain a new progression that still satisfies the condition, say, $a, a+d, \ldots$, where $(a, d)=1$ and $d \geq 1$. Since by the condition (for $n=1$) $a \mid a+d$, then $a \mid d$ and $a=1$. | k,2k,3k,\ldots,k\in\mathbb{N} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,899 |
4. (9th grade) Among the values of the function
$$
f(x)=\frac{\left(x^{2}+1\right)^{2}}{x\left(x^{2}-1\right)}
$$
find those that it takes the most number of times? | Answer: $(-\infty,-4) \cup(4,+\infty)$.
Solution. The desired values are precisely those $a$ for which the equation has the maximum number of roots:
$$
\begin{gathered}
f(x)=a \Longleftrightarrow\left(x^{2}+1\right)^{2}-a x\left(x^{2}-1\right)=0(\Rightarrow x \neq 0, \pm 1) \Longleftrightarrow\left(x^{2}-1\right)^{2}... | (-\infty,-4)\cup(4,+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,900 |
1. A line parallel to the selected side of a triangle with an area of 16 cuts off a smaller triangle with an area of 9. Find the area of the quadrilateral, three vertices of which coincide with the vertices of the smaller triangle, and the fourth lies on the selected side. Choose the answer option with the number close... | Answer: 12. (C)
 | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,901 |
1. A line parallel to the selected side of a triangle with an area of 27 cuts off a smaller triangle with an area of 3. Find the area of the quadrilateral, three vertices of which coincide with the vertices of the smaller triangle, and the fourth lies on the selected side. Choose the answer option with the number close... | Answer: 9. (B)
 | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,902 |
1. A line parallel to the selected side of a triangle with an area of 27 cuts off a smaller triangle with an area of 12. Find the area of the quadrilateral, three vertices of which coincide with the vertices of the smaller triangle, and the fourth lies on the selected side. Choose the answer option with the number clos... | Answer: 18. (E)
 | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,904 |
2. Calculate
$$
\frac{y^{2}+x y-\sqrt[4]{x^{5} y^{3}}-\sqrt[4]{x y^{7}}}{\sqrt[4]{y^{5}}-\sqrt[4]{x^{2} y^{3}}} \cdot(\sqrt[4]{x}+\sqrt[4]{y})
$$
where $x=3, \underbrace{22 \ldots 2}_{2013} 3, y=4, \underbrace{77 \ldots .7}_{2014}$. Choose the answer option with the number closest to the one you found. | Answer: 8. (D)
Options.
 | 8 | Algebra | MCQ | Yes | Yes | olympiads | false | 4,905 |
2. Calculate
$$
(\sqrt[4]{x}+\sqrt[4]{y}) \cdot \frac{x^{2}+x y-\sqrt[4]{x^{7} y}-\sqrt[4]{x^{3} y^{5}}}{\sqrt[4]{x^{5}}-\sqrt[4]{x^{3} y^{2}}}
$$
where $x=1, \underbrace{11 \ldots 1}_{2013} 2, y=3, \underbrace{88 \ldots 8}_{2014}$. Choose the answer option with the number closest to the one you found. | Answer: 5. (B)
Options.
 | 5 | Algebra | MCQ | Yes | Yes | olympiads | false | 4,906 |
2. Calculate
$$
\frac{\sqrt[4]{x^{5} y^{3}}+\sqrt[4]{x y^{7}}-x y-y^{2}}{\sqrt[4]{x^{2} y^{3}}-\sqrt[4]{y^{5}}} \cdot(\sqrt[4]{x}+\sqrt[4]{y})
$$
where $x=2, \underbrace{44 \ldots .4}_{2013} 5, y=1, \underbrace{55 \ldots .5}_{2014}$. Choose the answer option with the number closest to the one you found. | Answer: 4. (E)
Options.
 | 4 | Algebra | MCQ | Yes | Yes | olympiads | false | 4,907 |
3. In how many ways can a coach form a hockey team consisting of one goalkeeper, two defenders, and three forwards if he has 2 goalkeepers, 5 defenders, and 8 forwards at his disposal? Among the proposed answer options, choose the one closest to the correct one. | Answer: 1120. (B)
$$
\begin{array}{|l|l|l|l|l|l|l|}
\hline \mathbf{A} & 915 & \mathbf{B} & 1120 & \mathbf{C} & 1400 & \mathbf{D} \\
1960 & \mathbf{E} & 2475 & \mathbf{F} \\
\hline
\end{array}
$$ | 1120 | Combinatorics | MCQ | Yes | Yes | olympiads | false | 4,909 |
3. In how many ways can a team be selected from a group consisting of 7 boys and 8 girls, so that the team has 4 boys and 3 girls? Among the proposed answer options, choose the one closest to the correct one. | Answer: 1960. (D)
$$
\begin{array}{|l|l|l|l|l|l|l|l|}
\hline \mathbf{A} 915 & \mathbf{B} & 1120 & \mathbf{C} & 1400 & \mathbf{D} & 1960 & \mathbf{E} \\
\hline
\end{array}
$$ | 1960 | Combinatorics | MCQ | Yes | Yes | olympiads | false | 4,910 |
3. In how many ways can a coach form a basketball team consisting of two guards and three forwards if he has 6 guards and 11 forwards at his disposal? Among the options provided, choose the one closest to the correct answer.
| Answer: 2475. (E)
 | 2475 | Combinatorics | MCQ | Yes | Yes | olympiads | false | 4,911 |
3. In how many ways can a team be assembled consisting of 3 painters and 4 plasterers, if there are 6 painters and 8 plasterers? Among the options provided, choose the one closest to the correct answer. | Answer: $1400 .(\mathrm{C})$
 | 1400 | Combinatorics | MCQ | Yes | Yes | olympiads | false | 4,912 |
4. Determine the number of different values of $a$ for which the equation
$$
\left(3-a^{2}\right) x^{2}-3 a x-1=0
$$
has a unique solution | Answer: 2. (C)
Options.
 | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,915 |
4. Determine how many different values of $a$ exist for which the equation
$$
\left(a^{2}-5\right) x^{2}-2 a x+1=0
$$
has a unique solution | Answer: 2. (C)
Options.
$$
\begin{array}{|l|l|l|l|l|l|l|l|l|}
\hline \mathbf{A} & 0 & \mathbf{B} & 1 & \mathbf{C} & 2 & \mathbf{D} & 3 & \mathbf{E} \\
\hline
\end{array}
$$ | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,916 |
5. A product that initially contained $98\%$ water dried out over time and began to contain $97\%$ water. By what factor did it shrink (i.e., reduce its weight)? | Answer: 1.5. (D)
 | 1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,918 |
5. A product that initially contained $98\%$ water dried over some time and began to contain $95\%$ water. By what factor did it shrink (i.e., reduce its weight)? | Answer: 2.5. (D)
 | 2.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,920 |
1. The vertices of a regular 222-gon are painted red and blue. We will call a side monochromatic if the vertices are painted the same color, and bichromatic if they are painted different colors. Is it possible to color the vertices so that there are an equal number of monochromatic and bichromatic sides? | Answer: No.
Solution: Suppose it was possible to paint in such a way that there are 111 differently colored sides. Starting from an arbitrary point (for example, a red one), when we complete a full circle, the color will have changed 111 times, meaning the point will be blue. Contradiction. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,921 |
2. Determine whether the number $N=7 \times 9 \times 13+2020 \times 2018 \times 2014$ is prime or composite. Justify your answer. | Answer: Composite.
Solution: Note that 7+2020=9+2018=13+2014=2027. Therefore, $N \equiv 7 \times 9 \times 13+(-7) \times(-9) \times$ $(-13)=0(\bmod 2027)$ | Composite | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,922 |
3. Ivan Semenovich leaves for work at the same time every day, drives at the same speed, and arrives exactly at 9:00. One day he overslept and left 40 minutes later than usual. To avoid being late, Ivan Semenovich drove at a speed 60% faster than usual and arrived at 8:35. By what percentage should he have increased hi... | Answer: By $30 \%$.
Solution: By increasing the speed by $60 \%$, i.e., by 1.6 times, Ivan Semenovich reduced the time by 1.6 times and gained 40+25=65 minutes. Denoting the usual travel time as $T$, we get $\frac{T}{1.6}=T-65$, from which $T=\frac{520}{3}$. To arrive in $T-40=\frac{400}{3}$, the speed needed to be in... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,923 |
4. Prove that the sum of 6-digit numbers, not containing the digits 0 and 9 in their decimal representation, is divisible by 37. | Solution: Let's break it down into pairs, each complementing the other to make 9 in each digit (for example, 123456+876543). We get a sum of 999999 in each pair, which is equal to $9 \times 1001 \times 111$. It is easy to verify that 111 is divisible by 37. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 4,924 |
5. The English club is attended by 20 gentlemen. Some of them are acquainted (acquaintances are mutual, i.e., if A knows B, then B knows A). It is known that there are no three gentlemen in the club who are pairwise acquainted.
One day, the gentlemen came to the club, and each pair of acquaintances shook hands with ea... | Answer: 100 handshakes.
Solution: Choose a gentleman with the maximum number of acquaintances (if there are several, choose any one). Suppose he has $n$ acquaintances. These acquaintances cannot be pairwise acquainted with each other. Consider the remaining $(20-n-1)$ gentlemen, each of whom has no more than $n$ acqua... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,925 |
6. Compare the numbers $\frac{100}{101} \times \frac{102}{103} \times \ldots \times \frac{1020}{1021} \times \frac{1022}{1023}$ and $\frac{5}{16}$ | Answer: $\frac{100}{101} \times \frac{102}{103} \times \ldots \times \frac{2020}{2021} \times \frac{2022}{2023}<\frac{5}{16}$
Solution: Notice that $\frac{n}{n+1}<\frac{n+1}{n+2}$. Therefore, $A=\frac{100}{101} \times \frac{102}{103} \times \ldots \times \frac{2020}{2021} \times \frac{2022}{2023}<\frac{101}{102} \time... | A<\frac{5}{16} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,926 |
7. Olga Ivanovna, the class teacher of 5B, is staging a "Mathematical Ballet." She wants to arrange the boys and girls so that exactly 2 boys are 5 meters away from each girl. What is the maximum number of girls that can participate in the ballet if it is known that 5 boys are participating?
Answer: 20 girls. | Solution: Let's choose and fix two arbitrary boys - $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$. Suppose they are 5 meters away from some girl - G. Then $\mathrm{M}_{1}, \mathrm{M}_{2}$, and G form an isosceles triangle with the legs being 5 meters. Given the fixed positions of $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$, there ar... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,927 |
1. Solve the inequality $\frac{\sqrt{\frac{x}{\gamma}+(\alpha+2)}-\frac{x}{\gamma}-\alpha}{x^{2}+a x+b} \geqslant 0$.
In your answer, specify the number equal to the number of integer roots of this inequality. If there are no integer roots, or if there are infinitely many roots, enter the digit 0 in the answer sheet.
... | # Answer: 7.
Solution. Given the numerical values, we solve the inequality:
$$
\frac{\sqrt{x+5}-x-3}{x^{2}-15 x+54} \geqslant 0 \Leftrightarrow \frac{\sqrt{x+5}-x-3}{(x-6)(x-9)} \geqslant 0
$$
Since $\sqrt{x+5} \geqslant x+3 \Leftrightarrow\left[\begin{array}{c}-5 \leqslant x \leqslant-3, \\ x+5 \geqslant(x+3)^{2}\e... | 7 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,928 |
2. Solve the equation $\cos 2 x+\cos 6 x+2 \sin ^{2} x=1$.
In the answer, specify the number equal to the sum of the roots of the equation that belong to the interval $A$, rounding this number to two decimal places if necessary.
$$
A=\left[\frac{m \pi}{6} ; \frac{(m+1) \pi}{6}\right], m=5
$$ | Answer: 2.88 (exact value: $\frac{11 \pi}{12}$ ).
Solution. Since $\cos 2 x+2 \sin ^{2} x=1$, the original equation is equivalent to $\cos 6 x=0$. From all the roots of the series $x=\frac{\pi}{12}+\frac{\pi n}{6}$, only one root $x=\frac{\pi}{12}+\frac{5 \pi}{6}=\frac{11 \pi}{12}$ falls within the interval $A=\left[\... | 2.88 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,929 |
3. From point $M$, lying inside triangle $A B C$, perpendiculars are drawn to the sides $B C, A C, A B$, with lengths $k, l$, and $m$ respectively. Find the area of triangle $A B C$, if $\angle C A B=\alpha$ and $\angle A B C=\beta$. If the answer is not an integer, round it to the nearest integer.
$$
\alpha=\frac{\pi... | Answer: 67.
Solution. Denoting the sides of the triangle by $a, b, c$, using the sine theorem we get $S=\frac{k a+l b+m c}{2}=R(k \sin \alpha+l \sin \beta+m \sin \gamma)$.
Since, in addition, $S=2 R^{2} \sin \alpha \sin \beta \sin \gamma$, we can express $R=\frac{k \sin \alpha+l \sin \beta+m \sin \gamma}{2 \sin \alph... | 67 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,930 |
4. Solve the system
$$
\left\{\begin{array}{l}
x^{3}+3 y^{3}=11 \\
x^{2} y+x y^{2}=6
\end{array}\right.
$$
Calculate the values of the expression $\frac{x_{k}}{y_{k}}$ for each solution $\left(x_{k}, y_{k}\right)$ of the system and find the minimum among them. Provide the found minimum value in your answer, rounding ... | Answer: $-1.31$ (exact value: $-\frac{1+\sqrt{217}}{12}$).
Solution. Multiply the first equation by 6, the second by (-11), and add them:
$$
6 x^{3}-11 x^{2} y-11 x y^{2}+18 y^{3}=0
$$
Dividing both sides by $y^{3}$ and letting $t=\frac{x}{y}$, we get:
$$
6 t^{3}-11 t^{2}-11 t+18=0 \Leftrightarrow(t-2)\left(6 t^{2}... | -1.31 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,931 |
5. There are two alloys. The first alloy contains $p \%$ impurities, and the second - respectively $q \%$ impurities. Determine the proportion in which these alloys should be combined to obtain a new alloy containing $r \%$ impurities. In your answer, indicate the ratio of the mass of the first alloy to the mass of the... | Answer: 1.17 (exact answer: $\frac{7}{6}$).
Solution. Let the mass of the second alloy be $a$ (kg, g, t, or any other unit of mass), and the mass of the first be $x \cdot a$ (of the same units). Then $x$ is the desired quantity.
According to the problem, the mass of all impurities is $\frac{x a \cdot p}{100}+\frac{a ... | 1.17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,932 |
6. Find all integer values of $a$, not exceeding 15 in absolute value, for each of which the inequality
$$
\frac{4 x-a-4}{6 x+a-12} \leqslant 0
$$
is satisfied for all $x$ in the interval $[2 ; 3]$. In your answer, specify the sum of all such $a$. | Answer: -7.
Solution. Solving the inequality using the interval method, we find that its left side equals 0 at $x=1+\frac{a}{4}$ and is undefined at $x=2-\frac{a}{6}$. These two values coincide when $a=\frac{12}{5}$. In this case, the inequality has no solutions.
For $a>\frac{12}{5}$, the solution to the inequality i... | -7 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,933 |
10. In the answer, indicate $n_{1}+n_{k-4}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Answer: 36.
II. Among all triangles inscribed in a circle of fixed radius, with a known sum of the squares of all angles $\left(\alpha^{2}+\beta^{2}+\gamma^{2}=\pi^{2} / 2\right)$, find all triangles with the maximum possible area. For each such triangle, find the smallest value of all pairwise products of angles. In ... | 0.27 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,934 |
2.15. $C=19 \pi^{2} / 32$. Answer: 0.15. Here $\alpha=\frac{3 \pi}{4}, \beta=\frac{\pi}{8}$. The second triangle with $\beta=\frac{27 \pi}{40}$ does not exist. The exact answer is $\pi^{2} / 64$.
III. Find all pairs of positive numbers $x, y$ that satisfy the equation
$$
\begin{aligned}
& \frac{4 x^{2} y+6 x^{2}+2 x ... | Solution. The proposed equation can be rewritten as
$$
f\left(\alpha_{1} u_{1}+\alpha_{2} u_{2}+\alpha_{3} u_{3}\right)=\alpha_{1} f\left(u_{1}\right)+\alpha_{2} f\left(u_{2}\right)+\alpha_{3} f\left(u_{3}\right),
$$
where
$$
\begin{gathered}
\alpha_{1}=\frac{x^{2}}{(x+y)^{2}}, \alpha_{2}=\frac{y^{2}}{(x+y)^{2}}, \a... | 4.33 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,935 |
1.1. Find the greatest negative root of the equation $\sin 2 \pi x=\sqrt{3} \sin \pi x$. | Solution. The given equation is equivalent to
$$
\sin \pi x(2 \cos \pi x-\sqrt{3})=0 \Longleftrightarrow\left[\begin{array} { l }
{ \pi x = \pi n , n \in \mathbb { Z } } \\
{ \pi x = \pm \frac { \pi } { 6 } + 2 \pi k , k \in \mathbb { Z } , }
\end{array} \Longleftrightarrow \left[\begin{array}{l}
x=n, n \in \mathbb{Z... | -\frac{1}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,936 |
1.2. Find the greatest negative root of the equation $\sin \pi x = -\sqrt{3} \cos \frac{\pi x}{2}$. | Solution. The roots of the equation are the following series of values: $x=1+2 n, x=-\frac{2}{3}+4 n, x=$ $-\frac{4}{3}+4 n, n \in \mathbb{Z}$. Therefore, the greatest negative root of the equation is $-\frac{2}{3}$.
Answer: $-\frac{2}{3} \cdot(\mathrm{D})$
$$
\mathbf{A}-\frac{4}{3} \quad \mathbf{B}-1 \quad \boxed{\m... | -\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,937 |
1.3. Find the greatest negative root of the equation $\sin \pi x = -\sqrt{2} \sin \frac{\pi x}{2}$. | Solution. The roots of the equation are the following series of values: $x=2 n, x= \pm \frac{3}{2}+4 n, n \in \mathbb{Z}$. Therefore, the greatest negative root of the equation is $-\frac{3}{2}$.
Answer: $-\frac{3}{2} \cdot(\mathrm{C})$
$$
\mathbf{A}-2 \quad \mathbf{B}-\frac{5}{2} \quad \mathbf{C}-\frac{3}{2} \quad \... | -\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,938 |
1.4. Find the greatest negative root of the equation $\sin 2 \pi x=\sqrt{2} \cos \pi x$. Solution. The roots of the equation are the following series of values: $x=\frac{1}{2}+n, x=\frac{1}{4}+2 n, x=$ $\frac{3}{4}+2 n, n \in \mathbb{Z}$. Therefore, the greatest negative root of the equation is $-\frac{1}{2}$. | Answer: $-\frac{1}{2} \cdot(\mathrm{A})$
$$
\mathbf{A}-\frac{1}{2} \mathbf{B}-1 \quad \mathbf{C}-\frac{5}{4} \quad \mathbf{D}-\frac{3}{2} \quad \mathbf{E}-\frac{5}{2} \quad \mathbf{F}
$$ | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,939 |
2.1. The master's day shift lasts $10 \%$ longer than the apprentice's shift. If the apprentice worked as long as the master, and the master worked as long as the apprentice, they would produce the same number of parts. By what percentage does the master produce more parts per day than the apprentice? | Solution. Let $p=10 \%$. Suppose the productivity of the apprentice is $a$ parts per hour, the master's productivity is $b$ parts per hour, the apprentice's workday is $n$ hours, and the master's workday is $m$ hours. Then from the condition it follows:
$$
m=n\left(1+\frac{p}{100}\right), \quad \text { and } \quad m a... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,940 |
3.1. Find the sum of all two-digit numbers for each of which the sum of the squares of the digits is 37 more than the product of the same digits. | Solution. 1st method. For a two-digit number $\overline{a b}$, the condition means that
$$
a^{2}+b^{2}-a b=37
$$
Since the equation is symmetric, i.e., with each solution $(a, b)$, the pair $(b, a)$ is also a solution, we can assume without loss of generality that $a \geqslant b$.
- Suppose $a \leqslant 6$. Then the... | 231 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,944 |
3.2. Find the sum of all two-digit numbers for each of which the sum of the squares of the digits is 57 more than the product of the same digits. | Solution. Such two-digit numbers are $18, 78, 81, 87$. Their sum is 264.
Answer: 264. (D)
\section*{| A | 165 | $\mathbf{B}$ | 198 | $\mathbf{C}$ | 231 | $\mathbf{D}$ | 264 | $\mathbf{E}$ | 297 | $\mathbf{F}$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | 264 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,945 |
3.3. Find the sum of all two-digit numbers for each of which the sum of the squares of the digits is 73 more than the product of the same digits. | Solution. Such two-digit numbers are $19, 89, 91, 98$. Their sum is 297.
Answer: 297. ( ( )
 | 297 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,946 |
3.4. Find the sum of all two-digit numbers for each of which the sum of the squares of the digits is 31 more than the product of the same digits. | Solution. Such two-digit numbers are $16, 56, 61, 65$. Their sum is 198.
Answer: 198. (B)
 | 198 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,947 |
4.1. The segment connecting the lateral sides of the trapezoid and parallel to its bases, which are 3 and 21, divides the trapezoid into two parts of equal area. Find the length of this segment. | Solution. 1st method. Let $a=3$ and $b=21$ be the lengths of the bases of the trapezoid. If $c$ is the length of the segment parallel to the bases, and $h_{1}$ and $h_{2}$ are the parts of the height of the trapezoid adjacent to the bases $a$ and $b$ respectively, then, by equating the areas, we get:
$$
\frac{a+c}{2} ... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,948 |
4.2. The segment connecting the lateral sides of the trapezoid and parallel to its bases, which are 7 and 17, divides the trapezoid into two parts of equal area. Find the length of this segment. | Solution. $c=\sqrt{\left(a^{2}+b^{2}\right) / 2}=\sqrt{(49+289) / 2}=13$.
Answer: 13. (E)
$$
\begin{array}{|l|l|l|l|l|l|l|}
\hline \mathbf{A} & 11 & \mathbf{B} & 11.5 & \mathbf{C} & 12 & \mathbf{D} \\
12.5 & \mathbf{E} & 13 & \mathbf{F} \\
\hline
\end{array}
$$ | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,949 |
5.1. Find the sum of all integer values of the argument $x$ for which the corresponding values of the function
$$
y=x^{2}+x\left(\log _{2} 18-\log _{3} 12\right)-\log _{3} 16-4 \log _{2} 3
$$
do not exceed 8. | Solution. Let $a=\log _{2} 3$. Then the condition of the problem will turn into the inequality
$$
x^{2}+2\left(a-\frac{1}{a}\right) x-4\left(a+\frac{1}{a}+2\right) \leqslant 0
$$
Considering that $a \in\left(\frac{3}{2}, 2\right)$, we get $x \in\left[-2 a-2, \frac{2}{a}+2\right]$. Since $-6<-2 a-2<-5.3<\frac{2}{a}+2<... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,953 |
5.2. Find the sum of all integer values of the argument $x$ for which the corresponding values of the function
$$
y=x^{2}+x\left(\log _{2} 20-\log _{5} 8\right)-\log _{2} 5-9 \log _{5} 2
$$
do not exceed 6. | Solution. Let $a=\log _{2} 5$. Then the condition of the problem will turn into the inequality
$$
x^{2}+\left(a-\frac{3}{a}+2\right) x-\left(a+\frac{9}{a}+6\right) \leqslant 0
$$
Considering that $a \in(2,3)$, we get $x \in\left[-a-3, \frac{3}{a}+1\right]$. Since $-6<-a-3<-5,2<\frac{3}{a}+1<3$, the integer solutions ... | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,954 |
5.3. Find the sum of all integer values of the argument $x$ for which the corresponding values of the function
$$
y=x^{2}+x\left(\log _{2} 36-\log _{3} 16\right)-\log _{2} 9-4 \log _{3} 8
$$
do not exceed 11. | Solution. Let $a=\log _{2} 3$. Then the condition of the problem will turn into the inequality
$$
x^{2}+2\left(a-\frac{2}{a}+1\right) x-\left(2 a+\frac{12}{a}+11\right) \leqslant 0
$$
Considering that $a \in\left(\frac{3}{2}, 2\right)$, we get $x \in\left[-2 a-3, \frac{4}{a}+1\right]$. Since $-7<-2 a-3<-6,3<\frac{4}{... | -15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,955 |
5.4. Find the sum of all integer values of the argument $x$ for which the corresponding values of the function
$$
y=x^{2}+x\left(\log _{5} 2-\log _{2} 10\right)-\log _{2} 25-3 \log _{5} 2
$$
do not exceed 7. | Solution. Let $a=\log _{2} 5$. Then the condition of the problem will turn into the inequality
$$
x^{2}-\left(a-\frac{1}{a}+1\right) x-\left(2 a+\frac{3}{a}+7\right) \leqslant 0 .
$$
Considering that $a \in(2,3)$, we get $x \in\left[-\frac{1}{a}-2, a+3\right]$. Since $-3<-\frac{1}{a}-2<-2,5<a+3<6$, the integer soluti... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,956 |
6.1. Three pirates, Joe, Bill, and Tom, found a treasure containing 70 identical gold coins, and they want to divide them so that each of them gets at least 10 coins. How many ways are there to do this? | Solution. Let the treasure consist of $n=70$ coins and each pirate should receive no less than $k=10$ coins.
Give each pirate $k-1$ coins, and lay out the remaining $n-3 k+3$ coins in a row. To divide the remaining coins among the pirates, it is sufficient to place two dividers in the $n-3 k+2$ spaces between the coin... | 861 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,957 |
6.2. Three pirates, Joe, Bill, and Tom, found a treasure containing 80 identical gold coins, and they want to divide them so that each of them gets at least 15 coins. How many ways are there to do this? | Solution. Since $n=80, k=15$, it results in $C_{37}^{2}=666$ ways.
Answer: 666. | 666 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,958 |
6.3. Three pirates, Joe, Bill, and Tom, found a treasure containing 100 identical gold coins, and they want to divide them so that each of them gets at least 25 coins. How many ways are there to do this? | Solution. Since $n=100, k=25$, it results in $C_{27}^{2}=351$ ways.
Answer: 351. | 351 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,959 |
6.4. Three pirates, Joe, Bill, and Tom, found a treasure containing 110 identical gold coins, and they want to divide them so that each of them gets at least 30 coins. How many ways are there to do this? | Solution. Since $n=110, k=30$, we get $C_{22}^{2}=231$ ways.
Answer: 231.
Solution. Let $A=\underbrace{11 \ldots 1}_{1007}$. Then
$$
\sqrt{\underbrace{111 \ldots 11}_{2014}-\underbrace{22 ... | 231 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,960 |
8.1. Specify the integer closest to the larger root of the equation
$$
\operatorname{arctg}\left(\left(\frac{2 x}{7}+\frac{7}{8 x}\right)^{2}\right)-\operatorname{arctg}\left(\left(\frac{2 x}{7}-\frac{7}{8 x}\right)^{2}\right)=\frac{\pi}{4}
$$ | Solution. Let $y=\left(\frac{2 x}{7}-\frac{7}{8 x}\right)^{2}$, then $\left(\frac{2 x}{7}+\frac{7}{8 x}\right)^{2}=y+1$ and the equation will take the form
$$
\operatorname{arctg}(y+1)-\operatorname{arctg} y=\frac{\pi}{4}
$$
Since $0 \leqslant \operatorname{arctg} y<\operatorname{arctg}(y+1)<\frac{\pi}{2}$, the last ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,961 |
8.2. Specify the integer closest to the smaller root of the equation
$$
\operatorname{arctg}\left(\left(\frac{5 x}{26}+\frac{13}{10 x}\right)^{2}\right)-\operatorname{arctg}\left(\left(\frac{5 x}{26}-\frac{13}{10 x}\right)^{2}\right)=\frac{\pi}{4}
$$ | Solution. The equation is equivalent to $\frac{5 x}{26}=\frac{13}{10 x} \Longleftrightarrow|x|=\frac{13}{5}$.
Answer: -3 . | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,962 |
8.3. Specify the integer closest to the smaller root of the equation
$$
\operatorname{arctg}\left(\left(\frac{7 x}{10}-\frac{5}{14 x}\right)^{2}\right)-\operatorname{arctg}\left(\left(\frac{7 x}{10}+\frac{5}{14 x}\right)^{2}\right)=-\frac{\pi}{4}
$$ | Solution. The equation is equivalent to $\frac{7 x}{10}=\frac{5}{14 x} \Longleftrightarrow |x|=\frac{5}{7}$.
Answer: -1. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,963 |
8.4. Indicate the integer closest to the larger root of the equation
$$
\operatorname{arctg}\left(\left(\frac{3 x}{22}-\frac{11}{6 x}\right)^{2}\right)-\operatorname{arctg}\left(\left(\frac{3 x}{22}+\frac{11}{6 x}\right)^{2}\right)=-\frac{\pi}{4}
$$ | Solution. The equation is equivalent to $\frac{3 x}{22}=\frac{11}{6 x} \Longleftrightarrow |x|=\frac{11}{3}$.
Answer: 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,964 |
8.5. Indicate the integer closest to the smaller root of the equation
$$
\operatorname{arcctg}\left(\left(\frac{2 x}{7}+\frac{7}{8 x}\right)^{2}\right)-\operatorname{arcctg}\left(\left(\frac{2 x}{7}-\frac{7}{8 x}\right)^{2}\right)=-\frac{\pi}{4}
$$ | Solution. The equation is equivalent to $\frac{2 x}{7}=\frac{7}{8 x} \Longleftrightarrow |x|=\frac{7}{4}$.
Answer: -2. | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,965 |
8.6. Indicate the integer closest to the larger root of the equation
$$
\operatorname{arcctg}\left(\left(\frac{5 x}{26}+\frac{13}{10 x}\right)^{2}\right)-\operatorname{arcctg}\left(\left(\frac{5 x}{26}-\frac{13}{10 x}\right)^{2}\right)=-\frac{\pi}{4}
$$ | Solution. The equation is equivalent to $\frac{5 x}{26}=\frac{13}{10 x} \Longleftrightarrow|x|=\frac{13}{5}$.
Answer: 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,966 |
8.7. Indicate the integer closest to the larger root of the equation
$$
\operatorname{arcctg}\left(\left(\frac{7 x}{10}-\frac{5}{14 x}\right)^{2}\right)-\operatorname{arcctg}\left(\left(\frac{7 x}{10}+\frac{5}{14 x}\right)^{2}\right)=\frac{\pi}{4}
$$ | Solution. The equation is equivalent to $\frac{7 x}{10}=\frac{5}{14 x} \Longleftrightarrow |x|=\frac{5}{7}$.
Answer: 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,967 |
8.8. Provide the integer closest to the smaller root of the equation
$$
\operatorname{arcctg}\left(\left(\frac{3 x}{22}-\frac{11}{6 x}\right)^{2}\right)-\operatorname{arcctg}\left(\left(\frac{3 x}{22}+\frac{11}{6 x}\right)^{2}\right)=\frac{\pi}{4}
$$ | Solution. The equation is equivalent to $\frac{3 x}{22}=\frac{11}{6 x} \Longleftrightarrow |x|=\frac{11}{3}$.
Answer: -4. | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,968 |
9.1. In the triangular pyramid $S A B C$, the edges $S B, A B$ are perpendicular and $\angle A B C=120^{\circ}$. Point $D$ on edge $A C$ is such that segment $S D$ is perpendicular to at least two medians of triangle $A B C$ and $C D=A B=44 \sqrt[3]{4}$. Find $A D$ (if the answer is not an integer, round it to the near... | Solution. Since segment $SD$ is perpendicular to two medians of triangle $ABC$, it is perpendicular to the plane $(ABC)$ (see Fig. 3). By the theorem of three perpendiculars, it follows that $DB \perp AB$.
=2013-8 x^{3}+12 x^{2}-14 x-a-\sin 2 \pi x$ find the number of integer values of $a$, for each of which the equation
$$
\underbrace{f(f(\ldots f}_{2013 \text { times }}(x) \ldots))=2 x-1
$$
on the interval $[50 ; 51]$ has a unique solution. | Solution. Since
$$
2013-8 x^{3}+12 x^{2}-14 x-a-\sin 2 \pi x=2008-(2 x-1)^{3}-4(2 x-1)-a+\sin \pi(2 x-1)
$$
then after the substitution of the variable $t=2 x-1$, we get a new problem: "For the function $F(t)=2008-t^{3}-4 t- a+\sin \pi t$, find the number of integer values of $a$, for each of which the equation
$$
\... | 60013 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,970 |
10.2 For the function $f(x)=2013-a+12 x^{2}-\cos 2 \pi x-8 x^{3}-16 x$ find the number of integer values of $a$, for each of which the equation
$$
\underbrace{f(f(\ldots f}_{2013 \text { times }}(x) \ldots))=2 x-1
$$
on the interval $[50 ; 51]$ has a unique solution. | Solution. After substituting $t=2 x-1$, we obtain a new problem for the function $F(t)=2007-t^{3}-5 t-a+\cos \pi t$. The number of integer values of $a$ is
$$
g(101)-g(99)+1=2+101^{3}-99^{3}+505-495+1=60015, \quad \text { where } g(t)=t-F(t) .
$$
## Answer 60015 . | 60015 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,971 |
10.3 For the function $f(x)=2013+\sin 2 \pi x-8 x^{3}-12 x^{2}-18 x-a$, find the number of integer values of $a$ for each of which the equation
$$
\underbrace{f(f(\ldots f}_{2013 \text { times }}(x) \ldots))=2 x+1
$$
has a unique solution on the interval $[49,50]$. | Solution. After substituting $t=2 x+1$, we obtain a new problem for the function $F(t)=2020-t^{3}-6 t-a-\sin \pi t$. The number of integer values of $a$ is
$$
g(101)-g(99)+1=2+101^{3}-99^{3}+606-594+1=60017, \quad \text { where } g(t)=t-F(t)
$$
Answer 60017. | 60017 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,972 |
10.4 For the function $f(x)=2013-a+\cos 2 \pi x-12 x^{2}-8 x^{3}-20 x$ find the number of integer values of $a$, for each of which the equation
$$
\underbrace{f(f(\ldots f}_{2013 \text { times }}(x) \ldots))=2 x+1
$$
has a unique solution on the interval $[49,50]$. | Solution. After substituting $t=2 x+1$, we obtain a new problem for the function $F(t)=2021-t^{3}-7 t-a-\cos \pi t$. The number of integer values of $a$ is
$$
g(101)-g(99)+1=2+101^{3}-99^{3}+707-693+1=60019, \quad \text { where } g(t)=t-F(t)
$$
Answer 60019. | 60019 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,973 |
Task 1. An apple, a pear, an orange, and a banana were placed in four boxes (one fruit per box). Inscriptions were made on the boxes:
On the 1st: Here lies an orange.
On the 2nd: Here lies a pear.
On the 3rd: If in the first box lies a banana, then here lies an apple or a pear.
On the 4th: Here lies an apple.
It i... | Answer: 2431
Solution: The inscription on the 3rd box is incorrect, so in the first box lies a banana, and in the third - not an apple and not a pear, therefore, an orange. From the inscription on the 4th box, it follows that there is no apple there, so since the banana is in the 1st, and the orange is in the 2nd, the... | 2431 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,974 |
Problem 2. Beginner millionaire Bill buys a bouquet of 7 roses for $20 for the entire bouquet. Then he can sell a bouquet of 5 roses for $20 per bouquet. How many bouquets does he need to buy to earn a difference of $1000? | Answer: 125
Solution. Let's call "operation" the purchase of 5 bouquets (= 35 roses) and the subsequent sale of 7 bouquets (= 35 roses). The purchase cost is $5 \cdot 20=\$ 100$, and the selling price is $7 \cdot 20=\$ 140$. The profit from one operation is $\$ 40$.
Since $\frac{1000}{40}=25$, 25 such operations are ... | 125 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,975 |
Task 3. Find a natural number $N(N>1)$, if the numbers 1743, 2019, and 3008 give the same remainder when divided by $N$. | Answer: 23.
Solution. From the condition, it follows that the numbers $2019-1743=276$ and $3008-2019=989$ are divisible by $N$. Since $276=2^{2} \cdot 3 \cdot 23$, and $989=23 \cdot 43$, then $N=23$. | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,976 |
Task 4. Find the smallest natural number $n$ such that $n^{2}$ and $(n+1)^{2}$ contain the digit 7. | Answer: 26.
Solution. There are no squares ending in the digit 7. There are no two-digit squares starting with 7. Therefore, $n \geq 10$. The first square containing 7 is $576=24^{2}$. Since $25^{2}=625,26^{2}=676,27^{2}=729$, the answer is $n=26$.
Problem 4a. Find the smallest natural number $n$ such that $n^{2}$ an... | 26 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,977 |
Problem 5. A square with an integer side length was cut into 2020 squares. It is known that the areas of 2019 squares are 1, and the area of the 2020th square is not equal to 1. Find all possible values that the area of the 2020th square can take. In your answer, provide the smallest of the obtained area values. | Answer: 112225.
 | 112225 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,978 |
Problem 6. Master Li Si Qing makes fans. Each fan consists of 6 sectors, painted on both sides in red and blue (see fig.). Moreover, if one side of a sector is painted red, the opposite side is painted blue and vice versa. Any two fans made by the master differ in coloring (if one coloring can be transformed into anoth... | Answer: 36.
## Solution:
The coloring of one side can be chosen in $2^{6}=64$ ways. It uniquely determines the coloring of the opposite side. However, some fans - those that transform into each other when flipped, we have counted twice. To find their number, let's see how many fans transform into themselves when flip... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,979 |
Problem 7. How many solutions in integers does the equation
$6 y^{2}+3 x y+x+2 y-72=0$ have? | Answer: 4.
## Solution:
Factorize:
$(3 y+1)(2 y+x)=72$.
The first factor must give a remainder of 1 when divided by 3. The number 72 has only 4 divisors that give such a remainder: $-8,-2,1,4$. They provide 4 solutions.
Problem 7a. How many solutions in integers does the equation
$6 y^{2}+3 x y+x+2 y+180=0$ have?... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,980 |
Problem 9. In a convex quadrilateral $A B C D$, side $A B$ is equal to diagonal $B D, \angle A=65^{\circ}$, $\angle B=80^{\circ}, \angle C=75^{\circ}$. What is $\angle C A D$ (in degrees $) ?$ | Answer: 15.
Solution. Since triangle $ABD$ is isosceles, then $\angle BDA = \angle BAD = 65^{\circ}$. Therefore, $\angle DBA = 180^{\circ} - 130^{\circ} = 50^{\circ}$. Hence, $\angle CBD = 80^{\circ} - 50^{\circ} = 30^{\circ}$, $\angle CDB = 180^{\circ} - 75^{\circ} - 30^{\circ} = 75^{\circ}$. This means that triangle... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,982 |
1. A trip to St. Petersburg is being organized for 30 schoolchildren along with their parents, some of whom will be driving cars. Each car can accommodate 5 people, including the driver. What is the minimum number of parents that need to be invited on the trip?
ANSWER: 10 | Solution: No more than 4 students can fit in a car, so 8 cars will be needed, i.e., 10 drivers. | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,983 |
2. Eight numbers stand in a row such that the sum of any three consecutive numbers equals 50. The first and last numbers of these eight are known. Fill in the six empty places:
11 12.
ANSWER: $11,12,27,11,12,27,11,12$ | Solution: It follows from the condition that the sequence of numbers is periodic with a period of 3. | 11,12,27,11,12,27,11,12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,984 |
4.
In the test, there are 4 sections, each containing the same number of questions. Andrey answered 20 questions correctly. The percentage of his correct answers was more than 60 but less than 70. How many questions were in the test? | Answer: 32.
Solution. According to the condition $\frac{60}{100}<\frac{20}{x}<\frac{70}{100}$, hence $28 \frac{4}{7}=\frac{200}{7}<x<\frac{100}{3}=33 \frac{1}{3}$, that is $29 \leq x \leq 33$. From the first condition of the problem, it follows that the number of questions must be divisible by 4. | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,985 |
5. In square $A B C D$, points $F$ and $E$ are the midpoints of sides $A B$ and $C D$, respectively. Point $E$ is connected to vertices $A$ and $B$, and point $F$ is connected to $C$ and $D$, as shown in the figure. Determine the area of the rhombus $F G E H$ formed in the center, given that the side of the square $A B... | Solution - A square can be cut into pieces and folded to form 3 more such rhombuses, i.e., the rhombus constitutes $1 / 4$ of the area of the square.
Lomonosov Moscow State University
## Sc... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,986 | |
1. A trip to Nizhny Novgorod is being organized for 50 schoolchildren along with their parents, some of whom will be driving cars. Each car can accommodate 6 people, including the driver. What is the minimum number of parents that need to be invited on the trip?
ANSWER: 10 | Solution: No more than 5 students can fit in a car, so 10 cars will be needed, i.e., 10 drivers. | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,987 |
2. In a row, there are 8 numbers such that the sum of any three consecutive numbers equals 100. The first and last numbers of these eight are known. Fill in the six empty places:
20, _,_,_,_,_, 16 .
ANSWER: $20,16,64,20,16,64,20,16$ | Solution: It follows from the condition that the sequence of numbers is periodic with a period of 3. | 20,16,64,20,16,64,20,16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,988 |
3. Insert arithmetic operation signs and parentheses into the expression consisting of three numbers, $\frac{1}{8} \ldots \frac{1}{9} \ldots \frac{1}{28}$, so that the result of the calculations is equal to $\frac{1}{2016}$.
ANSWER: $\frac{1}{8} \times \frac{1}{9} \times \frac{1}{28}$ or $\left(\frac{1}{8}-\frac{1}{9}... | Solution: factorize 2016 into prime factors and notice that 2016=8x9x28. | \frac{1}{8}\times\frac{1}{9}\times\frac{1}{28} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,989 |
4. The test consists of 5 sections, each containing the same number of questions. Pavel answered 32 questions correctly. The percentage of his correct answers turned out to be more than 70 but less than 77. How many questions were in the test?
ANSWER: 45. | Solution: from the condition $0.7<32 / x<0.77$ it follows that $41<x<46$, but $x$ is a multiple of 5, so $x=45$. | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,990 |
5. At the vertices of a regular 9-gon
(see fig.) place the numbers $2016, 2017, \ldots, 2024$ in such a way that for any three vertices forming an equilateral triangle, one of the numbers is equal to the arithmetic mean of the other two.
![](https://cdn.mathpix.com/cropped/2024_05_06_54c562d050fb6badb826g-04.jpg?heig... | Solution: notice that any three numbers, spaced at equal intervals, work
Lomonosov Moscow State University
## School Olympiad "Conquer Sparrow Hills" in Mathematics
Final Round Tasks for 2015/2016 Academic Year for Grades 5-6 | 2016,2017,2018,2019,2020,2021,2022, | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,991 |
2. An integer was increased by 2, as a result, its square decreased by 2016. What was the number initially (before the increase)?
ANSWER: -505. | Solution: Solve the equation $(x+2)^{2}=x^{2}-2016$. | -505 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,992 |
3. Find all irreducible positive fractions that increase by 3 times when both the numerator and the denominator are increased by 12.
ANSWER: $2 / 9$. | Solution: Note that if the numerator is not less than 6, then by adding 12 to it, it will not increase more than three times, so the fraction itself certainly cannot increase three times. By trying numerators $1,2,3,4,5$, we get the fractions 1/3.6, 2/9, 3/18, 4/36, 5/90, among which only $2 / 9$ is irreducible. | \frac{2}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,993 |
4. A small garden measuring $6 x 7$ meters has been divided into 5 square plots. All boundaries between the plots run parallel to the sides of the square, and the side of each plot is a whole number of meters. Find the total length of the resulting boundaries. Consider the boundaries as lines with no thickness.
ANSWER... | Solution: 42 can be broken down into 5 squares in only one way: $42=4^{2}+3^{2}+3^{2}+2^{2}+2^{2}$. Their total perimeter is $4 * 4+4^{*} 3+4^{*} 3+4^{*} 2+4^{*} 2=56$. But we need to subtract the external boundaries, their length is $6+6+7+7=26$, and also take into account that each boundary participates in the perime... | 15\mathrm{M} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,994 |
5. Find the largest natural number that cannot be represented as the sum of two composite numbers.
ANSWER: 11 | Solution: Even numbers greater than 8 can be represented as the sum of two even numbers greater than 2. Odd numbers greater than 12 can be represented as the sum of 9 and an even composite number. By direct verification, we can see that 11 cannot be represented in such a way.
Lomonosov Moscow State University
## Scho... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,995 |
5. Philatelist Andrey decided to distribute all his stamps equally into 3 envelopes, but it turned out that one stamp was extra. When he distributed them equally into 5 envelopes, 3 stamps were extra; finally, when he distributed them equally into 7 envelopes, 5 stamps remained. How many stamps does Andrey have in tota... | Solution. If the desired number is $x$, then the number $x+2$ must be divisible by 3, 5, and 7, i.e., it has the form $3 \cdot 5 \cdot 7 \cdot p$. Therefore, $x=105p-2$. Since according to the condition $150 < x < 300$ for mathematics
Final stage tasks for the 2015/2016 academic year for 5-6 grades | 208 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,996 |
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