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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 |
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 |
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 |
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 |
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. 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 |
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 |
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 |
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. 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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. 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. 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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})
$$
where $x=3, \underbrace{22 \ldots 2}_{2013} 3, y=4, \underbrace{77 \ldots .7}_{2014}$. Choose the answer option with the number closest to the one you found. | Answer: 8. (D)
Options.
 | 8 | Algebra | MCQ | Yes | Yes | olympiads | false |
2. Calculate
$$
(\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}}}
$$
where $x=1, \underbrace{11 \ldots 1}_{2013} 2, y=3, \underbrace{88 \ldots 8}_{2014}$. Choose the answer option with the number closest to the one you found. | Answer: 5. (B)
Options.
 | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
2. Calculate
$$
\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})
$$
where $x=2, \underbrace{44 \ldots .4}_{2013} 5, y=1, \underbrace{55 \ldots .5}_{2014}$. Choose the answer option with the number closest to the one you found. | Answer: 4. (E)
Options.
 | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
3. In how many ways can a coach form a hockey team consisting of one goalkeeper, two defenders, and three forwards if he has 2 goalkeepers, 5 defenders, and 8 forwards at his disposal? Among the proposed answer options, choose the one closest to the correct one. | Answer: 1120. (B)
$$
\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}
$$ | 1120 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
3. In how many ways can a team be selected from a group consisting of 7 boys and 8 girls, so that the team has 4 boys and 3 girls? Among the proposed answer options, choose the one closest to the correct one. | Answer: 1960. (D)
$$
\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}
$$ | 1960 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
3. In how many ways can a coach form a basketball team consisting of two guards and three forwards if he has 6 guards and 11 forwards at his disposal? Among the options provided, choose the one closest to the correct answer.
| Answer: 2475. (E)
 | 2475 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
3. In how many ways can a team be assembled consisting of 3 painters and 4 plasterers, if there are 6 painters and 8 plasterers? Among the options provided, choose the one closest to the correct answer. | Answer: $1400 .(\mathrm{C})$
 | 1400 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
4. Determine the number of different values of $a$ for which the equation
$$
\left(3-a^{2}\right) x^{2}-3 a x-1=0
$$
has a unique solution | Answer: 2. (C)
Options.
 | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Determine how many different values of $a$ exist for which the equation
$$
\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}
$$ | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Ivan Semenovich leaves for work at the same time every day, drives at the same speed, and arrives exactly at 9:00. One day he overslept and left 40 minutes later than usual. To avoid being late, Ivan Semenovich drove at a speed 60% faster than usual and arrived at 8:35. By what percentage should he have increased hi... | Answer: By $30 \%$.
Solution: By increasing the speed by $60 \%$, i.e., by 1.6 times, Ivan Semenovich reduced the time by 1.6 times and gained 40+25=65 minutes. Denoting the usual travel time as $T$, we get $\frac{T}{1.6}=T-65$, from which $T=\frac{520}{3}$. To arrive in $T-40=\frac{400}{3}$, the speed needed to be in... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The English club is attended by 20 gentlemen. Some of them are acquainted (acquaintances are mutual, i.e., if A knows B, then B knows A). It is known that there are no three gentlemen in the club who are pairwise acquainted.
One day, the gentlemen came to the club, and each pair of acquaintances shook hands with ea... | Answer: 100 handshakes.
Solution: Choose a gentleman with the maximum number of acquaintances (if there are several, choose any one). Suppose he has $n$ acquaintances. These acquaintances cannot be pairwise acquainted with each other. Consider the remaining $(20-n-1)$ gentlemen, each of whom has no more than $n$ acqua... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the inequality $\frac{\sqrt{\frac{x}{\gamma}+(\alpha+2)}-\frac{x}{\gamma}-\alpha}{x^{2}+a x+b} \geqslant 0$.
In your answer, specify the number equal to the number of integer roots of this inequality. If there are no integer roots, or if there are infinitely many roots, enter the digit 0 in the answer sheet.
... | # Answer: 7.
Solution. Given the numerical values, we solve the inequality:
$$
\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
$$
Since $\sqrt{x+5} \geqslant x+3 \Leftrightarrow\left[\begin{array}{c}-5 \leqslant x \leqslant-3, \\ x+5 \geqslant(x+3)^{2}\e... | 7 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. From point $M$, lying inside triangle $A B C$, perpendiculars are drawn to the sides $B C, A C, A B$, with lengths $k, l$, and $m$ respectively. Find the area of triangle $A B C$, if $\angle C A B=\alpha$ and $\angle A B C=\beta$. If the answer is not an integer, round it to the nearest integer.
$$
\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 |
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 |
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:
$$
m=n\left(1+\frac{p}{100}\right), \quad \text { and } \quad m a... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. Find the sum of all two-digit numbers for each of which the sum of the squares of the digits is 37 more than the product of the same digits. | Solution. 1st method. For a two-digit number $\overline{a b}$, the condition means that
$$
a^{2}+b^{2}-a b=37
$$
Since the equation is symmetric, i.e., with each solution $(a, b)$, the pair $(b, a)$ is also a solution, we can assume without loss of generality that $a \geqslant b$.
- Suppose $a \leqslant 6$. Then the... | 231 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.2. Find the sum of all two-digit numbers for each of which the sum of the squares of the digits is 57 more than the product of the same digits. | Solution. Such two-digit numbers are $18, 78, 81, 87$. Their sum is 264.
Answer: 264. (D)
\section*{| A | 165 | $\mathbf{B}$ | 198 | $\mathbf{C}$ | 231 | $\mathbf{D}$ | 264 | $\mathbf{E}$ | 297 | $\mathbf{F}$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | 264 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.3. Find the sum of all two-digit numbers for each of which the sum of the squares of the digits is 73 more than the product of the same digits. | Solution. Such two-digit numbers are $19, 89, 91, 98$. Their sum is 297.
Answer: 297. ( ( )
 | 297 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.4. Find the sum of all two-digit numbers for each of which the sum of the squares of the digits is 31 more than the product of the same digits. | Solution. Such two-digit numbers are $16, 56, 61, 65$. Their sum is 198.
Answer: 198. (B)
 | 198 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. The segment connecting the lateral sides of the trapezoid and parallel to its bases, which are 3 and 21, divides the trapezoid into two parts of equal area. Find the length of this segment. | Solution. 1st method. Let $a=3$ and $b=21$ be the lengths of the bases of the trapezoid. If $c$ is the length of the segment parallel to the bases, and $h_{1}$ and $h_{2}$ are the parts of the height of the trapezoid adjacent to the bases $a$ and $b$ respectively, then, by equating the areas, we get:
$$
\frac{a+c}{2} ... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.2. 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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
$$
\sqrt{\underbrace{111 \ldots 11}_{2014}-\underbrace{22 ... | 231 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. Specify the integer closest to the larger root of the equation
$$
\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}
$$
Since $0 \leqslant \operatorname{arctg} y<\operatorname{arctg}(y+1)<\frac{\pi}{2}$, the last ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2. Specify the integer closest to the smaller root of the equation
$$
\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}
$$ | Solution. The equation is equivalent to $\frac{5 x}{26}=\frac{13}{10 x} \Longleftrightarrow|x|=\frac{13}{5}$.
Answer: -3 . | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.3. Specify the integer closest to the smaller root of the equation
$$
\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}
$$ | Solution. The equation is equivalent to $\frac{7 x}{10}=\frac{5}{14 x} \Longleftrightarrow |x|=\frac{5}{7}$.
Answer: -1. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.4. Indicate the integer closest to the larger root of the equation
$$
\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}
$$ | Solution. The equation is equivalent to $\frac{3 x}{22}=\frac{11}{6 x} \Longleftrightarrow |x|=\frac{11}{3}$.
Answer: 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5. Indicate the integer closest to the smaller root of the equation
$$
\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}
$$ | Solution. The equation is equivalent to $\frac{2 x}{7}=\frac{7}{8 x} \Longleftrightarrow |x|=\frac{7}{4}$.
Answer: -2. | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.6. Indicate the integer closest to the larger root of the equation
$$
\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}
$$ | Solution. The equation is equivalent to $\frac{5 x}{26}=\frac{13}{10 x} \Longleftrightarrow|x|=\frac{13}{5}$.
Answer: 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.7. Indicate the integer closest to the larger root of the equation
$$
\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}
$$ | Solution. The equation is equivalent to $\frac{7 x}{10}=\frac{5}{14 x} \Longleftrightarrow |x|=\frac{5}{7}$.
Answer: 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.8. Provide the integer closest to the smaller root of the equation
$$
\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}
$$ | Solution. The equation is equivalent to $\frac{3 x}{22}=\frac{11}{6 x} \Longleftrightarrow |x|=\frac{11}{3}$.
Answer: -4. | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. In the triangular pyramid $S A B C$, the edges $S B, A B$ are perpendicular and $\angle A B C=120^{\circ}$. Point $D$ on edge $A C$ is such that segment $S D$ is perpendicular to at least two medians of triangle $A B C$ and $C D=A B=44 \sqrt[3]{4}$. Find $A D$ (if the answer is not an integer, round it to the near... | Solution. Since segment $SD$ is perpendicular to two medians of triangle $ABC$, it is perpendicular to the plane $(ABC)$ (see Fig. 3). By the theorem of three perpendiculars, it follows that $DB \perp AB$.
=2013-8 x^{3}+12 x^{2}-14 x-a-\sin 2 \pi x$ find the number of integer values of $a$, for each of which the equation
$$
\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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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.
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 |
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. 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 |
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 |
2. Find the smallest natural N such that N+2 is divisible (without remainder) by 2, N+3 by 3, ..., N+10 by 10.
ANSWER: 2520. | Solution: Note that $N$ must be divisible by $2,3,4, \ldots, 10$, therefore, $N=$ LCM $(2,3,4, . ., 10)=2^{3} \times 3^{2} \times 5 \times 7=2520$. | 2520 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Five runners ran a relay. If the first runner ran twice as fast, they would have spent $5 \%$ less time. If the second runner ran twice as fast, they would have spent $10 \%$ less time. If the third runner ran twice as fast, they would have spent $12 \%$ less time. If the fourth runner ran twice as fast, they would ... | Solution: If each ran twice as fast, they would run 50% faster. This means that if the 5th ran faster, the time would decrease by $50-5-10-12-15=8 \%$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation
$$
\log _{3}(x+2) \cdot \log _{3}(2 x+1) \cdot\left(3-\log _{3}\left(2 x^{2}+5 x+2\right)\right)=1
$$ | Problem 4.
Answer: $x=1$.
Solution. For admissible values $x>-\frac{1}{2}$ we have: $\log _{3}(x+2)>0$.
If $\log _{3}(2 x+1)0$ and the left side of the equation is negative.
Thus, all three factors on the left side of the equation are positive, and the roots should be sought only among those $x$ for which
$$
\left... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Petrov lists the odd numbers: $1,3,5, \ldots, 2013$, while Vasechkin lists the even numbers: $2,4, \ldots, 2012$. Each of them calculated the sum of all digits of all their numbers and told the excellent student Masha. Masha subtracted Vasechkin's result from Petrov's result. What did she get? | Answer: 1007.
Solution: Let's break down the numbers of Petrov and Vasechkin into pairs as follows: $(2,3),(4,5), \ldots,(98,99),(100,101), \ldots$ (2012,2013), with 1 left unpaired for Petrov. Notice that in each pair, the sum of the digits of the second number is 1 greater than that of the first (since they differ o... | 1007 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. There are no fewer than 150 boys studying at the school, and there are $15 \%$ more girls than boys. When the boys went on a trip, 6 buses were needed, and each bus had the same number of students. How many people in total study at the school, given that the total number of students is no more than 400? | Answer: 387.
Solution: The number of boys is a multiple of 6, let's denote it as $6n$, obviously, $n \geqslant 25$. Then the number of girls is $6n \times 1.15 = 6.9n$. The total number of students is $12.9n \leqslant 400$, so $n \leqslant 31$. Considering that $6.9n$ must be an integer, and therefore $n$ must be a mu... | 387 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In the school Spartakiad, teams from classes $8^{\mathrm{A}}$, $8^{\text {Б }}$, and $8^{\mathrm{B}}$ participated. In each of the competitions, one of these teams took 1st place, another took 2nd place, and another took 3rd place. After the Spartakiad, points were tallied: $x$ points were awarded for 1st place, $y$... | Answer: 5 competitions, $8^{\text {B }}$.
Solution: Let $n \geqslant 2$ be the number of competitions in the Spartakiad, then the total number of points scored by all teams is $n(x+y+z)=22+9+9=40$. But $z \geqslant 1, y \geqslant 2, x \geqslant 3$, so $x+y+z \geqslant 6$. Consider the possible cases: $x+y+z=8, n=5$; $... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. Famous skater Tony Hawk is riding a skateboard (segment $A B$) in a ramp, which is a semicircle with diameter $P Q$. Point $M$ is the midpoint of the skateboard, $C$ is the foot of the perpendicular dropped from point $A$ to the diameter $P Q$. What values can the angle $\angle A C M$ take if it is known that the an... | Answer: $12^{\circ}$.
Solution: Extend the line $A C$ to intersect the circle at point $D$ (see figure). The chord $A D$ is perpendicular to the diameter $P Q$, therefore, it is bisected by it. Thus, $C M$ is the midline of triangle $A B D$, so $C M \| B D$ and, therefore, $\angle A C M=\angle A D B$. The angle $\angl... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. Find the number of natural numbers from 1 to 100 that have exactly four natural divisors, at least three of which do not exceed 10. | Answer: 8.
Solution: A number has exactly 4 natural divisors if it is either the cube of a prime number or the product of two prime numbers. The cubes of prime numbers (satisfying the conditions) are: 8 and 27. Prime numbers not greater than 10 are - 2, 3, 5, and 7. All their pairwise products satisfy the conditions, ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Karlson filled a conical glass with lemonade and drank half of it by height (measuring from the surface of the liquid to the apex of the cone), and Little Man drank the second half. How many times more lemonade did Karlson drink compared to Little Man? | # Answer: 7 times.
Solution. Let $r$ and $h$ be the radius of the base and the height of the conical glass, respectively. Then the volume of lemonade in the entire glass is $V_{\text {glass}}=\frac{1}{3} \pi r^{2} h$. The volume of lemonade drunk by Little One is $V_{\text {Little}}=\frac{1}{3} \pi(r / 2)^{2}(h / 2)=\... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Malvina and Buratino play according to the following rules: Malvina writes six different numbers in a row on the board, and Buratino comes up with his own four numbers $x_{1}, x_{2}, x_{3}, x_{4}$ and writes under each of Malvina's numbers one of the sums $x_{1}+x_{2}, x_{1}+x_{3}, x_{1}+x_{4}$, $x_{2}+x_{3}, x_{2}+... | # Answer: 14.
## Solution.
Solution. Let Malvina write the numbers $a_{1}>a_{2}>a_{3}>a_{4}>a_{5}>a_{6}$. If Buratino comes up with the numbers $x_{1}=\left(a_{1}+a_{2}-a_{3}\right) / 2, x_{2}=\left(a_{1}+a_{3}-a_{2}\right) / 2, x_{3}=\left(a_{2}+a_{3}-a_{1}\right) / 2, x_{4}=a_{4}-x_{3}$, then by writing $x_{1}+x_{2... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Find all four-digit numbers $\overline{a b c d}$ (where $a, b, c, d$ are the digits of the decimal representation), each of which is a divisor of at least one of the three four-digit numbers $\overline{b c d a}, \overline{c d a b}, \overline{d a b c}$ formed from it. | Answer: All numbers of the form $\overline{a b a b}$, where $a$ and $b$ are any digits except zero (there are 81 such numbers).
## Solution.
From the problem statement, it follows that there exists $k \in \mathbb{N}$ such that at least one of the following equalities holds:
(I) $k \cdot \overline{a b c d}=\overline{b... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. For what values of $a$ does the equation
$$
[x]^{2}+2012 x+a=0
$$
(where $[x]$ is the integer part of $x$, i.e., the greatest integer not exceeding $x$) have the maximum number of solutions? What is this number? | Answer: 89 solutions at $1006^{2}-20120
\end{array}\right.
$
has the maximum number of integer solutions.
The solution to the first inequality is the interval $\left[-1006-\sqrt{1006^{2}-a} ;-1006+\sqrt{1006^{2}-a}\right]$ under the condition $a \leqslant 1006^{2}$.
The solution to the second inequality for $a>1006^... | 89 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Determine how many zeros the number $N!$ ends with! | Solution. Let $N=2014$. Among the first 2014 natural numbers, 402 numbers are divisible by 5, of which 80 numbers are divisible by 25. Among these 80 numbers, 16 numbers are divisible by 125, of which 3 numbers are divisible by 625. There are more even numbers among the first 2014 natural numbers than those divisible b... | 501 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. By how much is the sum of the squares of the first hundred even numbers greater than the sum of the squares of the first hundred odd numbers | Answer: 20100.
Solution: Group the terms as $\left(2^{2}-1^{2}\right)+\left(4^{2}-3^{2}\right)+\cdots+\left(200^{2}-199^{2}\right)=$ $(2-1) \cdot(2+1)+(4-3) \cdot(4+3)+\ldots+(200-199) \cdot(200+199)=1+2+\cdots+$ $199+200$. Divide the terms into pairs that give the same sum: $1+200=2+199=\ldots=100+101=201$. There wil... | 20100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Petrov and Vasechkin were repairing a fence. Each had to nail a certain number of boards (the same amount). Petrov nailed two nails into some boards, and three nails into the rest. Vasechkin nailed three nails into some boards, and five nails into the rest. Find out how many boards each of them nailed, if it is know... | Answer: 30.
Solution: If Petrov had nailed 2 nails into each board, he would have nailed 43 boards and had one extra nail. If he had nailed 3 nails into each board, he would have nailed 29 boards. Therefore, the desired number lies between 29 and 43 (inclusive). Similarly, if Vasechkin had nailed 3 nails into each boa... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Six natural numbers (possibly repeating) are written on the faces of a cube, such that the numbers on adjacent faces differ by more than 1. What is the smallest possible value of the sum of these six numbers? | Answer: 18.
Solution: Consider three faces that share a common vertex. The numbers on them differ pairwise by 2, so the smallest possible sum would be for $1+3+5=9$. The same can be said about the remaining three faces.
Thus, the sum cannot be less than 18. We will show that 18 can be achieved - place the number 1 on... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Find the smallest three-digit number with the property that if a number, which is 1 greater, is appended to it on the right, then the result (a six-digit number) will be a perfect square. | Answer: 183
Solution: Let the required number be a, then $1000a + a + 1 = n^2$. We can rewrite this as: $1001a = (n - 1)(n + 1)$. Factorize $1001 = 7 \times 11 \times 13$, meaning the product (n-1)(n+1) must be divisible by 7, 11, and 13. Additionally, for the square to be a six-digit number, $n$ must be in the interv... | 183 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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