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Problem 10.5. In each cell of a square table of size $200 \times 200$, a real number not exceeding 1 in absolute value was written. It turned out that the sum of all the numbers is zero. For what smallest $S$ can we assert that in some row or some column, the sum of the numbers will definitely not exceed $S$ in absolut... | Answer: 100.
Solution. First, we show that $S<40000$.
$$
This means that one of the numbers $A$ or $D$ in absolute value exceeds 10000. However, each of the corresponding squares contains only 10000 cells, and the numbers in them do not exceed 1 in absolute value. Contradiction.
## Criteria
Any correct solution to ... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.5. Through the point with coordinates $(2,2)$, lines (including two parallel to the coordinate axes) are drawn, dividing the plane into angles of $18^{\circ}$. Find the sum of the abscissas of the points of intersection of these lines with the line $y=2016-x$. | Answer: 10080. Solution. The picture is symmetric with respect to the line $y=x$, so the sum of the abscissas is equal to the sum of the ordinates. Through the point $(2,2)$, 10 lines are drawn, and the line $y=2016-x$ intersects all of them. For each point on the line $y=2016-x$, the sum of the coordinates is 2016, so... | 10080 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Positive numbers $a, b, c, d$ are greater than 1. Find the smallest possible value of the expression
$$
\log _{a}\left(a b^{2}\right)+\log _{b}\left(b^{2} c^{3}\right)+\log _{c}\left(c^{5} d^{6}\right)+\log _{d}\left(d^{35} a^{36}\right)
$$ | Answer: 67.
Solution. From the properties of logarithms, it follows that $\log _{a} b \cdot \log _{b} c \cdot \log _{c} d \cdot \log _{d} a=1$. Also, all these four factors are positive, since all numbers $a, b, c, d$ are greater than 1.
Transform and estimate the given expression
$$
\begin{gathered}
S=\log _{a}\lef... | 67 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Tetrahedron $ABCD$ with acute-angled faces is inscribed in a sphere with center $O$. A line passing through point $O$ perpendicular to the plane $ABC$ intersects the sphere at point $E$ such that $D$ and $E$ lie on opposite sides relative to the plane $ABC$. The line $DE$ intersects the plane $ABC$ at point $F$, whi... | Answer: $40^{\circ}$.
Note that point $E$ is equidistant from points $A, B, C$, so its projection onto the plane $A B C$ coincides with the projection of point $O$ onto this plane and is the center of the circumscribed circle of triangle $A B C$.
Consider triangles $A D E$ and $B D E$. They have a pair of equal sides... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Buses from Moscow to Voronezh depart every hour, at 00 minutes. Buses from Voronezh to Moscow depart every hour, at 30 minutes. The trip between the cities takes 8 hours. How many buses from Voronezh will the bus that left from Moscow meet on its way? | Answer: 16.
It is clear that all buses from Moscow will meet the same number of buses from Voronezh, and we can assume that the bus from Moscow departed at 12:00. It is easy to understand that it will meet buses that left Oryol at 4:30, 5:30, ..., 18:30, 19:30 and only them. There are 16 such buses.
$\pm$ Correct rea... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In the vertices of a regular 2019-gon, numbers are placed such that the sum of the numbers in any nine consecutive vertices is 300. It is known that the 19th vertex has the number 19, and the 20th vertex has the number 20. What number is in the 2019th vertex? | Answer: 61.
Solution. Let the numbers at the vertices be denoted as $x_{1}, x_{2}, \ldots, x_{2019}$. Since the sum of any nine consecutive numbers is the same, the numbers that are 8 apart are equal. Therefore, $x_{1}=x_{10}=x_{19}=\ldots=x_{1+9 k}=\ldots$. Since 2019 is not divisible by 9 but is divisible by 3, cont... | 61 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let's call a natural number an almost palindrome if it can be transformed into a palindrome by changing one of its digits. How many nine-digit almost palindromes exist? (20 points) | Solution: After replacing one digit in the almost palindrome, we get a number of the form $\overline{a b c d e d c b a}$. Let's divide the digits into pairs by place value: 1 and 9, 2 and 8, 3 and 7, 4 and 6. In any almost palindrome, the digits in three of the specified pairs of place values must be the same, and in e... | 3240000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. If in an acute scalene triangle three medians, three angle bisectors, and three altitudes are drawn, they will divide it into 34 parts. | Write the answer in digits in ascending order, without spaces (for example, 12345).
## Mathematics $8-9$ grade
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 2. (20 points)
Find the maximum possible value of the ratio of a three-digit number to the sum of its digits.
# | # Solution.
Let $N=\overline{a b c}$, where $a, b, c$ are the digits of the number. Clearly, for "round" numbers $N=$ $100, 200, \ldots, 900$, we have $\frac{N}{a+b+c}=100$. Furthermore, if the number $N$ is not round, then $b+c>0$ and $a+b+c \geq a+1$. Since the leading digit of the number $N$ is $a$, we have $N<(a+1... | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In the language of the "Tekimar" tribe, there are only 7 letters: A, E, I, K, M, R, T, but the order of these letters in the alphabet is unknown. A word is defined as any sequence of seven different letters from the alphabet, and no other words exist in the language. The chief of the tribe listed all existing words ... | Solution: The total number of words in the tribe's language is $7!=5040$. Note that the number of words starting with any particular letter is the same for any first letter, which is $7!\div 7=6!=720$. If a word starts with the first letter in the alphabet, the numbering of any such word starts from number 1 and ends a... | 3745 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Find the number of triples of natural numbers $m, n, k$, which are solutions to the equation $m+$ $\sqrt{n+\sqrt{k}}=2023$. (20 points) | Solution: For the left side to be an integer, the numbers $k$ and $n+\sqrt{k}$ must be perfect squares, and since $n+\sqrt{k} \geq 2$, it follows that $\sqrt{n+\sqrt{k}} \geq 2$ and thus $m \leq 2021$. Since $1 \leq m \leq$ 2021, $\sqrt{n+\sqrt{k}}$ can take any value from 2 to 2022 - this value uniquely determines the... | 27575680773 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. A printing house determines the cost of printing a book as follows: it adds the cost of the cover to the cost of each page, and then rounds the result up to the nearest whole number of rubles (for example, if the result is 202 rubles and 1 kopeck, it is rounded up to 203 rubles). It is known that the cost of a book ... | Solution: Let's convert all costs into kopecks. Let one page cost $x$ kopecks, and the cover cost $100 y$ kopecks. Then, according to the problem, we have
$$
\left\{\begin{array}{l}
13300<100 y+104 x \leq 13400 \\
18000<100 y+192 x \leq 18100
\end{array}\right.
$$
Subtracting the first inequality from the second, we ... | 77 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Maxim came up with a new way to divide numbers by a two-digit number $N$. To divide any number $A$ by the number $N$, the following steps need to be performed:
1) Divide $A$ by the sum of the digits of the number $N$;
2) Divide $A$ by the product of the digits of the number $N$;
3) Subtract the second result from th... | Solution: Let $N=\overline{x y}=10 x+y$, where $x, y$ are digits. We need to find all pairs $(x, y)$ such that $\frac{A}{10 x+y}=\frac{A}{x+y}-\frac{A}{x y}$, which means $\frac{1}{10 x+y}=\frac{1}{x+y}-\frac{1}{x y}$. Note that $y \neq 0$. Transform the equation to the form
$$
\begin{gathered}
x y(x+y)=x y(10 x+y)-(x... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task 1. (20 points)
It is known that the only solution to the equation
$$
\pi / 4=\operatorname{arcctg} 2+\operatorname{arcctg} 5+\operatorname{arcctg} 13+\operatorname{arcctg} 34+\operatorname{arcctg} 89+\operatorname{arcctg}(x / 14)
$$
is a natural number. Find it. | # Solution.
Consider the equation:
$$
\operatorname{arcctg} a-\operatorname{arcctg} b=\operatorname{arcctg} y \text {. }
$$
Let $\alpha=\operatorname{arcctg} a, \beta=\operatorname{arcctg} b$. Now $y=\operatorname{ctg}(\alpha-\beta)=\frac{1+\operatorname{ctg} \alpha \operatorname{ctg} \beta}{\operatorname{ctg} \beta... | 2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Before you is a segment-digit. For displaying time on electronic clocks, each digit uses seven segments, each of which can be lit or not; the lit segments form the digit as shown in the figure
## 8723456789
That is, to display zero, six segments are used, to display one - two segments, and so on. On electronic clo... | Solution: Let's write down the number of segments required to display each digit:
\section*{8723456789
$$
\begin{array}{llllllllll}
6 & 2 & 5 & 5 & 4 & 5 & 6 & 3 & 7 & 6
\end{array}
$$
Consider several types of displayed time:
1) The last digit of the minutes is not 9. Then the next displayed time differs by only t... | 630 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Masha chose a natural number $n$ and wrote down all natural numbers from 1 to 6 n on the board. Then, Masha halved half of these numbers, reduced a third of the numbers by a factor of three, and increased all the remaining numbers by a factor of six. Could the sum of all the resulting numbers match the sum of the or... | Solution: Yes, this could have happened. Let $n=2$, then the sum of all numbers from 1 to 12 is
$$
1+2+3+\ldots+12=78
$$
Suppose Masha halved the numbers 2, 3, 5, 9, 10, 11, reduced the numbers 4, 6, 8, 12 by a factor of three, and increased the numbers 1, 7 by a factor of six. Then the sum of the resulting numbers w... | 78 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In how many ways can the numbers from 1 to 9 be arranged in a $3 \times 3$ table (each number appearing exactly once) such that in each column from top to bottom and in each row from left to right, the numbers are in increasing order? (20 points) | Solution: Let's number the cells of the table as shown in the figure. It is clear that the number 1 is in the upper left cell, and the number 9 is in the lower right cell.
| 1 | $a_{2}$ | $a_{3}$ |
| :---: | :---: | :---: |
| $a_{4}$ | $a_{5}$ | $a_{6}$ |
| $a_{7}$ | $a_{8}$ | 9 |
By the condition, $a_{5}>a_{2}, a_{5... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. How many units are in the number $S=9+99+999+\cdots+\overbrace{9 \ldots 90}^{1000}$? | $$
\begin{aligned}
& S=10-1+100-1+1000-1+\cdots+10^{1000}-1=10\left(1+10+100+\cdots+10^{999}\right)- \\
& -100=10 \frac{10^{1000}-1}{9}-1000=\underbrace{111 \ldots 10}_{999 \text { ones }}-1000 \Rightarrow
\end{aligned}
$$
the number has 998 ones. Answer: $\{998\}$. | 998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $n+S(n)=1964$, where $S(n)$ is the sum of the digits of the number $n$. | 1. We have $S(n) \leq 9 * 4=36 \Rightarrow n \geq 1964-36=1928 \Rightarrow n=1900+10 k+l$, where $2 \leq k \leq 9$ $0 \leq l \leq 9 ; 1900+10 k+l+10+k+l=1964,11 k+2 l=54,2 \leq 2 l \leq 18,36 \leq 11 k \leq 54$ $3 \frac{3}{11} \leq k \leq 4 \frac{10}{11} \Rightarrow k=4, l=5, n=1945$. Answer: $\{1945\}$. | 1945 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The denominator of an irreducible fraction is less than the square of the numerator by one. If 2 is added to both the numerator and the denominator, the value of the fraction will be greater than $\frac{1}{3}$. If 3 is subtracted from both the numerator and the denominator, the fraction will be less than $\frac{1}{1... | 4. Let $\frac{m}{n}$ be the desired irreducible fraction, where $m, n$ are single-digit numbers. According to the problem, we have $n=m^{2}-1, \frac{m+2}{n+2}>\frac{1}{3} \Rightarrow 3 m+6>m^{2}+1, m^{2}-3 m-5<0$, then $\frac{3-\sqrt{29}}{2}<m<\frac{3+\sqrt{29}}{2} \Rightarrow$ $\Rightarrow 0<m<\frac{3+5.4}{2} \Rightar... | 38 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. By the property of absolute value, replacing $x$ with $-x$ does not change this relation. This means that the figure defined by the given inequality is symmetric with respect to the OY axis. Therefore, it is sufficient to find the area of half of the figure for $x \geq 0$. In this case, we obtain the inequality $\le... | Answer: $\{30\}$.
Fig. 2 | 30 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. Since you have to approach each apple and return to the basket, the number of meters walked will be equal to twice the sum of the first hundred numbers, or 101 taken a hundred times, i.e., $10100 \mathrm{M}$. | Answer: $\{10100$ meters $\}$ | 10100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the area of the region defined by the inequality: $|y-| x-2|+| x \mid \leq 4$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. Consider three intervals of change for $x$: 1) $x \leq 0$; 2) $0 \leq x \leq 2$; 3) $x \geq 2$.
1) $x \leq 0$; then $|y+x-2|-x \leq 4, |y+x-2| \leq 4+x \Rightarrow -4 \leq x \leq 0$. Squaring the inequality, we get $(y+x-2)^{2} \leq (4+x)^{2}$, from which $(y-6)(y+2x+2) \leq 0$
2) $0 \leq x \leq 2$, then $|y+x-2| \l... | 32 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. Find all roots of the equation
$1-\frac{x}{1}+\frac{x(x-1)}{2!}-\frac{x(x-1)(x-2)}{3!}+\frac{x(x-1)(x-2)(x-3)}{4!}-\frac{x(x-1)(x-2)(x-3)(x-4)}{5!}+$
$+\frac{x(x-1)(x-2)(x-3)(x-4)(x-5)}{6!}=0 . \quad($ (here $n!=1 \cdot 2 \cdot 3 . . . n)$
In the Answer, indicate the sum of the found roots. | 4. Note that by substituting the numbers $1,2,3,4,5,6$ sequentially into the equation, we will get the equality: $0=0$. This means these numbers are roots of the given equation. Since the equation is of the sixth degree, there are no other roots besides the numbers mentioned above. Answer: $\{21\}$ | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Two acute angles $\alpha$ and $\beta$ satisfy the condition $\operatorname{Sin}^{2} \alpha+\operatorname{Sin}^{2} \beta=\operatorname{Sin}(\alpha+\beta)$. Find the sum of the angles $\alpha+\beta$ in degrees. | 6. Using the formula for $\operatorname{Sin}(\alpha+\beta)$, we get:
$\operatorname{Sin}^{2} \alpha+\operatorname{Sin}^{2} \beta=\operatorname{Sin} \alpha \operatorname{Cos} \beta+\operatorname{Cos} \alpha \operatorname{Sin} \beta \Leftrightarrow \operatorname{Sin} \alpha(\operatorname{Sin} \alpha-\operatorname{Cos} \... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Find $\max x^{2} y^{2} z$ subject to the condition that $x, y, z \geq 0$ and $2 x+3 x y^{2}+2 z=36$. | 7. From the inequality for the arithmetic mean and geometric mean of three numbers, we have:
$\sqrt[3]{2 x \cdot 3 x y^{2} \cdot 2 z} \leq \frac{2 x+3 x y^{2}+2 z}{3}=12 \Rightarrow 3 x y^{2} \cdot 2 z \cdot 2 x \leq 12^{3}, x^{2} y^{2} z \leq 144$
This inequality becomes an equality if the numbers are equal, i.e., $... | 144 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Masha has an integer number of times more toys than Lena, and Lena has the same number of times more toys than Katya. Masha gave Lena 3 toys, and Katya gave Lena 2 toys. After that, the number of toys the girls had formed an arithmetic progression. How many toys did each girl originally have? Indicate the total numb... | 1. Let initially Katya has $a$ toys, then Lena has $k a$, and Masha has $\kappa^{2} a$, where $\kappa-$ is a natural number $\geq 2$. After the changes: Katya has $a-2$, Lena has $a \kappa+5$, and Masha has $a \kappa^{2}-3$. Then these numbers form an arithmetic progression in some order. Let's consider the possible ca... | 105 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Find the minimum value of the sum
$$
\left|x-1^{2}\right|+\left|x-2^{2}\right|+\left|x-3^{2}\right|+\ldots+\left|x-10^{2}\right|
$$ | 4. Grouping the terms, we get $\left(\left|x-1^{2}\right|+\left|x-10^{2}\right|\right)+\left(\left|x-2^{2}\right|+\left|x-9^{2}\right|\right)+\left(\left|x-3^{2}\right|+\left|x-8^{2}\right|\right)+\left(\left|x-4^{2}\right|+\left|x-7^{2}\right|\right)+\left(\left|x-5^{2}\right|+\left|x-6^{2}\right|\right)$. It is known... | 275 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Solve the system
$\left\{\begin{array}{l}a+c=4 \\ a d+b c=5 \\ a c+b+d=8 \\ b d=1\end{array}\right.$
In the answer, write the sum of all solutions of the given system. | 7. We have
$\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=x^{4}+(a+c) x^{3}+(a c+b+d) x^{2}+(a d+b c) x+b d=$
$=x^{4}+4 x^{3}+6 x^{2}+5 x+2=(x+1)(x+2)\left(x^{2}+x+1\right)$. Since the polynomial $x^{2}+x+1$ does not have
real roots, the identity $\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=\left(x^{2}+3 x+2... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 1.
Since the pedestrian covers 1 km in 10 minutes, his speed is 6 km/h. There are more oncoming trams than overtaking ones because, relative to the pedestrian, the speed of the former is greater than that of the latter. If we assume the pedestrian is standing still, the speed of the oncoming trams is the sum of... | Answer: 15 km $/$ hour.
# | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The sequence of polynomials is defined by the conditions:
$$
P_{0}(x)=1, P_{1}(x)=x, P_{n+1}(x)=x P_{n}(x)-P_{n-1}(x), n=1,2, \ldots
$$
How many distinct real roots does the polynomial $P_{2018}(x)$ have? | 3. The sequence of polynomials is defined by the conditions:
$$
P_{0}(x)=1, P_{1}(x)=x, P_{n+1}(x)=x P_{n}(x)-P_{n-1}(x), n=1,2, \ldots
$$
How many distinct real roots does the polynomial $P_{2018}(x)$ have?
$\square$ We will prove by induction that the polynomial $P_{n}(x)=x^{n}+\ldots$ has exactly $n$ distinct rea... | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Misha, over the course of a week, picked an apple each day and weighed it. Each apple weighed a different amount, but the weight of each apple was a whole number of grams and ranged from 221 grams to 230 grams (inclusive). Misha also calculated the average weight of all the apples he picked, and it was always a whol... | Answer: 230 grams.
Solution: Each apple weighed 220 grams plus an integer from 1 to 10. From the numbers 1 to 10, we need to choose 7 numbers such that their sum is divisible by 7. One of these numbers is 5, so we need to choose 6 from the numbers $1,2,3,4,6,7,8,9,10$. The smallest sum of six of these numbers is 23, a... | 230 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Senya thought of two numbers, then subtracted the smaller from the larger, added both numbers and the difference, and got 68. What was the larger of the numbers Senya thought of? | Answer: 34.
Solution. The subtrahend plus the difference equals the minuend. Therefore, the doubled minuend equals 68.
Comment. Correct solution - 20 points. The result is obtained based on examples and the pattern is noticed but not explained - 15 points. The result is obtained based on one example - 10 points. The ... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The little squirrel has several bags of nuts. In two bags, there are 2 nuts each, in three bags, there are 3 nuts each, in four bags, there are 4 nuts each, and in five bags, there are 5 nuts each. Help the little squirrel arrange the bags on two shelves so that there are an equal number of bags and nuts on each she... | Solution. For example, $5+5+5+4+4+2+2=27$ nuts in 7 bags - the first shelf, $5+5+4+4+3+3+3=27$ nuts in 7 bags - the second shelf.
Comment. Correct solution - 20 points. Solution started, some progress made - 5 points. Solution started, but progress insignificant - 1 point. Solution incorrect or absent - 0 points. | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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.
$$
\alpha=\frac{\pi... | Answer: 67.
Solution. Denoting the sides of the triangle by $a, b, c$, using the sine theorem we get $S=\frac{k a+l b+m c}{2}=R(k \sin \alpha+l \sin \beta+m \sin \gamma)$.
Since, in addition, $S=2 R^{2} \sin \alpha \sin \beta \sin \gamma$, we can express $R=\frac{k \sin \alpha+l \sin \beta+m \sin \gamma}{2 \sin \alph... | 67 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2.1. The master's day shift lasts $10 \%$ longer than the apprentice's shift. If the apprentice worked as long as the master, and the master worked as long as the apprentice, they would produce the same number of parts. By what percentage does the master produce more parts per day than the apprentice? | Solution. Let $p=10 \%$. Suppose the productivity of the apprentice is $a$ parts per hour, the master's productivity is $b$ parts per hour, the apprentice's workday is $n$ hours, and the master's workday is $m$ hours. Then from the condition it follows:
$$
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}
$$ | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.3. Find the sum of all integer values of the argument $x$ for which the corresponding values of the function
$$
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
$$
Considering that $a \in\left(\frac{3}{2}, 2\right)$, we get $x \in\left[-2 a-3, \frac{4}{a}+1\right]$. Since $-7<-2 a-3<-6,3<\frac{4}{... | -15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.1. Three pirates, Joe, Bill, and Tom, found a treasure containing 70 identical gold coins, and they want to divide them so that each of them gets at least 10 coins. How many ways are there to do this? | Solution. Let the treasure consist of $n=70$ coins and each pirate should receive no less than $k=10$ coins.
Give each pirate $k-1$ coins, and lay out the remaining $n-3 k+3$ coins in a row. To divide the remaining coins among the pirates, it is sufficient to place two dividers in the $n-3 k+2$ spaces between the coin... | 861 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.2. Three pirates, Joe, Bill, and Tom, found a treasure containing 80 identical gold coins, and they want to divide them so that each of them gets at least 15 coins. How many ways are there to do this? | Solution. Since $n=80, k=15$, it results in $C_{37}^{2}=666$ ways.
Answer: 666. | 666 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.3. Three pirates, Joe, Bill, and Tom, found a treasure containing 100 identical gold coins, and they want to divide them so that each of them gets at least 25 coins. How many ways are there to do this? | Solution. Since $n=100, k=25$, it results in $C_{27}^{2}=351$ ways.
Answer: 351. | 351 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.4. Three pirates, Joe, Bill, and Tom, found a treasure containing 110 identical gold coins, and they want to divide them so that each of them gets at least 30 coins. How many ways are there to do this? | Solution. Since $n=110, k=30$, we get $C_{22}^{2}=231$ ways.
Answer: 231.
Solution. Let $A=\underbrace{11 \ldots 1}_{1007}$. Then
$$
\sqrt{\underbrace{111 \ldots 11}_{2014}-\underbrace{22 ... | 231 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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 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.
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. 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 |
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. 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 |
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 |
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 |
6. Petya formed all possible natural numbers that can be formed from the digits $2,0,1$, 8 (each digit can be used no more than once). Find their sum. Answer: 78331 | Solution: First, consider the units place. Each of the digits 1, 2, 8 appears once in this place for single-digit numbers, twice for two-digit numbers, four times for three-digit numbers, and four times for four-digit numbers - a total of 11 times.
In the tens place, each of them appears 3 times for two-digit numbers,... | 78331 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (5-7,8,9) There are 2014 boxes on the table, some of which contain candies, while the others are empty.
On the first box, it is written: “All boxes are empty.”
On the second - “At least 2013 boxes are empty.”
On the third - “At least 2012 boxes are empty.”
...
On the 2014th - “At least one box is empty.”
It is... | Answer: 1007.
Solution: Suppose that $N$ boxes are empty, then $2014-N$ boxes contain candies. Note that on the box with number $k$, it is written that there are at least $2015-k$ empty boxes. Therefore, the inscriptions on the boxes with numbers $1,2, \ldots, 2014-N$ are false. Consequently, $N=2014-N$, from which $N... | 1007 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. $(5-7,8)$ The phone PIN code consists of 4 digits (and can start with zero, for example, 0951). Petya calls "lucky" those PIN codes where the sum of the outer digits equals the sum of the middle digits, for example $1357: 1+7=3+5$. In his phone, he uses only "lucky" PIN codes. Petya says that even if he forgets one ... | Answer: a) $10 ;$ b) 670 .
Solution: a) Obviously, in the worst case, Petya tries all possible values for one digit and reconstructs the PIN code from them. Thus, no more than 10 combinations need to be tried. Using the example of the PIN code 0099, we can see that if you forget the two middle digits, you need to try ... | 670 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (5-7,8) There are 10 segments, the length of each of which is expressed as an integer not exceeding some $N$. a) Let $N=100$. Provide an example of a set of 10 segments such that no three of them can form a triangle. b) Find the maximum $N$ for which it can be guaranteed that there will be three segments that can fo... | Answer: a) $1,1,2,3,5,8,13,21,34,55 ;$ b) $N=54$.
Solution: a) If each new segment is chosen to be equal to the sum of the two largest of the remaining ones, then it is impossible to form a triangle with its participation.
b) From the previous point, it is clear that for $N=55$ such a sequence can be constructed.
We... | 54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. $(8,9)$ What is the maximum possible area of quadrilateral $A B C D$, the sides of which are $A B=1, B C=8, C D=7$ and $D A=4$? | Answer: 18.
Solution: Note that $1^{2}+8^{2}=7^{2}+4^{2}=65$. With fixed lengths of $A B$ and $B C$, the area of $\triangle A B C$ will be maximized if $\angle A B C=$ $90^{\circ}$. In this case, $A C=\sqrt{65}$, and thus $\angle B C D=$ $90^{\circ}$ as well, and the area of $\triangle B C D$ is also maximized. | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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