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8.5. For a natural number $N$, all its divisors were listed, then the sum of digits for each of these divisors was calculated. It turned out that among these sums, all numbers from 1 to 9 were found. Find the smallest value of $N$. | Answer: 288.
Solution. Note that the number 288 has divisors $1,2,3,4,32,6,16,8,9$. Therefore, this number satisfies the condition of the problem. We will prove that there is no smaller number that satisfies the condition.
Indeed, since $N$ must have a divisor with the sum of digits 9, $N$ is divisible by 9. Now cons... | 288 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.5. On an $8 \times 8$ chessboard, $k$ rooks and $k$ knights are placed such that no figure attacks any other. What is the largest $k$ for which this is possible? | Answer: 5.
Solution: From the condition, it follows that in one row (column) with a rook, no other figure can stand.
Suppose 6 rooks were placed on the board. Then they stand in 6 rows and 6 columns. Therefore, only 4 unpicked cells will remain (located at the intersection of two empty rows and two empty columns). It... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task 7.2
There are 30 logs, the lengths of which are 3 or 4 meters, and their total length is one hundred meters. How many cuts are needed to saw all these logs into pieces 1 meter long? (Each cut saws exactly one log).
Points 7
# | # Answer:
70
## Solution
## First Method
The total length of the logs is 100 meters. If it were a single log, 99 cuts would be needed. Since there are 30 logs, 29 cuts have already been made. Therefore, another $99-29=70$ cuts are needed.
## Second Method
Let's find the number of logs of each type. If all the log... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 5. CONDITION
A right triangle $ABC$ with hypotenuse $AB$ is inscribed in a circle. A point $D$ is taken on the larger leg $BC$ such that $AC = BD$, and point $E$ is the midpoint of the arc $AB$ containing point $C$. Find the angle $DEC$. | Solution. Point $E$ is the midpoint of arc $AB$, so $AE = BE$. Moreover, inscribed angles $CAE$ and $EBD$, subtending the same arc, are equal. Given that $AC = BD$, triangles $ACE$ and $BDE$ are congruent, which implies that angle $CEA$ is equal to angle $BED$. Therefore, angle $DEC$ is equal to angle $BEA$ and both ar... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. The numbers $\sqrt{2}$ and $\sqrt{5}$ are written on the board. You can add to the board the sum, difference, or product of any two different numbers already written on the board. Prove that you can write the number 1 on the board. | Solution: For example, we get $\sqrt{5}-\sqrt{2}$, then $\sqrt{5}+\sqrt{2}$ and $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=5-2=3$, then $\sqrt{2} \cdot \sqrt{5}=\sqrt{10}$, then $\sqrt{10}-3$ and $\sqrt{10}+3$ and finally $(\sqrt{10}-3)(\sqrt{10}+3)=10-9=1$.
Criteria. The goal is achieved if the same numbers are used in ... | 1 | Number Theory | proof | Yes | Yes | olympiads | false |
3. From 80 identical Lego parts, several figures were assembled, with the number of parts used in all figures being different. For the manufacture of the three smallest figures, 14 parts were used, and in the three largest, 43 were used in total. How many figures were assembled? How many parts are in the largest figure... | Answer: 8 figurines, 16 parts.
Solution. Let the number of parts in the figurines be denoted by $a_{1}43$, so $a_{n-2} \leqslant 13$.
Remove the three largest and three smallest figurines. In the remaining figurines, there will be $80-14-$ $43=23$ parts, and each will have between 7 and 12 parts. One figurine is clea... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. The diagonals of the circumscribed trapezoid $A B C D$ with bases $A D$ and $B C$ intersect at point O. The radii of the inscribed circles of triangles $A O D, A O B, B O C$ are 6, 2, and $3 / 2$ respectively. Find the radius of the inscribed circle of triangle $C O D$. | Answer: 3
Solution. We will prove a more general statement, that $\frac{1}{r_{1}}+\frac{1}{r_{3}}=\frac{1}{r_{2}}+\frac{1}{r_{4}}$, where $r_{1}, r_{2}, r_{3}$ and $r_{4}$ are the radii of the inscribed circles of triangles $A O D, A O B, B O C$ and $C O D$ respectively.
Let $A B=a, B C=b, C D=c, A D=d, O A=x, O D=y,... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.1. In the product of three natural numbers, each factor was decreased by 3. Could the product have increased by exactly $2016$?
(N. Agakhanov, I. Bogdanov) | Answer. Yes, it could.
Solution. The product $1 \cdot 1 \cdot 676$ serves as an example. After the specified operation, it becomes $(-2) \cdot(-2) \cdot 673 = 2692 = 676 + 2016$.
Remark. The given example is the only one. Here is how to come up with it. Suppose two of the factors were 1, and the third was $-a$. Their... | 676 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.3. How many rectangles exist on this picture with sides running along the grid lines? (A square is also a rectangle.)
 | Answer: 24.
Solution. In a horizontal strip $1 \times 5$, there are 1 five-cell, 2 four-cell, 3 three-cell, 4 two-cell, and 5 one-cell rectangles. In total, $1+2+3+4+5=15$ rectangles.
In a vertical strip $1 \times 4$, there are 1 four-cell, 2 three-cell, 3 two-cell, and 4 one-cell rectangles. In total, $1+2+3+4=10$ r... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.3. A cuckoo clock is hanging on the wall. When a new hour begins, the cuckoo says "cuckoo" a number of times equal to the number the hour hand points to (for example, at 19:00, "cuckoo" sounds 7 times). One morning, Maxim approached the clock when it was 9:05. He started turning the minute hand until he advan... | Answer: 43.
Solution. The cuckoo will say "cuckoo" from 9:05 to 16:05. At 10:00 it will say "cuckoo" 10 times, at 11:00 - 11 times, at 12:00 - 12 times. At 13:00 (when the hand points to the number 1) "cuckoo" will sound 1 time. Similarly, at 14:00 - 2 times, at 15:00 - 3 times, at 16:00 - 4 times. In total
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. From a $6 \times 6$ grid square, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then,
$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA... | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. Given an isosceles triangle $A B C$, where $A B=A C$ and $\angle A B C=53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $A K$. Point $M$ is chosen such that:
- $B$ and $M$ are on the same side of line $A C$;
- $K M=A B$
- angle $M A K$ is the maximum possible.
How many degrees does angl... | Answer: 44.
Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
It is clear that the number of rhombuses of each orientation will be the same, so let's consider ... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.5. A circle $\omega$ is inscribed in trapezoid $A B C D$, and $L$ is the point of tangency of $\omega$ and side $C D$. It is known that $C L: L D=1: 4$. Find the area of trapezoid $A B C D$, if $B C=9$, $C D=30$.
![](https://cdn.mathpix.com/cropped/2024_05_06_899cf5197845501d962eg-35.jpg?height=305&width=40... | Answer: 972.
Solution. Let's mark the center of the circle $I$, as well as the points of tangency $P, Q, K$ with the sides $B C$, $A D, A B$ respectively. Note that $I P \perp B C, I Q \perp A D$, i.e., points $P, I, Q$ lie on the same line, and $P Q$ is the height of the given trapezoid, equal to the diameter of its ... | 972 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-1. A beginner gardener planted daisies, buttercups, and marguerites on their plot. When they sprouted, it turned out that there were 5 times more daisies than non-daisies, and 5 times fewer buttercups than non-buttercups. What fraction of the sprouted plants are marguerites? | Answer. Zero. They did not germinate.
Solution. Daisies make up $5 / 6$ of all the flowers, and dandelions make up $1/6$. Therefore, their total number equals the total number of flowers.
Criteria. Only the answer - 0 points. Complete solution - 7 points. | 0 | Other | math-word-problem | Yes | Yes | olympiads | false |
8-3. Vika has been recording her grades since the beginning of the year. At the beginning of the second quarter, she received a five, after which the proportion of fives increased by 0.15. After another grade, the proportion of fives increased by another 0.1. How many more fives does she need to get to increase their p... | Answer: 4.
Solution: Let's say Vika had $n$ grades in the first quarter, of which $k$ were fives. Then, after the first five in the second quarter, the proportion of fives increased by $\frac{k+1}{n+1}-\frac{k}{n}=0.15$. Similarly, after the second five, the increase was $\frac{k+2}{n+2}-\frac{k+1}{n+1}=0.1$. Simplify... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A new series of "Kinder Surprises" - chocolate eggs, each containing a toy car - was delivered to the store. The seller told Pete that there are only five different types of cars in the new series, and it is impossible to determine which car is inside by the appearance of the egg. What is the minimum number of "Kind... | Solution. If Petya buys 10 "Kinder Surprises," in the least favorable situation for him, he will get two cars of each type. If he buys 11 "Kinder Surprises," he will get three cars of one type. Let's prove this. Suppose Petya bought 11 "Kinder Surprises" but did not get three cars of one type. This means the number of ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. A $3 \times 3$ table is initially filled with zeros. In one move, any $2 \times 2$ square in the table is selected, and all zeros in it are replaced with crosses, and all crosses with zeros. We will call any arrangement of crosses and zeros in the table a "pattern." How many different patterns can be obtained as a r... | Solution. First method. Different drawings differ by at least one cell in the table, either plus or minus. The symbol in a given specific cell is determined by the parity (even or odd) of the number of swaps performed in each $2 \times 2$ square containing that cell. Each $2 \times 2$ square contains exactly one corner... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Two cyclists are training on a circular stadium. In the first two hours, Ivanov lapped Petrov by 3 laps. Then Ivanov increased his speed by 10 km/h, and as a result, after 3 hours from the start, he lapped Petrov by 7 laps. Find the length of the lap. | 4. Answer: 4 km. Solution: let I be Ivanov's speed (initial), Π be Petrov's speed, K be the length of the circle. Then 2I - 2Π = 3K and 2I - 2(Ι - Π) = 3K and 3(Ι - Π) = 3K - 10, express Ι - Π from both equations and equate, we get 14K - 20 = 9K, from which K = 4.
Criteria: correct solution - 7 points; the system is s... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.2. How many rectangular trapezoids $A B C D$ exist, where angles $A$ and $B$ are right angles, $A D=2$, $C D=B C$, the sides have integer lengths, and the perimeter is less than 100? | Answer: 5.
Solution. Let $C D=B C=a, A B=b$. Drop a perpendicular from point $D$ to $B C$, and apply the Pythagorean theorem: $(a-2)^{2}+b^{2}=a^{2}$. From this, $b^{2}=4(a-1), a=\frac{b^{2}}{4}+1$. Suppose $A D=2$ is the smaller base of the trapezoid. The perimeter $P=2 a+b+2a$, so $b>2$. Considering that $a=\frac{b^... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.5. What is the maximum number of figures consisting of 4 1x1 squares, as shown in the diagram, that can be cut out from a $6 \times 6$ table, if cutting can only be done along the grid lines?
# | # Solution:
Example:
The diagram shows that 8 figures can be cut out.
Evaluation:
. Since $\mathrm{BK} \| \mathrm{AD}$, then $\angle \mathrm{KBE}=\angle \mathrm{DAE}$. Moreover, $\angle \mathrm{KEB}=\angle \mathrm{DEA}$ and $\mathrm{AE}=\mathrm{BE}$, so triangles $\mathrm{BKE}$ and $\mathrm{ADE}$ are congruent. Therefore, $\mathr... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.1. Solve the equation $2 \sin ^{2} x+1=\cos (\sqrt{2} x)$. | Answer: $x=0$. Solution. The left side of the equation $\geq 1$, and the right side $\leq 1$. Therefore, the equation is equivalent to the system: $\sin x=0, \cos \sqrt{2} x=1$. We have: $x=\pi n, \sqrt{2} x=2 \pi k$ ( $n, k-$ integers). From this, $n=k \cdot \sqrt{2}$. Since $\sqrt{2}$ is an irrational number, the las... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. At a round table, 12 people are sitting. Some of them are knights, who always tell the truth, and the rest are liars, who always lie. Each person declared their left neighbor to be a liar. Can we definitely state how many knights and how many liars are at the table? | 1. Answer: Yes, it is possible.
If a knight is sitting in some place, then he told the truth, and to his left should sit a liar. Conversely, if a liar is sitting in some place, then to his left sits the one who was incorrectly called a liar, that is, a knight. This means that knights and liars alternate around the tab... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Find the smallest natural number that ends in 56, is divisible by 56, and has the sum of its digits equal to 56. | 2. Answer: 29899856.
The number has the form $100 A+56$, where $A$ is divisible by 7 and is even. The sum of the digits of $A$ is 45. It is sufficient to indicate the smallest number with these properties. Clearly, it must be at least six digits. The smallest even number with a digit sum of 45 is 199998, but it is not... | 29899856 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.1. The student was given an assignment consisting of 20 problems. For each correctly solved problem, they receive 8 points, for each incorrectly solved problem - minus 5 points, and for a problem they did not attempt - 0 points. The student scored a total of 13 points. How many problems did the student attempt? | # Solution.
Let $x$ be the number of problems solved correctly, $y$ be the number of problems solved incorrectly. Then we get the equation $8x - 5y = 13$. This equation can be solved in two ways.
1st method. Rewrite the equation as $8(x + y) = 13(1 + y)$. We see that the number $x + y$ is divisible by 13. On the othe... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.2. Solve the equation:
$1+\frac{3}{x+3}\left(1+\frac{2}{x+2}\left(1+\frac{1}{x+1}\right)\right)=x$. | # Solution.
$1+\frac{1}{x+1}=\frac{x+2}{x+1}$, therefore the given equation is equivalent to the equation $1+\frac{3}{x+3}\left(1+\frac{2}{x+1}\right)=x$ under the condition that $\mathrm{x} \neq-2$. Proceeding similarly, we get $1+\frac{3}{x+3}=x$, where $\mathrm{x} \neq-2$ and $\mathrm{x} \neq-3$. The roots of this ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.5. Three families of parallel lines have been drawn, with 10 lines in each. What is the maximum number of triangles they can cut out of the plane? | Solution. Consider 100 nodes - the intersection points of lines from the first and second directions. Divide them into 10 sectors: the first sector - nodes lying on the first lines of the first and second directions. The second sector - nodes lying on the second lines (excluding points lying in the first sector) and so... | 150 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Denis housed chameleons that can change color only to two colors: red and brown. Initially, the number of red chameleons was five times the number of brown chameleons. After two brown chameleons turned red, the number of red chameleons became eight times the number of brown chameleons. Find out how many chameleons D... | Solution. Let $t$ be the number of brown chameleons Denis had. Then the number of red chameleons was $5t$. From the problem statement, we get the equation $5 t+2=8(t-2)$. Solving this, we find $t=6$. Therefore, the total number of chameleons is $6 t$, which is 36.
Answer. 36
Recommendations for checking. Only the cor... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. For various positive numbers $a$ and $b$, it is known that
$$
a^{3}-b^{3}=3\left(2 a^{2} b-3 a b^{2}+b^{3}\right)
$$
By how many times is the larger number greater than the smaller one? | Solution. Let's consider and transform the difference:
$$
\begin{aligned}
& 0=a^{3}-b^{3}-3\left(2 a^{2} b-3 a b^{2}+b^{3}\right)= \\
& (a-b)\left(a^{2}+a b+b^{2}\right)-3\left(2 a b(a-b)-b^{2}(a-b)\right)= \\
& (a-b)\left(a^{2}+a b+b^{2}-6 a b+3 b^{2}\right)= \\
& (a-b)\left(a^{2}-5 a b+4 b^{2}\right)=(a-b)(a-4 b)(a-... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Next to the number 2022 on the board, an unknown positive number less than 2022 was written. Then one of the numbers on the board was replaced by their arithmetic mean. This replacement was performed 9 more times, and the arithmetic mean was always an integer. Find the smaller of the numbers originally written on th... | Solution. Let at some point the numbers $a$ and $b$ be written on the board, with $a > b$. Then notice that after the specified operation, the difference between the numbers will become twice as small, regardless of which number we erase, since
$$
a - b = 2\left(a - \frac{a + b}{2}\right) = 2\left(\frac{a + b}{2} - b\... | 998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The numbers from 1 to 2019 were written in a row. Which digit was written more: ones or twos, and by how many? | 2. Answer. There are 990 more ones. Solution. From 1 to 999, their quantity is the same. From 1000 to 1999, there are 1000 more ones. From 2000 to 2009, there are 10 more twos. From 2010 to 2019, it is the same again. In total, $1000-10=990$.
Grading criteria. Full solution - 7 points. In other cases $-\mathbf{0}$ poi... | 990 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The length of the road from the Capital to city O is 2020 kilometers. Along the road are kilometer markers. On the first marker, on the front side, 1 is written, and on the back - 2019; on the second - 2 and 2018 ..., on the last - 2019 and 1. A marker is called good if the two written numbers have a common divisor ... | 3. Answer: 800. Solution. Note that the sum of two numbers on a pillar is 2020. If both numbers are divisible by some common divisor, then $2020=4 * 5 * 101$ is also divisible by this divisor. All even pillars are good, all divisible by 5 are good, all divisible by 101 are good. In total, there are odd pillars $2020 / ... | 800 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. On November 15, a tournament of the game of dodgeball was held. In each game, two teams competed. A win was worth 15 points, a draw 11, and a loss earned no points. Each team played against each other once. At the end of the tournament, the total number of points scored was 1151. How many teams were there? | 6. Answer: 12 teams. Solution. Let there be $\mathrm{N}$ teams. Then the number of games was $\mathrm{N}(\mathrm{N}-1) / 2$. For each game, a total of 15 or 22 points are earned. Therefore, the number of games was no less than $53(1151 / 22)$ and no more than $76(1151 / 15)$.
Note that if there were no more than 10 te... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. How many positive numbers are there among the first 100 terms of the sequence: $\sin 1^{\circ}, \sin 10^{\circ}, \sin 100^{\circ}, \sin 1000^{\circ}, \ldots ?$ | 2. Note that all members of the sequence, starting from $\sin 1000^{\circ}$, are equal to each other, since the difference between the numbers $10^{k+1}$ and $10^{k}$ for natural $k>2$ is a multiple of 360. Indeed, $10^{k+1}-10^{k}=10^{k}(10-1)=9 \cdot 10^{k}=$ $9 \cdot 4 \cdot 10 \cdot 25 \cdot 10^{k-3} \vdots 360$. B... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In triangle $A B C \quad A C=18 \text{ cm}, B C=21$ cm. Point $K$ is the midpoint of side $B C$, and point $M$ is the midpoint of side $A B$, point $N$ lies on side $A C$ and $A N=6 \text{ cm}$. Given that $M N=K N$. Find the length of side $A B$. | 3. Since $M$ and $K$ are midpoints of the sides, we will extend the segments $NM$ and $NK$ beyond the specified points by the same distance and connect the points $L, B, A, N$; as well as $F, B, N, C$. Then the quadrilaterals $ALBN$ and $NBFC$ become parallelograms. Since in a parallelogram the sum of the squares of th... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. We will call a number greater than 25 semi-prime if it is the sum of some two different prime numbers. What is the maximum number of consecutive semi-prime numbers that can be semi-prime? | 5. Note that an odd semiprime number can only be the sum of two and an odd prime number.
Let's show that three consecutive odd numbers $2n+1, 2n+3, 2n+5$, greater than 25, cannot all be semiprimes simultaneously. Assume the opposite. Then we get that the numbers $2n-1, 2n+1, 2n+3$ are prime, and all of them are greate... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Out of four classes, $28\%$ of the students received a "5" for the final math test, "4" - $35\%$, "3" - $25\%$, and "2" - $12\%$. How many students took the test if there are no more than 30 students in each class? | 1. Answer: 100.
From the condition, it follows that the number of schoolchildren must be divisible by 25, 20, and 4. The smallest suitable number is 100, the next one is 200, but it would not work for us since, according to the condition, there cannot be more than 120 people in 4 classes. | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In the office, each computer was connected by wires to exactly 5 other computers. After some computers were infected by a virus, all wires from the infected computers were disconnected (a total of 26 wires were disconnected). Now, each of the uninfected computers is connected by wires to only 3 others. How many comp... | # 5. Answer: 8.
Let $\mathrm{m}$ be the number of infected computers, and $\mathrm{n}$ be the number of uninfected computers. Then, before the infection, there were $5(\mathrm{~m}+\mathrm{n}) / 2$ cables, and after the disconnection, there were $3 \mathrm{n} / 2$ cables (from which it follows that $\mathrm{n}$ is even... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8-1. Kolya made a figure from four identical blocks as shown in the picture. What is the surface area of this figure? Express your answer in square centimeters.
 | Answer: 64.
Solution. The surface area of one block is 18 cm². Out of this area, 2 cm² is "lost" at the joints with other blocks, leaving a total area of $18-2=16$ cm². Since there are 4 blocks, the answer is $4 \cdot 16=64$ cm². | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-3. Olya bought three gifts and packed them in three rectangular boxes: blue, red, and green. She tried to place these gifts in different ways: one on the table, and two on top of each other on the floor. Some distances are given in the diagram. Find the height of the table $h$. Express your answer in centimeters.
, adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.8. Inside a large triangle with a perimeter of 120, several segments were drawn, dividing it into nine smaller triangles, as shown in the figure. It turned out that the perimeters of all nine small triangles are equal to each other. What can they be equal to? List all possible options.
The perimeter of a fig... | Answer: 40.
Solution. Let's add the perimeters of the six small triangles marked in gray in the following figure:
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.7. Anya places pebbles on the sand. First, she placed one stone, then added pebbles to form a pentagon, then made a larger outer pentagon with pebbles, then another outer pentagon, and so on, as shown in the picture. The number of stones she had arranged on the first four pictures: 1, 5, 12, and 22. If she co... | Answer: 145.
Solution. On the second picture, there are 5 stones. To get the third picture from it, you need to add three segments with three stones on each. The corner stones will be counted twice, so the total number of stones in the third picture will be $5+3 \cdot 3-2=12$.
![](https://cdn.mathpix.com/cropped/2024_... | 145 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. In the chat of students from one of the schools, a vote was held: "On which day to hold the disco: October 22 or October 29?"
The graph shows how the votes were distributed an hour after the start of the voting.
Then, 80 more people participated in the voting, voting only for October 22. After that, the ... | Answer: 260.
Solution. Let $x$ be the number of people who voted an hour after the start. From the left chart, it is clear that $0.35 x$ people voted for October 22, and $-0.65 x$ people voted for October 29.
In total, $x+80$ people voted, of which $45\%$ voted for October 29. Since there are still $0.65 x$ of them, ... | 260 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. A line $\ell$ is drawn through vertex $A$ of rectangle $ABCD$, as shown in the figure. Perpendiculars $BX$ and $DY$ are dropped from points $B$ and $D$ to line $\ell$. Find the length of segment $XY$, given that $BX=4$, $DY=10$, and $BC=2AB$.
Fig. 5: to the solution of problem 9.4
Solution. Note that since $\angle Y A D=90^{\circ}-\angle X A B$ (Fig. 5), right triangles $X A B$ and $Y D A$ are similar by the acute an... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. In triangle $ABC$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $AC$, point $K$ is marked, and on side $BC$, points $L$ and $M$ are marked such that $KL=KM$ (point $L$ lies on segment $BM$).
Find the length of segment $LM$, if it is known that $AK=4$, $BL=31$, and $MC=3$.
... | Answer: 14.
Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). Since this height is also a median, then $M H=H L=... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.2. Points $A, B, C, D, E, F, G$ are located clockwise on a circle, as shown in the figure. It is known that $A E$ is the diameter of the circle. Also, it is known that $\angle A B F=81^{\circ}, \angle E D G=76^{\circ}$. How many degrees does the angle $F C G$ measure?
=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.2. A square was cut into five rectangles of equal area, as shown in the figure. The width of one of the rectangles is 5. Find the area of the square.
 | Answer: 400.
Solution. The central rectangle and the rectangle below it have a common horizontal side, and their areas are equal. Therefore, the vertical sides of these rectangles are equal, let's denote them by $x$ (Fig. 13). The vertical side of the lower left rectangle is $2x$, and we will denote its horizontal sid... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.3. In a football tournament, 15 teams participated, each playing against each other exactly once. For a win, 3 points were awarded, for a draw - 1 point, and for a loss - 0 points.
After the tournament ended, it turned out that some 6 teams scored at least $N$ points each. What is the largest integer value ... | # Answer: 34.
Solution. Let's call these 6 teams successful, and the remaining 9 teams unsuccessful. We will call a game between two successful teams an internal game, and a game between a successful and an unsuccessful team an external game.
First, note that for each game, the participating teams collectively earn n... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.8. Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. A point $X$ is chosen on the edge $A_{1} D_{1}$, and a point $Y$ is chosen on the edge $B C$. It is known that $A_{1} X=5, B Y=3, B_{1} C_{1}=14$. The plane $C_{1} X Y$ intersects the ray $D A$ at point $Z$. Find $D Z$.
, if: 1) a pawn cannot be placed on the $e4$ square; 2) no two pawns can stand on squares that are symmetric with respect to the $e4$ square?
Answer: 39 pawns. | Solution. All fields of the board except for the vertical $a$, the horizontal 8, and the field e4 can be divided into pairs that are symmetrical relative to e4. Such pairs form 24. According to the condition, no more than one pawn can be placed on the fields of each pair. In addition, no more than one pawn can be place... | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Borya and Vova are playing the following game on an initially white $8 \times 8$ board. Borya moves first and on each of his turns, he colors any four white cells black. After each of his moves, Vova colors an entire row (row or column) completely white. Borya aims to color as many cells black as possible, while Vov... | Solution. Let Vova make white the row with the most black cells on each of his moves. Then, as soon as Borya achieves a row of no less than four black cells (we will call such a row "rich"), Vova will remove at least four cells, meaning that Borya will not be able to increase the number of black cells compared to his p... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. The password consists of four different digits, the sum of which is 27. How many password options exist | Answer: 72.
Solution. Among the digits of the password, there is a 9. Otherwise, since all digits of the password are different, the maximum sum of the digits will not exceed $8+7+6+5=26$. If there are 9 and 8, then the other two digits are no more than 7, and their sum is 10. There are two possible cases: 7,3 and 6,4... | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Eight cards have the numbers $1,1,2,2,3,3,4,4$ written on them. Can these cards be placed in a row so that there is one card between the ones, two cards between the twos, three cards between the threes, and exactly four cards between the fours? | Answer: Yes.
Solution: For example, this: 41312432.
Instructions for checking:
The score can only be one of two: 0 points or 7 points. | 41312432 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 4. Masha wrote a three-digit number on the board, and Vera wrote the same number next to it, but she swapped the last two digits. After that, Polina added the obtained numbers and got a four-digit sum, the first three digits of which are 195. What is the last digit of this sum? (The answer needs to be justified.)
... | Solution. Let Masha write the number $100 x+10 y+z$. Then Vera wrote the number $100 x+10 z+y$, and the sum of these numbers is $200 x+11 y+11 z$. For $x \leqslant 8$ this expression does not exceed 1798, and therefore cannot start with 195. Thus, $x=9$. Then $11(y+z)$ is a three-digit number starting with 15. Among th... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The function $f(x)$ is such that for all values of $x$, the equality $f(x+1)-f(x)=x+1$ holds. It is known that $f(0)=4$. Find $f(62)$.
# | # Solution.
$$
f(62)-f(61)=61+1=62 \rightarrow f(62)-f(0)=31 * 63=1953
$$
Therefore, $f(62)=1957$. | 1957 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.1.1. Let $A, B, C, D, E, F, G, H$ - be different digits from 0 to 7 - satisfy the equation
$$
\overline{A B C}+\overline{D E}=\overline{F G H}
$$
Find $\overline{D E}$, if $\overline{A B C}=146$.
(The notation $\overline{A B C}$ represents a three-digit number consisting of digits $A, B, C$, similarly cons... | Answer: 57.
Solution. Substitute 146 for $\overline{A B C}$ and write the example in a column:
$$
\begin{array}{r}
146 \\
+\quad D E \\
\hline F G H
\end{array}
$$
$D, E, F, G, H$ can only be the digits $0,2,3,5,7$. Let's consider $E$.
- If $E=0$, then $H=6$ - but this contradicts the fact that $C=6$.
- If $E=2$, t... | 57 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.2.1. In the figure, rays $O A, O B, O C, O D, O E, O F$ are such that:
- $O B$ is the bisector of angle $A O C$
- $O E$ is the bisector of angle $D O F$;
- $\angle A O F=146^{\circ}, \angle C O D=42^{\circ}$.
How many degrees does angle $B O E$ measure?
, there are no fewer balls than in the previous one. It is also known that ther... | Answer: 1 red ball, 3 blue balls.
Solution. Among all the boxes, there can be:
- a maximum of 2 boxes with one ball (one with a red ball, the other with a blue ball),
- a maximum of 3 boxes with two balls (one with two red balls, another with one red and one blue, and the third with two blue balls),
- a maximum of 4 ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7.1. A seven-digit natural number $N$ is called interesting if:
- it consists of non-zero digits;
- it is divisible by 4;
- any number obtained from the number $N$ by permuting its digits is divisible by 4.
How many interesting numbers exist? | Answer: 128.
Solution. We will use the divisibility rule for 4: a number is divisible by 4 if and only if the number formed by its last two digits is divisible by 4.
In an interesting number, the digits can be rearranged in any way, and the divisibility by 4 should remain unchanged. This implies that there are no odd... | 128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.4 Given a triangle $A B C$. Point $P$ is the incenter. Find the angle $B$, if it is known that $R_{A B C}=R_{A P C}$, where $R_{A B C}, R_{A P C}$ are the radii of the circumcircles of triangles $ABC$ and $APC$ respectively. | 10.4. Given a triangle $A B C$. Point $P$ is the incenter. Find the angle $B$, if it is known that $R_{A B C}=R_{A P C}$, where $R_{A B C}, R_{A P C}$ are the radii of the circumcircles of triangles $ABC$ and $APC$ respectively.
Answer: $60^{\circ}$. Hint: Let $\alpha=\angle A B C$. Then $\angle A P C=90^{\circ}+\frac... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. The distance between points A and B is 90 km. At 9:00, a bus left point A for point B at a speed of 60 km/h. Starting from 9:00, every 15 minutes, buses leave point B towards point A at a speed of 80 km/h. The bus that left point A, after traveling 45 km, reduces its speed to 20 km/h due to a breakdown and continues... | 2. A bus leaving point A will travel 45 km in $\frac{45}{60}$ hours. During this time, a bus leaving point B at 9:00 will travel $\frac{45}{60} \cdot 80=60$ km and will be closer to point A than the bus leaving point A. A bus leaving point B at 9:15 will travel $\left(\frac{45}{60}-\frac{15}{60}\right) \cdot 80=40$ km ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In the country, there are 20 cities. An airline wants to organize two-way flights between them so that from any city, it is possible to reach any other city with no more than $\mathrm{k}$ transfers. At the same time, the number of air routes should not exceed four. What is the smallest $\mathrm{k}$ for which this is... | 5. $\mathrm{k}=2$. At least two transfers will be required. From an arbitrary city A, one can reach no more than four cities without a transfer, and with one transfer - no more than $4 \times 3=12$ cities. That is, if using no more than one transfer, one can fly to no more than 16 other cities, but 19 are required. | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3.1. Semyon was solving the quadratic equation $4 x^{2}+b x+c=0$ and found that its two roots are the numbers $\operatorname{tg} \alpha$ and $3 \operatorname{ctg} \alpha$ for some $\alpha$. Find $c$. | Answer: 12
Solution. By Vieta's theorem, $c / 4$ equals the product of the roots. Considering that $\operatorname{tg} \alpha \cdot \operatorname{ctg} \alpha=1$, we get $c / 4=3$. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. Given an arithmetic progression $a_{1}, a_{2}, \ldots, a_{100}$. It is known that $a_{3}=9.5$, and the common difference of the progression $d=0.6$. Find the sum $\left\{a_{1}\right\}+\left\{a_{2}\right\}+\ldots+\left\{a_{100}\right\}$. The notation $\{x\}$ represents the fractional part of the number $x$, i.e., t... | Answer: 50
Solution. Let's look at the first digits after the decimal point in the decimal representation of the terms of the progression. Since $d=0.6$ and $a_{3}=9.5$, the sequence of these digits is: $3,9,5,1,7,3,9,5,1,7, \ldots$ It is periodic with a period of 5. Then the sum of the fractional parts of any five co... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. A random number sensor outputs a number $a$ - one of the natural numbers $1,2, \ldots, 100$ (with equal probability). For this value of $a$, we find the maximum possible value $M$ of the function
$$
f(x)=\frac{700}{x^{2}-2 x+2 a}
$$
The probability that $M>10$ is $n$ percent. What is $n$? | Answer: 35
Solution. The minimum value of the quadratic trinomial $x^{2}-2 x+2 a$ is $2 a-1$. Since for the considered values of $a$ we have $2 a-1>0$ (which is true for $a>1 / 2$), the maximum value of $f(x)$ is $M=\frac{700}{2 a-1}$. Next, the condition $M>10$ (for $2 a-1>0$) is equivalent to the following condition... | 35 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.1. In a row, the numbers are written: $100^{100}, 101^{101}, 102^{102}, \ldots, 876^{876}$ (i.e., the numbers of the form $n^{n}$ for natural numbers n from 100 to 876.) How many of the written numbers are perfect cubes? (A perfect cube is the cube of an integer.) | # Answer: 262
Solution. Consider a number of the form $m^{k}$, where $m$ and $k$ are natural numbers. If $k$ is divisible by 3, then $m^{k}$ is a perfect cube. Otherwise, $m^{k}$ is a perfect cube if and only if $m$ is a perfect cube. Thus, the answer to our problem is the total number of numbers that are divisible by... | 262 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. In a triangular pyramid $A B C D$, it is known that: $A B=C D=6, A D=B C=10, \angle A B C=120^{\circ}$. Find $R^{2}$, where $R$ is the radius of the smallest sphere that can contain such a pyramid. | Answer: 49
Solution. Since the segment $AC$ fits inside the sphere, $2R \geqslant AC$. On the other hand, the sphere constructed with $AC$ as its diameter covers both the triangle $ABC$, since $\angle ABC > 90^{\circ}$, and the congruent (by three sides) triangle $BAD$, and thus covers the entire tetrahedron. Therefor... | 49 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.1. On the coordinate plane, a triangle $O A B$ is drawn, where the point of intersection of the medians is at $\left(\frac{19}{3}, \frac{11}{3}\right)$, and points $A$ and $B$ have natural coordinates. Find the number of such triangles. (Here, $O$ denotes the origin - the point $(0,0)$; two triangles with the same se... | Answer: 90.
Let $M$ be the midpoint of $AB$. Then, by the property of the median, $OM = \frac{3}{2} OG$, where $G$ is the centroid. Therefore, $M$ has coordinates $\left(\frac{19}{2}, \frac{11}{2}\right)$.
:
- X202020;
- 2 X 02020;
- $20 \times 2020$
- $202 \times 020$
- $2020 \times 20$
- $20202 \mathrm{X} 0$
- $202020 \mathrm{X}$.
Case 1. If the first... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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