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# 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 |
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
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. From a $6 \times 6$ grid square, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then,
$$
\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 |
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. Point E is the midpoint of side AB of parallelogram ABCD. On segment DE, there is a point F such that $\mathrm{AD}=\mathbf{B F}$. Find the measure of angle CFD. | Solution: Extend DE until it intersects line BC at point K (see figure). 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 |
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 |
10.5. 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 |
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 |
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 |
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 |
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.
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 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?
 | 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 |
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}
$$
$D, E, F, G, H$ can only be the digits $0,2,3,5,7$. Let's consider $E$.
- If $E=0$, then $H=6$ - but this contradicts the fact that $C=6$.
- If $E=2$, t... | 57 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.2.1. In the figure, rays $O A, O B, O C, O D, O E, O F$ are such that:
- $O B$ is the bisector of angle $A O C$
- $O E$ is the bisector of angle $D O F$;
- $\angle A O F=146^{\circ}, \angle C O D=42^{\circ}$.
How many degrees does angle $B O E$ measure?
. For this value of $a$, we find the maximum possible value $M$ of the function
$$
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 |
Problem 6.7. The Amur and Bengal tigers started running in a circle at 12:00, each at their own constant speed. By 14:00, the Amur tiger had run 6 more laps than the Bengal tiger. Then the Amur tiger increased its speed by 10 km/h, and by 15:00, it had run a total of 17 more laps than the Bengal tiger. How many meters ... | Answer: 1250.
Solution. In the first 2 hours, the Amur tiger ran 6 more laps, i.e., in 1 hour, it ran 3 more laps. If it had not increased its speed, in the first 3 hours, it would have run 9 more laps. However, the increase in speed resulted in an additional $17-9=8$ laps in the third hour. Since it increased its spe... | 1250 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.2 There were 25 sparrows sitting on two bushes. After 5 sparrows flew from the first bush to the second, and 7 sparrows flew away from the second bush, it turned out that there were twice as many sparrows left on the first bush as on the second. How many sparrows were there on each bush initially? | 7.2 On the first bush, there were 17 sparrows, and on the second, there were 8.
After 7 sparrows flew away, 18 remained. At this point, there were twice as many sparrows on the first bush as on the second. This means there were 12 sparrows on the first bush and 6 on the second. If we return 7 sparrows to the second bu... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.4. What angle do the hour and minute hands form at twenty minutes past one? (Don't forget to justify your answer). | 7.4 $110^{\circ}$.
At 12:00, the angle between the hour and minute hands is $0^{\circ}$. The minute hand completes a full circle of $360^{\circ}$ in 60 minutes, which is $6^{\circ}$ per minute, and in 20 minutes it will cover $120^{\circ}$. The hour hand moves 12 times slower than the minute hand. Therefore, in 20 min... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.5. Find all three-digit numbers $\mathrm{N}$ such that the sum of the digits of the number $\mathrm{N}$ is 11 times smaller than the number $\mathrm{N}$ itself (do not forget to justify your answer). | 7.5. The number 198 is unique.
From the condition, we get the relation $11 \cdot(a+b+c)=100 a+10 b+c$, or $10 c+b=89 a$. In this relation, the left side is a number less than 100. If $a$ is greater than 1, then the right side will be a number greater than 100. Therefore, $a=1$, $c=8$, $b=9$. All-Russian School Olympia... | 198 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.7. A green chameleon always tells the truth, while a brown chameleon lies and immediately turns green after lying. In a company of 2019 chameleons (green and brown), each in turn answered the question of how many of them are currently green. The answers were the numbers $1,2,3, \ldots, 2019$ (in some order, not neces... | Answer: 1010.
Solution. Consider two chameleons who spoke in a row. One of them was brown at the moment of speaking; indeed, if both were green, the number of green chameleons would not have changed after the first one spoke, and the second one would have named the same number as the first. We can divide all the chame... | 1010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.8. In an acute-angled triangle $A B C$, the bisector $B L$ is drawn. The circumcircle of triangle $A B L$ intersects side $B C$ at point $D$. It turns out that point $S$, symmetric to point $C$ with respect to the line $D L$, lies on side $A B$ and does not coincide with its endpoints. What values can $\angle A B C$ ... | Answer: $60^{\circ}$.
First solution. From the symmetry, triangles $C L D$ and $S L D$ are equal, so $D S = D C$, $\angle C D L = \angle S D L$, and $\angle D L C = \angle D L S$. Since the quadrilateral $A L D B$ is inscribed in a circle, we have $\angle B A L = \angle L D C$ (see Fig. 1). The chords $A L$ and $D L$ ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Variant 1 Petya wrote 9 different positive integers on the board. It turned out that the arithmetic mean of these numbers is 16. What is the greatest value that the largest number on the board can take | Answer: 108.
Solution: The sum of the given numbers is $9 \cdot 16=144$. Since all the numbers are distinct, the sum of the 8 smallest of them is no less than $1+2+\cdots+8=36$. Therefore, the largest number cannot be greater than $144-36=108$. This is possible: $(1+2+\cdots+8+108): 9=16$.
Variant 2 Petya wrote 9 dif... | 108 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 7. Option 1.
Unit cubes were used to assemble a large parallelepiped with sides greater than 3. Two cubes will be called adjacent if they touch by faces. Thus, one cube can have up to 6 neighbors. It is known that the number of cubes that have exactly 6 neighbors is 429. Find the number of cubes that have exactly 4 ... | Answer: 108.
Solution: Let $a, b$ and $c$ be the lengths of the sides of the large parallelepiped. Then, the number of cubes with exactly 6 neighbors is: $(a-2)(b-2)(c-2)$. Since each of the factors $a-2, b-2$, and $c-2$ is greater than 1 and their product equals the product of the three prime numbers 3, 11, and 13, t... | 108 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.5. The figure shows 4 squares. It is known that the length of segment $A B$ is 11, the length of segment $F E$ is 13, and the length of segment $C D$ is 5. What is the length of segment $G H$?
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
. Therefore, it must contain the num... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
 | Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =... | 103 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers located inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
We get 8 rectangles $1 \... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Find all positive solutions of the equation
$$
x^{101} + 100^{99} = x^{99} + 100^{101}
$$
Do not forget to justify your answer. | Solution. The equation is reduced to the form
$$
x^{99}\left(x^{2}-1\right)=100^{99}\left(100^{2}-1\right)
$$
Obviously, the solution is $x=100$. For $0<x<1$, the left side is less than the right side. For $x>1$, the left side is an increasing function, so the required value is taken no more than once. Therefore, the... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the box, there are balls of seven colors. One tenth of the balls are red, one eighth are orange, and one third are yellow. There are 9 more green balls than red ones, and 10 more blue balls than orange ones. There are 8 blue balls in the box. The remaining balls are purple. What is the smallest possible number of... | Answer: 25 balls.
Solution. Let the total number of balls be $x$, and the number of violet balls be $y$. Then
$$
\frac{x}{10}+\frac{x}{8}+\frac{x}{3}+\frac{x}{10}+9+\frac{x}{8}+10+8+y=x
$$
from which $\frac{47 x}{60}+27+y=x$, that is, $y=\frac{13 x}{60}-27$, and the smallest value of $y$ is achieved at the smallest ... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Vanya and Petya decided to mow the lawn for a football field. Vanya could do it alone in 5 hours, and Petya in 6 hours. They started at 11 o'clock, and stopped mowing simultaneously when their parents called them, but Petya took a one-hour lunch break, and Vanya's lunch break lasted two hours. One tenth of the lawn ... | Answer: At 15 o'clock.
Solution: Let $t$ be the time from 11 o'clock to the end, then Petya worked $(t-1)$ hours, and Vanya worked $(t-2)$ hours. In one hour, Petya completed $\frac{1}{6}$ of the work, and Vanya completed $\frac{1}{5}$ of the work. We get the equation: $\frac{1}{6} \cdot(t-1)+\frac{1}{5} \cdot (t-2)=\... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A company produces a lemon drink by diluting lemon juice with water. Initially, the company produced a drink containing $15\%$ lemon juice. After some time, the CEO ordered to reduce the lemon juice content to $10\%$. By what percentage will the amount of lemon drink produced increase with the same volumes of lemon ... | Answer: By $50 \%$.
Solution. Method 1. The content of lemon juice in the drink after the CEO's directive decreased by one and a half times. This means that from the same lemons, one and a half times more lemon drink can be made. In other words, the amount of lemon drink produced will increase by one and a half times ... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On a standard graph paper, an angle is drawn (see figure). Find its measure without using measuring instruments. Justify your answer. | Answer: $45^{\circ}$.
Solution. Connect the two "extreme" points with a segment (as shown in the figure). The resulting triangle is isosceles because two of its sides, $A B$ and $B C$, are diagonals of three-cell rectangles. Diagonal $A B$ divides the angle of the rectangle at vertex $B$ into two angles that complemen... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.3. Numbers $1,2,3,4,5,6,7$ and 8 are placed at the vertices of a cube, such that the sum of any three numbers belonging to any face is not less than 10. Find the minimum possible sum of four numbers belonging to one face. | Answer: 16. Evaluation. Consider an arbitrary face. If the largest number written at the vertex of this face is no more than 5, then the sum of the remaining numbers is no more than $4+3+2=9$. Therefore, the largest number written at the vertex of any face is at least 6, and the minimum possible sum of four numbers bel... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Given a two-digit number, where the first digit is greater than the second. Its digits were rearranged, and a new two-digit number was obtained. The difference between the first and the second number turned out to be a perfect square. How many possible two-digit numbers satisfy this condition? | 1. Answer: 13 numbers.
Let the first number be $10a + b$. Then the second number is $10b + a$, where $a > b > 0$. The difference is $9(a - b)$, which implies that the difference $a - b$ is a perfect square. The difference can be 1 or 4. For the first case, we have 8 values for the larger number $(21, 32, \ldots, 98)$,... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. On a plane, there are 10 different points. We considered the midpoints of all segments connecting all pairs of points. What is the smallest number of midpoints that could have arisen? | 4. Answer: 17 midpoints.
Consider the case when the points lie on a single line. Introduce coordinates on it, and let the points be $a_{1}<\cdots<a_{10}$. Consider the midpoints of segments involving $a_{1}$. There are nine of them, and they form an increasing sequence $\frac{a_{1}+a_{2}}{2}<\frac{a_{1}+a_{3}}{2}<\cdo... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. There is a $10 \times 10$ grid for playing "Battleship". In this case, the game follows unusual rules, and only ships of size $1 \times 2$ are allowed, which can be placed both horizontally and vertically. What is the maximum number of these ships that can be placed on such a field, if the ships must not extend beyo... | 5. Answer: 13.
One of the examples of placing 13 ships is shown in the figure:
| | | | | | | | | | |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| | $\mathrm{x}$ | $\mathrm{x}$ | | $\mathrm{x}$ | $\mathrm{x}$ | | $\mathrm{x}$ | $\mathrm{x}$ | |
| | | | | | | | | | |... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.5. On each of ten cards, a real number is written. For each non-empty set of these cards, the sum of all numbers written on the cards in this set was found. It is known that not all of the obtained sums are integers. What is the maximum possible number of integer sums that could have resulted? | Answer: 511.
Solution: It is clear that at least one of the cards has a non-integer number, otherwise the sum of the numbers in any set would be an integer. We select one of these cards and call it $a$, and the number on it $-x$. For each set $C$ that does not contain $a$ (including the empty set), we associate the se... | 511 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.1. In the store, Sasha bought two pens, three notebooks, and one pencil and paid 33 rubles, Dima bought one pen, one notebook, and two pencils and paid 20 rubles. How much did Tanya pay for four pens, five notebooks, and five pencils? | Answer: 73 rubles.
Let $a, b, c$ be the price of a pen, a notebook, and a pencil, respectively. Then, according to the condition, we have $2a + 3b + c = 33$ and $a + b + 2c = 20$. Multiplying the second relation by 2 and adding it to the first equation, we get the answer. | 73 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.5. The figure shows 4 squares. It is known that the length of segment $A B$ is 11, the length of segment $F E$ is 13, and the length of segment $C D$ is 5. What is the length of segment $G H$?
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
. Therefore, it must contain the num... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
 | Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =... | 103 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers located inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
We get 8 rectangles $1 \t... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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