problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
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 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | 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 inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
This results in 8 rectan... | 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 |
8.3. On the side $AC$ of triangle $ABC$, a point $M$ is taken. It turns out that $AM = BM + MC$ and $\angle BMA = \angle MBC + \angle BAC$. Find $\angle BMA$. | Answer: $60^{\circ}$. Solution. First, we show that triangle $A B C$ is isosceles. Indeed, this follows from the condition $\angle B M A=\angle M B C+\angle B A C$ and the property of the exterior angle: $\angle B M A=\angle M B C+\angle B C A$. From these two equalities, we have $\angle B C A=\angle B A C$, and thus, ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) The first term of the sequence is 934. Each subsequent term is equal to the sum of the digits of the previous term, multiplied by 13. Find the $2019-$th term of the sequence. | Solution. Let's find the first few terms of the sequence:
$a_{1}=934 ; a_{2}=16 \cdot 13=208 ; a_{3}=10 \cdot 13=130 ; a_{4}=4 \cdot 13=52 ;$
$a_{5}=7 \cdot 13=91 ; a_{6}=10 \cdot 13=130=a_{3}$.
Since each subsequent number is calculated using only the previous number, the terms of the sequence will repeat with a pe... | 130 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) The function $f$ is such that for any $x>0, y>0$ the equality $f(x y)=f(x)+f(y)$ holds. Find $f(2019)$, if $f\left(\frac{1}{2019}\right)=1$. | Solution. When $y=1 \quad f(x)=f(x)+f(1), f(1)=0$.
When $x=2019 \quad y=\frac{1}{2019} f(1)=f(2019)+f\left(\frac{1}{2019}\right)$,
$f(2019)=f(1)-f\left(\frac{1}{2019}\right)$
$f(2019)=-f\left(\frac{1}{2019}\right)=-1$.
Answer. $f(2019)=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) On side $AB$ of triangle $ABC$, a point $K$ is marked, and on side $AC$ - a point $M$. Segments $BM$ and $CK$ intersect at point $P$. It turns out that angles $APB$, $BPC$, and $CPA$ are each $120^{\circ}$, and the area of quadrilateral $AKPM$ is equal to the area of triangle $BPC$. Find the angle $BAC$.
... | # Solution.
Add the area of triangle $B P K$ to both sides of the equality $S_{\text {AKPM }}=S_{\text {BPC }}$:
$$
\begin{aligned}
& S_{A K P M}+S_{B P K}=S_{B P C}+S_{B P K}, \\
& S_{A B ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1 Four princesses each thought of a two-digit number, and Ivan thought of a four-digit number. After they wrote down their numbers in a row in some order, the result was 132040530321. Find Ivan's number.
Answer: 5303 | Solution: Let's go through the options. Option 1320 is not suitable because the remaining part of the long number is divided into segments of two adjacent digits: $40,53,03,21$, and the segment 03 is impossible as it is not a two-digit number. Option 3204 is impossible due to the invalid segment 05 (or the segment 1 fr... | 5303 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.3 It is known that for some natural number $n$ each of the numbers $3 n-1$ and $n-10$ is divisible by a prime number p. Find the number p. | Answer: 29
Solution. The number $3 n-1-3(n-10)=29$ is divisible by p. Since $29$ is a prime number, then $\mathrm{p}=29$. | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.4 In the class, there are more than 30 people but less than 40. Any boy is friends with three girls, and any girl is friends with five boys. How many people are in the class? | Answer: 32
Solution. Let $\mathrm{m}$ be the number of boys, $\mathrm{d}$ be the number of girls, and $\mathrm{r}$ be the number of friendly pairs "boy-girl". According to the problem, $\mathrm{r}=3 \mathrm{~m}$ and $\mathrm{r}=5 \mathrm{~d}$. Therefore, r is divisible by 3 and 5, and thus by 15: $\mathrm{r}=15 \mathr... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.1. In the first grade, there are more than 10 but fewer than 15 children. On New Year's morning, Father Frost came to them with a bag of candies and distributed them equally to everyone, and found that he had 9 candies left. How many candies did each child receive if there were initially 207 candies in the bag? | Solution. 207-9=198 candies were given out by Father Frost. 198=18$\cdot$11 - factorization taking into account that one of the factors meets the condition of being between 10 and 15. Therefore, 11 children received 18 candies each.
Answer: 18 candies.
## Criteria:
7 points - complete solution;
6 points - factoriza... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2. Find the GCD of all numbers obtained by all possible permutations of the digits of the number 202120222023
Solution. By the divisibility rule, all these numbers are divisible by 9 (the sum of the digits is 18). A sufficient condition to prove that there are no other numbers is that the difference between any two ... | Answer: 9.
## Criteria
7 points - complete solution;
3 points - GCD found, but not proven that there can be no other divisors. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. On December 31, 2011, Yevgeny Alexandrovich's age coincided with the sum of the digits of his birth year. How old was Yevgeny Alexandrovich on December 31, 2014? Prove the uniqueness of the answer.
Solution. The maximum sum of the digits of the birth year can be $1+9+9+9=28$, and the minimum - 2. Therefore, Y.A. ... | Answer: 23 years.
## Criteria:
7 points - complete solution;
4 points - equations are correctly set up;
3 points - the earliest and latest year of birth are estimated;
2 points - noted that in each decade there is no more than one suitable year. | 23 | 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
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. From a $6 \times 6$ square grid, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
.
For convenience, let's introduce some notations. The top-left corner cell of the $5 \times 5$ table will be called $A$, and the bottom-right corner c... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\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 $B C=9$, $C D=30$.
$\angle KFC$ is the angle between the chord and the tangent, which is equal to half the arc $FC$, just like the inscribed angle $\angle FDC$ ... | 61 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.1. Among fifty consecutive natural numbers, exactly 8 are divisible by 7 without a remainder. What is the remainder when the eleventh number in the sequence is divided by 7?
# | # Answer: 3
Solution. The closest numbers that give the same remainder when divided by 7 differ by 7. Therefore, among fifty consecutive numbers, 6 groups of numbers with the same remainder are formed, each containing 7 numbers, and only one group containing 8 numbers, with the first and last numbers among them. Thus,... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.1. Four chess players - Ivanov, Petrov, Vasiliev, and Kuznetsov - played a round-robin tournament (each played one game against each other). A win earns 1 point, a draw earns 0.5 points each. It turned out that the player who took first place had 3 points, and the player who took last place had 0.5 points. How many v... | Answer: 36
Solution. First, let's determine what quantities of points the four participants can have under the given conditions. Since the one who took first place has 3 points, he won against everyone, and the one who took second place could not have scored more than 2, as he lost to the first. There were a total of ... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.1. People stand in a circle - liars who always lie, and ryiars who always tell the truth. And each of them said that among the people standing next to them, there are as many liars as ryiars. How many people are there in total if there are 48 ryiars? | Answer: 72
Solution. Let's denote a knight as $\mathrm{P}$, and a liar as L. Notice that each knight stands between a knight and a liar, otherwise he would have told a lie. Therefore, knights stand in groups of two. Liars cannot stand in groups larger than one, as in such a case, the liar standing at the edge of the g... | 72 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. A merchant bought several bags of salt in Tver and sold them in Moscow with a profit of 100 rubles. With all the money earned, he again bought salt in Tver (at the Tver price) and sold it in Moscow (at the Moscow price). This time the profit was 120 rubles. How much money did he spend on the first purchase? | Solution. Let the merchant pay $x$ rubles for the salt during the first purchase in Tver. Then he sold it in Moscow for $x+100$ rubles. The second time, he spent $x+100$ rubles in Tver and received $x+100+120=x+220$ rubles in Moscow. Since the ratio of Moscow and Tver prices did not change, we can set up the proportion... | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In a triangle, two medians are perpendicular to each other, their lengths are 18 cm and 24 cm. Calculate the area of the triangle. | Solution. $S=\frac{1}{2} A C \cdot B M$ (see the figure). By the property of medians, we have $A O=\frac{2}{3} \cdot A E=16, O C=\frac{2}{3} \cdot C D=12$. From $\triangle A O C$, we find $A C=\sqrt{A E^{2}+O C^{2}}=20$. Let $D F \perp A C, O K \perp A C$,
![](https://cdn.mathpix.com/cropped/2024_05_06_b4606a2fddb2d21... | 288 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. We will build a sequence of numbers in the following way. We will place the number 7 in the first position, and after each number, we will place the sum of the digits of its square, increased by one. For example, the second position will hold the number 14, since $7^{2}=49$, and $4+9+1=14$. The third position will h... | Solution Let's continue finding the first few terms of the sequence: $7 ; 14 ; 17 ; 20 ; 5 ; 8 ; 11 ; 5 ; \ldots$ - the number 5 is repeated. This means the sequence has a period of length 3: the numbers 5; 8; 11 will repeat subsequently. The number 8 is in the sixth position, so for any $\mathrm{k}>0$, the number 8 wi... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Thirty students from five classes came up with 40 tasks, and students from one class came up with the same number of tasks, while students from different classes came up with a different number. How many students came up with one task? | Solution: We will choose 5 students, one from each class. Each of them came up with a different number of problems. Therefore, the total number of problems they came up with is no less than $1+2+3+4+5=$ 15. The remaining 25 students came up with no more than $40-15=25$ problems. It is clear that each of them came up wi... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.5. Hooligan Dima made a construction in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the figure.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire construc... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
, 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 $A B C$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $A C$, point $K$ is marked, and on side $B C$, points $L$ and $M$ are marked such that $K L=K M$ (point $L$ lies on segment $B M$).
Find the length of segment $L M$, if it is known that $A K=4, B L=31, M C=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 greatest 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 $D Z$.
Each small arc between adjacent vertices is $\frac{1}{\mathrm{n}} \cdot 360^{\circ}$. The angle $\mathrm{ACD}$ is an inscribed angle and s... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.3 Let a, b, c be non-negative integers such that $28a + 30b + 31c = 365$. Prove that $a + b + c = 12$. | Solution. If it were $\mathrm{a}+\mathrm{b}+\mathrm{c} \leq 11$, then we would have $28 \mathrm{a}+30 \mathrm{~b}+31 \mathrm{c} \leq 31(\mathrm{a}+\mathrm{b}+\mathrm{c}) \leq 31 \cdot 11 < 365$. Therefore, $\mathrm{a}+\mathrm{b}+\mathrm{c} \geq 12$. We will show that the inequality $\mathrm{a}+\mathrm{b}+\mathrm{c} \ge... | 12 | Number Theory | proof | Yes | Yes | olympiads | false |
Problem 7.1. In class 7 "B", there are more than 3 but less than 15 children. On New Year's, Santa Claus came to them with a bag containing 195 candies. After distributing the candies equally among all the children in the class, Santa Claus found that there were 8 candies left in the bag. How many candies did each chil... | Answer: 17.
Solution. The children received $195-8=187=11 \cdot 17$ candies. This number must be divisible by the number of children in the class, which is more than 3 and less than 15. Therefore, there are 11 children in the class, and each of them received $187: 11=17$ candies. | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.3. In the example with fractions, some two-digit natural numbers have been replaced by the letters $A$ and $B$.
$$
\frac{A-5}{A}+\frac{4}{B}=1
$$
(a) (2 points) What is the smallest value that $A$ can take?
(b) (2 points) What is the largest value that $B$ can take? | # Answer:
(a) (2 points) 15.
(b) (2 points) 76.
Solution. According to the condition,
\[
1=\frac{A-5}{A}+\frac{4}{B}=1-\frac{5}{A}+\frac{4}{B}
\]
from which we obtain \(\frac{A}{5}=\frac{B}{4}\) and \(4A=5B\). It follows that for some integer \(k\), the equalities \(A=5k\) and \(B=4k\) hold.
Since \(B=4k \geqslan... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. The figure shows 5 lines intersecting at one point. One of the resulting angles is $34^{\circ}$. How many degrees is the sum of the four angles shaded in gray?
 | Answer: $146^{\circ}$.
Solution. Replace the two "upper" gray angles with their equal vertical angles, as shown in the picture:
Now it is clear that the gray angles, together with the $34^{\... | 146 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. On an island, there live knights who always tell the truth, and liars who always lie. One day, 35 inhabitants of the island sat down at 7 tables, with 5 people at each table. Each of these 35 inhabitants was asked: "Are there more than three tables where at least 3 knights are sitting?"
(a) (1 point) What... | Answer:
(a) (1 point) 35.
(b) (3 points) 23.
Solution. (a) It is clear that all 35 people could have answered "Yes" if they were knights. More than 35 could not be.
(b) Consider two cases.
First case. The statement "There are more than three tables where at least 3 knights are sitting" is true.
Then there are at l... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.8. In a chess tournament, 30 chess players participated, each playing against each other exactly once. A win was awarded 1 point, a draw - $1 / 2$, and a loss - 0. What is the maximum number of chess players who could have ended up with exactly 5 points by the end of the tournament? | Answer: 11.
Solution. Let $N-$ be the number of chess players with 5 points.
Assume $N \geqslant 12$. In each game between two playing chess players, 1 point is played, so the sum of points of all chess players who scored 5 points at the end of the tournament is no less than $\frac{N \cdot(N-1)}{2}$. Then, by the pig... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.3. In each of the three chests, Ali-Baba found gold and silver coins; in total, there were 40 gold and 40 silver coins. In the first chest, there were 7 more gold coins than silver coins, in the second chest, there were 15 fewer silver coins than gold coins. Which type of coin is more in the third chest and by how ma... | Answer: There are 22 more silver coins.
Solution. In the first two chests, the total number of gold coins is $7+15=22$ more than the total number of silver coins. Since initially there were an equal number of gold and silver coins, there are 22 fewer gold coins than silver coins in the third chest.
## Evaluation crit... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.4. The frame for three square photographs has the same width everywhere (see figure). The perimeter of one opening is 60 cm, the perimeter of the entire frame is 180 cm. What is the width of the frame? | Answer: 5 cm.
Solution. From the condition of the problem, it follows that the length of one hole's side is 15 cm. Let $d$ cm be the width of the frame, then the perimeter of the rectangle is $8 \cdot 15 + 12 d = 120 + 12 d$ (cm). According to the condition, $120 + 12 d = 180$, which means $d = 5$.
Similar reasoning ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.5. The arithmetic mean of four numbers is 10. If one of these numbers is erased, the arithmetic mean of the remaining three increases by 1; if instead another number is erased, the arithmetic mean of the remaining numbers increases by 2; and if only the third number is erased, the arithmetic mean of the remaining inc... | Answer: will decrease by 6.
Solution. From the fact that the arithmetic mean of four numbers is 10, it follows that the sum of these numbers is 40. Similarly, the sum of three numbers (excluding the first) is 33, the sum of three numbers (excluding the second) is 36, and the sum of three numbers (excluding the third) ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.6. What is the smallest natural number $a$ for which there exist integers $b$ and $c$ such that the quadratic trinomial $a x^{2}+b x+c$ has two distinct positive roots, each not exceeding $\frac{1}{1000} ?$
(A. Khryabrov) | Answer. $a=1001000$.
First solution. We will prove that $a \geqslant 1001000$. Notice that if $y$ is a root of the quadratic polynomial $a x^{2}+b x+c$, then $1 / y$ is a root of the quadratic polynomial $c x^{2}+b x+a$. Therefore, in the problem, we need to find the smallest natural $a$ for which the roots $x_{1}$ an... | 1001000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) 45 candies cost as many rubles as can be bought for 20 rubles. How many candies can be bought for 50 rubles?
Answer: 75 candies. | Solution. Let $x$ be the cost of one candy in rubles. Then $45 x=\frac{20}{x}$, from which $x=\frac{2}{3}$. Then with 50 rubles, one can buy $\frac{50}{x}=75$ candies.
Criteria. Any correct solution: 7 points.
The equation $45 x=\frac{20}{x}$ is correctly set up, but an arithmetic error is made in solving it or in su... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.1. When composing the options for the district mathematics olympiad for grades $7, 8, 9, 10, 11$, the jury aims to ensure that in the option for each grade, there are exactly 7 problems, of which exactly 4 do not appear in any other option. What is the maximum number of problems that can be included in the olympiad? | Answer: 27.
Solution. The number of non-repeating tasks will be 20, the number of repeating tasks does not exceed $3 \cdot 5 / 2<8$. We will provide an example of variants with exactly 7 repeating tasks: 7th grade: $1,2,3$; 8th grade: $1,2,3$; 9th grade: $4,5,6$; 10th grade: $4,5,7$; 11th grade: $4,6,7$. | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.5. One hundred integers are written in a circle. Each number is greater than the sum of the two numbers following it in a clockwise direction. What is the maximum number of positive numbers that can be among the written ones
(S. Berlov) | Answer: 49.
Solution: Suppose that two non-negative numbers stand next to each other. Then the number preceding them is greater than their sum, meaning it is positive. Similarly, the number before it is also positive, and so on. In the end, we get that all numbers are non-negative; but then the smallest of them cannot... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.6. The field is a square grid of $41 \times 41$ cells, one of which conceals a tank. The fighter aircraft shoots at one cell per shot. If a hit occurs, the tank moves to an adjacent cell along a side; if not, it remains in place. After each shot, the pilot does not know whether a hit occurred. To destroy the tank, it... | Answer. $\frac{3 \cdot 41^{2}-1}{2}=2521$ shots.
Solution. Color the cells in a checkerboard pattern so that the corners of the field are black. Suppose the pilot first shoots at all the white cells, then at all the black cells, and then again at all the white cells. If the tank was on a white cell, the pilot will des... | 2521 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.8. On the board, there are $N \geqslant 9$ different non-negative numbers, each less than one. It turns out that for any eight different numbers on the board, there is a ninth, different from them, such that the sum of these nine numbers is an integer. For which $N$ is this possible?
(F. Nilov) | Answer. Only for $N=9$.
Solution. It is clear that for $N=9$ the required is possible - it is sufficient to write 9 different positive numbers on the board with a unit sum. We will show that for $N>9$ the required is impossible. Suppose the opposite; let $S$ be the sum of all numbers on the board.
Choose arbitrary nu... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 1. CONDITION
How many integers between 100 and 10000 are there such that their representation contains exactly 3 identical digits? | Solution. If the digit that repeats three times is zero, then by adding one of the other 9 digits in front, we get 9 numbers. If the digit $a \neq 0$ repeats in the number, then the number of numbers with different $b, b \neq a$, will be $9 \cdot 4 \cdot 9$. Therefore, in total, there are 333 numbers.
Answer: 333. | 333 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.1. The price for a ride on the "Voskhod" carousel in February 2020 was 300 rubles per person. In March, the price was reduced, and the number of visitors increased by $50 \%$, while the revenue increased by $25 \%$. By how many rubles was the price reduced? | Answer: 50.
Solution. The entry fee for two people in February 2020 was 600 rubles. In March, instead of every two people, the stadium is visited by three people, as the number of visitors increased by $50 \%$. Since the total collection increased by $25 \%$, they will pay $600+0.25 \cdot 600=750$ (rub.), and thus one... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.2. The external angles of a triangle are proportional to the numbers $5: 7: 8$. Find the angle between the altitudes of this triangle drawn from the vertices of its smaller angles. | Answer: $90^{0}$.
Solution. The sum of the exterior angles of a triangle is
$$
180^{\circ} \cdot 3-180^{\circ}=360^{\circ}
$$
From this, we get that one of the exterior angles of the triangle is $\frac{360^{\circ}}{5+7+8} \cdot 5=90^{0}$, and therefore the adjacent interior angle of the triangle is $90^{0}$. Consequ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. On the island, there are 217 residents, 17 of whom are knights, and the remaining 200 are tricksters. One day, a private detective arrived on the island and decided to find out who is who. For this, he asked each resident to write a list of 200 people they consider to be tricksters. The detective does not know who w... | Answer: Yes, it can always.
Solution. We will combine identical lists into groups. Let's call a group important if it contains at least 17 lists. Note that the number of important groups is no more than 12 (otherwise, the number of residents is at least $13 * 17 = 221$, which is more than 217).
Since all the knights ... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.1. Does there exist a ten-digit number, divisible by 11, in which all digits from 0 to 9 appear? | Answer. Yes, for example, 9576843210.
Solution. Let's consider a possible way to find the required number. Note that a number is divisible by 11 if the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions is divisible by 11. If we write all ten digits in descend... | 9576843210 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. There is a balance scale without weights and 11 visually identical coins, among which one may be counterfeit, and it is unknown whether it is lighter or heavier than the genuine coins (genuine coins have the same weight). How can you find at least 8 genuine coins in two weighings? | Solution. Let's divide the coins into three piles of three coins each. Compare pile 1 and pile 2, and then compare pile 2 and pile 3. If all three piles weigh the same, then all the coins in them are genuine, and we have found 9 genuine coins. Otherwise, one of the piles differs in weight from the others, and the count... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.5. Find all numbers of the form $22 \ldots 2$ that can be represented as the sum of two perfect squares. | Answer: 2.
Solution. Let $22 \ldots 2=a^{2}+b^{2}$ for some integers $a$ and $b$.
If the numbers $a$ and $b$ are even, then the sum of their squares is divisible by 4, but the number $22 \ldots 2$ is not.
Thus, the numbers $a$ and $b$ can only be odd:
$$
a=2 k+1, b=2 l+1(k, l \in Z)
$$
Therefore, the sum
$$
a^{2}... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 10.1. Find any natural $x$ such that the value of the expression $2^{x}+2^{8}+2^{11}$ is a square of a natural number. | Answer: 12.
Solution. Note that $x=12$ works: $2^{12}+2^{8}+2^{11}=\left(2^{6}\right)^{2}+\left(2^{4}\right)^{2}+2 \cdot 2^{6} \cdot 2^{4}=\left(2^{6}+2^{4}\right)^{2}$. Remark. The number $x=12$ is the only one that fits the condition of the problem. | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.2. There are 36 balls lying in a circle, each of which is either red or blue (balls of each of these colors are present). It is known that:
- for any red ball, there is exactly one red ball such that there is exactly one ball between them;
- for any red ball, there is exactly one red ball such that there ar... | Answer:
(a) (2 points) 12.
(b) (2 points) 24.
Solution. By the condition, there will be two red balls, between which lies exactly one ball. Number the balls clockwise with numbers \(1, 2, \ldots, 36\) such that the 1st and 3rd balls are red. From the condition, it follows that the 35th ball is blue (otherwise, the f... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.3. Quadrilateral $ABCD$ is such that $\angle ABD = \angle ACD = 90^\circ$, $\angle CAD = 42^\circ$. Rays $CB$ and $DA$ intersect at point $K$. It is known that $BK = 3$, $AD = 6$.
... | # Answer:
(a) (2 points) $28^{\circ}$.
(b) (2 points) $34^{\circ}$.
Solution. From the condition, it follows that points $A, B, C, D$ lie on a circle with diameter $A D$. Connect point $B$ to the midpoint $O$ of segment $A D$, which is the center of this circle (Fig. 6). Clearly, $A O=O D=O B=B K=3$.
Triangles $K B... | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.4. Given a rectangular grid of cells. It is known that:
- there are exactly 940 ways to cut out a $1 \times 2$ rectangle from it along the grid lines;
- there are exactly 894 ways to cut out a $1 \times 3$ rectangle from it along the grid lines.
How many ways are there to cut out a $1 \times 5$ rectangle f... | Answer: 802.
Solution. Let the table have $a$ rows and $b$ columns. Without loss of generality, $a \leqslant b$. It is clear that at least one of the numbers $a$ and $b$ is not less than 2.
- Suppose $a=1$. Then from the table $1 \times b$, a rectangle $1 \times 2$ can be cut in $b-1=940$ ways, and a rectangle $1 \ti... | 802 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.5. Several sweet-tooths participated in a candy eating contest. Each participant ate an integer number of candies, and any two participants ate a different number of candies. Summarizing the contest, the jury ranked all the people in descending order of the number of candies eaten (for example, the winner at... | Answer: 21.
Solution. Let there be $n$ sweet-tooth participants in total, and they ate $S$ candies in total. From the condition, it follows that the winner ate $\frac{S}{15}$ candies, the person in third place $-\frac{S}{21}$ candies, and the person in last place $-\frac{S}{22}$ candies.
All participants, except the ... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. Given trapezoid $ABCD$. Segments $AB$ and $CD$ - its lateral sides - are equal to 24 and 10, respectively. Points $X$ and $Y$ are marked on side $AB$ such that $AX=6, XY=8, YB=10$. It is known that the distances from points $X$ and $Y$ to line $CD$ are 23 and 27, respectively.
 (1 point) 100.
(b) (3 points) 260.
Fig. 7: to the solution of problem 10.6
Solution. We will prove that the distances from points $A$ and $B$ to the line $C D$ are 20 and 32,... | 260 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. On an island, there are 23 knights and 200 liars; all residents have different names. A tourist who knows this asked each of the 223 residents to write down 200 names of liars on a piece of paper. Each knight correctly wrote down 200 names of liars, while each liar wrote down an arbitrary list of 200 name... | Answer: 16.
Solution. First, let's outline a strategy for the liars that allows identifying no more than 16 of them. Let's call the knights $R_{1}, R_{2}, \ldots, R_{23}$, and the liars $-L_{1}, L_{2}, \ldots, L_{200}$. All knights will list $L_{1}, L_{2}, \ldots, L_{200}$ in their lists. For each integer $0 \leqslant... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.5. Hooligan Dima made a construction in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the figure.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire construc... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
, 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), the right triangles $X A B$ and $Y D A$ are similar by the acut... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. In triangle $A B C$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $A C$, point $K$ is marked, and on side $B C$, points $L$ and $M$ are marked such that $K L=K M$ (point $L$ lies on segment $B M$).
Find the length of segment $L M$, if it is known that $A K=4, B L=31, M C=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 as $x$ (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 greatest 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 $D Z$.
. Find the sum of all angles with vertex O, whose sides are rays OA, OC, OD, OE, OF, OB, if $\angle \mathrm{COF}=97^{\circ}, \angle \mathrm{DOE}=35^{\c... | Answer: $1226^{\circ}$.
Instructions. There are four pairs of adjacent angles and a straight angle, which sum up to $180^{\circ}$ or are equal to $180^{\circ}$. There are two pairs of angles whose sum equals $\angle \mathrm{COF}$, and the angle itself. And there is still $\angle \mathrm{DOE}$. Therefore, the sum of th... | 1226 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In a room, 15 chairs are arranged in a circle. Three jewelers, when no one is watching, sit on three adjacent chairs, and the one sitting in the middle chair hides a diamond in the chair he is sitting on. The inspector has several detectors that show whether someone has sat on a chair or not. What is the minimum num... | Answer: 9 detectors.
Instructions. Evaluation. Consider five chairs such that between any two nearest ones there are two other chairs. If there is no detector in two nearest ones, then by adding to the two chairs between them any of them, we get two possible seating arrangements of jewelers, in which the detector read... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. What is the total number of emeralds in the boxes?
![](htt... | # Answer: 12.
Solution. The total number of rubies is no more than $13+8=21$, and the number of diamonds is no less than $2+4=6$. According to the condition, their quantities differ by 15. This is only possible if the rubies are in the boxes with 13 and 8 stones, and the diamonds are in the boxes with 2 and 4 stones. ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
![](https://cdn.mathpix.co... | Answer: 22.
Solution. Note that there are a total of 91 coins, so after all moves, each box should have exactly 13 coins. At least 7 coins need to be moved from the box with 20 coins. Now consider the boxes adjacent to the box with 20 coins. Initially, they have a total of 25 coins, and at least 7 more coins will be t... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.