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4. How many four-digit numbers are there for which the sum of the digits is a multiple of ten? | 4. Answer: 900.
To any three-digit number, you can append exactly one decimal digit on the right so that the sum of the digits becomes a multiple of 10. For example, to 780, we append 5; to 202, we append 6; to 334, we append 0, and so on. Since three-digit numbers range from 100 to 999 inclusive, there are exactly $9... | 900 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. On the side $B C$ of rectangle $A B C D$, a point $K$ is marked. Point $H$ on segment $A K$ is such that $\angle A H D=90^{\circ}$. It turns out that $A K=B C$. How many degrees does angle $A D H$ measure if $\angle C K D=71^{\circ}$?
Fig. 5: to the solution of problem 8.3
Solution. Since $B C \| A D$, we get $\angle A D K=\angle C K D=71^{\circ}$ (Fig. 5).
Since $A K=B C=A D$, triangle $A K D$ is isosceles a... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4. Thirty-six children are standing in a circle, each dressed in a red or blue sweater. It is known that next to each boy stands a girl, and next to each girl stands a person in a blue sweater. Find the maximum possible number of girls in red sweaters.
# | # Answer: 24.
Solution. Note that there will not be 3 girls in red sweaters standing in a row (otherwise, the condition would not be satisfied for the middle one). By dividing 36 children into 12 triplets, we get that in each of them there are no more than 2 girls in red sweaters, and the total number of girls in red ... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. A car left the city for the village at the same time a cyclist left the village for the city. When the car and the cyclist met, the car immediately turned around and headed back to the city. As a result, the cyclist arrived in the city 35 minutes after the car. How many minutes did the cyclist spend on the... | Answer: 55.
Fig. 7: to the solution of problem 8.5
Solution. In Fig. 7, let's mark village $A$, city $B$, point $P$ where the car and the cyclist meet, and point $Q$, where the cyclist was ... | 55 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. In an acute-angled triangle $A B C$, the altitude $B H$ is drawn. It turns out that $C H=A B+A H$. How many degrees does the angle $B A C$ measure if $\angle A B C=84^{\circ}$ ?
 | Answer: 64.
Fig. 8: to the solution of problem 8.7
Solution. Mark a point \( K \) on the segment \( CH \) such that \( AH = HK \). Then, from the condition, it follows that \( AB = CK \) (Fig... | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) The king decided to test his hundred sages and announced that the next day he would line them up with their eyes blindfolded and put a black or white hat on each of them. After their eyes are uncovered, each, starting from the last in line, will name the supposed color of their hat. If he fails to guess c... | Solution. Let's describe the strategy that the sages should adhere to. The last in line looks ahead, counts the number of black hats, and says "black" if this number is even. In doing so, he cannot save himself for sure. However, the 99th, 98th, ..., 1st in line receive very important information. Thus, the 99th counts... | 99 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. The parabola $y=x^{2}-20 x+c$, where $c \neq 0$, intersects the $O x$ axis at points $A$ and $B$, and the $O y$ axis at point $C$. It is known that points $A$ and $C$ are symmetric with respect to the line $y=-x$. Find the area of triangle $A B C$. | Answer: 231.
Solution. Since $y(0)=c$, we have $C(0, c)$ and $A(-c, 0)$. Therefore, one of the roots of the equation $x^{2}-20 x+c=0$ is $-c$, i.e., $c^{2}+20 c+c=0$, from which $c=-21, x_{1}=21$. By Vieta's theorem, $x_{1}+x_{2}=20$. Thus, $x_{2}=-1$. The length of the base of the triangle is $x_{1}-x_{2}=22$, and th... | 231 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. Petya had a large wooden cube with dimensions $30 \times 30 \times 30$ cm $^{3}$. Petya decided to paint the entire cube, which took 100 grams of paint. Later, Petya needed smaller cubes, so he cut the large cube with 6 cuts, parallel to the faces of the cube (2 cuts parallel to each pair of faces), into 27 cubes ... | Answer: 200 grams.
First solution. Each cut increases the unpainted surface area by an amount equal to twice the area of the cut. It is clear that the area of the cut is the same as the area of a face. Therefore, it is necessary to paint an additional surface area equal to \(6 \cdot 2 = 12\) faces, which is twice the ... | 200 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.4. Every day, from Monday to Friday, the old man went to the blue sea and cast his net into the water. Each day, the net caught no more fish than the previous day. In total, over the five days, the old man caught exactly 100 fish. What is the smallest total number of fish he could have caught on the three days - Mon... | Answer: 50.
Solution. If the old man caught 25 fish each of the first four days and caught nothing on Friday, the conditions of the problem are met, and exactly 50 fish were caught over the specified three days.
Let's prove that in the specified days, fewer than 50 fish could not have been caught. Indeed, suppose few... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. In triangle $A B C$, points $M$ and $N$ are the midpoints of sides $A C$ and $B C$ respectively. It is known that the point of intersection of the medians of triangle $A M N$ is the point of intersection of the altitudes of triangle $A B C$. Find the angle $A B C$. | Answer: $45^{\circ}$.
Solution. Let $H$ be the orthocenter (the point of intersection of the altitudes) of triangle $ABC$. Then the altitude $AT$ of triangle $ABC$ contains the median of triangle $AMN$, that is, it intersects segment $MN$ at its midpoint - point $E$ (see Fig. 10.5a, b). We can reason in different ways... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In the drawer, there are 23 socks: 8 white and 15 black. Every minute, Marina approaches the drawer and pulls out a sock. If at any moment Marina pulls out more black socks than white ones, she exclaims: "Finally!" - and ends the process.
What is the maximum number of socks Marina can pull out before she exclaims: ... | Answer: 17.
Solution: If Marina takes out 17 socks, then among them there will be no more than 8 white ones, which means there will be at least 9 black ones. Therefore, at this moment, she will definitely exclaim, "Finally!"
On the other hand, if she takes out only 16 socks, she might be unlucky: if she sequentially ... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Circle $k_{1}$ with a radius of 8 cm lies inside circle $k$. Both circles intersect circle $k_{2}$ with a radius of 15 cm, as shown in the figure. What is the radius of $k$ if the shaded area inside $k$ but outside $k_{1}$ is equal to the total shaded area inside $k_{2}$?
 \Leftrightarrow\left(2^{x}-2^{-x}\right)^{2}=4 \cos ^{2 a x} / 2\right) \Leftrightarrow$
$$
\left[\begin{array} { l }
{ 4 ^ { x / 2 } - 4 ^ { - x / 2 } = 2 \operatorname { cos } \frac { a x } { 2 } ... | 4042 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.2. For the angles of triangle $ABC$, it is known that $\sin \angle A + \cos \angle B = \sqrt{2}$ and $\cos \angle A + \sin \angle B = \sqrt{2}$. Find the measure of angle $C$. | Answer: $90^{\circ}$.
Solution. Square both equalities and add them. Then use the fundamental trigonometric identity, and after transformations, we get $2 \sin \angle A \cos \angle \mathrm{B}+2 \sin \angle B \cos \angle \mathrm{A}=2 \sin (\angle A+\angle \mathrm{B})=2 \quad, \quad$ that is, $\quad \sin (\angle A+\angl... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.6. Points $M$ and $N$ are the midpoints of sides $B C$ and $A D$ of quadrilateral $A B C D$. It is known that $\angle B=$ $150^{\circ}, \angle C=90^{\circ}$ and $A B=C D$. Find the angle between the lines $M N$ and $B C$. | Answer: $60^{\circ}$.
Solution. First method. Construct parallelogram $A B M K$ and rectangle $C D L M$ (see Fig. 8.6a). Since $A K\|B C\| L D$ and $A K=B M=$ $M C=L D$, then $A K D L$ is also a parallelogram. Therefore, the midpoint $N$ of its diagonal $A D$ is also the midpoint of diagonal $K L$.
. How many steps are there on the escalator if Petya counted 20 steps with his feet (i.e., before falling) and 30 steps with his sides (after fallin... | Answer: 80
Solution. Let the escalator have a length of $2 \mathrm{~L}$ (steps), the speed of the escalator be $u$, and Petya runs with a speed of $\mathrm{x}$, and flies with a speed of $3 \mathrm{x}$. Then the time until the fall is $\frac{L}{u+x}$ and during this time Petya will count $\frac{L x}{u+x}$ steps. From ... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. We will call a right-angled triangle elegant if one of its legs is 10 times longer than the other. Is it possible to cut a square into 2020 identical elegant triangles? | Answer: Yes
Solution. An elegant triangle with legs of 10 and 100 can be cut into 100 triangles with legs of 1 and 10 (the cuts are straight lines parallel to its sides). From four such large triangles, we can form a square (its sides will be the hypotenuses of these triangles) with a hole (in the shape of a square wi... | 2020 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. A three-digit number, all digits of which are different and non-zero, will be called balanced if it is equal to the sum of all possible two-digit numbers formed from the different digits of this number. Find the smallest balanced number. | Answer: 132.
Solution: Let the desired number be of the form $\overline{a b c}$. Then it can be represented as the sum of six different two-digit numbers: $\overline{a b c}=\overline{a b}+\overline{b a}+\overline{a c}+\overline{c a}+\overline{b c}+\overline{c b}$.
From the last equality, we get $100 a+10 b+c=22 a+22 ... | 132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.1 In a competition of meaningless activity, a participant recorded 2022 numbers in a circle such that each number is equal to the product of its two neighbors. What is the maximum number of different numbers that could have been used? | Solution: If there is a zero among the numbers, then its neighbors are also zeros, so all the recorded numbers will be equal to zero. We will assume that there are no zeros among the numbers. Let $a$ and $b$ be two adjacent numbers. Then on the other side of $a$ stands $a / b$, and on the other side of $b$ stands $b / ... | 337 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. Each student in class 7B ate the same number of chocolate bars during math lessons over the week. Nine of them together ate fewer than 288 chocolate bars in a week, while ten of them together ate more than 300 chocolate bars. How many chocolate bars did each student in class 7B eat? Explain your answer. | Answer: 31 chocolates each
Solution 1: Since ten students ate more than 300 chocolates, nine students ate more than (310:10)$\cdot$9 = 270 chocolates. It is also known that nine students together ate less than 288 chocolates. The only number divisible by 9 in the range from 271 to 287 is 279. Therefore, nine students ... | 31 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
7.2. Find the largest natural number in which each digit, starting from the third, is equal to the sum of all the previous digits of the number. | Answer: 101248.
Solution: Let the first digit of the number be $a$, the second digit be $b$. Then the third digit is $(a+b)$, the fourth digit is $(2a+2b)$, the fifth digit is $(4a+4b)$, and the sixth digit is $(8a+8b)$. There cannot be a seventh digit, because if it exists, it would be equal to $(16a+16b)$, but this ... | 101248 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. Given an equilateral triangle $\mathrm{ABC}$. On the sides $\mathrm{AB}$ and $\mathrm{BC}$, isosceles right triangles ABP and BCQ are constructed externally with right angles $\angle \mathrm{ABP}$ and $\angle \mathrm{BCQ}$. Find the angle $\angle \mathrm{PAQ}$. | Answer: 90.
Solution: Consider triangle ACQ. It is isosceles because $\mathrm{AC}=\mathrm{BC}$ (by the condition of the equilateral triangle $\mathrm{ABC}$), and $\mathrm{BC}=\mathrm{CQ}$ (by the condition of the isosceles triangle BCQ). The angle at the vertex $\angle A C Q=60 \circ+90 \circ=150$ . Then $\angle \math... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. How many five-digit numbers exist that are not divisible by 1000, and have the first, third, and last digits even?
Otvet: 9960. | Solution. The first digit of the number can be any of the four (2, 4, 6, or 8), the second and fourth can be any of ten each, and the third and fifth, if we abandon the condition "not divisible by a thousand," can be any of five (0, 2, 4, 6, or 8). Therefore, there are $4 \times 10 \times 5 \times 10 \times 5=10000$ fi... | 9960 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.4. On each of 2013 cards, a number is written, and all these 2013 numbers are distinct. The cards are face down. In one move, it is allowed to point to ten cards, and in response, one of the numbers written on them will be reported (it is unknown which one). For what largest $t$ can it be guaranteed to find $t$ card... | Answer. $t=1986=2013-27$.
Solution. 1. First, we will show that it is impossible to guess 1987 cards. Number the cards $A_{1}, \ldots, A_{2013}$; we will demonstrate how to arrange the answers so that none of the numbers on the cards $A_{1}, \ldots, A_{27}$ can be determined.
For each $i=1, \ldots, 9$, combine the ca... | 1986 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 1. Option 1.
A confectionery factory received 5 rolls of ribbon, each 50 m long, for packaging cakes. How many cuts need to be made to get pieces of ribbon 2 m long? | Answer: 120.
Solution. From one roll, 25 pieces of ribbon, each 2 m long, can be obtained. For this, 24 cuts are needed. Therefore, a total of $5 \cdot 24=120$ cuts are required. | 120 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 4. Option 1.
In a box, there are red, blue, and green pencils. It is known that if you remove two red pencils, the number of pencils of all colors will be equal. And if after that you add 10 red pencils to the box, the number of red pencils will be exactly half of all the pencils. How many pencils were in the box in... | Answer: 32.
Solution. Before adding pencils, it could be considered that there were as many red ones as green ones. Then, the 10 added pencils are the number of blue ones. So, at the moment of adding, there were 10 red, blue, and green pencils each.
# | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 5. Option 1.
The chocolate bar has the shape of a square with a side of 100 cm, divided into pieces with a side of 1 cm. The sweet tooth ate all the pieces along each of the four sides, as shown in the figure. How many pieces did the sweet tooth eat in total?
. The chocolate bar consists of $100 \times 100=10000$ pieces in total. If 2 rows of pieces are removed from each side, a square with a side length of 96 cm remains. Thus, the sweet tooth will eat $10000-96 \cdot 96=784$ pieces.
 is
$$
\frac{a^{3}-b^{3}}{a^{2}+b^{2}+a \cdot b}=\frac{(a-b)\left(a^{2}+a b+b^{2}\right)}{a^{2}+a b+b^{2}}=a-b=149^{2}-199=22002
$$
Thus, his pension will be enough for one ticket.
Answer: It will be enough.
Recommendations for check... | 22002 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5. The first two digits of a natural four-digit number are either each less than 5, or each greater than 5. The same can be said about the last two digits. How many such numbers are there? Justify your answer. | Solution: There are $4^{2}=16$ two-digit numbers where both digits are greater than 5, and there are $4 \cdot 5=20$ (the first digit is not zero) where both digits are less than 5. In total, there are $16+20=36$ such numbers. This is the conclusion about the first two digits of the number. The last two digits will prov... | 1476 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. There is an unlimited number of chips in six colors. What is the smallest number of chips that need to be arranged in a row so that for any two different colors, there are two adjacent chips of these colors in the row. | 5. It follows from the condition that for each fixed color A, a chip of this color must be paired with a chip of each of the other 5 colors. In a row, a chip has no more than two neighbors, so a chip of color A must appear at least 3 times. Similarly for each other color. Thus, there should be no fewer than 3*6=18 chip... | 18 | 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 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 |
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 |
7.5. 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
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. From a $6 \times 6$ square grid, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
.
For convenience, let's introduce some notations. The top-left corner cell of the $5 \times 5$ table will be called $A$, and the bottom-right corner c... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\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 |
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 |
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).
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 $A B C$, 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?
 | 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 $D Z$.
![](https://cdn.mathpix.com... | Answer: 20.
Solution. Lines $C_{1} Y$ and $Z X$ lie in parallel planes $B B_{1} C_{1} C$ and $A A_{1} D_{1} D$, so they do not intersect. Since these two lines also lie in the same plane $C_{1} X Z Y$, they are parallel. Similarly, lines $Y Z$ and $C_{1} X$ are parallel. Therefore, quadrilateral $C_{1} X Z Y$ is a par... | 20 | Geometry | math-word-problem | 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 |
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 |
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 |
# 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 |
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 |
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