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5. Can the 8 natural numbers $1,2,3,4,5,6,7,8$ be divided into two groups of 4 numbers each such that the sum of the numbers in the first group equals the sum of the numbers in the second group, and the sum of the squares of the numbers in the first group equals the sum of the squares of the numbers in the second group | Solution. The sum of the numbers from 1 to 8 is 36, and the sum of their squares is 204. Therefore, we need to find a quartet of numbers from $1, \ldots, 8$, the sum of which is 18, and the sum of their squares is 102. Assume that this quartet contains 8. Then the sum of the squares of the other three numbers in the qu... | {2,3,5,8},{1,4,6,7} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,552 |
Problem 10.1. The smallest divisor of a number, different from 1, will be called minimal. The largest divisor of a number, different from the number itself, will be called maximal. Find a four-digit number for which the maximal divisor is 49 times greater than the minimal one. It is sufficient to provide an example of ... | Answer: 1225 or 2401.
Solution. Note that any divisor $d$ of a number $n$ corresponds to a divisor $n / d$, and the smaller $d$ is, the larger $n / d$ is, and this is a one-to-one correspondence of divisors with themselves.
Let $n$ be our four-digit number, and $p$ be its smallest divisor. Then the largest divisor of... | 1225or2401 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,554 |
Problem 10.2. On a sheet of paper, three intersecting circles are drawn, forming 7 regions. We will call two regions adjacent if they share a common boundary. Regions that border at exactly one point are not considered adjacent.
Two regions already have numbers written in them. Write integers in the remaining 5 region... | Answer: -8.
Solution. Let the unknown number be $x$. Then we can place the numbers in the other regions as follows:
The sum of the numbers in the regions surrounding $x$ should be equal to $... | -8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,555 |
Problem 10.3. Petya and Daniil are playing the following game. Petya has 36 candies. He lays out these candies in the cells of a $3 \times 3$ square (some cells may remain empty). After this, Daniil chooses four cells forming a $2 \times 2$ square and takes all the candies from there. What is the maximum number of cand... | # Answer: 9.
Solution. If Petya places 9 candies in each corner cell (and does not place any candies in the other cells), then in any $2 \times 2$ square there will be exactly 9 candies. After this, Daniil will be able to take only 9 candies.
Let's prove that Daniil can get at least 9 candies. Suppose the opposite: l... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,556 |
Problem 10.4. Roma thought of a natural number, the sum of the digits of which is divisible by 8. Then he added 2 to the thought number and again got a number, the sum of the digits of which is divisible by 8. Find the smallest number that Roma could have thought of. | Answer: 699.
Solution. If both numbers are divisible by 8, then their difference is also divisible by 8. If there was no carry-over when adding, the sum of the digits would differ by 2, which is not divisible by 8. If there was a carry-over but not into the hundreds place, the sum of the digits would differ by $9-2=7$... | 699 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,557 |
Problem 10.5. On the side $AB$ of the rectangle $ABCD$, a circle $\omega$ is constructed with $AB$ as its diameter. Let $P$ be the second intersection point of the segment $AC$ and the circle $\omega$. The tangent to $\omega$ at point $P$ intersects the segment $BC$ at point $K$ and passes through point $D$. Find $AD$,... | Answer: 24.
Solution. Note that triangles $A D P$ and $C K P$ are similar (the equality $\angle D A P=\angle K C P$ and $\angle A D P=\angle C K P$ is ensured by the parallelism $A D \| B C$). Moreover, triangle $A D P$ is isosceles, as $A D$ and $D P$ are segments of tangents. Therefore, triangle $C K P$ is also isos... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,558 |
Problem 10.7. Nikita schematically drew the graph of the quadratic polynomial $y=a x^{2}+b x+c$. It turned out that $A B=C D=1$. Consider four numbers $-a, b, c$ and the discriminant of the quadratic polynomial. It is known that three of them are equal in some order to $1 / 4, -1, -3 / 2$. Find what the fourth number i... | Answer: $-1 / 2$.
Solution. From the drawn graph, it follows that $a$, $b$, and $c$ are negative. Indeed, $a$ is negative because the branches of the parabola are directed downwards. The free term $c$ is negative because it equals the value of the trinomial at $x=0$. Finally, $b$ coincides with the slope of the line $... | \frac{-1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,560 |
Problem 10.8. On the face $ABC$ of the tetrahedron $ABCD$, a point $P$ is marked. Points $A_1, B_1, C_1$ are the projections of point $P$ onto the faces $BCD, ACD, ABD$ respectively. It turns out that $PA_1 = PB_1 = PC_1$. Find $\angle BA_1C$, given that $\angle BC_1D = 136^\circ$ and $\angle CB_1D = 109^\circ$.
$ and the hypotenuse (common $P D$). From this, we get that $D C_{1}=D A_{1}$.
Considering triangles $B P C_{1}$ and $B P A_{1}$, we similarly obtain $B C_{1}=B A_{1}$. Then triangles $B D C_{... | 115 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,561 |
1. (7 points) Winnie-the-Pooh eats 3 cans of condensed milk and a jar of honey in 25 minutes, while Piglet takes 55 minutes. One can of condensed milk and 3 jars of honey, Pooh eats in 35 minutes, while Piglet takes 1 hour 25 minutes. How long will it take them to eat 6 cans of condensed milk together? | # Solution.
1st method. From the condition, it follows that Winnie-the-Pooh eats 4 cans of condensed milk and 4 jars of honey in 1 hour, while Piglet does so in 2 hours and 20 minutes. Therefore, one can of condensed milk and one jar of honey are eaten by Winnie-the-Pooh in 15 minutes, and by Piglet in 35 minutes.
Us... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,566 |
2. (7 points) It is known that $a+b+c<0$ and that the equation $a x^{2}+b x+c=0$ has no real roots. Determine the sign of the coefficient $c$.
| Solution. Let $y=a x^{2}+b x+c$, then $y(1)=a+b+c, y(1)<0$.
Since the equation $a x^{2}+b x+c=0$ has no real roots, the branches of the parabola are directed downwards and then $c<0$.
Answer. $c<0$. | <0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,567 |
3. (7 points) Prove that if the sides of a right triangle form an arithmetic progression, then its common difference is equal to the radius of the inscribed circle. | Solution. Let $a$, $b$, $c$ be the sides of a right triangle, where $b=a+d, c=a+2d$.
Then on one hand, $c^{2}=a^{2}+b^{2}=2a^{2}+2ad+d^{2}$.
On the other hand, $c^{2}=(a+2d)^{2}=a^{2}+4ad+4d^{2}$.
We obtain that $2a^{2}+2ad+d^{2}=a^{2}+4ad+4d^{2}, a^{2}-2ad-3d^{2}=0$.
Solving the quadratic equation for $a$, we get:... | d | Geometry | proof | Yes | Yes | olympiads | false | 13,568 |
4. (7 points) Natural numbers $m, n$ are such that the fraction $\frac{m}{n}$ is irreducible. Is the fraction $\frac{n}{2 m+3 n}$ reducible? | Solution. Let $n$ be divisible by $d$ and $2 m+3 n$ be divisible by $d, d \neq 1$. Then $3 n$ is divisible by $d$ and $2 m$ is divisible by $d$.
From the fact that $2 m$ is divisible by $d$ it follows that 2 is divisible by $d$, i.e., $d=2$ (since $m$ is not divisible by $d$, as the fraction $\frac{m}{n}$ is irreducib... | Reducibleforevenn | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,569 |
5. (7 points) Inside triangle $ABC$, there is a point $M$. Prove that
$$
\frac{S_{A C M}+S_{B C M}}{S_{A B M}}=\frac{C M}{C_{1} M}
$$
where $C_{1}$ is the point of intersection of the lines $AB$ and $CM$. | # Solution.
Draw the altitude $C H$ in triangle $A B C$ and the altitude $M H_{1}$ in triangle $A M B$. Then
$$
\begin{gathered}
S_{A B C}=\frac{1}{2} \cdot A B \cdot C H=\frac{1}{2} \cdot ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,570 |
9.5. Different real numbers $a$ and $b$ are such that the equation
$$
\left(x^{2}+20 a x+10 b\right)\left(x^{2}+20 b x+10 a\right)=0
$$
has no roots. Prove that the number $20(b-a)$ is not an integer.
(P. Kozlov) | 9.5. Suppose the opposite. Let, for definiteness, $b>a$; then $20(b-a) \geqslant 1$, that is, $b-a \geqslant \frac{1}{20}$.
The discriminant of the second quadratic polynomial in the original equation is negative, hence $100 b^{2}-10 a<0$. Therefore, $10 b^{2}<a \leqslant b-\frac{1}{20}$, which means $10 b^{2}-b+\frac... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,571 |
9.6. Each of the 1000 gnomes has a hat, blue on the outside and red on the inside (or vice versa). If a gnome is wearing a red hat, he can only lie, and if it is blue, he can only tell the truth. During one day, each gnome said to each other, "You have a red hat!" (during the day, some gnomes turned their hats inside o... | 9.6. Answer. 998 turnovers.
Let's call a gnome red or blue if he is wearing a cap of the corresponding color. Note that one gnome can say the required phrase to another if and only if these gnomes are of different colors: a blue gnome will tell the truth in this case, and a red one will lie. Now, if any three gnomes h... | 998 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,572 |
9.7. We will call a natural number $n$ unsuccessful if it cannot be represented as $n=\begin{aligned} & x^{2}-1 \\ & y^{2}-1\end{aligned}$ with natural numbers $x, y>1$. Is the number of unsuccessful numbers finite or infinite? $\quad$ (V. Senderov) | 9.7. Answer. Infinitely.
We will prove that any number of the form $n=p^{2}$, where $p$ is an odd prime number, is unsuccessful. Suppose the opposite, i.e.,
$$
\left(y^{2}-1\right) p^{2}=x^{2}-1
$$
for some natural numbers $x, y \neq 1$. Then either $x+1$ or $x-1$ is divisible by $p$.
Let $x+1 \vdots p$. Then $x-1=... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,573 |
9.8. In an acute-angled triangle $A B C$, the median $A M$ is longer than the side $A B$. Prove that triangle $A B C$ can be cut into three parts from which a rhombus can be formed.
(S. Volchenkov) | 9.8. Let $N$ be the midpoint of side $AC$, and $K$ be a point on the line $MN$ such that $MK = MN$. Then triangles $MNC$ and $MKB$ are symmetric with respect to $M$ and therefore equal. We make a cut along the midline $MN$; by flipping triangle $MNC$ so that it coincides with $\triangle MKB$, we obtain parallelogram $A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,574 |
Task 1.3. A $4 \times 4$ table is divided into four $2 \times 2$ squares.
Vika wrote 4 ones, 4 twos, 4 threes, and 4 fours in the cells of the table such that in each column, each row, and each $2 \times 2$ square, all numbers are different.
The hooligan Andrey erased some of the numbers. Help Vika restore which cell... | Answer: $A-3, B-2, C-4, D-1$. | A-3,B-2,C-4,D-1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,576 |
Problem 2.1. Points $A, B, C, D$ are marked on a line, in that exact order. Point $M$ is the midpoint of segment $A C$, and point $N$ is the midpoint of segment $B D$. Find the length of segment $M N$, given that $A D=68$ and $B C=20$.
}{2} = 44 - \frac{x}{2}
$$
Now it is not difficult to calculate $MN$:
$$
MN = AD - AM - ND = 68 - \frac{x}{2} - \left(44 - \frac{x}{2}\right) = 24
$$ | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,577 |
Problem 6.1. Given a quadrilateral $A B C D$, in which $A D \| B C$. The bisector of angle $A$ intersects side $C D$ at point $X$, and the extension of side $B C$ beyond point $C$ - at point $Y$. It turns out that $\angle A X C=90^{\circ}$. Find the length of segment $A B$, if it is known that $A D=16$ and $C Y=13$.
!... | Answer: 14.5.
Solution. $\quad$ Let $AB=a$. Note that $\angle BYA=\angle YAD=\angle YAB$. We get that triangle $ABY$ is isosceles $(AB=BY)$, and $BC=BY-CY=AB-CY=a-13$ (Fig. 1).
Fig. 1: to t... | 14.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,579 |
# 2. CONDITION
"Once, during a chess game, I had three times fewer pieces and pawns than my opponent, and six times fewer than the free squares on the board, but I still won the game!" - Ivan Ivanov once told. Should we believe him? | Solution. Suppose Ivan has $n$ pawns and pieces left. If we believe Ivan, at this moment his opponent had $3 n$ pawns and pieces, and there were $-6 n$ empty squares on the board. But then it turns out that there are $n+3 n+6 n=10 n$ squares on the board, while in fact there are 64 - not divisible by 10.
Answer. Do no... | Donotbelieve | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,584 |
# 3. CONDITION
Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with edge 1. A line $l$ passes through point $E$, the midpoint of edge $C_{1} D_{1}$, and intersects lines $A D_{1}$ and $A_{1} B$. Find the distance from point $E$ to the point of intersection of line $l$ with line $A_{1} B$. | Solution. Consider the plane $A D_{1} C_{1} B$. Since the line $l$ passes through the midpoint of the edge $C_{1} D_{1}$ and intersects the line $A D_{1}$, it has two common points with this plane, which means it entirely belongs to this plane. At the same time, the line $A_{1} B$ lies in the plane $A B A_{1} B_{1}$. T... | 1.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,585 |
# 4. CONDITION
Two two-digit numbers, written one after the other, form a four-digit number that is divisible by their product. Find these numbers. | Solution. Let $a$ and $b$ be two-digit numbers, then $100a + b$ is a four-digit number. According to the condition, $100a + b = k \cdot ab$, hence $b = a(kb - 100)$, i.e., $b$ is divisible by $a$. Therefore, $b = ma$, but $a$ and $b$ are two-digit numbers, so $m$ is a single-digit number. Since $100a + b = 100a + ma = ... | 1734;1352 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,586 |
# 6. CONDITION
Can 2005 non-zero vectors be drawn on a plane such that from any ten of them, three can be chosen with a zero sum? | Solution. Suppose we managed to draw 2005 non-zero vectors on a plane such that from any 10 of them, we can choose 3 with a zero sum. Choose a line that is not perpendicular to any of the 2005 drawn vectors. Consider the projections of the drawn vectors onto the chosen line. None of these projections are zero vectors. ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,587 |
Problem 5.2. During a physical education class, 25 students from 5B class lined up. Each of the students is either an excellent student who always tells the truth, or a troublemaker who always lies.
Excellent student Vlad stood in the 13th place. Everyone except Vlad stated: "There are exactly 6 troublemakers between ... | Answer: 12.
Solution. Note that students in places $7-12$ are troublemakers, since there are fewer than 6 people between each of them and Vlad. Therefore, the student with number 6 is an excellent student. The same can be said about the student in the 5th place, then about the 4th, the 3rd, the 2nd, and the 1st.
Thus... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,589 |
Problem 5.3. Petya and Vasya were playing with toy soldiers. Petya arranged his knights in a "rectangle" - some number of columns and some number of rows. When all the knights from the first and second rows went on reconnaissance, 24 knights remained. Then Vasya's archers drove away all the knights who were left in the... | Answer: 40.
Solution. The archers routed $24-18=6$ knights. This is two columns. So, at that moment, there were 3 knights in each column. That means 24 knights stood in 8 columns, with 3 knights in each. It follows that 16 knights (two rows, 8 knights in each) went on reconnaissance (Fig. 1).
![](https://cdn.mathpix.... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,590 |
Task 5.4. Masha drew two little people in her notebook. The area of each cell is 1.
Which of the little people has a larger area?
What is the difference? If the areas are the same, write "0" in the answer.
,
- 2 large triangl... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,591 |
Problem 5.5. Denis has identical ten-ruble coins, identical two-ruble coins, and identical one-ruble coins (more than 20 coins of each type). In how many ways can Denis pay exactly 16 rubles for a pie without receiving change? It is not necessary to use coins of each type. | Answer: 13.
Solution. If Denis uses a ten-ruble coin, he will need to collect 6 rubles using two-ruble and one-ruble coins. There are 4 ways to do this: using from 0 to 3 two-ruble coins.
If Denis does not use the ten-ruble coin, he will need to collect 16 rubles using two-ruble and one-ruble coins. There are 9 ways ... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,592 |
Problem 5.6. There are 4 absolutely identical cubes, each of which has 6 dots on one face, 5 on another, ..., and 1 on the remaining face. It is known that on any two opposite faces of the cube, the total number of dots is 7.
These 4 cubes were glued together to form the figure shown in the image. It is known that on ... | Answer: $A=2, B=2, C=6$.
Solution. First, we use the condition that the sum of the number of dots on a pair of opposite faces of the die is 7, to understand how the die, three faces of which we know, looks. Its hidden faces are shown in Fig. 2 (up to the arrangement of dots on the face with 2 dots, which cannot be uni... | A=2,B=2,C=6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,593 |
Problem 5.7. In a class, there are 31 students. Three of them have exactly three friends each, the next three have six each, the next three have nine each, ..., and the next three have thirty each. How many friends does the 31st student have? (Friendship between people is mutual.)
![](https://cdn.mathpix.com/cropped/2... | # Answer: 15.
Solution. Let the 31st student have $x$ friends.
Consider three people, each of whom has 30 friends in the class. There are a total of 31 students in the class, so they are friends with all their classmates.
Let's expel them from the class. Then the number of friends for each person will decrease by 3:... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,594 |
Problem 5.8. In the large family of Ivanovs, there are no twins. A reporter came to the Ivanovs to interview them.
During the interview, each of the children said: "I have an older brother." After thinking for a moment, the reporter was very surprised. But the head of the family explained that some of the children wer... | Answer: 8, 10.
Solution. Let the Ivanov family have $x$ boys. Then the oldest boy joked, and the remaining $x-1$ boys told the truth.
If $x \geqslant 8$, then at least 7 children told the truth, which contradicts the problem's condition.
If $x \leqslant 5$, then there are no more than 1 girl in the family. But then ... | 8,10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,595 |
Problem 2. On his way home, the hunter first walked through the swamp at a speed of 2 km/h, then through the forest at a speed of 4 km/h, and finally along the highway at a speed of 6 km/h. In 4 hours, he walked 15 kilometers. Did he spend more time walking through the swamp or walking along the highway? | Answer. For walking through the swamp. Solution. Let the hunter walked through the swamp for $a$ hours, through the forest for $-b$ hours, and along the highway for $-c$ hours. In 4 hours, he walked
$$
2 a+4 b+6 c=4(a+b+c)+2(c-a)=4 \cdot 4+2(c-a)=15 \text { kilometers. }
$$
It follows that $2(c-a)=-1$, hence $c<a$.
... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,596 | |
Task 3. Can positive numbers (not necessarily integers) be placed in a $6 \times 6$ table so that the product of all numbers in any row, any column, and any $2 \times 2$ square equals two? | Answer: No. Solution: Suppose it is possible to arrange the numbers in such a way. Since the table can be divided into 6 rows, the product of all numbers in it should be $2^{6}$. On the other hand, the table can be divided into 9 $2 \times 2$ squares, and therefore the product of all numbers in it should be $2^{2}$. Co... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,597 |
Problem 5. Do there exist 100 distinct natural numbers such that the product of any fifty-two of them is divisible by the product of the remaining forty-eight? | Answer. They exist. The first solution. The numbers 100!, 100!/2, ..., 100!/100 will work. The product of any 52 of them is divisible by (100!)^51, and the product of any 48 of them is a divisor of (100!)^48, and thus of (100!)^51. The second solution. The numbers 2^n, 2^(n+1), ..., 2^(n+99), where n ≥ 575, will work. ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,599 |
5.1. 20 schoolchildren came to the mathematics Olympiad. Everyone who brought a pencil also brought a pen. 12 people forgot their pencils at home, and 2 schoolchildren forgot their pen. By how many fewer schoolchildren brought a pencil than those who brought a pen but forgot a pencil? | Answer: 2.
Solution: 8 students brought a pencil, which means they also brought a pen. 18 students brought a pen. Therefore, 10 students brought a pen without a pencil. Then 10-8=2. | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,600 |
5.2. Three students, Anna, Vika, and Masha, study at the Literature, History, and Biology faculties. If Anna is a literature student, then Masha is not a historian. If Vika is not a historian, then Anna is a literature student. If Masha is not a literature student, then Vika is a biologist. Which girl studies at which ... | Answer: Vika is a historian, Anna is a biologist, Masha is a philologist.
Solution. Suppose Vika is not a historian, then (by condition 2) Anna is a philologist, but if Anna is a philologist, then Masha is not a historian - this is an obvious contradiction. Therefore, Vika is a historian. Then Masha is a philologist -... | Vika | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,601 |
5.3. A counterfeiter made fake coins on order. In the end, he got three piles: 15, 19, and 25 pieces, one of which accidentally included a real coin. All the coins look the same, the fake coins weigh the same, but differ in weight from the real one. Can the counterfeiter find the pile with all fake coins in one weighin... | Answer: It can.
Solution: Place all the coins from the first pile on one scale pan, and 15 coins from the second (or third) pile on the other. If the scales balance, then all the coins in the first pile are counterfeit. If they do not balance, then the pile whose coins are not involved in the weighing contains only co... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,602 |
5.5. Four dolls and five robots cost 4100 rubles, while five dolls and four robots cost 4000. How much does one doll cost? | Answer: 400 rubles.
Solution. Let's briefly write down the condition of the problem:
$$
\begin{aligned}
& 4 \kappa + 5 p = 4100 \\
& 5 \kappa + 4 p = 4000 .
\end{aligned}
$$
Then $9 \kappa + 9 p = 8100$ and the cost of one doll together with one robot will be 900 rubles. Therefore, 4 dolls and 4 robots cost 3600 rub... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,604 |
1. 1.1. An online electric scooter rental service charges a fixed amount for a ride and a certain amount for each minute of rental. Katya paid 78 rubles for a ride lasting 3 minutes. Lena paid 108 rubles for a ride lasting 8 minutes. It takes Kolya 5 minutes to get from home to work. How much will he pay for this ride? | Answer: 90.
## Solution.
1st method. If the rental of a scooter costs $x$ rubles, and a minute of use costs $y$ rubles, then $x+3y=78$, and $x+8y=108$. Solving this system, we get that $y=6, x=60$. Then Kolya will pay $60+30=90$ rubles.
2nd method. Notice that Lena's trip is 30 rubles more expensive than Katya's tri... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,605 |
2. 2.1. Dealer Dima bought a "LADA Kalina" car from the manufacturer and increased its price by $A \%$. There was no demand, and Dima had to sell the car at a sale with a $20 \%$ discount. As a result, his profit was $20 \%$. Find $A$. | Answer: 50.
## Solution.
1st method. Let's assume Dima bought a car from the manufacturer for $X$ rubles. After raising the price, the car cost $X(1+A / 100)$ rubles. After reducing the price by $20 \%$, the car cost $0.8 X(1+A / 100)$. According to the condition, this is the same as $1.2 X$. Therefore, $1+A / 100=1.... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,606 |
3. 3.1. Petya marked 5 points on a face of a cube, turned it and marked 6 points on an adjacent face, then turned it again and marked 7 points, and so on. He marked points on each face this way. What is the maximum number of points that can be on two opposite faces? | Answer: 18.
## Solution.
Since a cube has only 6 faces, the maximum number of points marked by Petya is 10. Then the number 9 will be on an adjacent face to the number 10 and cannot be on the opposite face. Therefore, the maximum sum will not exceed 18. The sum of 18 can be achieved, for example, as follows:
 = a x^{2} + b x + c \), where \( a > 0 \), the condition \( |f(1)| = |f(2)| = |f(3)| = 1 \) is satisfied. What can the coefficients be? | Answer: $a=2, b=-8, c=7$ or $a=1, b=-3, c=1$ or $a=1, b=-5, c=5$.
## Solution.
Note that a quadratic trinomial can take the same value at no more than two different points. Therefore, the numbers $f(1), f(2), f(3)$ cannot all have the same sign. Thus, exactly two of these numbers must take the same value. If a quadra... | =2,b=-8,=7or=1,b=-3,=1or=1,b=-5,=5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,608 |
4.2. For the quadratic trinomial $f(x)=a x^{2}+b x+c$, where $a>0$, the condition $|f(1)|=|f(2)|=|f(3)|=2$ is satisfied. What can the coefficients be? | Answer: $a=4, b=-16, c=14$ or $a=2, b=-6, c=2$ or $a=2, b=-10, c=10$. | =4,b=-16,=14or=2,b=-6,=2or=2,b=-10,=10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,609 |
4.3. For the quadratic trinomial $f(x)=a x^{2}+b x+c$, where $a>0$, the condition $|f(1)|=|f(2)|=|f(3)|=3$ is satisfied. What can the coefficients be? | Answer: $a=6, b=-24, c=21$ or $a=3, b=-15, c=15$ or $a=3, b=-9, c=3$ | =6,b=-24,=21or=3,b=-15,=15or=3,b=-9,=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,610 |
4.4. For the quadratic trinomial $f(x)=a x^{2}+b x+c$, where $a>0$, the condition $|f(1)|=|f(2)|=|f(3)|=4$ is satisfied. What can the coefficients be? | Answer: $a=8, b=-32, c=28$ or $a=4, b=-20, c=20$ or $a=4, b=-12, c=4$ | =8,b=-32,=28or=4,b=-20,=20or=4,b=-12,=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,611 |
5. 5.1. An isosceles trapezoid \(ABCD\) is inscribed in a circle with diameter \(AD\) and center at point \(O\). A circle is inscribed in triangle \(BOC\) with center at point \(I\). Find the ratio of the areas of triangles \(AID\) and \(BIC\), given that \(AD=15, BC=5\). | Answer: 9.
## Solution.
From the fact that $A D$ is the diameter of the circle circumscribed around the trapezoid, it follows that $A O = B O = C O = D O$ as radii. Therefore, triangle $B O D$ is isosceles. Hence, $\angle O B D = \angle O D B = \angle D B C$ (the last equality is due to $A D \parallel B C$). Thus, $B... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,612 |
6. Find the largest natural number in which all digits are different and any two adjacent digits differ by 6 or 7. | Answer: 60718293.
## Solution.
We will map each digit from 0 to 9 to a vertex in a graph and connect the vertices with an edge if the corresponding digits differ by 6 or 7.
We see that the ... | 60718293 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,613 |
7. Find the smallest positive integer $n$ such that $A_{n}=1+11+111+\ldots+1 \ldots 1$ (the last term contains $n$ ones) is divisible by 45. | Answer: 35.
## Solution.
For the sum $A_{n}$ to be divisible by 45, it must be divisible by 5 and by 9. By the divisibility rule for 5, we get that $n$ must be a multiple of 5. Since any natural number gives the same remainder when divided by 9 as the sum of its digits, $A_{n}$ gives the same remainder when divided b... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,614 |
8. 8.1. The vertices of a regular 22-sided polygon are numbered. In how many ways can four of its vertices be chosen to form a trapezoid? (A trapezoid is a quadrilateral with one pair of parallel sides and the other two sides not parallel). | Answer: 990.
## Solution.
Note that all trapezoids we obtain will be isosceles. We will count the number of inscribed quadrilaterals that have an axis of symmetry that does not pass through the vertices. Thus, we will count all trapezoids once (since each isosceles trapezoid has only one axis of symmetry), and count ... | 990 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,615 |
1.1. The product of 3 integer numbers is -5. What values can their sum take? Be sure to list all options. | Answer: 5 or -3 or -7
Solution. Since 5 is not divisible by any number other than 1 and itself, the only three factors that give a product of 5 are 1, 1, and 5. Since the sign of the product is negative, either one or all three numbers are negative. We get the following options: $1+1-5=-3 ; 1-1+5=5$; $-1-1-5=-7$. | 5or-3or-7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,616 |
2.1. Petya writes on the board such different three-digit natural numbers that each of them is divisible by 3, and the first two digits differ by 2. What is the maximum number of such numbers he can write if they end in 6 or 7? | Answer: 9
Solution. A number is divisible by 3 if the sum of its digits is a multiple of 3. If the number ends in 6, then the sum of the other two digits leaves a remainder of 0 when divided by 3. Such numbers are: 2,4 and 4,2; 5,7 and 7,5. If the number ends in 7, then the sum of the other two digits leaves a remaind... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,620 |
3.1. Every hour, between two adjacent nettle bushes in a row, two more of the same grow. How many bushes do you need to plant initially so that after three hours, the total number of bushes together is 190? | Answer: 8
Solution. If at the moment there are $\mathrm{n}$ bushes, then on the next move their number increases by $2(n-1)$. Thus, if after 3 hours the total number of bushes should be 190, then one hour before that, there should be $\frac{190-1}{3}+1=64$. One more hour back, $\frac{64-1}{3}+1=22$. And on the next mo... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,621 |
5.1. 40 people came into a room where there were 40 chairs, black and white, and sat on them. All of them said they were sitting on black chairs. Then they somehow resat, and exactly 16 claimed they were sitting on white chairs. Each of those sitting either lied both times or told the truth both times. How many of them... | Answer: 8
Solution. Initially, everyone who told the truth sat on black chairs, and everyone who lied sat on white ones. After some of them switched chairs, 16 claimed they were sitting on white chairs. Obviously, this group includes those who told the truth and were sitting on white chairs, and those who switched wit... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,623 |
6.1. On a circular road, there are three cities: $A, B$ and $C$. It is known that the path from $A$ to $C$ is 3 times longer along the arc not containing $B$ than through $B$. The path from $B$ to $C$ is 4 times shorter along the arc not containing $A$ than through $A$. How many times shorter is the path from $A$ to $B... | Answer: 19
Solution. From the condition of the problem, the following relationship follows:
$$
\begin{aligned}
& x=3(y+z) \\
& 4 z=x+y \\
& 3 y+3 z=4 z-y \\
& 4 y=z \\
& x=15 y \\
& x+z=19 y
\end{aligned}
$$
Let the first two volumes be Pushkin's, $s_{1}$ - the number of them read by Fyodor; the last two - Lermontov... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,626 |
1. Two boys, Petya and Vasya, are walking from the same house to the same school at the same time. Petya's step is $25\%$ shorter than Vasya's, but in the same amount of time, he manages to take $25\%$ more steps than Vasya. Who will arrive at the school first? | # Solution
Let's take Vasya's step as 1, then Petya's step is 0.75. For every 100 steps Vasya takes, Petya takes 125 steps. After 100 steps, Vasya will have walked a distance of 100, while Petya will have walked $0.75 * 125 = 75$ in the same time. Therefore, Vasya will arrive faster.
## Grading Criteria
Correct answ... | Vasya | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,627 |
2. When adding two integers, a student mistakenly added an extra zero at the end of the second addend and got a sum of 7182 instead of 3132. Determine the addends.
# | # Solution
By subtracting the second sum from the first, we find the second addend multiplied by 9. Thus, the required numbers are 2682 and 450.
## Grading Criteria
The numbers are correct, but the method of obtaining them is not specified - 1 point.
The correct answer is obtained with a valid explanation - 7 point... | 2682450 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,628 |
4. Is it possible to arrange 10 black and several white chips in a circle such that each black chip has a diametrically opposite white chip and no two white chips are adjacent? | Solution
Since each black chip corresponds to a diametrically opposite white chip and no two white chips are adjacent, the chips must alternate and there must be an equal number of each.
There are nine chips between a black and a white chip on the semicircle, so the outermost chips are of the same color, which means ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,630 |
# Problem 10.2 (7 points)
Vasya, Petya, and 2020 other people stood in a circle, with Vasya and Petya not standing next to each other. Then Vasya chooses one of his two neighbors and "spots" them (taps them on the shoulder). Next, Petya does the same, followed by Vasya again, and so on. The person who is spotted leave... | # Solution:
Neither player wants to create a gap of length 0 between them, as this would result in an immediate loss. Therefore, at some point, both gaps will contain only one person. By this point, $2020-2=2018$ moves will have been made (the last move was by the second player, Petya). Thus, Vasya is forced to create... | Petya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,633 |
# Problem 10.3 (7 points)
In opposite corners of a $2020 \times 2021$ grid, two grasshoppers are sitting. On each move, they simultaneously jump over a side to an adjacent cell. Will the grasshoppers ever be able to end up in the same cell? | # Solution:
Let's color the cells of the table in a checkerboard pattern. The grasshoppers will end up in cells of different colors. After each move, the grasshoppers will simultaneously change the color of their cells and again end up in cells of different colors. Therefore, they will never be able to end up in the s... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,634 |
# Problem 10.4 (7 points)
In triangle $\mathrm{ABC}$, the angle bisectors $\mathrm{AA}_{1}$ and $\mathrm{CC}_{1}$ are drawn. The radii of the circles inscribed in triangles $\mathrm{AA}_{1} \mathrm{C}$ and $\mathrm{CC}_{1} \mathrm{~A}$ are equal. Prove that triangle $\mathrm{ABC}$ is isosceles. | # Solution:
Proof.
If $\mathrm{O}_{1}$ and $\mathrm{O}_{2}$ are the centers of the circles inscribed in triangles $\mathrm{CC}_{1}A$ and $\mathrm{AA}_{1}C$, and $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ are the points of tangency of these circles with side $\mathrm{AC}$, then $\mathrm{O}_{1} \in \mathrm{AA}_{1}, \mathrm{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,635 |
6.1. Replace each asterisk in the example of adding decimal fractions
$$
0, * *+0, * *+0, * *+0, * *=1
$$
with the digit 2 or the digit 3 so that the equation is correct. | Solution: The sum of the four digits in the hundredths place must be a multiple of 10. Since each of the digits is either 2 or 3, there must be exactly two 2s and two 3s among these digits. Then, the sum of the digits in the tens place should end in the digit 9, so among the tens digits, there are three 2s and one 3. I... | 0.32+0.22+0.23+0.23=1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,636 |
6.2. The sixth-grade students went to a party. Each boy had 5 balloons, and each girl had 4 balloons. On the way, the children started to play and pop each other's balloons. (Of course, they did not pop their own balloons.) In the end, each girl popped exactly one balloon, and each boy popped exactly two balloons. Dima... | Solution: Let's assume that the student who punctured the balloon compensates the victim by giving one of their own. This is equivalent to children only puncturing their own balloons. Then, each girl will part with one balloon, and each boy with two. Each sixth-grader will be left with three balloons, so the total numb... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,637 |
6.3. Petya, Kolya, and Vasya started simultaneously in a 100-meter race, and Petya came in first. When Petya had run half the distance, Kolya and Vasya had run a total of 85 meters. It is known that the speed of each of the three boys was constant throughout the distance. How many meters in total did Kolya and Vasya ha... | Solution: Since the speeds of each of the boys are constant, during the time it takes Petya to run the second half of the distance, Kolya and Vasya together will run the same 85 meters, and in total from the start $2 \cdot 85=170$. They should run 200. Therefore, they have $200-170=30$ meters left to run together.
Ans... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,638 |
6.4. A square (see figure) was cut into rectangles. It turned out that the areas of all six cut-off rectangles are equal to each other. Find how many times the long side of the shaded rectangle is greater than its short side. Justify your answer. | Solution: Since the areas of all hexagons are equal, the area of each is $1 / 6$ of the area of the square, so the three left rectangles make up half of the square. Therefore, the horizontal side of the shaded rectangle is also equal to half the side of the square. Since its area is $1 / 6$ of the area of the square, t... | 1.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,639 |
6.5. There are nuts in three boxes. In the first box, there are six fewer nuts than in the other two boxes combined, and in the second box, there are ten fewer nuts than in the other two boxes combined. How many nuts are in the third box? Justify your answer. | Solution: Let there be $x$ nuts in the first box, $y$ and $z$ in the second and third boxes, respectively. Then the condition of the problem is given by the equations $x+6=y+z$ and $x+z=y+10$. From the first equation, $x-y=z-6$, and from the second, $x-y=10-z$. Therefore, $z-6=10-z$, from which $z=8$.
Answer: 8 nuts.
... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,640 |
10.1. The first 20 numbers of the natural number sequence were written on the board. When one of the numbers was erased, it turned out that among the remaining numbers, one is the arithmetic mean of all the others. Find all the numbers that could have been erased. | Answer: 1 and 20.
Solution. The sum of the first twenty numbers in the natural number sequence is $(1+20) \cdot 10=210$. Therefore, if one of them is erased, the sum $S$ of the remaining numbers satisfies the inequality $190 \leq S \leq 209$. The arithmetic mean of the remaining numbers is $S / 19$. According to the c... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,641 |
10.2. To transport 80 tons of cargo, a truck required 5 more trips than to transport 40 tons of the same cargo. On all trips, except possibly the last one in each transportation, the truck was fully loaded. What values can the truck's carrying capacity take? | Answer: $7 \frac{3}{11} \leq x<8 \frac{8}{9}$.
Solution. Let $x$ be the load capacity of the truck, and $n$ be the number of trips required to transport 40 tons of cargo.
Consider 2 cases:
When transporting 40 tons, in the last trip the truck was loaded to no more than half, i.e., in the $(n-1)$ trips the truck was ... | 7\frac{3}{11}\leqx<8\frac{8}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,642 |
10.3. In the quadrilateral $A B C D$, $A B=2, B C=4, C D=5$. Find its area, given that it is both inscribed and circumscribed. | Answer: $2 \sqrt{30}$.
Solution. Since the quadrilateral $A B C D$ is circumscribed, then $A B+C D=$ $B C+A D$, from which $A D=3$. By the cosine theorem from triangle $A B D$ we get $B D^{2}=A B^{2}+A D^{2}-2 A B \cdot A D \cos A$, and from triangle $B C D$
$$
B D^{2}=B C^{2}+C D^{2}-2 B C \cdot C D \cos C
$$
Since... | 2\sqrt{30} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,643 |
10.4. Find $f(2021)$ if for any real $x$ and $y$ the equality $f(x+f(y))=x+y$ holds. | Answer: 2021.
Solution. Let $f(0)=b$, then for $y=0$ we get $f(x+b)=x$, from which $f(x)=x-b$. Thus, $f(x+f(y))=f(x+(y-b))=f(x+y-b)=x+y-b-b=x+y-2b$.
Since $f(x+f(y))=x+y$ for any real $x$ and $y$, then $b=0$, so $f(x)=x$, and $f(2021)=2021$.
Comment. Only the example of the function $f(x)=x-1$ gets a point. | 2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,644 |
10.5. The cells of a table are filled with the numbers 0 and 1, as shown in the figure.
| 0 | 1 | 0 | 0 |
| :--- | :--- | :--- | :--- |
| 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 0 |
The following operations are allowed:
- add one to all numbers in the same row;
- add one to all numbers in the same column;
- ... | Answer: No.
Solution. Let's mark 8 cells in the table (see the figure).
Each row and each column contain 2 marked cells, and each diagonal contains 0 or 2. Therefore, the parity of the numb... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,645 |
11.1. The quadratic trinomial $f(x)=a x^{2}+b x+c$, which has no roots, is such that the coefficient $b$ is rational, and among the numbers $c$ and $f(c)$, exactly one is irrational. Can the discriminant of the trinomial $f(x)$ be rational?
(Zhukov G.) | Answer. No, it cannot.
Solution. Since the quadratic polynomial $f(x)$ has no roots, then $c=$ $=f(0) \neq 0$ and $f(c) \neq 0$. Therefore, the expression $\frac{f(c)}{c}$ is irrational as the ratio of a rational and an irrational number. But $\frac{f(c)}{c}=$ $=\frac{a c^{2}+b c+c}{c}=a c+b+1$. Since $b+1$ is rationa... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,646 |
11.2. Positive numbers $x, y$, and $z$ satisfy the condition $x y z \geqslant$ $\geqslant x y + y z + z x$. Prove the inequality
$$
\sqrt{x y z} \geqslant \sqrt{x} + \sqrt{y} + \sqrt{z}
$$
(A. Khryabrov) | Solution. By the inequality of means, we have $x y+x z \geqslant 2 \sqrt{x y \cdot x z}, \quad x y+y z \geqslant 2 \sqrt{x y \cdot y z}, \quad x z+y z \geqslant 2 \sqrt{x z \cdot y z}$. Adding these three inequalities and dividing the result by 2, and considering the condition, we get
$$
x y z \geqslant x y+x z+y z \g... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 13,647 |
11.3. In triangle $A B C$, the bisector $B L$ is drawn. A point $M$ is chosen on the segment $C L$. The tangent at point $B$ to the circumcircle $\Omega$ of triangle $A B C$ intersects the ray $C A$ at point $P$. The tangents at points $B$ and $M$ to the circumcircle $\Gamma$ of triangle $B L M$ intersect at point $Q$.... | Solution. Since $BL$ is the bisector of $\angle ABC$, we have $\angle ABL = \angle LBC$. Since $PB$ is tangent to $\Omega$, we have $\angle PBA = \angle BCA$ (see Fig. 4). Additionally, $\angle PBL = \angle PBA + \angle ABL = \angle BCA + \angle LBC = \angle BLP$, so $\angle BPM = 180^\circ - (\angle PBL + \angle BLP) ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,648 |
11.4. There is a grid board $2015 \times 2015$. Dima places a detector in $k$ cells. Then Kolya places a grid ship in the shape of a square $1500 \times 1500$ on the board. The detector in a cell reports to Dima whether this cell is covered by the ship or not. For what smallest $k$ can Dima place the detectors in such ... | Answer. $k=2(2015-1500)=1030$.
Solution. We will show that 1030 detectors are enough for Dima. Let him place 515 detectors in the 515 leftmost cells of the middle row of the square, and the remaining 515 detectors in the 515 top cells of the middle column. Note that for any position of the ship, its left column lies i... | 1030 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,649 |
4-1. Katya attached a square with a perimeter of 40 cm to a square with a perimeter of 100 cm as shown in the figure. What is the perimeter of the resulting figure in centimeters?
 | Answer: 120.
Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,650 |
4-5. Given a figure consisting of 33 circles. You need to choose three circles that are consecutive in one of the directions. In how many ways can this be done? The image shows three of the desired ways.
. When Petya was at the 22nd... | Answer. At the 64th lamppost.
Solution. There are a total of 99 intervals between the lampposts. From the condition, it follows that while Petya walks 21 intervals, Vasya walks 12 intervals. This is exactly three times less than the length of the alley. Therefore, Petya should walk three times more to the meeting poin... | 64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,654 |
5-1. A square with a side of 100 was cut into two equal rectangles. They were placed next to each other as shown in the figure. Find the perimeter of the resulting figure.
 | Answer: 500.
Solution. The perimeter of the figure consists of 3 segments of length 100 and 4 segments of length 50. Therefore, the length of the perimeter is
$$
3 \cdot 100 + 4 \cdot 50 = 500
$$ | 500 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,655 |
5-5. Along a straight alley, 400 lamps are placed at equal intervals, numbered in order from 1 to 400. At the same time, from different ends of the alley, Alla and Boris started walking towards each other at different constant speeds (Alla from the first lamp, Boris from the four hundredth). When Alla was at the 55th l... | Answer. At the 163rd lamppost.
Solution. There are a total of 399 intervals between the lampposts. According to the condition, while Allа walks 54 intervals, Boris walks 79 intervals. Note that $54+79=133$, which is exactly three times less than the length of the alley. Therefore, Allа should walk three times more to ... | 163 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,656 |
5-6. A rectangular table of size $x$ cm $\times 80$ cm is covered with identical sheets of paper of size 5 cm $\times 8$ cm. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right... | Answer: 77.
Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
Let's call such ... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,657 |
5-7. On the faces of a die, the numbers $6,7,8,9,10,11$ are written. The die was rolled twice. The first time, the sum of the numbers on the four "vertical" (that is, excluding the bottom and top) faces was 33, and the second time - 35. What number can be written on the face opposite the face with the number 7? Find al... | Answer: 9 or 11.
Solution. The total sum of the numbers on the faces is $6+7+8+9+10+11=51$. Since the sum of the numbers on four faces the first time is 33, the sum of the numbers on the two remaining faces is $51-33=18$. Similarly, the sum of the numbers on two other opposite faces is $51-35=16$. Then, the sum on the... | 9or11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,658 |
6-3. The red segments in the figure have equal length. They overlap by equal segments of length $x$ cm. What is $x$ in centimeters?
 | Answer: 2.5.
Solution. Adding up the lengths of all the red segments, we get 98 cm. Why is this more than 83 cm - the distance from edge to edge? Because all overlapping parts of the red segments have been counted twice. There are 6 overlapping parts, each with a length of $x$. Therefore, the difference $98-83=15$ equ... | 2.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,660 |
8-1. Two rectangles $8 \times 10$ and $12 \times 9$ are overlaid as shown in the figure. The area of the black part is 37. What is the area of the gray part? If necessary, round the answer to 0.01 or write the answer as a common fraction.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,663 |
9-1. Segment $P Q$ is divided into several smaller segments. On each of them, a square is constructed (see figure).
What is the length of the path along the arrows if the length of segment ... | Answer: 219.
Solution. Note that in each square, instead of going along one side, we go along three sides. Therefore, the length of the path along the arrows is 3 times the length of the path along the segment, hence the answer $73 \cdot 3=219$. | 219 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,664 |
10-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,666 |
11-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 58.
Solution. Angle $ACD$ is a right angle since it subtends the diameter of the circle.
Therefore, $\angle CAD = 90^{\circ} - \angle CDA = 48^{\circ}$. Also, $AO = BO = CO$ as they... | 58 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,667 |
10.5. Find the largest natural number $n$, for which the product of the numbers $n, n+1, n+2, \ldots, n+20$ is divisible by the square of one of them. | Answer: 20!.
Solution. For $n=20$ ! we have $\frac{n(n+1)(n+2) \ldots(n+20)}{n^{2}}=$ $=\frac{(n+1)(n+2) \ldots(n+20)}{20!}=C_{n+20}^{20}-$ an integer.
Now let $n>20$ ! and suppose $P=n(n+1)(n+2) \ldots(n+20)$ is divisible by $k^{2}$, where $k=n+i, i \in\{0,1,2, \ldots, 20\}$. We have $P / k=$ $=(k-i)(k-i+1) \ldots(k... | 20! | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,668 |
10.6. A square $100 \times 100$ is divided into squares $2 \times 2$. Then it is divided into dominoes (rectangles $1 \times 2$ and $2 \times 1$). What is the smallest number of dominoes that could end up inside the squares of the division?
(C. Berlov)
# | # Answer. 100.
Solution. Example. We will divide the top and bottom horizontals into horizontal dominoes - they will end up in $2 \times 2$ squares. The remaining rectangle $98 \times 100$ will be divided into vertical dominoes - they will not end up in $2 \times 2$ squares.
Estimate. Consider the squares $A_{1}, A_{... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,669 |
10.7. Given a trapezoid $A B C D$, where $A D \| B C$, and rays $A B$ and $D C$ intersect at point $G$. The common external tangents to the circumcircles of triangles $A B C$ and $A C D$ intersect at point $E$. The common external tangents to the circumcircles of triangles $A B D$ and $B C D$ intersect at point $F$. Pr... | Solution. Let the line $EC$ intersect the circle $(ABC)$ again at point $X$, and the line $EA$ intersect the circle $(ACD)$ again at point $Y$ (we will consider the arrangement of points as shown in Fig. 5; other cases are treated similarly).
Consider the homothety with center $E$ that maps $(ABC)$ to $(ACD)$. Under t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,670 |
10.8. Given a number $a \in(0,1)$. Positive numbers $x_{0}, x_{1}, \ldots, x_{n}$ satisfy the conditions $x_{0}+x_{1}+\ldots+x_{n}=n+a$ and $\frac{1}{x_{0}}+\frac{1}{x_{1}}+$ $+\ldots+\frac{1}{x_{n}}=n+\frac{1}{a}$. Find the minimum value of the expression $x_{0}^{2}+x_{1}^{2}+\ldots+x_{n}^{2}$.
(A. Khryabrov) | Answer: $n+a^{2}$.
Solution: Note that equality is achieved when $x_{0}=a$ and $x_{1}=\ldots=x_{n}=1$.
Write $\sum x_{k}^{2}=\sum\left(1-x_{k}\right)^{2}+2 \sum x_{k}-(n+1)=\sum\left(1-x_{k}\right)^{2}+$
$+n-1+2 a$. It is sufficient to prove that $\sum\left(1-x_{k}\right)^{2} \geqslant(1-a)^{2}$. Let $x_{0}$ be the s... | n+^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,671 |
1. Find the smallest 10-digit number, the sum of whose digits is not less than that of any smaller number. | Answer: 1899999999.
Solution. Among 9-digit numbers, the largest sum of digits is for the number 999999999, which is 81. Since the desired 10-digit number is greater than 999999999, we need to find the smallest number with a sum of digits no less than 81. The sum of the last eight digits of the desired number is no mo... | 1899999999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,672 |
2. In a chess tournament, everyone played against each other once. The winner won half of the games and drew the other half. It turned out that he scored 13 times fewer points than all the others. (1 point for a win, 0.5 for a draw, 0 for a loss.) How many chess players were there in the tournament? | Answer: 21 chess players.
Solution. If the number of participants is $n$, then each played $n-1$ games. The winner won half of the games and scored $\frac{1}{2}(n-1)$ points. The winner drew half of the games and scored another $\frac{1}{4}(n-1)$ points. In total, the winner scored $\frac{1}{2}(n-1)+\frac{1}{4}(n-1)=\... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,673 |
3. The length of the diagonal of a rectangular parallelepiped is 3. What is the maximum possible value of the surface area of such a parallelepiped? | Answer: 18.
Solution. Let the edges of the rectangular parallelepiped be $-a, b$, and $c$. According to the problem, its diagonal is $\sqrt{a^{2}+b^{2}+c^{2}}=3$, and thus, $a^{2}+b^{2}+c^{2}=9$. The surface area of the parallelepiped is $f=2(a b+b c+c a)$. We will prove the inequality $a b+b c+c a \leqslant a^{2}+b^{... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,674 |
4. All values of the quadratic trinomial $f(x)=a x^{2}+b x+c$ on the interval $[0 ; 2]$ do not exceed 1 in absolute value. What is the greatest value that the quantity $|a|+|b|+|c|$ can have under these conditions? For which function $f(x)$ is this value achieved? | Answer: the maximum value is 7; for example, it is achieved for $f(x)=2 x^{2}-4 x+1$.
Solution. By the condition, the values of $f(x)=a x^{2}+b x+c$ on the interval $[0 ; 2]$ do not exceed one in absolute value. In particular, $|f(0)| \leqslant 1,|f(1)| \leqslant 1,|f(2)| \leqslant 1$, which is equivalent to the syste... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,675 |
5. In triangle $A B C$, the bisectors $B B_{1}$ and $B B_{2}$ of the internal and external angles at vertex $B$ are drawn. From point $H$ of intersection of the altitudes, perpendiculars $H H_{1}$ and $H H_{2}$ are dropped to lines $B B_{1}$ and $B B_{2}$. In what ratio does the line $H_{1} H_{2}$ divide the side $A C$... | Answer: The line $H_{1} H_{2}$ bisects $A C$.
Solution. (Fig. 4.) Let $A D$ and $C E$ be the altitudes of triangle $A B C$. Since the bisectors of the internal and external angles are perpendicular, the quadrilateral $H H_{2} B H_{1}$ is a rectangle. The diagonals $H_{1} H_{2}$ and $B H$ are diameters of the circumscr... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,676 |
10.1. It is known that the quadratic equations in $x$, $2017 x^{2} + p x + q = 0$ and $u p x^{2} + q x + 2017 = 0$ (where $p$ and $q$ are given real numbers) have one common root. Find all possible values of this common root and prove that there are no others. | Solution: Let $x_{0}$ be the common root of these equations, that is, the equalities $2017 x_{0}^{2} + p x_{0} + q = 0$ and $p x_{0}^{2} + q x_{0} + 2017 = 0$ are satisfied. Multiply both sides of the first equation by the number $x_{0}$ and subtract the second equation. We get the consequential equation: $2017 x_{0}^{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,677 |
10.2. On a plane, a semicircle with diameter $A B=36$ cm was constructed; inside it, a semicircle with diameter $O B=18$ cm was constructed ($O-$ the center of the larger semicircle). Then, a circle was constructed that touches both semicircles and the segment АО. Find the radius of this circle. Justify your answer. | Solution: Let the desired radius be $x$. Let points $C$ and $Q$ be the centers of the smaller semicircle and the inscribed circle, respectively, and $D$ be the point of tangency of the inscribed circle with the diameter $AB$. Due to the tangency of the circle and the larger semicircle, the ray $OQ$ passes through the p... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,678 |
10.3. Vasya chose some real number $x$ and wrote down an infinite sequence: $a_{1}=1+x^{2}+x^{3}, a_{2}=1+x^{3}+x^{4}, a_{3}=1+x^{4}+x^{5}, \ldots$, $a_{n}=1+x^{n+1}+x^{n+2}, \ldots$ It turned out that $a_{2}^{2}=a_{1} \cdot a_{3}$. Prove that then for all natural $n \geqslant 3$ the equality $a_{n}^{2}=a_{n-1} \cdot a... | Solution: We will carry out equivalent transformations.
$$
\begin{gathered}
a_{2}^{2}=a_{1} \cdot a_{3} \\
\left(1+x^{3}+x^{4}\right)^{2}=\left(1+x^{2}+x^{3}\right)\left(1+x^{4}+x^{5}\right)
\end{gathered}
$$
$$
\begin{gathered}
1+x^{6}+x^{8}+2 x^{3}+2 x^{4}+2 x^{7}=1+x^{2}+x^{3}+x^{4}+x^{6}+x^{7}+x^{5}+x^{7}+x^{8} \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,679 |
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