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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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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:
 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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.
 | 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
10.5. We will call a natural number semi-prime if it is greater than 25 and is the sum of two distinct prime numbers. What is the maximum number of consecutive natural numbers that can be semi-prime? Justify your answer. | Solution: There are many sets of five consecutive semiprime numbers, for example $30(=13+17), 31(=2+29), 32(=3+29), 33(=2+31), 34$ $(=5+29)$ or $102(=5+97), 103(=2+101), 104(=31+73), 105(=2+103)$, $106(=47+59)$. We will show that there are no six consecutive semiprime numbers.
Indeed, among any six consecutive numbers... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.6. Ali-baba came to a cave where there was gold and diamonds. Ali-baba had one large bag with him. It is known that a full bag of gold weighs 200 kg, and if the entire bag is filled with diamonds, it will weigh 40 kg (the empty bag weighs nothing). A kilogram of gold costs 20 dinars, and a kilogram of diamonds costs... | # Solution:
Method 1. Let Ali-Baba have placed a set of gold and diamonds that gives him the maximum possible revenue - we will call such a bag optimal. Then either the bag is filled to the brim, or it weighs exactly 100 kg: otherwise, more diamonds could be added to it and the revenue would increase. Consider the fir... | 3000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.1. Inside a convex pentagon, a point is marked and connected to all vertices. What is the maximum number of the ten segments drawn (five sides and five segments connecting the marked point to the vertices of the pentagon) that can have a length of 1? (A. Kuznetsov) | Answer: 9 segments.
Solution. First, we prove that all 10 segments cannot have a length of 1. Assume the opposite. Let $A B C D E$ be a pentagon, $O$ be a point inside it, and all 10 drawn segments have a length of 1 (see Fig. 6). Then triangles $O A B, O B C$, $O C D, O D E$, and $O E A$ are equilateral, so $\angle A... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.5. Let's call a trapezoid with bases 1 and 3 a "boat," which is obtained by gluing two triangular half-cells to the opposite sides of a unit square. In a $100 \times 100$ square, an invisible boat is placed (it can be rotated, it does not go beyond the boundaries of the square, its middle cell is entirely on one of ... | Answer: 4000 shots.
First solution. We will call a boat horizontal or vertical depending on whether its parallel sides are horizontal or vertical.
First, we will show that 4000 shots are enough. We divide the $100 \times 100$ square into 400 squares of size $5 \times 5$, and in each square, we make 10 shots as shown ... | 4000 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.1. Masha lives in apartment No. 290, which is located in the 4th entrance of a 17-story building. On which floor does Masha live? (The number of apartments is the same in all entrances of the building on all 17 floors; apartment numbers start from 1.$)$ | # Answer: 7.
Solution. Let $x$ be the number of apartments per floor, then there are $17 x$ apartments in each entrance. Thus, in the first three entrances, there are $51 x$ apartments, and in the first four $68 x$.
If $x \geqslant 6$, then in the first three entrances there are at least 306 apartments, so apartment ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 11.2. On the table, there are 30 coins: 23 ten-ruble coins and 7 five-ruble coins, with 20 of these coins lying heads up and the remaining 10 tails up. What is the smallest $k$ such that among any $k$ randomly selected coins, there will definitely be a ten-ruble coin lying heads up? | Answer: 18.
Solution. If you choose 18 coins, then among them there will be no more than 10 lying heads down, so at least 8 coins will be lying heads up. Among these coins, no more than 7 will be five-ruble coins, so at least one will be a ten-ruble coin, which is what we need.
On the other hand, if initially on the ... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.3. The product of positive numbers $a$ and $b$ is 1. It is known that
$$
(3 a+2 b)(3 b+2 a)=295
$$
Find $a+b$. | Answer: 7.
Solution. Expanding the brackets, we get
$$
295=6 a^{2}+6 b^{2}+13 a b=6\left(a^{2}+b^{2}\right)+13
$$
from which $a^{2}+b^{2}=47$. Then
$$
(a+b)^{2}=a^{2}+b^{2}+2 a b=47+2=49=7^{2}
$$
which gives $a+b=7$ (note that $a+b>0$, since $a>0$ and $b>0$). | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.4. A convex pentagon $A B C D E$ is such that $\angle A B C=128^{\circ}, \angle C D E=9^{\circ}$ $104^{\circ}, A B=B C, A E=E D$. How many degrees does the angle $A D B$ measure?
 | Answer: 26.
Fig. 14: to the solution of problem 11.4
Solution. The angles at the base of the isosceles triangle $ABC$ are equal to $\frac{1}{2}\left(180^{\circ}-128^{\circ}\right)=26^{\circ}... | 26 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. For what least natural $n$ can the numbers from 1 to $n$ be arranged in a circle so that each number is either greater than all 40 following it in the clockwise direction, or less than all 30 following it in the clockwise direction? | Answer: 70.
Solution. If $n \leqslant 39$, then for the number $n$ the condition cannot be satisfied: it cannot be greater than the next 40 numbers (since it is not greater than itself), nor can it be less than the next 30 numbers (since it is the largest).
If $40 \leqslant n \leqslant 69$, then for the number 40 the... | 70 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. The polynomial $P(x)$ has all coefficients as non-negative integers. It is known that $P(1)=4$ and $P(5)=152$. What is $P(11) ?$ | Answer: 1454.
Solution. Suppose the degree of the polynomial $P$ is not less than 4, then its leading coefficient is not less than 1. Since all other coefficients of $P(x)$ are non-negative, then $P(5) \geqslant 5^{4}=625$, which contradicts the condition $P(5)=152$.
Therefore, the degree of the polynomial $P$ is not... | 1454 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.8. In a $28 \times 35$ table, some $k$ cells are painted red, another $r-$ in pink, and the remaining $s-$ in blue. It is known that
- $k \geqslant r \geqslant s$
- each boundary cell has at least 2 neighbors of the same color;
- each non-boundary cell has at least 3 neighbors of the same color.
What is th... | Answer: 28.
Solution. From the condition, it is easy to understand that each cell can have no more than one neighbor of a different color.
We will prove that the coloring of the table must be "striped," meaning that either each row or each column is completely colored in one color. To do this, it is sufficient to sho... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. At a round table, 30 people are sitting, each of whom is either a knight, who always tells the truth, or a liar, who always lies. Each person was asked: "How many knights are among your neighbors?" (Two people are called neighbors of each other if there is no one else sitting between them.) 10 people answered "one,"... | Answer: 22 knights.
Solution. From the condition of the problem, it follows that the number of knights who have two knight-neighbors does not exceed 10; the same can be said about the number of knights who have one knight-neighbor. Therefore, the number of pairs of knight-neighbors does not exceed $(2 \cdot 10 + 1 \cd... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In triangle $A B C$, angle $A$ is $75^{\circ}$, and angle $C$ is $60^{\circ}$. On the extension of side $A C$ beyond point $C$, segment $C D$ is laid off, equal to half of side $A C$. Find angle $B D C$. | Answer: $\angle B D C=45^{\circ}$.
Solution. In this triangle, angle $A B C$ is obviously equal to $180^{\circ}-75^{\circ}-60^{\circ}=45^{\circ}$. Draw the altitude $A H$ and note that triangle $A H B$ is isosceles, since $\angle B A H=90^{\circ}-45^{\circ}=\angle A B H$ (Fig. 1). Therefore, $A H=B H$. In the right tr... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 10.5
The field is a $41 \times 41$ grid, in one of the cells of which a tank is hidden. The fighter aircraft shoots 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 ta... | # Answer:
2521 shots
## Solution
## Example.
Let's color the cells in a checkerboard pattern so that the corner cells are black. Suppose the pilot first shoots at all the white cells, then at all the black ones, and then again at all the white ones. If the tank was on a white cell, the pilot will destroy it in the ... | 2521 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Real numbers $a$ and $b$ are such that $a^{5}+b^{5}=3, a^{15}+b^{15}=9$. Find the value of the expression $a^{10}+b^{10}$. | Answer: 5.
Solution. Let $x=a^{5}, y=b^{5}$. Then $x+y=3, x^{3}+y^{3}=9 ;(x+y)\left(x^{2}-x y+y^{2}\right)=9$; $3\left(x^{2}-x y+y^{2}\right)=9 ; x^{2}-x y+y^{2}=3 ; x^{2}+2 x y+y^{2}=9 ; 3 x y=6 ; x y=2 ; x^{2}+y^{2}=9-2 x y=9-$ $2 \cdot 2=5 ; a^{10}+b^{10}=x^{2}+y^{2}=5$
Criteria. Simplification using substitution w... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In a company of 8 people, each person is acquainted with exactly 6 others. In how many ways can four people be chosen such that any two of them are acquainted? (We assume that if A is acquainted with B, then B is also acquainted with A, and that a person is not acquainted with themselves, as the concept of acquainta... | Answer: 16.
Solution. Each person in the company does not know $8-1-6=1$ person in this company. This means the company can be divided into four pairs of strangers. From each pair, we can only take one person to form a company of four people who are all acquainted with each other. The selection of such a company consi... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.1. In each cell of a $6 \times 6$ table, numbers are written. All the numbers in the top row and all the numbers in the left column are the same. Each of the other numbers in the table is equal to the sum of the numbers written in the two adjacent cells - the cell to the left and the cell above. What number can be w... | Answer: 8. Let $a$ be the number written in the corner cell at the top left. Then the table is filled in sequentially. For the bottom right corner, we get the relation $252 a = 2016$, from which it follows that $a = 8$.
Comment. An answer with a chain of calculations - 7 points. An answer obtained using combinatorial ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.4. A set of composite numbers from the set $\{1,2,3,4, \ldots, 2016\}$ is called good if any two numbers in this set do not have common divisors (other than 1). What is the maximum number of numbers that a good set can have? | Answer: 14 numbers.
Example. $2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}, 17^{2}, 19^{2}, 23^{2}, 29^{2}, 31^{2}, 37^{2}, 41^{2}, 43^{2}$. All the presented numbers are composite, as they are squares of natural numbers greater than 1. It is also clear that different prime numbers do not have common divisors greater th... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. Write the number 2013 several times in a row so that the resulting number is divisible by 9. Explain your answer. | Answer: for example, 201320132013.
Solution. We will provide several ways to justify this.
First method. The sum of the digits of the written number is $(2+0+1+3) \cdot 3=18$, so it is divisible by 9 (by the divisibility rule for 9).
Second method. The sum of the digits of the number 2013 is divisible by 3, so if we... | 201320132013 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. In the sum $+1+3+9+27+81+243+729$, any addends can be crossed out and the signs before some of the remaining numbers can be changed from “+” to “-”. Masha wants to use this method to first obtain an expression whose value is 1, then, starting over, obtain an expression whose value is 2, then (starting over again) ... | Answer: up to the number 1093 (inclusive).
Solution. The number 1 is obtained by erasing all addends except the first. Then Masha can get the numbers $2=-1+3, 3=+3$, and $4=+1+3$.
We will show that by adding the addend 9, one can obtain any integer from $5=9-4$ to $13=9+4$. Indeed, $5=-1-3+9; 6=-3+9; 7=+1-3+9; 8=-1+9... | 1093 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. On January 1, 2013, a little boy was given a bag of chocolate candies, containing 300 candies. Each day, the little boy ate one candy. On Sundays, Karlson would fly over, and the little boy would treat him to a couple of candies. How many candies did Karlson eat? (January 1, 2013, was a Tuesday).
2. Petya can swap ... | 1. Answer: 66.
Solution. In a full week, Little One eats 7 candies (one each day), and Karlson eats 2 (on Sunday), so together they eat 9 candies. Dividing 300 by 9 with a remainder, we get the quotient 33 and the remainder 3 (33$\cdot$9+3=300). In the last incomplete week, Little One will eat 3 candies: on Wednesday,... | 66 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. A $5 \times 5$ square is to be cut into two types of rectangles: $1 \times 4$ and $1 \times 3$. How many rectangles can result from the cutting? Justify your answer. | 5. Answer: 7.
Solution. Example:
If there are no more than 6 rectangles, then they occupy no more than $6 \cdot 4 = 24$ cells, while there are 25 cells. Therefore, there must be at least 7 r... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. 50 students from fifth to ninth grade published a total of 60 photos on Instagram, each not less than one. All students of the same grade (same parallel) published an equal number of photos, while students of different grades (different parallels) published a different number. How many students published only one ph... | Solution. Let each of the students publish one photo first, during which 50 photos out of 60 will be published. There are 10 photos left to be published, which we will call additional photos. We have a total of five different classes (parallels), and these 10 additional photos must be published by different students fr... | 46 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Find the value of the expression
$$
2^{2}+4^{2}+6^{2}+\ldots+2018^{2}+2020^{2}-1^{2}-3^{2}-5^{2}-\ldots-2017^{2}-2019^{2}
$$ | # Solution.
$$
\begin{aligned}
& 2020^{2}-2019^{2}+2018^{2}-2017^{2}+\ldots+6^{2}-5^{2}+4^{2}-3^{2}+2^{2}-1^{2}= \\
& =(2020-2019)(2020+2019)+(2018-2017)(2018+2017)+\ldots+ \\
& +(6-5)(6+5)+(4-3)(4+3)+(2-1)(2+1)= \\
& =2020+2019+2018+2017+\ldots+6+5+4+3+2+1= \\
& =\frac{2020+1}{2} \cdot 2020=2041210
\end{aligned}
$$
... | 2041210 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Given the natural numbers $1,2,3 \ldots, 10,11,12$. Divide them into two groups such that the quotient of the product of all numbers in the first group divided by the product of all numbers in the second group is an integer, and takes the smallest possible value. What is this quotient? | Solution. Let's factorize the given numbers into prime factors and find their product.
$$
1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 10 \cdot 11 \cdot 12=2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7 \cdot 11
$$
The factors 7 and 11 do not have pairs. One of the factors 3 does not have a pair. Therefore, to make the quoti... | 231 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. For what value of the parameter m is the sum of the squares of the roots of the equation
$$
x^{2}-(m+1) x+m-1=0
$$
the smallest? | 3. $\mathrm{m}=3$.
Note that the discriminant $\mathrm{D}$ of the equation is equal to
$$
\mathrm{D}=(\mathrm{m}+1)^{2}-4(\mathrm{~m}-1)=(\mathrm{m}-1)^{2}+4>0
$$
therefore, the equation has two roots for any $\mathrm{m}$. Let's represent the sum of the squares of the roots $\mathrm{x}_{1}$ and $\mathrm{x}_{2}$ as
... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.1. Arina wrote down a number and the number of the month of her birthday, multiplied them and got 248. In which month was Arina born? Write the number of the month in your answer.
 | Answer: 8
Solution. $248=2 * 2 * 2 * 31$. Obviously, 31 is the number, then 8 is the month number. | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3.1. A number was multiplied by its first digit and the result was 494, by its second digit - 988, by its third digit - 1729. Find this number. | Answer: 247
Solution. It can be noticed that all three obtained numbers are divisible by 247, from which we get the answer. | 247 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. It is known that in the past chess tournament, in each round all players were paired, the loser was eliminated (there were no draws). It is known that the winner played 6 games. How many participants in the tournament won at least 2 games more than they lost? | Answer: 8
Solution. Since the winner played 6 games, there were a total of $2^{6}=64$ players. The win-loss ratio for those eliminated in the 1st round is -1; for those who lost in the 2nd round, it is 0; for those who lost in the 3rd round, it is +1; for those who lost in the 4th round and beyond, it is at least +2. ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.1. In a basketball team, there is a certain number of players. The coach added up all their heights and divided by the number of players (let's call this the average height), getting 190 cm. After the first game, the coach removed Nikolai, who is 197 cm tall, and replaced him with Petr, who is 181 cm tall, after whic... | Answer: 8
Solution. The change in the total height of the players by $197-181=16$ cm leads to a change in the average height of the team by $190-188=2$ cm, so the team has $\frac{16}{2}=8$ people. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. On the counter lie 10 weights with masses $n, n+1, \ldots, n+9$. The seller took one of them, after which the total weight of all the remaining ones was 1457. Which weight did the seller take? | # Answer: 158
Solution. The total weight of all 10 weights is $10 n+45$. Suppose the weight taken has a weight of $n+x$, where $x<10$. Then
$10 n+45-n-x=1457$.
$9 n+45-x=1457$.
We can see that 1457, when divided by 9, gives a remainder of 8, so $x=1$. Then
$9 n+45-1=1457$
$9 n=1413$.
$n=157$.
$n+x=158$. | 158 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. In a circle, 58 balls of two colors - red and blue - are arranged. It is known that the number of triples of consecutive balls, among which there are more red ones, is the same as the number of triples with a majority of blue ones. What is the smallest number of red balls that could be present?
, the elements of each row and each column form arithmetic progressions. What is the number $x$ in the central cell?
 | Answer: 111.
Solution. The middle term of an arithmetic progression is equal to the half-sum of the extreme terms, so in the middle cell of the first row, the number is $(3+143) / 2=73$, and in the middle cell of the last row, the number is $(82+216) / 2=149$.
$ points. According to the problem, the ratio of the number of points scored by girls to the number of points scored by boys is 1:4. Therefore, the gir... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find all natural $\boldsymbol{n}$ such that the value of the expression $\frac{2 n^{2}-2}{n^{3}-n}$ is a natural number. | Solution. $\frac{2 n^{2}-2}{n^{3}-n}=\frac{2(n-1)(n+1)}{n(n-1)(n+1)}$. For $n \neq \pm 1$, this expression is identically equal to $\frac{2}{n}$. Since $n \neq 1$, the value of the last expression is a natural number only when $n=2$.
Answer: $n=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) The number 20 is written on the board. In one move, it is allowed to either double the number or erase its last digit. Is it possible to get the number 25 in several moves? | Answer. Possible.
Solution. The number 25 can be obtained by erasing the last digit of the number 256, which is a power of two. Thus, the required chain of transformations can look like this:
$$
20 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 16 \rightarrow 32 \rightarrow 64 \rightarrow 128 \rightarrow 256 \... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. How many emeralds are there in total in the boxes?
Fig. 1: to the solution of problem 8.4
Notice that $\angle A B K=\angle C B L$, since they both complement $\angle... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
 standing consecutively. Find the number of chains, the sum of the numbers in which is positive.
# | # Answer: 4951
Solution. Let's divide all chains (except the chain consisting of all numbers) into pairs that complement each other. If the sum of the numbers in one chain of the pair is $s$, and in the second is $t$, then $s+t=1$. Since $s$ and $t$ are integers, exactly one of them is positive. Therefore, exactly hal... | 4951 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.5 The square of a natural number a, when divided by a natural number n, gives a remainder of 8. The cube of the number a, when divided by n, gives a remainder of 25. Find n. | Answer: $n=113$
Solution. Note that the number $x=a^{6}-8^{3}=(a^{2})^{3}-8^{3}=(a^{2}-8)(a^{4}+8a^{2}+64)$ is divisible by $n$. Also note that the number $y=a^{6}-25^{2}=(a^{3})^{2}-25^{2}=(a^{3}-25)(a^{3}+25)$ is divisible by $n$. Then the difference $x-y=(a^{6}-8^{3})-(a^{6}-25^{2})=625-512=113$ is divisible by $n$... | 113 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The number 1232 is divisible by the sum of its digits $1+2+3+2=8$. What is the next number that is divisible by the sum of its digits? | Answer: 1233.
Solution: One can notice that the next number fits, as 1233 is divisible by the sum of its digits $1+2+3+3=9$. Indeed, $1233: 9=137$.
Comment: One could have avoided performing the division and instead used the divisibility rule for 9. | 1233 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. A farmer wants to start growing watermelons. He wants to sell at least 10,000 watermelons every year. Watermelons are grown from seeds (one seed grows into one watermelon). Each watermelon can produce 250 good seeds that can be planted the following year, but then this watermelon cannot be sold. What is the minimum ... | Answer: 10041.
Solution: Let the smallest number of seeds the farmer needs to buy be $10000+x$. Then he will be able to sell 10000 watermelons, and the remaining $x$ watermelons can be used for seeds. For the seeds to be sufficient for the next season, the inequality $250 x \geqslant 10000+x$ must hold. From this, $x ... | 10041 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. For the function $f(x)$, it is known that it is odd, i.e., $f(-x) = -f(x)$ for every real $x$. Additionally, it is known that for every $x$, $f(x+5) = f(x)$, and also $f(1 / 3) = 2022$, and $f(1 / 2) = 17$. What is the value of
$$
f(-7) + f(12) + f\left(\frac{16}{3}\right) + f\left(\frac{9}{2}\right) ?
$$ | Answer: 2005.
Solution. Let's find the values of $f(-7)+f(12)$, $f\left(\frac{16}{3}\right)$, and $f\left(\frac{9}{2}\right)$ separately.
- $f(-7)+f(12)=f(-7+5)+f(12-2 \cdot 5)=f(-2)+f(2)=0$;
- $f\left(\frac{16}{3}\right)=f\left(5+\frac{1}{3}\right)=f\left(\frac{1}{3}\right)=2022$.
- $f\left(\frac{9}{2}\right)=f\left... | 2005 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. For the numbers $1000^{2}, 1001^{2}, 1002^{2}, \ldots$, the last three digits are discarded. How many of the first terms of the resulting sequence form an arithmetic progression? | Answer: 32.
Solution. $(1000+k)^{2}=1000000+2000 k+k^{2}$. As long as $k^{2}<1000$, after discarding the last three digits, the number $1000+2 k$ will remain, i.e., each term of the sequence will be 2 more than the previous one. When $k=31: k^{2}=961<1000$, and when $k=32: k^{2}=1024$. Therefore, the element obtained ... | 32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. While tidying up the children's room before the guests arrived, mom found 9 socks. Among any four socks, at least two belong to the same owner. And among any five socks, no more than three belong to the same owner. How many children scattered the socks, and how many socks does each child own?
Answer. There are thre... | Solution. No child owned more than three socks, as otherwise the condition "among any five socks, no more than three belonged to one owner" would not be met. There are a total of 9 socks, so there are no fewer than three children. On the other hand, among any four socks, there are two socks belonging to one child, so t... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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