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5. A positive integer $N$ and $N^{2}$ end with the same sequence of digits $\overline{a b c d}$, where $a-$ is a non-zero digit. Find $\overline{a b c}$. | Answer: 937.
Solution. Let's represent the number $N$ as $10000 M+k$, where $k$ and $M$ are natural numbers, and $1000 \leq k \leq 9999$. Since $N$ and $N^{2}$ end with the same sequence of digits, the difference
$$
N^{2}-N=\left(10^{4} M+k\right)^{2}-\left(10^{4} M+k\right)=10^{4}\left(10^{4} M^{2}+2 M k-M\right)+k^... | 937 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.2. In the city of Perpendicularinsk, it was decided to build new multi-story houses (some of them may be single-story), but in such a way that the total number of floors would be 30. The city architect, Parallelnikov, proposed a project according to which, if after construction one climbs to the roof of each new hou... | Solution: 1) We will show that the project does not involve building houses with more than two floors. Assume the opposite, that such houses are planned. Take the lowest of them and reduce it by one floor, building an additional one-story house as a result. The sum of the numbers in question will decrease by the number... | 112 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.9. In the company, there are 100 children, some of whom are friends (friendship is always mutual). It is known that by selecting any child, the remaining 99 children can be divided into 33 groups of three such that in each group, all three are pairwise friends. Find the minimum possible number of pairs of friends.
... | Answer: 198.
Solution: Let's translate the problem into the language of graphs, associating each child with a vertex and each friendship with an edge. Then we know that in this graph with 100 vertices, after removing any vertex, the remaining vertices can be divided into 33 triples such that the vertices in each tripl... | 198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Will Katya be able to write a ten-digit number on the board where all digits are different and all differences between two adjacent digits are different (when finding the difference, the larger is subtracted from the smaller)? | Answer: will be able to.
Example: 9081726354.
Criteria. 7 points for any correct example. | 9081726354 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. There is a paper rectangle $3 \times 100$, divided into 300 cells $1 \times 1$. What is the maximum number of pairs consisting of one corner and one $2 \times 2$ square that can be cut out along the grid lines? (A corner is obtained from a $2 \times 2$ square by removing one of its corner cells). | Answer: 33.
Sketch of the solution. The square in the middle row occupies two cells, and the corner - at least one, so the pair occupies at least three cells in the middle row. If there are no fewer than 34 pairs, then they occupy at least $34 \times 3=102$ cells in the middle row, while there are only 100 there.
Exa... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.1. Parallelogram $A B C D$ is such that $\angle B<90^{\circ}$ and $A B<B C$. Points $E$ and $F$ are chosen on the circle $\omega$ circumscribed around triangle $A B C$ such that the tangents to $\omega$ at these points pass through $D$. It turns out that $\angle E D A=\angle F D C$. Find the angle $A B C$.
(A. Yaku... | Answer: $60^{\circ}$.
Solution. Let $\ell$ be the bisector of angle $E D F$. Since $D E$ and $D F$ are tangents to $\omega$, the line $\ell$ passes through the center $O$ of the circle $\omega$.
.
The product of the remaining factors ends in 6. Indeed, since the product $1 \times 2 \cdot 3 \cdot 4 \cdot 6 \... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. Does there exist a ten-digit number, divisible by 11, in which all digits from 0 to 9 appear? | Answer. Yes, for example, 9576843210.
Solution. Let's consider a possible way to find the required number. Note that a number is divisible by 11 if the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions is divisible by 11. If we write all ten digits in descend... | 9576843210 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. From the set of numbers $1,2,3,4, \ldots, 2021$, one number was removed, after which it turned out that the sum of the remaining numbers is divisible by 2022. Which number was removed? | Answer: 1011.
Solution. Let's write out the sum and perform grouping:
$1+2+3+\cdots+2019+2020+2021=(1+2021)+(2+2020)+$ $\cdots+(1012+1010)+1011$.
All the terms enclosed in parentheses are divisible by 2022. If we remove the last ungrouped term from the sum, the sum will be a multiple of 2022.
Comment. Answer only -... | 1011 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. Petya came home from school today at $16:45$, looked at the clock and wondered: after what time will the hands of the clock be in the same position for the seventh time since he came home from school? | Answer: 435 minutes.
Solution. The speed of the minute hand is 12 divisions/hour (one division here refers to the distance between adjacent numbers on the clock face), and the hour hand is 1 division/hour. Before the seventh meeting of the minute and hour hands, the minute hand must first "lap" the hour hand 6 times, ... | 435 | Other | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) There are apples in five boxes, with an equal number of apples in each. When 60 apples were taken out of each box, after that, the total number of apples left was the same as the number of apples that were originally in two boxes. How many apples were in each box?
Answer: 100. | Solution. In total, $60 \cdot 5=300$ apples were taken out, and this is equal to the number of apples that were in three boxes. Therefore, there were 100 apples in each box.
Criteria. Any correct solution: 7 points.
If it is not justified that there were 100 apples in each box, but it is verified that the condition i... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) A gnome in shoes weighs 2 kg more than a gnome without shoes. If you put five identical gnomes in shoes and five such gnomes without shoes on the scales, the scales will show 330 kg. How much does a gnome in shoes weigh? | Answer: 34 kg.
Solution. Let's put boots on five gnomes, then the weight will increase by 10 kg. It turns out that ten gnomes in boots weigh 340 kg. Therefore, one gnome in boots weighs 34 kg.
Criteria. Correctly found the weight of a gnome in boots by any method: 7 points.
Correctly found the weight of a gnome with... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Two runners, starting simultaneously at constant speeds, run on a circular track in opposite directions. One of them runs the loop in 5 minutes, while the other takes 8 minutes. Find the number of different meeting points of the runners on the track, if they ran for at least an hour. | Answer: 13 points.
Solution.
Let the length of the track be $\mathrm{S}$ meters. Then the speeds of the runners are $\mathrm{S} / 5$ and S/8 meters per minute, respectively. | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Three families of parallel lines have been drawn, with 10 lines in each. What is the maximum number of triangles they can cut out from the plane?
Answer: 150 triangles
# | # Solution.
Consider 100 nodes - the intersection points of lines in the first and second directions. Divide them into 10 sectors: the first sector - nodes lying on the first lines of the first and second
... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.1. What is the sum of the digits of the number $A=10^{50}-10^{40}+10^{30}-10^{20}+10^{10}-1$? | # Answer: 270.
Solution. The number is the sum of three numbers: a number composed of 10 nines followed by 40 zeros, a number composed of 10 nines followed by 20 zeros, and finally, a number composed of 10 nines. All the nines fall on the zeros in the other addends, so there is no carry-over, and the answer is $90+90+... | 270 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. Find the largest natural number with all distinct digits such that the sum of any two of its digits is a prime number. | Answer: 520.
Solution: If the desired number is at least a four-digit number, then it either has three digits of the same parity or two pairs of digits of the same parity. In each of these cases, we get that two of the sums of the digits are even. Therefore, they must equal 2. The number 2 can be represented as the su... | 520 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.1. In the class, there are more than 20 but fewer than 30 students. In the class, those who attend the chess club are half the number of those who do not attend. And those who attend the checkers club are one-third the number of those who do not attend. How many students are in the class? Provide all possible answers... | Solution: Let $n$ be the number of students in the class who attend the chess club, then $2n$ students do not attend, and the total number of students in the class is $3n$, meaning the total number of students in the class is divisible by 3. Similarly, from the fact that the number of people attending the checkers club... | 24 | Other | math-word-problem | Yes | Yes | olympiads | false |
9.1 How many five-digit natural numbers are there that are divisible by 9, and for which the last digit is 2 more than the second last digit? | Answer: 800. Hint The second last digit (the tens digit) can be any from 0 to 7 (so that after adding two, the last digit makes sense). The third and second digits can be any (from 0 to 9). After choosing the specified three digits (the second, third, and fourth), the last digit is uniquely determined by the second las... | 800 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1-0. The number 111123445678 is written on the board, and several digits (not all) need to be erased to get a number that is a multiple of 5. In how many ways can this be done? | Answer: 60
Solution. The digits 6, 7, and 8 must be crossed out, and 5 must be kept (otherwise, the number will not be divisible by 5). Each digit before the five can be crossed out or not. There are two options for each of the digits 2 and 3 (cross out or not), three options for the digit 4 (do not cross out, cross o... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4-0. Kisa and Busya came to the cafeteria during the break, where only muffins and doughnuts were sold, costing a whole number of rubles. Kisa bought 8 muffins and 3 doughnuts, spending less than 200 rubles, while Busya bought 4 muffins and 5 doughnuts, spending more than 150 rubles. Name the highest possible price of ... | Answer: 19
Solution. Let the price of a cake be $k$, and the price of a bun be $p$ rubles. Then $8 k+3 p<150$. Multiplying the first inequality by 5, and the second by 3, we get $40 k+15 p<1000, -12 k-15 p<-450$. Adding these inequalities: $28 k<550$, from which, taking into account the integer nature, $k \leqslant 19... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7-0. The number $n$ has exactly six divisors (including 1 and itself). They were arranged in ascending order. It turned out that the third divisor is seven times greater than the second, and the fourth is 10 more than the third. What is $n$? | Answer: 2891
Solution. If $n$ has six divisors, then either $n=p^{5}$ or $n=p \cdot q^{2}$ (where $p$ and $q$ are prime numbers). In the first case, $p=7$ (since the third divisor is seven times the second), but then the second condition is not satisfied.
Therefore, $n$ has two prime divisors (one of which is squared... | 2891 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8-0. Once, King Shahryar said to Scheherazade: "Here is a paper circle with 1001 points on its boundary. Each night, you must cut the figure you have along a straight line containing any two of the marked points, keeping only one fragment and discarding the other. And make sure that the figure you keep is not a polygon... | # Answer: 1999
Solution. First, let's describe the strategy that Scheherazade will use to meet Shahryar's conditions for 1998 nights. Initially, she will cut along the lines connecting adjacent vertices and discard the smaller part (let's call such a part a segment). She will do this for 1000 nights. After this, a 100... | 1999 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. A right triangle ABC is inscribed in a circle with hypotenuse AB. On the larger leg BC, a point D is taken such that AC = BD, and point E is the midpoint of the arc AB containing point C. Find the angle DEC. | 3. Point $\mathrm{E}$ is the midpoint of arc $\mathrm{AB}$, so $\mathrm{AE}=\mathrm{BE}$. Moreover, the inscribed angles $\mathrm{CAE}$ and $\mathrm{EBC}$, which subtend the same arc, are equal. Also, by the given condition, $\mathrm{AC}=\mathrm{BD}$. Therefore, triangles $\mathrm{ACE}$ and $\mathrm{BDE}$ are congruent... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. There are 40 visually identical coins, among which 3 are counterfeit - they weigh the same and are lighter than the genuine ones (the genuine coins also weigh the same). How can you use three weighings on a balance scale without weights to select 16 genuine coins? | 5. First solution: Divide all the coins into two parts of 20 coins each and weigh them. Since the number of counterfeit coins is odd, one of the piles will weigh more. This means that there is no more than one counterfeit coin in it. Divide it into two piles of 10 coins and weigh them. If the scales are in balance, the... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Find the value of the expression $a^{3}+b^{3}+12 a b$, given that $a+b=4$. | 1. $a^{3}+b^{3}+12 a b=(a+b)\left(a^{2}-a b+b^{2}\right)+12 a b=4\left(a^{2}-\right.$ $\left.a b+b^{2}\right)+12 a b=4 a^{2}+4 b^{2}+8 a b=4(a+b)^{2}=4 \cdot 16=64$ | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The boy went to the shooting range with his father. The father bought him 10 bullets. Later, the father took away one bullet for each miss and gave one additional bullet for each hit. The son fired 55 times, after which he ran out of bullets. How many times did he hit the target? | 2. Each time the boy hit the target, the number of bullets he had remained the same (he used one and received one from his father). Each time the boy missed, the number of bullets he had decreased by 2 (he used one and his father took one). This means that the son missed $10: 2=5$ times out of 55 shots, so he hit 55 - ... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.2. From a square with a side of 10, a green square with a side of 2, a blue square, and a yellow rectangle were cut out (see figure). What is the perimeter of the remaining figure?
The perimeter of a figure is the sum of the lengths of all its sides.
. It is known that among any 10 parrots, there is definitely a red one, and among any 12 parrots, there is definitely a yellow one. What is the ma... | Answer: 19.
Solution. Let there be $x$ red, $y$ yellow, and $z$ green parrots in the zoo.
Since among any 10 parrots there is a red one, the number of non-red parrots does not exceed 9, that is, $y+z \leqslant 9$. By similar reasoning, we get that the number of non-yellow parrots does not exceed 11, that is, $x+z \le... | 19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.2. In a convex quadrilateral $\mathrm{ABCD}$, the bisector of angle $\mathrm{B}$ passes through the midpoint of side $\mathrm{AD}$, and $\angle \mathrm{C}=\angle \mathrm{A}+\angle \mathrm{D}$. Find the angle $\mathrm{ACD}$.
$, from which $\angle \mathrm{AEB}=180-\angle \mathrm{A}-\angle \mathrm{B} / 2=\angle \mathrm{D}$. Therefor... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.3. On the board, 100 pairwise distinct natural numbers $a_{1}, a_{2}, \ldots, a_{100}$ were written. Then, under each number $a_{i}$, a number $b_{i}$ was written, obtained by adding to $a_{i}$ the greatest common divisor of the remaining 99 original numbers. What is the smallest number of pairwise distinct numbers t... | # Answer. 99.
First solution. If we set $a_{100}=1$ and $a_{i}=2 i$ for $i=1,2, \ldots, 99$, then $b_{1}=b_{100}=3$, so there will be no more than 99 different numbers among the numbers $b_{i}$. It remains to prove that there will always be 99 different numbers among the numbers $b_{i}$.
Without loss of generality, w... | 99 | Number Theory | 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 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
![](https://cdn.mathpix.co... | Answer: 22.
Solution. Note that there are a total of 91 coins, so after all moves, each box should have exactly 13 coins. At least 7 coins need to be moved from the box with 20 coins. Now consider the boxes adjacent to the box with 20 coins. Initially, they have a total of 25 coins, and at least 7 more coins will be t... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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?
=1$, that is, $A+B=90^{\circ}$.
Second method. Adding the original equalities, we get $\sin \left(A+45^{\circ}\right)+\sin \left(B+45^{\circ}\right)=2$, from which $\sin \left(A+45^{\circ}\right)=\si... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Two parks with a total area of 110 hectares are divided into the same number of plots, and in each park, the plots have the same area, but differ from those in the other park. If the first park were divided into plots of the same area as the second, it would result in 75 plots. If the second park were divided into p... | 2. Answer. 50 and 60 hectares. Solution. Let x be the area of the first park. According to the condition, we will construct a table (for the new division):
| Park | In the new division | | |
| :---: | :---: | :---: | :---: |
| | Area of the park | Number of plots | Area of the plot |
| First | x | 75 | x/75 |
| Sec... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.3. Given a triangle $A B C$. It is known that $\angle B=60^{\circ}, \angle C=75^{\circ}$. On side $B C$ as the hypotenuse, an isosceles right triangle $B D C$ is constructed inside triangle $A B C$. What is the measure of angle $D A C$? | Answer: $30^{0}$.
Solution 1: From the problem statement, it follows that $\angle A=45^{0}$. Draw a circle with center $\mathrm{M}$ and radius $\mathrm{MB}=\mathrm{MC}$. Since $\angle \mathrm{BDC}=90^{\circ}$, the major arc BC is seen at an angle of $45^{0}$. Therefore, vertex A lies on this circle. This means that tr... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. The altitudes of an acute-angled, non-isosceles triangle \(ABC\) intersect at point \(H\). \(O\) is the center of the circumcircle of triangle \(BHC\). The center \(I\) of the inscribed circle of triangle \(ABC\) lies on the segment \(OA\). Find the angle \(BAC\). | Answer: $60^{\circ}$.
Solution. From the problem statement, it follows that point O lies at the intersection of the angle bisector of angle $A$ and the perpendicular bisector of side $BC$. Since these lines intersect on the circumcircle of triangle $ABC$, point O lies on this circle and is the midpoint of arc $BC$ (se... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.1. Given two five-digit numbers without the digits 0 and 1 in their notation. The absolute value of their difference is a four-digit number \( S \). It is known that if each digit of one of the original numbers is decreased by 1, then the absolute value of the difference becomes 10002. What values can the number \( ... | Answer: 1109.
Solution: Let $A$ and $B$ be the two given numbers, and $C$ be the number obtained from $B$ by decreasing each of its digits by 1, that is, $C = B - 11111$. If $A10000 > B - A$ (this number is four-digit), then $C > B$. This is a contradiction. Therefore, $A > C$. Also, the case $A > B$ is impossible (th... | 1109 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Little One gave a big box of candies to Karlson. Karlson ate all the candies in three days. On the first day, he ate 0.2 of the entire box and 16 more candies. On the second day, he ate -0.3 of the remainder and 20 more candies. On the third day, he ate -0.75 of the remainder and the last 30 candies. How many candie... | Solution. Let $x$ be the number of candies in the box. On the first day, $(0.2x + 16)$ candies were eaten; on the second and third days, $(0.8x - 16)$ candies were eaten. On the second day, $(0.3(0.8x - 16) + 20) = (0.24x + 15.2)$ candies were eaten; on the third day, $(0.56x - 31.2)$ candies remained. Since on the thi... | 270 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The figure "archer" on a grid board attacks along a ray - along cells upwards, downwards, to the right, or to the left (exactly one of the four directions; the directions for different archers are independent). What is the maximum number of non-attacking archers that can be placed on an $8 \times 8$ chessboard? | Solution. Answer: 28. We will prove that it is impossible to place more than 28 archers. Consider any arrangement with the maximum possible number of archers. Perform the following two operations sequentially:
1) turn those archers who are standing at the edge of the board so that they shoot "outward" (this will not "... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.2. In the parliament of the island state of Promenade-and-Tornado, only the indigenous inhabitants of the island can be elected, who are divided into knights and liars: knights always tell the truth, liars always lie. In the last convocation, 2020 deputies were elected to the parliament. At the first plenary session ... | Solution. Note that the statement was made by more than half of the deputies. If all those who made the statement are knights, then their statements turn out to be false; if all those who made the statement are liars, then their statements turn out to be true - both are impossible. Therefore, among those who made the s... | 1010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.3. A trading organization wholesale purchased exotic fruits, the moisture content of which was $99\%$ of their mass. After delivering the fruits to the market, the moisture content dropped to $98\%$. By what percentage should the trading organization increase the retail price of the fruits (the price at which it will... | Solution. In 100 kg of fresh fruits, there was 99 kg of water and 1 kg of solid mass. After drying, 1 kg of solid mass constituted $2 \%$ of the mass of 50 kg, i.e., every 100 kg of fresh fruits dried down to 50 kg, i.e., by half, and the retail price, compared to the wholesale price, should be increased by two times.
... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.4. Find the smallest natural number ending in the digit 6 (on the right) which, when this digit is moved to the beginning (to the left), increases exactly four times. | Solution. Multiplication "in column" is performed from the end, so we can start the process of multiplication by sequentially finding all the digits: *****6 ( $6 \times 4=24$ - "4 we write, 2 in mind" - the 4 written under the line is the 4th digit that stood before 6; by writing 4 in the top line before 6, we will con... | 153846 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. What is the smallest identical number of pencils that need to be placed in each of 6 boxes so that any 4 boxes contain pencils of any of the 26 pre-specified colors (there are enough pencils available)? Prove that fewer is impossible. | Solution. Let's assume that we have fewer than 3 pencils of some color. Then, if we take 4 boxes in which such pencils are not present (and such boxes can be found, since there are no more than $2 y x$ pencils of that color), the condition of the problem will not be met. This means that there are at least 3 pencils of ... | 78 | Combinatorics | proof | Yes | Yes | olympiads | false |
1. Real numbers $a, b, c$ are such that $a+1 / b=9, b+1 / c=10$, $c+1 / a=11$. Find the value of the expression $a b c+1 /(a b c)$. | Answer: 960.
Sketch of the solution. By multiplying the equations, expanding the brackets, and grouping, we get: $a b c + 1/(a b c) + a + 1/b + b + 1/c + c + 1/a = 990$. From this, $a b c + 1/(a b c) = 990 - 9 - 10 - 11 = 960$. | 960 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Several cells on a $14 \times 14$ board are marked. It is known that no two marked cells are in the same row and the same column, and also, that a knight can, starting from some marked cell, visit all marked cells in several jumps, visiting each exactly once. What is the maximum possible number of marked cells? | Answer: 13.
Since there is no more than one marked cell (field) in each row, there are no more than 14 marked cells. Suppose there are 14 marked cells. Number the rows and columns from 1 to 14 from bottom to top and from left to right, and color the cells in a checkerboard pattern, where the sum of the row and column ... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 8.1. Can a cube be cut into 71 smaller cubes (of any non-zero size)? | Answer: Yes, it is possible.
Solution: Let's divide each edge of the cube in half and cut the cube into 8 smaller cubes. Now, take one of these smaller cubes and divide it into 8 even smaller cubes. One cube disappears, but 8 new ones appear, increasing the total number of parts by \(8-1=7\) to 15. Repeating this oper... | 71 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.3. In triangle $A B C$ with angle $C$ equal to $30^{\circ}$, median $A D$ is drawn. Angle $A D B$ is equal to $45^{\circ}$. Find angle $B A D$.
# | # Answer: $30^{\circ}$.
Solution. Draw the height $B H$ (see the figure). In the right triangle $B H C$, the leg $B H$ lies opposite the angle $30^{\circ}$, so $B H=\frac{B C}{2}=B D$. The angle $H B C$ is $180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}$. Triangle $B H D$ is isosceles with an angle of $60^{\circ}$, so i... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.5. What is the maximum number of members that can be in a sequence of non-zero integers, for which the sum of any seven consecutive numbers is positive, and the sum of any eleven consecutive numbers is negative? | Answer: 16.
Solution: Estimation. Assume there are no fewer than 17 numbers. Construct a table with 7 columns and 11 rows, in which the first 17 numbers are arranged.
| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $a_{2}$ | $a_{3}$ |... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.5. Given a $5 \times 5$ square grid of cells. In one move, you can write a number in any cell, equal to the number of cells adjacent to it by side that already contain numbers. After 25 moves, each cell will contain a number. Prove that the value of the sum of all the resulting numbers does not depend on the order in... | Solution. Consider all unit segments that are common sides for two cells. There are exactly forty such segments - 20 vertical and 20 horizontal. If a segment separates two filled cells, we will say that it is "painted." Note that when we write a number in a cell, it indicates the number of segments that were not painte... | 40 | Combinatorics | proof | Yes | Yes | olympiads | false |
3. Find the smallest natural number $n$ such that the sum of the digits of each of the numbers $n$ and $n+1$ is divisible by 17. | Answer: 8899.
Solution. If $n$ does not end in 9, then the sums of the digits of the numbers $n$ and $n+1$ differ by 1 and cannot both be divisible by 17. Let $n=\overline{m 99 \ldots 9}$, where the end consists of $k$ nines, and the number $m$ has a sum of digits $s$ and does not end in 9. Then the sum of the digits ... | 8899 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In the Rhind Papyrus (Ancient Egypt), among other information, there are decompositions of fractions into the sum of fractions with a numerator of 1, for example,
$\frac{2}{73}=\frac{1}{60}+\frac{1}{219}+\frac{1}{292}+\frac{1}{x}$
One of the denominators here is replaced by the letter $x$. Find this denominator. | # Solution:
First, find $\stackrel{1}{-.}$ from the equation:
$\frac{2}{73}=\frac{1}{60}+\frac{1}{219}+\frac{1}{292}+\frac{1}{x}$
$\frac{2}{73}-\frac{1}{219}-\frac{1}{292}-\frac{1}{60}=\frac{1}{x}$
$\frac{2}{73}-\frac{1}{73 \cdot 3}-\frac{1}{73 \cdot 4}-\frac{1}{60}=\frac{1}{x}$
$\frac{2 \cdot 3 \cdot 4-4-3}{73 \c... | 365 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. At a certain moment, Anna measured the angle between the hour and minute hands of her clock. Exactly one hour later, she measured the angle between the hands again. The angle turned out to be the same. What could this angle be? (Consider all cases).
# | # Solution:
After 1 hour, the minute hand remains in its place. During this time, the hour hand has turned $30^{\circ}$. Since the angle has not changed, the minute hand must be dividing one of the
, the number of desired numbers can be obtained by erasing one digit to the right of the " " "middle digit of the final number" in the number 12343210. There are 4 such numbers.
If the middle digit of a seven-digit number is 5, the number... | 7608 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the following equation for positive $x$.
$$
x^{2014} + 2014^{2013} = x^{2013} + 2014^{2014}
$$ | Answer: 2014.
Solution. Rewrite the equation in the following form $x^{2013}(x-1)=2014^{2013}(2014-1)$. The solution is $x=2014$. Since the function on the left, for $x>1$ this function is increasing, as the product of two positive increasing functions, the intersection with a constant function can be no more than one... | 2014 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 9.2 (7 points)
In a bag, there are 70 balls that differ only in color: 20 red, 20 blue, 20 yellow, and the rest are black and white.
What is the minimum number of balls that need to be drawn from the bag, without seeing them, to ensure that among them there are at least 10 balls of the same color? | # Solution:
By drawing 37 balls, we risk getting 9 red, 9 blue, and 9 yellow balls, and we won't have ten balls of one color. If we draw 38 balls, the total number of red, blue, and yellow balls among them will be at least 28, and the number of balls of one of these colors will be at least ten (since $28 > 3 \cdot 9$)... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 9.3 (7 points)
$H$ is the point of intersection of the altitudes of an acute-angled triangle $ABC$. It is known that $HC = BA$. Find the angle $ACB$.
# | # Solution:
Let $\mathrm{D}$ be the foot of the altitude dropped from vertex A to side BC. Angles НСВ and DAB are equal as acute angles with mutually perpendicular sides. Therefore, $\triangle \mathrm{CHD}=\Delta \mathrm{ABD}$ (by hypotenuse and acute angle). Hence, $=\mathrm{AD}$, i.e., triangle $\mathrm{ACD}$ is iso... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 9.4 (7 points)
A palindrome is a number, letter combination, word, or text that reads the same in both directions. How much time in a day do palindromes appear on the clock display, if the clock shows time from 00.00.00 to 23.59.59? | # Solution:
If the digits on the display are ab.cd.mn, then $\mathrm{a}=0,1,2,0 \leq \mathrm{b} \leq 9,0 \leq \mathrm{c} \leq 5,0 \leq \mathrm{d} \leq 9,0 \leq \mathrm{m} \leq 5$, $0 \leq \mathrm{n} \leq 9$. Therefore, if $\mathrm{a}=\mathrm{n}, \mathrm{b}=\mathrm{m}, \mathrm{c}=\mathrm{d}$, then the symmetric number ... | 96 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 9.5 (7 points)
In a store, there are buttons of six colors. What is the smallest number of buttons that need to be bought so that they can be sewn in a row, such that for any two different colors in the row, there are two adjacent buttons of these colors? | # Solution:
From the condition, it follows that for each fixed color A, a button of this color must appear in a pair with a button of each of the other 5 colors. In a row, a button has no more than two neighbors, so a button of color A must appear at least 3 times. The same applies to each other color. Thus, there sho... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2.4. From sticks of the same length, a row of 800 hexagons was laid out as shown in the figure. How many sticks were used in total?
Answer, option 1. 5001.
Answer, option 2. 2501.
Answer, option 3. 3001.
Answer, option 4. 4001. | Solution option 1. In the first hexagon, there are 6 sticks. Building each subsequent hexagon adds 5 sticks. In total, it will be $6+5 \cdot 999=5001$ sticks. | 5001 | Geometry | MCQ | Yes | Yes | olympiads | false |
7.4. In isosceles triangle $ABC$, the base $AC$ is equal to $x$, and the lateral side is 12. On the ray $AC$, point $D$ is marked such that $AD=24$. A perpendicular $DE$ is dropped from point $D$ to line $AB$. Find $x$ if it is known that $BE=6$.
 | 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 |
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 |
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.
We will call suc... | 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.
.
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 |
# Problem №2
In an $8 \times 8$ frame that is 2 cells wide (see figure), there are a total of 48 cells.
How many cells are in a $254 \times 254$ frame that is 2 cells wide? | Answer: 2016.
## Solution
First method. Cut the frame into four identical rectangles as shown in the figure. The width of the rectangles is equal to the width of the frame, i.e., 2 cells. The length of each
.
Since $\mathrm{BK} \| \mathrm{AD}$, then $\angle \mathrm{KBE}=\angle \mathrm{DAE}$.
Moreover, $\angle \mathrm{KEB}=\angle \mathrm{DEA}$ and $\mathrm{AE}=\mathrm{BE}$, thus triangles $\mathrm{BKE}$ ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Four princesses each thought of a two-digit number, and Ivan thought of a four-digit number. After they wrote their numbers in a row in some order, the result was 132040530321. Find Ivan's number. | Solution. Let's go through the options. Option 1320 is not suitable because the remaining part of the long number is divided into fragments of two adjacent digits: 40, 53, 03, 21, and the fragment 03 is impossible, as it is not a two-digit number. Option 3204 is impossible due to the invalid fragment 05 (or the fragmen... | 5303 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. There are more than 30 people in the class, but less than 40. Any boy is friends with three girls, and any girl is friends with five boys. How many people are in the class | Solution. Let $\mathrm{m}$ be the number of boys, $\mathrm{d}$ be the number of girls, and $\mathrm{r}$ be the number of friendly pairs "boy-girl". According to the problem, $\mathrm{r}=3 \mathrm{~m}$ and $\mathrm{r}=5 \mathrm{~d}$. Therefore, $\mathrm{r}$ is divisible by 3 and 5, and thus by 15: $\mathrm{r}=15 \mathrm... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.2. Little kids were eating candies. Each one ate 11 candies less than all the others together, but still more than one candy. How many candies were eaten in total? | Answer: 33 candies.
Solution: Let $S$ be the total number of candies eaten by the children. If one of the children ate $a$ candies, then according to the condition, all the others ate $a+11$ candies, and thus all together ate $S=a+(a+11)=2a+11$ candies. This reasoning is valid for each child, so all the children ate t... | 33 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.4. The number 49 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 50 in several moves? | Answer: Yes.
Solution: The number 50 can be obtained by doubling 25, and 25 can be obtained by erasing the last digit of the number 256, which is a power of two. Thus, the necessary chain of transformations can look like this: $49 \rightarrow 4 \rightarrow 8 \rightarrow 16 \rightarrow 32 \rightarrow 64 \rightarrow 128... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.3. Given an acute-angled triangle $A B C$. Point $M$ is the intersection point of its altitudes. Find the angle $A$, if it is known that $A M=B C$.
---
The text has been translated while preserving the original formatting and line breaks. | 83. Given an acute-angled triangle $A B C$. Point $M$ is the intersection point of its altitudes. Find the angle $A$, if it is known that $A M=B C$.
Answer: $45^{\circ}$. Hint Let $\mathrm{K}$ be the foot of the altitude from point В. We will prove that triangles АМ К and BKC are equal. Indeed, we have right triangles... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 3. The numbers $2^{2021}$ and $5^{2021}$ are written one after another. How many digits are written in total?
# | # Solution
Let the number $2^{2021}$ have $\mathrm{k}$ digits, and the number $5^{2021}$ have $\mathrm{m}$ digits, then the number of digits in the desired number is $\mathrm{k}+\mathrm{m}$. $10^{k-1}<2^{2021}<10^{k}, 10^{m-1}<5^{2021}<10^{m}$, therefore, $10^{k+m-2}<$ $10^{2021}<10^{m+k}$ and $\mathrm{k}+\mathrm{m}=2... | 2022 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.1. At first, there were natural numbers, from 1 to 2021. And they were all white. Then the underachiever Borya painted every third number blue. Then the top student Vova came and painted every fifth number red. How many numbers remained white? (7 points)
# | # Solution
Borya repainted the numbers divisible by 3, a total of [2021:3]=673 numbers. Vova repainted [2021:5]=404 numbers. 673+404=1077. However, numbers divisible by 15 were counted twice. [2021:15]=134. Therefore, the number of white numbers remaining is 2021-1077+134=1078.
Answer: 1078 white numbers
| criteria ... | 1078 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.2. Doctor Vaccinov and Doctor Injectionov vaccinated all the residents of the village of Covido. Vaccinov thought: if I had given 40% more vaccinations, then Injectionov's share would have decreased by 60%. And how would Injectionov's share change if Vaccinov had given 50% more vaccinations? (7 points)
# | # Solution
$40 \%$ of the injections given by Privevkin equals $60 \%$ of the number of injections given by Ukolkin, so Privevkin gave 1.5 times more injections. Therefore, an increase in Privevkin's share by $n \%$ would decrease Ukolkin's share by $1.5 n \%$.
Answer: It would decrease by $75 \%$.
| criteria | poin... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. On an $8 \times 8$ chessboard, tokens are placed according to the following rule. Initially, the board is empty. A move consists of placing a token on any free square. With this move, exactly one of the tokens that ends up adjacent to it is removed from the board (if there is such an adjacent token). What is the max... | Answer: No more than 61 chips can be placed. | 61 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. In a certain country, there are 47 cities. Each city has a bus station from which buses run to other cities in the country and possibly abroad. A traveler studied the schedule and determined the number of internal bus routes departing from each city. It turned out that if the city of Lake is not considered, then for... | Solution.
Note that external lines are not considered in this problem.
There are a total of 47 variants of the number of internal lines - from 0 to 46. Note that the existence of a city with 46 lines excludes the existence of a city with 0 lines and vice versa.
Suppose there is a city with 46 lines. Then the smalles... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.1. The teacher suggested that his students - Kolya and Seryozha - solve the same number of problems during the lesson. After some time from the start of the lesson, it turned out that Kolya had solved a third of what Seryozha had left to solve, and Seryozha had left to solve half of what he had already completed. Ser... | # 8.1. Answer: 16.
Solution. Let Tanya have solved x problems, then she has $\frac{x}{2}$ problems left to solve.
Let $\mathrm{t}_{1}$ be the time interval after which Tanya and Kolya evaluated the shares of solved and remaining problems, and $\mathrm{t}_{2}$ be the remaining time. Since Tanya's problem-solving speed... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.8. In the vertices of a regular 100-gon, 100 chips numbered $1, 2, \ldots, 100$ were placed, in exactly that order clockwise. In one move, it is allowed to swap two adjacent chips if their numbers differ by no more than $k$. For what smallest $k$ can a series of such moves result in a configuration where each chip i... | # Answer. 50.
Solution. Example. The chip 50 is sequentially exchanged 99 times with the next one counterclockwise. We get the required arrangement.
There are several ways to prove the estimate, below we provide two of them.
The first way. Suppose that for some $k<50$ the required arrangement is obtained.
At any mo... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.4. Point $E$ is the midpoint of side $A B$ of parallelogram $A B C D$. On segment $D E$, there is a point $F$ such that $A D = B F$. Find the measure of angle $C F D$. | Answer: $90^{\circ}$.
Solution. Extend $D E$ to intersect line $B C$ at point $K$ (see Fig. 8.4). Since $B K \| A D$, then $\angle K B E = \angle D A E$. Moreover, $\angle K E B = \angle D E A$ and $A E = B E$, therefore, triangles $B K E$ and $A D E$ are equal. Thus, $B K = A D = B C$.
Therefore, in triangle $CFK$, ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.6. A triangle is divided into triangular cells as shown in the figure. A natural number is written in each cell. For each side of the triangle, there are four layers parallel to this side, containing seven, five, three, and one cell, respectively. It turns out that the sum of the numbers in each of these twelve layer... | Answer: 22.
Solution. Example. In each of the three corner cells, we write the number 3, and in each of the others, we write the number 1. Then the sum of the written numbers is $3 \cdot 3 + 13 \cdot 1 = 22$, and the sums of the numbers in the layers are: 11, 5, 3, and 3, respectively.
Estimation. Any corner cell is ... | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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