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# Problem 5. (3 points)
In trapezoid $A B C D$, the lateral side $A B$ is equal to the diagonal $A C$. On the smaller arc $A D$ of the circumscribed circle of triangle $A B D$, a point $E$ is chosen such that $A B=A E$. Find the angle $\angle C E D$. | Answer: $90^{\circ}$.
Solution:
From the isosceles property of triangle $ABC$ and the parallelism of $BC$ and $AD$, we get $\angle ABC = \angle ACB = \angle CAD = \alpha$.
Let the line $BC$ intersect the circumcircle of triangle $ABD$ at point $F$. Then $ABFD$ is an inscribed, i.e., isosceles, trapezoid, from which ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8. (5 points)
Girl Katya doesn't like the number 239. She listed several different numbers, none of which contain the sequence of digits 239 (in a row and exactly in that order). Prove that the sum of the reciprocals of these numbers is no more than 30000.
# | # Solution:
The number of suitable $3 n+1$-digit numbers is no more than $9 \cdot 999^{n}: 9$ options for the first digit and no more than 999 options for each subsequent triplet of digits. Each of them is no less than $10^{3 n}$.
The number of suitable $3 n+2$-digit numbers is no more than $90 \cdot 999^{n}$. Each o... | 30000 | Number Theory | proof | Yes | Yes | olympiads | false |
# Problem 5. (3 points)
In trapezoid $A B C D$, the lateral side $B C$ is equal to the diagonal $B D$. On the smaller arc $A B$ of the circumscribed circle of triangle $A B C$, a point $E$ is chosen such that $B C=B E$. Find the angle $\angle A E D$. | Answer: $90^{\circ}$.
## Solution:
From the isosceles triangle $B C D$ and the parallelism of $A B$ and $C D$, we get $\angle B C D = \angle B D C = \angle D B A = \alpha$.
Let the line $C D$ intersect the circumcircle of triangle $A B C$ at point $F$. Then $B C F A$ is an inscribed, i.e., isosceles, trapezoid, from... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8. (5 points)
Boy Kolya doesn't like the number 1234. He listed several different numbers, none of which contain the sequence of digits 1234 (in a row and exactly in that order). Prove that the sum of the reciprocals of these numbers is no more than 400000. | Solution:
The number of suitable $4 n+1$-digit numbers is no more than $9 \cdot 9999^{n}: 9$ options for the first digit and no more than 9999 options for each subsequent quartet of digits. Each of them is no less than $10^{4 n}$.
The number of suitable $4 n+2$-digit numbers is no more than $90 \cdot 9999^{n}$. Each ... | 400000 | Number Theory | proof | Yes | Yes | olympiads | false |
# Problem 5. (3 points)
In trapezoid $A B C D$, the lateral side $C D$ is equal to the diagonal $A C$. On the smaller arc $B C$ of the circumscribed circle of triangle $B C D$, a point $E$ is chosen such that $C D=C E$. Find the angle $\angle A E B$. | Answer: $90^{\circ}$.
Solution:
From the isosceles triangle $A C D$ and the parallelism of $B C$ and $A D$, we get $\angle A D C = \angle C A D = \angle A C B = \alpha$.
Let the line $A D$ intersect the circumcircle of triangle $B C D$ at point $F$. Then $C D F B$ is an inscribed, i.e., isosceles, trapezoid, from wh... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8. (5 points)
Girl Masha doesn't like the number 729. She listed several different numbers, none of which contain the sequence of digits 729 (in a row and exactly in that order). Prove that the sum of the reciprocals of these numbers is no more than 30000.
# | # Solution:
The number of suitable $3 n+1$-digit numbers is no more than $9 \cdot 999^{n}: 9$ options for the first digit and no more than 999 options for each subsequent triplet of digits. Each of them is no less than $10^{3 n}$.
The number of suitable $3 n+2$-digit numbers is no more than $90 \cdot 999^{n}$. Each o... | 30000 | Number Theory | proof | Yes | Yes | olympiads | false |
# Problem 5. (3 points)
In trapezoid $A B C D$, the lateral side $A D$ is equal to the diagonal $B D$. On the smaller arc $C D$ of the circumscribed circle of triangle $A C D$, a point $E$ is chosen such that $A D=D E$. Find the angle $\angle B E C$. | Answer: $90^{\circ}$.
## Solution:
From the isosceles triangle $B A D$ and the parallelism of $A B$ and $C D$, we get $\angle B A D = \angle D B A = \angle B D C = \alpha$.
Let the line $A B$ intersect the circumcircle of triangle $A C D$ at point $F$. Then $D A F C$ is an inscribed, i.e., isosceles, trapezoid, from... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8. (5 points)
Boy Anton does not like the number 2048. He listed several different numbers, none of which contain the sequence of digits 2048 (in a row and exactly in this order). Prove that the sum of the reciprocals of these numbers is no more than 400000.
# | # Solution:
The number of suitable $4 n+1$-digit numbers is no more than $9 \cdot 9999^{n}: 9$ options for the first digit and no more than 9999 options for each subsequent quartet of digits. Each of them is no less than $10^{4 n}$.
The number of suitable $4 n+2$-digit numbers is no more than $90 \cdot 9999^{n}$. Eac... | 400000 | Number Theory | proof | Yes | Yes | olympiads | false |
3. $\mathrm{ABCD}$ is an isosceles trapezoid, $\mathrm{AB}=\mathrm{CD}=25, \mathrm{BC}=40, \mathrm{AD}=60$. $\mathrm{BCDE}$ is also an isosceles trapezoid. Find AE. (Points A and E do not coincide) If there are multiple possible values, list them in any order separated by a semicolon. | Answer: 44
## Examples of answer recording:
45
$45 ; 56$
Problem 10 (4 points). | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (2 points) Does there exist a four-digit natural number with the sum of its digits being 21, which is divisible by $14 ?$ | Answer: Yes, for example 6384. | 6384 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (3 points) Given the rebus: MIMIMI + NYANYANYA = OLALAOY. Identical letters represent identical digits, different letters represent different digits. Find MI + NY.
# | # Answer: 119
Solution: The rebus can be rewritten as OLAOLOY = (MI + NYA) $\cdot$10101. First, this means that the last digit of MI + NYA is Y. Second, if MI + NYA $<100$, the result will be a four-digit number. Let MI + NYA $=1 Z$ Y, where $Z$ is some digit.
Then OLAOLOY $=1 Z$ Y $0000+1 Z$ Y $00+1 Z$ Y. If Y $<9$,... | 119 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) A quadrilateral is divided into 1000 triangles. What is the maximum number of different points at which the vertices of these triangles can be located? | Answer: 1002
Solution: The sum of the angles of a quadrilateral is $360^{\circ}$, and for thousands of triangles, it is $180000^{\circ}$. Where did the extra $1796400^{\circ}$ come from? Each internal vertex, where only the angles of triangles meet, adds $360^{\circ}$. Each vertex on the side of the original triangle ... | 1002 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (2 points) Does there exist a four-digit natural number with the sum of its digits being 14, which is divisible by $14 ?$ | Answer: Yes, for example 6314. | 6314 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (3 points) Given the puzzle: ЛЯЛЯЛЯ + ФУФУФУ = ГГЫГЫЫР. Identical letters represent identical digits, different letters represent different digits. Find ЛЯ + ФУ. | Answer: 109
Solution: The rebus can be rewritten as GGUGUUU $=(\text{LYA} + \text{FU}) 10101$. First, this means that the last digit of $\text{LYA} + \text{FU}$ is R. Second, if $\text{LYA} + \text{FU} < 100$, the result will be a four-digit number. Let $\text{LYA} + \text{FU} = 1ZP$, where $Z$ is some digit.
Then OL... | 109 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Boy Zhenya lives in a building with a total of 100 floors. The elevator takes 1 second to travel between adjacent floors. Unfortunately, only two buttons in the elevator work: “+13 floors” and “-7 floors”. The buttons are pressed instantly.
How many seconds will it take Zhenya to get home from the 1st floor to the ... | Answer: 107
Examples of writing answers:
45 | 107 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. A 103-gon is inscribed in a circle with diameter $\mathrm{XY}=4$ and has an axis of symmetry perpendicular to this diameter. Find the sum of the squares of the distances from the vertices of the 103-gon to the point $X$. | Answer: 824
## Examples of how to write the answer:
1,7
$1 / 7$
17
# | 824 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Given 50 quadratic trinomials. The graphs of these trinomials have their vertices on the $x$-axis. No three graphs intersect at the same point. What is the minimum number of intersection points they can form? | Answer: 600
## Examples of how to write the answer:
## 17
# | 600 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. From the digits $1,2,3,4,5,6,7$, all possible seven-digit numbers with all different digits were formed. The obtained numbers were written in ascending order. Find which number will be in the 1972nd place. | Answer: 3641572
## Examples of how to write the answer:
1234567
# | 3641572 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On Misfortune Island, a table tennis tournament was held in a round-robin format (i.e., everyone played against everyone else once). After each match, both participants separately approached the Chief Referee and reported the result. Among the participants, there were only three types of people: knights, who always ... | Answer: 16
## Examples of answer recording:
17
# | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Mitya wrote on the board the sum of the digits of each of the numbers from 1 to 800 inclusive. What is the sum of the numbers on the board | Answer: 10008
## Examples of answer notation:
12345
# | 10008 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Boy Zhenya lives in a building with a total of 100 floors. The elevator takes 1 second to travel between adjacent floors. Unfortunately, only two buttons in the elevator work: “+13 floors” and “-7 floors”. The buttons are pressed instantly.
How many seconds will it take Zhenya to get home from the 1st floor to the ... | Answer: 107
Examples of writing answers:
45 | 107 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11. A lot of snow fell, and the children decided to build snowmen. For this, they rolled 99 snowballs with masses of 1 kg, 2 kg, 3 kg, ..., 99 kg. A snowman consists of three snowballs stacked on top of each other, and one snowball can be placed on another if and only if the mass of the first is at least twice as small... | 11. Answer: 24.
Two snowballs with masses from 50 to 99 kg cannot end up in the same snowman, so in each snowman, there will be at least two snowballs with masses from 1 to 49 kg. Therefore, there cannot be more than 24 snowmen.
An example can be constructed greedily: first, take three snowballs with masses, for exam... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
18. Among 65 residents of the village, two are tricksters, and the rest are knights. A knight always tells the truth, while a trickster can either tell the truth or lie. You can show any resident a list of a group of residents (including a group of one person) and ask if they are all knights. How can you find both tric... | 18. In fact, 16 questions are enough (and this is not the minimum).
In the first four questions, we will find a knight. We will do this as follows: ask the first and second residents about the third. If both answered that he is a knight, then he is a knight (otherwise, all three are liars, which is impossible).
If no... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
21. Baron Münchhausen told a story. "A whole crowd gathered. We reached a crossroads. Then half of our group turned left, a third - right, and a fifth - straight ahead." "But excuse me," the duke noticed, - the sum of a half, a third, and a fifth is not equal to one, so you are lying!" The baron objected: "I am not lyi... | 21. Answer: 37 (18 to the left, 12 to the right, 7 straight).
Evaluation: $\frac{1}{2}+\frac{1}{3}+\frac{1}{5}=\frac{31}{30}$. If there were a total of people, then $\frac{1}{30} \cdot$ is the error. Let's calculate the maximum error (in fractions of a person): $\frac{1}{2}+\frac{1}{3}+\frac{2}{5}=\frac{37}{30}$. Ther... | 37 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
32. In the country, there are 1000000 people, each of whom is acquainted with at least one resident. After surveys, a paradoxical situation has emerged: exactly $90 \%$ of the population admitted that they believe in Santa Claus, however, each resident can claim that among their acquaintances, exactly $10 \%$ believe i... | 32. Each person knows at least 10 people. In total, we have 900,000 people who believe in Santa Claus, and they have at least 8,100,000 acquaintances with those who do not believe in him. In total, 100,000 people do not believe in Santa Claus, so among them, there will be someone who has at least 81 acquaintances who b... | 810 | Combinatorics | proof | Yes | Yes | olympiads | false |
35. There is a pile of 660 stones on the table. In one move, you can split any of the existing piles into two smaller ones. At the same time, the sizes of any two piles on the table at the same time must differ by less than a factor of two. What is the maximum number of piles that can result?
## 9th Grade
## Plot 1
... | 35. Answer: 30
Example: Let's go in reverse - we will combine two piles into one without violating the conditions of the problem. Let's say we initially had piles (two of each kind): $15,15,16,16, \ldots, 29,29$. It is easy to see that the condition is not violated, and the sum is 660. We will combine the two smallest... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. There are 1000 crows sitting on a wire. At the end of each minute, every third (third, sixth, ninth, and so on) crow flies away.
a) Which crows, by their initial count, will remain on the wire in the end?
b) How many minutes will pass before the crows stop flying away? | Solution. a) It is clear that at the end there will be no more than two ravens left, and the first and second never fly away. Therefore, it will be precisely these two. Answer: 1 and 2.
b) It is easy to see that the number of ravens flying away in the next minute is one-third of their number, rounded down. Thus, it on... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. On a 5 by 5 board, a certain number of bishops and knights were placed. Then, they counted how many times each bishop is attacked, and all the obtained numbers were added together. What is the maximum sum that could have been obtained?
# | # Solution:
Answer: 80.
The example is constructed through a chessboard coloring - bishops on one color, rooks on the other.
Evaluation: First, place bishops everywhere. The total sum will be 64. Now, in each cell, write the number of points that can be gained by replacing a bishop with a knight. We will get the fol... | 80 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3.3. Let $N=15$ and only the second operation is allowed. Prove that from any arrangement, less than $10^{4}$ others can be obtained. | Solution. We will paint the positions in three colors; the parity of the number of blacks on all positions changes simultaneously. With a fixed set of parities, there are exactly $2^{12}$ arrangements, and with a fixed set up to inversion $-2^{13}<10000$. | 10000 | Combinatorics | proof | Yes | Yes | olympiads | false |
1. Timur thought of a three-digit number and told it to Anton, who then wrote down three numbers that result from replacing one of the digits of the original number with 1 (for example, if the number 215 was thought of, Anton would write down 115, 215, 211). The sum of all the numbers written down by Anton turned out t... | Solution: Notice that from the number abc, we get the sum $1 b c + a 1 c + a b 1 = 111 + 2^{\star} a b c$. From the equation $1243 = 111 + 2 * a b c$, it follows that $a b c = 566$. | 566 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. In each cell of a $10 \times 10$ board, there sits a rabbit. Between rabbits in adjacent cells, there are partitions that can be removed. What is the minimum number of partitions that need to be removed so that any rabbit can visit any other rabbit, traveling through no more than 17 cells (not counting the starting ... | Solution. An example for 100 partitions is given in the figure. The correctness of the example follows from the fact that, as can be easily seen, any rabbit can reach the central 4-cell square by moving no more than 8 cells. From the central square to any cell, it can also be reached in no more than 8 moves. No more th... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. In the company, several employees have a total monthly salary of 10000 dollars. A kind manager proposes to triple the salary for everyone earning up to 500 dollars, and increase the salary by 1000 dollars for the rest, so the total salary will become 24000 dollars. A mean manager proposes to reduce the salary to 500... | Solution. Note that the increase proposed by the kind manager is twice as large as the salary proposed by the evil manager (this is true for both poor and rich employees). The increase according to the kind manager's proposal is 14000, so the salary according to the evil manager's proposal is 7000. | 7000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. There are 30 people in the bar. The bartender knows that among them, there are 10 knights (who always tell the truth), 10 liars (who always lie), and 10 troublemakers. The bartender can ask person $X$ about person $Y$: "Is it true that $Y$ is a troublemaker?" If $X$ is not a troublemaker, they will answer the questi... | Answer: 19. Indeed, with the first question, a peaceful client may be identified, so the bartender cannot guarantee to keep all peaceful clients.
Let's show how the bartender can leave 19 peaceful clients.
Solution 1. First, he asks everyone about client $A$ until someone throws him out of the bar. Let's say $B$ thre... | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. In a $100 \times 100$ square, 10000 cuts were made along the grid lines (each cut one cell long), and it split into 2500 four-cell figures. How many of them are $2 \times 2$ squares? | Solution:
Answer: 2300.
Solution. Note that the perimeter of a $2 \times 2$ square is 8, while for the other four-cell figures (rectangle $1 \times 4$, T-shape, L-shape, or S-shape) it is 10. Let the number of squares be $x$, and the number of other figures be $2500-x$. Then their total perimeter is $8 x + 10 \cdot (... | 2300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Milla was writing the letters M and L in uppercase. At the end, she counted that the letter matched the previous one 59 times, and did not match 40 times. Determine the maximum number of letters M that Milla could have written, and prove that it is indeed the maximum.
| Solution. Answer: 80 letters.
For each letter except the first, it is known whether it matches the previous letter or not. Therefore, Milla wrote $1+59+40=100$ letters.
Divide all the written letters into groups of consecutive identical letters. Then, groups of M (M-groups) will alternate with groups of L (L-groups).... | 80 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1.4. Prove that Shakti will always be able to find a strange sociable set of no more than 82 troglodytes.
# | # Solution.
Consider a communicative set $A$ of the smallest size. Note that we can choose $A$ such that for each troglodyte $a$ in $A$, there exists a troglodyte $v_{a}$ whose only friend in the set $A$ is the troglodyte $a$. Indeed, otherwise, if $a$ is friends with someone in $A$, it can simply be removed, and if i... | 82 | Combinatorics | proof | Yes | Yes | olympiads | false |
4. The number 100 is represented as the sum of several two-digit numbers, and in each addend, the digits are swapped. What is the largest number that could result from the new sum? | Solution. If the digits in the two-digit number $\overline{b a}$ are swapped, the number increases by $9(a-b)$. Therefore, the new sum is $S=100+9 U-9 D$, where $D$ is the sum of the tens digits, and $U$ is the sum of the units digits in the original addends. Since $10 D+U=100$, then $S=1000-99 D$. Thus, we need to min... | 406 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Pompous Vova has an iPhone XXX, and on that iPhone there is a voice-command calculator: "Multiply my number by two and subtract two from the result," "Be so kind as to multiply my number by three and then add four," and finally, "Add seven to my number!" The iPhone knows that Vova initially had the number 1. How man... | Answer: 9000 (or 18000, if negative numbers are considered).
We will prove that we can obtain all four-digit numbers. Note that the command +7 allows us to obtain from the current number all larger numbers with the same remainder when divided by seven. Therefore, it is sufficient to use the first two buttons to obtain... | 9000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Winnie-the-Pooh decided to give Eeyore a pot of honey. On the way to Eeyore, he tried the honey from the pot several times. When he tried the honey for the first time, the pot became half as heavy. And after the second time, the pot became half as heavy again. And after the third! And after the fourth! Indeed, after... | Solution. $\quad$ Answer: 3000 g. Indeed, $3000=(((200 \cdot 2) \cdot 2) \cdot 2) \cdot 2)-200$. | 3000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Petya has several 5-ruble coins and several 2-ruble coins. Vanya has as many 5-ruble coins as Petya has 2-ruble coins, and as many 2-ruble coins as Petya has 5-ruble coins. Petya has 60 rubles more than Vanya. Which coins does Petya have more of - 5-ruble or 2-ruble? By how many? | Solution. Answer: Petya has 20 more 5-ruble coins.
Suppose Petya has more 2-ruble coins than 5-ruble coins. Let Petya remove one 2-ruble coin, and Vanya remove one 5-ruble coin. Then the difference in the sums of Petya's and Vanya's coins will increase by 3. If they repeat this operation until Petya has an equal numbe... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The Elector George has 100 coins, some of which are counterfeit (possibly all or none). George can show the expert from 10 to 20 coins, and the expert will tell him how many of them are counterfeit. The problem is that the only expert in the entire region is Baron Münchhausen, and he exaggerates: the result given by... | Solution. It will work. We will submit $X$ random coins for expertise, and then the same $X$ plus 1 more. If the baron says the same number both times, the added coin is genuine; otherwise, it is counterfeit. This way, we can divide the coins into groups of 10, and check each group for counterfeits in 11 queries: first... | 110 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. There are several cards. On each of them, on each of the two sides, a circle is drawn: red, blue, or yellow. Among any 30 cards, there is a card with a red circle, among any 40 cards, there is a yellow circle, and among any 50, there is a blue one. In total, there are 20 cards with circles of different colors. Prove... | # Solution:
There are no more than 29 cards without red circles, no more than 39 cards without yellow, and no more than 49 cards without blue. Adding all these numbers, we get no more than 117. In this process, single-colored cards have been counted twice, and two-colored cards have been counted once. Therefore, the d... | 48 | Combinatorics | proof | Yes | Yes | olympiads | false |
7. Kopyatych took some natural number, raised it to the 1st, 2nd, 3rd, 4th, and 5th powers. Then he encrypted the numbers by replacing the same digits with the same letters and different digits with different letters. He wrote each encrypted number on a separate piece of paper. But Nyusha left only a fragment of each p... | # Solution:
It is clear that the letters К, Ё, Ж, И are of the same parity, and among them there are no 0 or 5.
If К, Ё, Ж, И are even, then all powers except the first one must be divisible by 4, and then О, Ш are also even digits, which is not good.
The last digit of the number in the first and fifth power is the ... | 189 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. We remind you that the sum of the angles in a triangle equals 180 degrees. In triangle $A B C$, angle $A$ is a right angle. Let $B M$ be the median of the triangle, and $D$ be the midpoint of $B M$. It turns out that $\angle A B D = \angle A C D$. What are these angles? | # Solution.
Draw $A D$. Since triangle $A B M$ is a right triangle, its median $A D$ equals half the hypotenuse $B M$. Therefore, $\angle A B D = \angle B A D = \alpha$, and $\angle A D M$ equals $2 \alpha$, as the exterior angle of triangle $A B D$. However, since $D M = A D$, $\angle D A M = \angle D M A = 90^{\circ... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. In the Magic and Wizardry club, all first and second-year students wear red robes, third-year students wear blue, and fourth-year students wear black.
Last year, at the general assembly of students, there were 15 red, 7 blue, and several black robes, while this year - blue and black robes are equal in number, and r... | # Solution.
(a) 7 mantles. Since 7 blue mantles from last year will turn black this year and there are an equal number of blue and black mantles this year, there will also be 7 blue mantles this year, which will turn black next year.
(b) 6 students. This year, there are 7 blue, 7 black, and $14=2 \times 7$ red mantle... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Each of five friends multiplied several consecutive numbers starting from 1. It turned out that one of the products is equal to the sum of the other four. Find all possible values of this product and show that there are no other values. | Solution. The product of all consecutive numbers from 1 to $x$ is called the factorial of the number $x$ and is denoted by $x!$. Thus, we need to solve the equation $x!=a!+b!+c!+d!$. Let the numbers $a, b, c, d$ be ordered in ascending order. Then $x>d$, i.e., $x! \geqslant x \cdot d!$. But we know that $x! \leqslant 4... | 24 | Number Theory | proof | Yes | Yes | olympiads | false |
1. Captain Billy the pirate plundered 1010 gold doubloons and set sail on his ship to a deserted island to bury them as treasure. Every evening of the voyage, he paid each of his pirates one doubloon. On the eighth day of the voyage, the pirates plundered a Spanish caravel, and Billy's treasure doubled, while the numbe... | Solution. Answer: 30
Before the pirates looted the caravel, Billy managed to pay the pirates their daily wages 7 times. After this, his fortune doubled. This is equivalent to Billy having 2020 doubloons before the voyage, and he paid the wages 14 times. After this, Billy paid the wages 40 times to half of the remainin... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The length of the escalator is 200 steps. When Petya walks down the escalator, he manages to count 50 steps. How many steps will he count if he runs twice as fast? | Solution. 80 steps. Let's call the step of the escalator from which Petya begins his descent the first step. When Petya walks down, he passes 50 steps. During this time, the first step manages to descend $200-50=150$ steps. Therefore, the escalator moves three times faster than Petya walks. When Petya runs, the speed r... | 80 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. A number is written on the board. In one move, you can either increase or decrease any of its digits by three (if it results in a digit), or swap two adjacent digits. Show how to transform the number 123456 into 654321 in 11 moves. | Solution. In the first 5 moves, we move 6 to the beginning: 612345. Then we swap 1 and 2, 4 and 5, getting 621354. Now we increase 2 and 1 by three, and decrease 5 and 4 by three - resulting in 654321. | 654321 | Logic and Puzzles | proof | Yes | Yes | olympiads | false |
3. A group of toddlers in a kindergarten has 90 teeth in total. Any two toddlers together do not have more than 9 teeth. What is the minimum number of toddlers that can be in the group? | Solution. If all children have fewer than 5 teeth, then there are no fewer than $90 / 4$, i.e., no fewer than 23 children.
If one child has exactly 5 teeth, then the others have no more than 4, and there are no fewer than $1+85 / 4$, i.e., also no fewer than 23 children.
If any of the children have between 6 and 9 te... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. A chessboard $(8 \times 8)$ was cut into several equal parts in such a way that all white cells remained uncut, while each black cell was cut. How many parts could have been obtained? | Solution. Note that there are 32 white cells, and each part contains an integer number of white cells, so the answer must be a divisor of 32.
Obviously, it cannot be 1. The answers $2, 4, 8, 16$, and 32 are possible. To achieve them,
 - are standing in a row. Each of them said one of two phrases: "To the left of me, there are twice as many knights as liars" or "To the left of me, there are as many knights as liars." In reality, there were more knights than liars... | Answer: 49.
Solution:
There are more knights than half, so:
1) Either they alternate like this: KRK...LK, but this option does not work: the phrase "there are twice as many knights as liars" cannot be said by any knight, but there are more than 50 such phrases.
2) Or some two knights stand next to each other. Two k... | 49 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. A group of toddlers in a kindergarten has 90 teeth in total. Any two toddlers together do not have more than 9 teeth. What is the minimum number of toddlers that can be in the group | Solution. If all children have fewer than 5 teeth, then there are no fewer than $90 / 4$, i.e., no fewer than 23 children.
If one child has exactly 5 teeth, then the others have no more than 4, and there are no fewer than $1 + 85 / 4$, i.e., also no fewer than 23 children.
If any of the children have between 6 and 9 ... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. From head to tail of the zebra Hippotigris - 360 stripes of the same width. Flea Masha and flea Dasha started crawling from the head of the zebra to its tail. At the same time, flea Sasha started crawling from the tail to the head. Flea Dasha crawls twice as fast as flea Masha. Before meeting flea Sasha, Masha overc... | Solution: 240 stripes.
Mashka crawled half of the zebra. Let her speed be $v$, then the closing speed of Mashka and Sashka is $2v$, and the closing speed of Dashka and Sashka is $3v$. Therefore, Sashka will crawl $3/2$ times fewer stripes before meeting Dashka than before meeting Mashka, i.e., 120 stripes. The remaini... | 240 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. In a row, 100 knights and 100 liars are standing (in some order). The first person was asked: "Are you a knight?", and the rest were asked in turn: "Is it true that the previous person answered 'Yes'?" What is the maximum number of people who could have said "Yes"? Knights always tell the truth, liars always lie. | Solution. 150, for example, if there are 100 knights followed by 100 liars. We will now prove that this is the maximum.
Consider any liar, except possibly the first one. Either they said "no," or the previous person said "no." Thus, to each of the 99 liars, we can associate at least one "no" response (either theirs or... | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In a certain city, the fare scheme for traveling by metro with a card is as follows: the first trip costs 50 rubles, and each subsequent trip costs either the same as the previous one or one ruble less. Petya spent 345 rubles on several trips, and then on several subsequent trips - another 365 rubles. How many trips... | Solution. A total of 710 rubles was spent. The total number of trips could not have been 14 or less (for 14 trips, a maximum of 700 rubles could be spent), but it also could not have been 17 or more $(50+49+48+\cdots+35+$ $34=714$, and this is the minimum that could be spent). Therefore, the choice is between two optio... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. In a company, several employees have a total monthly salary of 10000 dollars. A kind manager proposes to triple the salary for those earning up to 500 dollars, and increase the salary by 1000 dollars for the rest, so the total salary will become 24000 dollars. A mean manager proposes to reduce the salary to 500 doll... | Solution. Note that the increase proposed by the kind manager is twice as large as the salary proposed by the evil manager (this is true for both poor and rich employees). The increase according to the kind manager's proposal is 14000, so the salary according to the evil manager's proposal is 7000. | 7000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. 31 cars started simultaneously from one point on a circular track: the first car at a speed of 61 km/h, the second at 62 km/h, and so on (the 31st at 91 km/h). The track is narrow, and if one car overtakes another by a full lap, they crash into each other, both fly off the track, and are eliminated from the race. In... | Solution. First, the fastest car collides with the slowest, then the second fastest collides with the second slowest, and so on. In the end, the car with the median speed remains, i.e., the 16th. It travels at a speed of $76 \mathrm{Km} /$ h.
Criteria. Full solution - 3 points.
1 point if only the answer 76 is writte... | 76 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Five brothers were dividing their father's inheritance equally. The inheritance included three houses. Since the houses couldn't be divided, the three older brothers took them, and the younger brothers were given money: each of the three older brothers paid $2000. How much did one house cost?
## Solution: | Answer: $5000.
Since each of the brothers who received money got $3000, the total inheritance was estimated at $15000. Therefore, each house was worth a third of this amount. | 5000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the Parliament of the Emerald City, 5 parties are represented, which together developed 100 laws over the year (each law was developed by exactly one of the parties). It is known that any three parties together developed no fewer than 50 laws. What is the maximum number of laws that the Green Lenses party could h... | Solution:
Answer: 33.
Evaluation. Let this quantity be $x$. Since any three parties together have developed no less than 50, any party together with the Green Lenses party has developed no more than 50. Therefore, each of the four other parties has developed no more than $50-x$, and all of them together have develope... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Andrey placed chips of a hundred different colors in the cells of a $10 \times 10$ board. Each minute, one of the chips changes color, and only a chip that was unique (i.e., differed in color from all others) in its row or column before this operation can change color. After $N$ minutes, it turned out that no chip c... | Solution. Let's look at the moment when there are no moves left. Suppose we have $k$ colors left, then we have repainted at least $100-k$ cells. Note that if nothing can be repainted, then there are at least 4 cells of each color. In total, there are at least 25 colors left after this process. Therefore, we will repain... | 75 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Masha bought 2021 weights of pairwise distinct masses. Now Masha places one weight on each pan of a two-pan balance (weights placed on the balance previously are not removed). Each time the balance is in equilibrium, Masha rejoices. What is the maximum number of times she can find a reason to be happy | Solution. Answer: 673 times.
Example. Let Masha have bought 673 triples of the form $x, y, x+y$ (we will choose the weights of the new triples so that they do not duplicate the old ones). The last two weights are any. Masha puts $x$ on the left pan, then $y$ on the same left pan, and finally $x+y$ on the right pan and... | 673 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Seven fishermen stand in a circle. The fishermen have a professional habit of exaggerating numbers. Each fisherman has a measure of lying (each has their own, an integer) - how many times the number mentioned by the fisherman is greater than the true value. For example, if a fisherman with a lying measure of 3 catch... | Solution. Note that the product of all the numbers named by the fishermen in each of the surveys is the product of the number of fish caught, multiplied by the product of all the measures of lying of these fishermen. Therefore, the seventh answered $\frac{12 \cdot 12 \cdot 20 \cdot 24 \cdot 32 \cdot 42 \cdot 56}{12 \cd... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. The gnomes have gone to work, and Snow White is feeling lonely. She laid out a pile of fifteen stones on the table. Every minute, Snow White splits one pile into two non-empty piles and adds a stone to one of them. How can Snow White use these actions to get seven identical piles? | Solution. First, let's understand how many stones are in the piles. With each action, the number of piles increases by one, as does the number of stones. Therefore, 7 piles will arise after 6 actions, with the total number of stones being $15+6=21$. This means each pile should have three stones. From these consideratio... | 3333333 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In the company, several employees have a total monthly salary of 10000 dollars. A kind manager proposes to double the salary for everyone earning up to 500 dollars, and increase the salary by 500 dollars for the rest, so the total salary will become 17000 dollars. A mean manager proposes to reduce the salary to 500 ... | Solution. Answer: 7000
Let's look at the difference between the total salary from the kind manager and the current one, and we will understand that this is the total salary from the evil manager:
up to 500 - from the doubled we subtract the current, we get the current
over 500 - from the increased by 500 we subtract... | 7000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Cheburashka bought as many mirrors from Galina in the store as Gen bought from Shapoklyak. If Gen had bought from Galina, he would have 27 mirrors, and if Cheburashka had bought from Shapoklyak, he would have 3 mirrors. How many mirrors would Gen and Cheburashka buy together if Galina and Shapoklyak agreed and set t... | Solution. Let Gena have $x$ times more money than Cheburashka. If we swap Gena's and Cheburashka's money, then with the second method of purchase, the number of mirrors should be equal. Therefore, $3 x=\frac{27}{x}$, from which $3 x^{2}=27$, and $x=3$. So, Gena initially had three times more money than Cheburashka, and... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Let $N=400$. What is the maximum number of balls that the magician can guarantee are not in their correct vessels?
# | # Solution:
Answer: 533.
Let's imagine that inside the vessels, the balls are divided into cells, and each cell contains one ball. Then we can imagine that (regardless of whether they are in the same vessel or different ones) balls $i$ and $j$ simply swap cells upon the command ( $i j$ ). Thus, any sequence of comman... | 533 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. 31 cars started simultaneously from one point on a circular track: the first car at a speed of 61 km/h, the second at 62 km/h, and so on (the 31st at 91 km/h). The track is narrow, and if one car overtakes another by a full lap, they crash into each other, both fly off the track, and are eliminated from the race. In... | Solution. First, the fastest car collides with the slowest, then the second fastest collides with the second slowest, and so on. In the end, the car with the median speed remains, i.e., the 16th. It travels at a speed of $76 \mathrm{Km} /$ h.
Criteria. Full solution - 7 points.
Solution that correctly describes the o... | 76 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Each cell of a $10 \times 10$ board is painted black or white. A cell is said to be out of place if it has at least seven neighbors of a different color than itself. (Neighbors are cells that share a common side or corner.) What is the maximum number of cells on the board that can be out of place at the same time? | # Solution.
Answer 26. The example consists of 13 dominoes, which do not share any cells and do not touch the border of the board (see right).
Estimation. Let's call a cell that is not in its place an NVCT-cell. Obviously, NVCT-cells cannot be adjacent to the border. Let's examine what form a connected component of N... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Each cell of a $10 \times 10$ board is painted black or white. A cell is said to be out of place if it has at least seven neighbors of a different color than itself. (Neighbors are cells that share a common side or corner.) What is the maximum number of white cells on the board that can be out of place at the same t... | Solution. Answer: 26
Example: 13 dominoes, not touching each other or the edge of the board.
Estimate. Let's call a cell that is not in its own place an OOP cell. For each OOP cell, we draw ... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Masha bought 2021 weights of pairwise distinct masses. Now Masha places one weight on each pan of a two-pan balance (weights placed on the balance previously are not removed). Each time the balance is in equilibrium, Masha rejoices. What is the maximum number of times she can find a reason to be happy | Solution. Answer: 673 times.
Example. Let Masha have bought 673 triples of the form $x, y, x+y$ (we will choose the weights of the new triples so that they do not duplicate the old ones). The last two weights are any. Masha puts $x$ on the left pan, then $-y$ on the same left pan, and finally $x+y$ on the right pan an... | 673 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Seven fishermen stand in a circle. The fishermen have a professional habit of exaggerating numbers. Each fisherman has a measure of lying (each has their own, an integer) - how many times the number mentioned by the fisherman is greater than the true value. For example, if a fisherman with a lying measure of 3 catch... | Solution. Note that the product of all the numbers named by the fishermen in each of the surveys is the product of the number of fish caught, multiplied by the product of all the measures of lying of these fishermen. Therefore, the seventh answered $\frac{12 \cdot 12 \cdot 20 \cdot 24 \cdot 32 \cdot 42 \cdot 56}{12 \cd... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. The lye trader Matti wrote several three-digit palindromic numbers on the board. It turned out that no two of them add up to a palindrome. Could he have written more than half of all three-digit palindromes? Recall that a palindrome is a number that remains the same when its digits are reversed. | Solution. No. There are $9 \cdot 10$ three-digit palindromes in total (9 options for the first digit, 10 for the second, and the third is uniquely determined by the first). We will prove that at least 45 palindromes are missing from the board.
If $a, b, c, d$ are such digits that $a+c=b+d=11$, then
$$
\overline{a b a... | 47 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Andrey named all natural numbers from 180 to 220 inclusive ("one hundred eighty", "one hundred eighty one", etc.). How many words did he say? | Solution. A total of 41 numbers are named. In each, the hundreds place is named (a total of 41 words). In addition to the hundreds place, the numbers 180, 190, 201-220 each contain one more word (this is 22 words). The number 200 does not contribute any more words, while the remaining 18 numbers each contain two more w... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. What is the smallest area that a figure on the plane $Oxy$ can have, located between the lines $x=-26$ and $x=2$, bounded below by the $x$-axis, and above by the tangent to the graph of the function $y=5+\sqrt{4-x}$ at the point of tangency with abscissa $x_{0}$, where $x_{0}$ lies in the interval $-26 \leq x_{0} \l... | Solution. Solution. We will write the equation of the tangent line to the graph of the function $y=5+\sqrt{4-x}$ at the point with abscissa $x_{0}$
$$
y_{\text {tan }}=-\frac{1}{2 \sqrt{4-x_{0}}}\left(x-x_{0}\right)+5+\sqrt{4-x_{0}}
$$
^{3}$ after combining like terms. | Solution: $\left(1+x+x^{2}+\cdots+x^{80}\right)^{3}=\left(1+x+x^{2}+\cdots+x^{80}\right)\left(1+x+x^{2}+\cdots+x^{80}\right)\left(1+x+x^{2}+\cdots+x^{80}\right)$
Let after expanding the brackets, the term $x^{p} \cdot x^{q} \cdot x^{r}$ is obtained if the factor $x^{p}$ is chosen from the first bracket, $x^{q}$ - from... | 3321 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. The resistance to bending of a beam with a rectangular cross-section is proportional to the product of the width of this section and the square of its height. What should be the width of the cross-section of a beam cut from a round log with a diameter of $15 \sqrt{3}$ so that its resistance to bending is maximized?
... | # Solution.
$F(a)=k a b^{2}, a-$ width of the beam, $b$ - height of the beam. Since $b^{2}=225 \cdot 3-a^{2}$, then
$F(a)=k a\left(225 \cdot 3-a^{2}\right)=k\left(225 \cdot 3 a-a^{3}\right), F^{\prime}(a)=k\left(225 \cdot 3-3 a^{2}\right)=3 k\left(225-a^{2}\right)$.
The derivative equals zero and changes sign from p... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find all pairs of integers $(x, y)$ that satisfy the equation $x y=20-3 x+y$. For each found pair $(x, y)$, calculate the product $x y$. In the answer, write the sum of these products. | Solution. $\quad x y=20-3 x+y, \quad x y=17+3(1-x)+y, \quad(x-1) y+3(x-1)=17$, $(x-1)(y+3)=17$. Since x and y are integers, we have four cases:
1) $\left\{\begin{array}{c}y+3=17, \\ x-1=1,\end{array} \Leftrightarrow\left\{\begin{array}{c}y=14 \\ x=2\end{array}\right.\right.$
2) $\left\{\begin{array}{c}y+3=-17, \\ x-1=... | 56 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find all natural solutions to the inequality
$$
\frac{4}{5}+\frac{4}{45}+\frac{4}{117}+\cdots+\frac{4}{16 n^{2}-8 n-3}>n-5
$$
In your answer, write the sum of all found solutions. | Solution. We have $16 n^{2}-8 n-3=(4 n-3)(4 n+1), \quad \frac{1}{4 n-3}-\frac{1}{4 n+1}=\frac{4}{16 n^{2}-8 n-3}$,
$\frac{4}{1 \cdot 5}+\frac{4}{5 \cdot 9}+\frac{4}{9 \cdot 13}+\cdots+\frac{4}{(4 n-3)(4 n+1)}>n-5$,
$1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\cdots+\frac{1}{4 n-3}-\frac{1}{4 n+1}>n-5,1-\frac{1}{4 n+1}>n-5... | 15 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. There is an unlimited supply of square glasses in 10 colors. In how many ways can 4 glasses be inserted into a $2 \times 2$ window frame so that some color appears in both the upper and lower halves of the window. | Solution. If the upper half of the glass is of one color (one of 10), then the colors of the lower glasses can be chosen in $10^{2}-9^{2}=19$ ways. If the upper glasses are different $\left(10 * 9=90\right.$ ways to choose), then the colors of the lower ones can be chosen in $10^{2}-8^{2}=36$ ways.
In total $10 * 19+9... | 3430 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Let $x, y, z$ be the roots of the equation $t^{3}-5 t-3=0$. Find $x^{3} y^{3}+x^{3} z^{3}+y^{3} z^{3}$. | Solution. The polynomial has 3 different real roots, since $\mathrm{P}(-100)0, \mathrm{P}(0)0$. By Vieta's theorem, $\mathrm{x}+\mathrm{y}+\mathrm{z}=0, \mathrm{xy}+\mathrm{xz}+\mathrm{yz}=-5, \mathrm{xyz}=3$.
$$
\begin{gathered}
x^{3} y^{3}+x^{3} z^{3}+y^{3} z^{3}=(5 x+3)(5 y+3)+(5 x+3)(5 z+3)+(5 y+3)(5 z+3) \\
=25(x... | -98 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. There are 5 pieces of transparent glass of the same square shape and size. Each piece of glass is conditionally divided into 4 equal parts (right triangles) by its diagonals, and one of these triangles is painted with an opaque paint of its individual color, different from the colors of the painted parts of the othe... | Solution. First, consider some fixed vertical order of laying the glasses (from bottom to top). It is clear that by rotating the entire layout by some angle, we will not change the layout (we will not get a new layout). Therefore, we can assume that the bottom glass in the layout is always fixed (does not rotate). Then... | 7200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A student wrote a program to recolor a pixel into one of 128 different colors. These colors he numbered with natural numbers from 1 to 128, and the primary colors received the following numbers: white color - number 1, red - 5, orange - 13, yellow - 19, green - 23, blue - 53, blue - 55, purple - 83, black - 128. If ... | Solution. The final pixel color number is equal to $f^{[2019]}(5)$, where $f^{[k]}(n)=\underbrace{f(f(f(\ldots(f}_{k \text { times }}(n) \ldots)-k$ is the $k$-fold composition of the function $f(n)$, which is equal to $3 n-2$ for $n \leq 17$, and equal to $|129-2 n|$ for $n \geq 18$. Let's compute and write down the fi... | 55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The wolf saw a roe deer several meters away from him and chased after her along a straight forest path. The wolf's jump is $22\%$ shorter than the roe deer's jump. Both animals jump at a constant speed. All the roe deer's jumps are of the same length, and the wolf's jumps are also equal to each other. There is a per... | Solution: Let x be the length of the roe deer's jump, then $0.78x$ is the length of the wolf's jump; y - the number of jumps the roe deer makes in a unit of time specified in the condition, $y\left(1+\frac{t}{100}\right)$ - the number of jumps the wolf makes in the same unit of time. The wolf will not be able to catch ... | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the arithmetic progression $\left(a_{n}\right) a_{1000}=75, d=0.25$.
Calculate: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cdot a_{1581}}+\frac{1}{a_{1581} \cdot a_{1582}}+\ldots+\frac{1}{a_{2019} \cdot a_{2020}}\right)$. | Solution: The expression in parentheses consists of several terms of the form $\frac{1}{x \cdot(x+d)}$, which can be decomposed into the sum of simpler fractions: $\frac{1}{x \cdot(x+d)}=\frac{1}{d}\left(\frac{1}{x}-\frac{1}{x+d}\right)$. Transform the original expression: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cd... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In $\triangle A B C$ with $\angle B=120^{0}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn.
Segment $A_{1} B_{1}$ intersects the bisector $C C_{1}$ at point M. Find the angle $\Delta B_{1} C_{1} M$.
# | # Solution.
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$.
Considering that point $A_{1}$ lies on the bisector of $\angle B A C$, and thus is equidistant from its sides, we get that $A_{1}$ is equid... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. What is the minimum number of cells that need to be painted in a square with a side of 65 cells (65x65 - there are a total of 4225 cells in the square), so that among any four of its cells forming a corner (an "L" shape), there is at least one painted cell.
# | # Solution.
Shading should be done diagonally every 3rd (see fig.). Thus, $\left[\frac{N^{2}}{3}\right]$ cells will be shaded. This is the minimum possible number, as within any $3 \times 3$ square, at least three cells need to be shaded. $\left[\frac{65^{2}}{3}\right]=1408$.
=10|\mathrm{p}-3|+|2 \mathrm{p}-| \mathrm{p}+\mathrm{c}||-6 \mathrm{p}$. The coefficient of the first modulus in absolute value is greater than the sum of the other coefficients of $\mathrm{p}$. $10>2+1+6$. Therefore, on all intervals up to $\mathrm{p}=3$, the c... | -147 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the arithmetic progression $\left(a_{n}\right) a_{1}=1, d=4$.
$$
\text { Calculate } A=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{1579}}+\sqrt{a_{1580}}}
$$
In the answer, write the smallest integer greater than $A$. | Solution: We transform the expression by multiplying the numerator and denominator of each fraction by the expression conjugate to the denominator:
$$
\begin{aligned}
& A=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{1579}}+\sqrt{a_{1580}}}= \\
& =\frac{\sqrt{a_{2}}-\... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Right triangles $\triangle M D C$ and $\triangle A D K$ have a common right angle $\angle D$. Point $K$ lies on $C D$ and divides it in the ratio $2: 3$ from point $C$. Point $M$ is the midpoint of side $A D$. Find the sum of $\angle A K D$ and $\angle M C D$, if $A D: C D=2: 5$.
$ satisfies the condition $f(x+1)+1=f(x)+\frac{20}{(x+1)(x+2)}$. Calculate $\frac{2019}{f(2019)}$, if $\quad f(0)=2019$. | # Solution.
Notice that
$f(x+2019)-f(x)=(f(x+2019)-f(x+2018))+(f(x+2018)-f(x+2017))+\cdots+(f(x+$ 1) $-f(x))=\frac{20}{(x+2019)(x+2020)}-1+\frac{20}{(x+2018)(x+2019)}-1+\cdots+\frac{20}{(x+1)(x+2)}-1$. Therefore, $f(2019)-f(0)=20\left(\frac{1}{2019}-\frac{1}{2020}+\ldots+1-\frac{1}{2}\right)-2019=20\left(1-\frac{1}{2... | 101 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. What is the minimum number of cells that need to be painted in a square with a side of 65 cells (a total of $65 \times 65$ cells, which is 4225 cells in the square), so that from any unpainted cell it is impossible to move to another unpainted cell with a knight's move in chess? | # Solution
Cells should be colored in a checkerboard pattern. Thus, $\left[\frac{N^{2}}{2}\right]$ cells will be colored. Since any "knight's move" lands on a cell of a different color, there is no move to a cell of the same color. A "knight's move" can traverse any square table larger than 4x4 such that the knight vi... | 2112 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. Find the sum of all integer values of $\mathrm{h}$ for which the equation ||$r+h|-r|-4 r=9|r-3|$ in terms of $r$ has no more than one solution.
# | # Solution
Consider the function $\mathrm{f}(\mathrm{r})=9|\mathrm{r}-3|-|| \mathrm{r}+\mathrm{h}|-\mathrm{r}|+4 \mathrm{r}$. The coefficient of the first modulus in absolute value is greater than the sum of the other coefficients of $\mathrm{r}$. $9>1+1+4$. Therefore, on all intervals up to $r=3$, the coefficient of ... | -93 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. A table consisting of 1861 rows and 1861 columns is filled with natural numbers from 1 to 1861 such that each row contains all numbers from 1 to 1861. Find the sum of the numbers on the diagonal that connects the top left and bottom right corners of the table, if the table is filled symmetrically with respect to thi... | # Solution:
We will show that all numbers from 1 to 1861 are present on the diagonal. Suppose the number $a \in\{1,2,3 \ldots, 1861\}$ is not on the diagonal. Then, due to the symmetry of the table, the number $a$ appears an even number of times. On the other hand, since the number $a$ appears once in each row, the to... | 1732591 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Ilya takes a triplet of numbers and transforms it according to the rule: at each step, each number changes to the sum of the other two. What is the difference between the largest and the smallest numbers in the triplet after 1989 applications of this rule, if the initial triplet of numbers was $\{70 ; 61 ; 20\}$? If... | # Solution:
Let's denote the 3 numbers as $\{x ; x+a ; x+b\}$, where $0<a<b$. Then the difference between the largest and the smallest number at any step, starting from the zeroth step, will be an invariant, that is, unchanged and equal to $b . \quad b=70-20=50$.
Answer: 50. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Two points are moving along a circle. When moving in the same direction, the faster point catches up with the slower point every 16 seconds. If they move in opposite directions at the same speeds, they meet every 4 seconds. It is known that when moving towards each other along the circle, the distance between the ap... | # Solution:
Let $v-$ be the speed of the faster point, and $u-$ be the speed of the slower point. According to the problem, we set up the system of equations: $\left\{\begin{array}{l}(u+v) \cdot 4=(v-u) \cdot 16 \\ (u+v) \cdot 2=64\end{array} ;\left\{\begin{array}{l}3 v=5 u \\ u+v=32\end{array} ;\left\{\begin{array}{l... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In triangle $A B C$, side $B C$ is 18 cm. The perpendicular $D F$, drawn from the midpoint of side $A B$ - point $D$, intersects side $B C$ at point $F$. Find the perimeter of triangle $A F C$, if side $A C$ is $9 \, \text{cm}$. | # Solution:
Triangle $ABF (BF = AF)$ is isosceles, since $DF \perp AB$, and $D$ is the midpoint of $AB$. $P_{AFC} = AF + FC + AC = BF + FC + AC = BC + AC = 27 \text{ cm}$.
Answer: 27 cm. | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. There are two types of containers: 27 kg and 65 kg. How many containers of the first and second type were there in total, if the cargo in the first type of containers exceeds the cargo in the second type of containers by 34 kg, and the number of 65 kg containers does not exceed 44 units?
# | # Solution:
Let $x$ be the number of containers weighing 27 kg, and $y$ be the number of containers weighing 65 kg. We get the equation $27 x - 65 y = 34$.
$27(x - 2y) - 11y = 34$, let $x - 2y = k$
$27k - 11y = 34$
$11(2k - y) + 5k = 34$, let $2k - y = t(2)$
$11t + 5k = 34$
$5(2t + k) + t = 34$, let $2t + k = n(3... | 66 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. A farmer presented 6 types of sour cream in barrels of $9,13,17,19,20,38$ liters at the market. On the first day, he sold sour cream from three barrels completely, and on the second day, from two more barrels completely. The volume of sour cream sold on the first day was twice the volume of sour cream sold on the se... | # Solution:
A total of 116 liters were delivered; $116=3 \cdot 38+2$; therefore, the unsold barrel, when divided by 3, gives a remainder of 2.
1) If it is 17 liters, then $116-17=99$, then on the second day, one third of 99 liters were sold, which is $33=13+20$, then on the first day, barrels with capacities of $9, 1... | 66 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. In triangle $A B C$ with $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Segment $A_{1} B_{1}$ intersects the angle bisector $C C_{1}$ at point M. Find $\angle C B M$.
# | # Solution:
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$.
Considering that point $A_{1}$ lies on the bisector of $\angle B A C$, and therefore is equidistant from its sides.
We obtain that $A_{1}$ i... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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