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10.4 Find the smallest positive integer $k$, for which in any coloring of the numbers of the set $M=\{1,2,3, \ldots, k\}$ in two colors, there will be ten not necessarily distinct numbers of the same color from the set $M$ | Answer: 109.
Solution. Let $k \geq 100$. Suppose that for any 10 numbers of one color, their sum is either a number of the other color or does not belong to M. Let the number 1 be of the first color, and the number 2 be of the second color. Then the numbers 10 and 20 are of the second and first colors, respectively. N... | 109 | Combinatorics | 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 marked is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of gl... | 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?
It remains to show that after removing any $n=5099$ stones, there will still be many stones le... | 5099 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. A circle is inscribed in trapezoid $A B C D$, touching the lateral side $A D$ at point $K$. Find the area of the trapezoid if $A K=16, D K=4$ and $C D=6$. | Answer: 432.
Solution. Let $L, M, N$ be the points of tangency of the inscribed circle with the sides $BC, AB, CD$ respectively; let $I$ be the center of the inscribed circle. Denote the radius of the circle by $r$. Immediately note that $DN = DK = 4$ (the first equality follows from the equality of the segments of ta... | 432 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Let's call a natural number "remarkable" if all its digits are different, it does not start with the digit 2, and by erasing some of its digits, the number 2018 can be obtained. How many different seven-digit "remarkable" numbers exist? | Answer: 1800.
Solution: To correctly count the number of options, it is necessary to follow the rule: before the digit 2, there must be one of the six digits $-3,4,5,6,7$ or 9. Let's assume for definiteness that this is 3, then two different digits from the remaining five ( $\frac{5 \cdot 4}{2}=10$ options) can be in ... | 1800 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Lena downloaded a new game for her smartphone, where it is allowed to conduct alchemical reactions of two types. If she combines one "fire" element and one "stone" element, she gets one "metal" element. And if she combines one "metal" element and one "stone" element, she gets three "stone" elements. Lena has 50 "fir... | Answer: 14 elements.
Solution. Consider the expression $S=2 x+y+z$, where $x$ is the number of "metal" elements, $y$ is the number of "fire" elements, and $z$ is the number of "stone" elements. It is easy to see that this expression does not change with each of the two alchemical operations:
$$
\begin{aligned}
& 2(x+... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.1. A three-digit number is 56 times greater than its last digit. How many times greater is it than its first digit? Justify your answer.
# | # Solution:
Method 1. The last digit is such that when multiplied by 6, the resulting number ends with the same digit. By exhaustive search, we confirm that this can be any even digit (and only it). Therefore, this three-digit number is either 112, 224, 336, or 448 (the option with the last digit 0 is not valid, as it... | 112 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.3. On the way from city $A$ to city $B$, there are kilometer markers every kilometer. On each marker, one side shows the distance to $A$, and the other side shows the distance to $B$. In the morning, a tourist passed by a marker where one number was twice the other. After walking another 10 km, the tourist saw a mark... | Solution: Let $C_{1}$ and $C_{2}$ be the poles mentioned in the problem ($C_{1}$ - the pole near which the tourist was in the morning). Without loss of generality, assume the tourist was walking from $A$ to $B$. Then there are two possible situations: 1) $C_{1} A = 2 C_{1} B$ or 2) $C_{1} B = 2 C_{1} A$. Since $C_{2} A... | 120 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.5. 101 people bought 212 balloons of four colors, and each of them bought at least one balloon, but no one had two balloons of the same color. The number of people who bought 4 balloons is 13 more than the number of people who bought 2 balloons. How many people bought only one balloon? Provide all possible answers an... | # Solution:
Method 1. First, exclude the 13 people who bought 4 balloons. There will be 88 people left, who bought a total of 160 balloons, with each person buying between 1 and 4 balloons, and the number of people who bought 2 and 4 balloons is now equal. Suppose now that each person who bought 4 balloons gives one b... | 52 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3.1. Every day, a sweet tooth buys one more candy than the previous day. In one week, on Monday, Tuesday, and Wednesday, he bought a total of 504 candies. How many candies did he buy on Thursday, Friday, and Saturday in total for the same week? | Answer: 513
Solution. On Thursday, 3 more candies were bought than on Monday, on Friday - 3 more than on Tuesday, and on Saturday - 3 more than on Wednesday. In total, 9 more candies were bought on Thursday, Friday, and Saturday compared to Monday, Tuesday, and Wednesday, i.e., $504+9=513$ candies. | 513 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. In an $11 \times 11$ square, the central cell is painted black. Maxim found a rectangular grid of the largest area that is entirely within the square and does not contain the black cell. How many cells does it have? | Answer: 55
Solution. If both sides of the rectangle are not less than 6, then it must contain the central cell, since the distance from the painted cell to the sides is 5 cells. If one of its sides is not more than 5, then the other is definitely not more than 11, and therefore its area is not more than 55. | 55 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.1. In a rectangle of size $5 \times 18$, numbers from 1 to 90 were placed. This resulted in five rows and eighteen columns. In each column, the median number was selected, and from these median numbers, the largest one was chosen. What is the smallest value that this largest number can take?
Recall that among 99 num... | Answer: 54
Solution. The largest of the average numbers $N$ is not less than each of the eighteen selected averages. Each of the eighteen averages, in turn, is not less than three numbers from its column (including itself). In total, $N$ is not less than some 54 numbers in the table. Therefore, $N$ is at least 54.
To... | 54 | Other | math-word-problem | Yes | Yes | olympiads | false |
# Task 8.2
For a natural number $N$, all its divisors were listed, and then the sum of digits for each of these divisors was calculated. It turned out that among these sums, all numbers from 1 to 9 were found. Find the smallest value of $\mathrm{N}$.
## Number of points 7 | Answer:
288
## Solution
Note that the number 288 has divisors $1,2,3,4,32,6,16,8,9$. Therefore, this number satisfies the condition of the problem. We will prove that there is no smaller number that satisfies the condition.
Indeed, since $\mathrm{N}$ must have a divisor with the sum of digits 9, $\mathrm{N}$ is div... | 288 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task 8.5
On the Island of Liars and Knights, a circular arrangement is called correct if each person standing in the circle can say that among their two neighbors, there is a representative of their tribe. Once, 2019 natives formed a correct arrangement in a circle. A liar approached them and said: "Now we can also ... | Answer:
1346
## Solution
We will prove that a correct arrangement around a circle is possible if and only if the number of knights is at least twice the number of liars.
Indeed, from the problem's condition, it follows that in such an arrangement, each liar has two knights as neighbors, and among the neighbors of a... | 1346 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Find the largest even three-digit number $x$ that gives a remainder of 2 when divided by 5 and satisfies the condition $\operatorname{GCD}(30, \operatorname{GCD}(x, 15))=3$. | Solution. From the condition, we get that there exist such $a, b \in \mathbb{N}$ that $3a=30$, GCD $(x, 15)=3b$, and GCD $(a, b)=1$.
Consider the equality GCD $(x, 15)=3b$. This means that $\exists c, d \in \mathbb{N}$ such that $x=3bc$ and $15=3bd$, and GCD $(c, d)=1$.
From the equality $15=3bd$ it follows that $bd=... | 972 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. A hundred people are standing in a circle, each of whom is either a knight or a liar (liars always lie, and knights always tell the truth). Each of those standing said: “I have a liar as a neighbor.” Find the minimum possible number of liars among these 100 people. | Answer: 34.
Solution: Note that 3 knights cannot stand next to each other, as in this case, the middle knight would be lying. Therefore, among any 3 standing next to each other, there is a liar. Take any liar, and divide the remaining 99 people into 33 groups of three standing next to each other. Since there is at lea... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. On the Island of Misfortune, only knights, who always tell the truth, and liars, who always lie, live. The island is governed by a group of 101 people. At the last meeting, it was decided to reduce this group by 1 person. But each member of the group stated that if they were removed from the group, the majority of t... | 2. 50 knights and 51 liars. Let there be $k$ knights in the group, then there are $101-k$ liars. Each one said that if they were removed, there would be no less than 51 liars among the remaining 100 people. Since the knights told the truth, then $k-1 \leq 49$ and
, then $\angle A C B=2 \alpha=\angle C A D$ (alternate interior angles). Triangle ACB is isosceles by condition, so $... | 22 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.1. In a six-digit number, the first digit, which is 2, was moved to the last place, leaving the other digits in the same order. The resulting number turned out to be three times larger than the original. Find the original number. | Answer: 285714.
Solution. According to the condition, the desired number has the form $\overline{2 a b c d e}$, then we have: $\overline{a b c d e 2}=$ $3 \cdot \overline{2 a b c d e}$, or $\overline{a b c d e} \cdot 10+2=3 \cdot(200000+\overline{a b c d e})$. Let $\overline{a b c d e}=X$ - a five-digit number, then $... | 285714 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 1. Clone 1
The figure is divided into 7 equal squares and several rectangles. The perimeter of rectangle A is 112 cm. What is the perimeter of rectangle B? Express your answer in centimeters.
![](https://cdn.mathpix.com/cropped/2024_05_06_d0dc74dfba1714359e24g-01.jpg?height=485&width=483&top_left_y=1008&top_left_x=... | # Answer: 168
## Solution
1st method. Rectangles A and B are composed of identical squares: Rectangle A consists of three, and Rectangle B consists of four. Let's call the side of such a square a "stick" and count the perimeter in sticks. The perimeter of Rectangle B is four sticks more than the perimeter of Rectangl... | 168 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 3. Clone 1
A rope was divided into 19 equal parts and arranged in a snake-like pattern. After that, a cut was made along the dotted line. The rope split into 20 pieces: the longest of them is 8 meters, and the shortest is 2 meters. What was the length of the rope before it was cut? Express your answer in meters.
-4x=32
$$ | 32 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.2. Each of the natural numbers $1,2,3, \ldots, 377$ is painted either red or blue (both colors are present). It is known that the number of red numbers is equal to the smallest red number, and the number of blue numbers is equal to the largest blue number. What is the smallest red number? | Answer: 189.
Solution. Let $N$ be the largest blue number. Then only numbers from 1 to $N$ can be painted blue. Since there are a total of $N$ blue numbers, we get that all numbers from 1 to $N$ are blue. Accordingly, all numbers from $N+1$ to 377 are red. Since the number of red numbers equals the smallest red number... | 189 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.3. Krosh and Yozhik decided to check who would run faster along a straight road from Kopyatych's house to Losyash's house. When Krosh had run 20 meters, Yozhik had run only 16 meters. And when Krosh had 30 meters left, Yozhik had 60 meters left. How many meters is the length of the road from Kopyatych's house... | Answer: 180.
Solution. When Krosh ran 20 meters, Yozhik ran only 16 meters, so their speeds are in the ratio of $5: 4$.
When Krosh had 30 meters left to run, let him have already run $5 x$ meters (where $x$ is not necessarily an integer). Then by this point, Yozhik had run $4 x$ meters. Therefore, the total length of... | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. Consider seven-digit natural numbers, in the decimal representation of which each of the digits $1,2,3,4,5,6,7$ appears exactly once.
(a) (1 point) How many of them have the digits from the first to the sixth in ascending order, and from the sixth to the seventh in descending order?
(b) (3 points) How ma... | Answer: (a) 6. (b) 15.
Solution. (a) From the condition, it follows that the sixth digit is the largest, so it is 7. The last digit can be any digit from 1 to 6, and this uniquely determines the entire number (since the first five digits must be in ascending order). Therefore, there are exactly 6 such seven-digit numb... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. In the forest, there live elves and gnomes. One day, 60 inhabitants of this forest lined up facing the same direction, at which point some of them might have been wearing hats. (There could have been from 0 to 60 elves, and inhabitants wearing hats could also have been from 0 to 60 inclusive.)
Each of the... | Answer: (a) 59. (b) 30.
Solution. (a) Note that the rightmost resident cannot be telling the truth, since there is no one to the right of him. This means he cannot be a hatless elf, so the total number of hatless elves is no more than 59.
Now let's provide an example where there are exactly 59 hatless elves. Suppose ... | 59 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.8. The cells of a $50 \times 50$ table are colored in $n$ colors such that for any cell, the union of its row and column contains cells of all $n$ colors. Find the maximum possible number of blue cells if
(a) (1 point) $n=2$
(b) (3 points) $n=25$. | Answer: (a) 2450. (b) 1300.
Solution. We will prove that the number of cells of any color $A$ present in the coloring is no less than 50. Suppose this is not the case, and the number of cells of color $A$ is no more than 49. Then there exists a row without cells of color $A$, and there also exists a column without cel... | 1300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 7.8.1. The cells of a $50 \times 50$ table are colored in $n$ colors such that for any cell, the union of its row and column contains cells of all $n$ colors. Find the maximum possible number of blue cells if
(a) (1 point) $n=2$;
(b) (3 points) $n=25$. | Answer: (a) 2450. (b) 1300. | 1300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 7.8.2. The cells of a $40 \times 40$ table are colored in $n$ colors such that for any cell, the union of its row and column contains cells of all $n$ colors. Find the maximum possible number of blue cells if
(a) (1 point) $n=2$;
(b) (3 points) $n=20$. | Answer: (a) 1560. (b) 840. | 840 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 7.8.3. The cells of a $30 \times 30$ table are colored in $n$ colors such that for any cell, the union of its row and column contains cells of all $n$ colors. Find the maximum possible number of blue cells if
(a) $(1$ point) $n=2$;
(b) $(3$ points) $n=15$. | Answer: (a) 870. (b) 480. | 480 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 7.8.4. The cells of a $20 \times 20$ table are colored in $n$ colors such that for any cell, the union of its row and column contains cells of all $n$ colors. Find the maximum possible number of blue cells if
(a) (1 point) $n=2$;
(b) (3 points) $n=10$. | Answer: (a) 380. (b) 220. | 380 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.4. Roma thought of a natural number, the sum of the digits of which is divisible by 8. Then he added 2 to the thought number and again got a number, the sum of the digits of which is divisible by 8. Find the smallest number that Roma could have thought of. | Answer: 699.
Solution. If both numbers are divisible by 8, then their difference is also divisible by 8. If there was no carry-over when adding, the sum of the digits would differ by 2, which is not divisible by 8. If there was a carry-over but not into the hundreds place, the sum of the digits would differ by $9-2=7$... | 699 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.5. On the side $AB$ of the rectangle $ABCD$, a circle $\omega$ is constructed with $AB$ as its diameter. Let $P$ be the second intersection point of the segment $AC$ and the circle $\omega$. The tangent to $\omega$ at point $P$ intersects the segment $BC$ at point $K$ and passes through point $D$. Find $AD$,... | Answer: 24.
Solution. Note that triangles $A D P$ and $C K P$ are similar (the equality $\angle D A P=\angle K C P$ and $\angle A D P=\angle C K P$ is ensured by the parallelism $A D \| B C$). Moreover, triangle $A D P$ is isosceles, as $A D$ and $D P$ are segments of tangents. Therefore, triangle $C K P$ is also isos... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.8. On the face $ABC$ of the tetrahedron $ABCD$, a point $P$ is marked. Points $A_1, B_1, C_1$ are the projections of point $P$ onto the faces $BCD, ACD, ABD$ respectively. It turns out that $PA_1 = PB_1 = PC_1$. Find $\angle BA_1C$, given that $\angle BC_1D = 136^\circ$ and $\angle CB_1D = 109^\circ$.
$ and the hypotenuse (common $P D$). From this, we get that $D C_{1}=D A_{1}$.
Considering triangles $B P C_{1}$ and $B P A_{1}$, we similarly obtain $B C_{1}=B A_{1}$. Then triangles $B D C_{... | 115 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) Winnie-the-Pooh eats 3 cans of condensed milk and a jar of honey in 25 minutes, while Piglet takes 55 minutes. One can of condensed milk and 3 jars of honey, Pooh eats in 35 minutes, while Piglet takes 1 hour 25 minutes. How long will it take them to eat 6 cans of condensed milk together? | # Solution.
1st method. From the condition, it follows that Winnie-the-Pooh eats 4 cans of condensed milk and 4 jars of honey in 1 hour, while Piglet does so in 2 hours and 20 minutes. Therefore, one can of condensed milk and one jar of honey are eaten by Winnie-the-Pooh in 15 minutes, and by Piglet in 35 minutes.
Us... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.6. Each of the 1000 gnomes has a hat, blue on the outside and red on the inside (or vice versa). If a gnome is wearing a red hat, he can only lie, and if it is blue, he can only tell the truth. During one day, each gnome said to each other, "You have a red hat!" (during the day, some gnomes turned their hats inside o... | 9.6. Answer. 998 turnovers.
Let's call a gnome red or blue if he is wearing a cap of the corresponding color. Note that one gnome can say the required phrase to another if and only if these gnomes are of different colors: a blue gnome will tell the truth in this case, and a red one will lie. Now, if any three gnomes h... | 998 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 2.1. Points $A, B, C, D$ are marked on a line, in that exact order. Point $M$ is the midpoint of segment $A C$, and point $N$ is the midpoint of segment $B D$. Find the length of segment $M N$, given that $A D=68$ and $B C=20$.
}{2} = 44 - \frac{x}{2}
$$
Now it is not difficult to calculate $MN$:
$$
MN = AD - AM - ND = 68 - \frac{x}{2} - \left(44 - \frac{x}{2}\right) = 24
$$ | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. Denis has identical ten-ruble coins, identical two-ruble coins, and identical one-ruble coins (more than 20 coins of each type). In how many ways can Denis pay exactly 16 rubles for a pie without receiving change? It is not necessary to use coins of each type. | Answer: 13.
Solution. If Denis uses a ten-ruble coin, he will need to collect 6 rubles using two-ruble and one-ruble coins. There are 4 ways to do this: using from 0 to 3 two-ruble coins.
If Denis does not use the ten-ruble coin, he will need to collect 16 rubles using two-ruble and one-ruble coins. There are 9 ways ... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.7. In a class, there are 31 students. Three of them have exactly three friends each, the next three have six each, the next three have nine each, ..., and the next three have thirty each. How many friends does the 31st student have? (Friendship between people is mutual.)
![](https://cdn.mathpix.com/cropped/2... | # Answer: 15.
Solution. Let the 31st student have $x$ friends.
Consider three people, each of whom has 30 friends in the class. There are a total of 31 students in the class, so they are friends with all their classmates.
Let's expel them from the class. Then the number of friends for each person will decrease by 3:... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5.5. Four dolls and five robots cost 4100 rubles, while five dolls and four robots cost 4000. How much does one doll cost? | Answer: 400 rubles.
Solution. Let's briefly write down the condition of the problem:
$$
\begin{aligned}
& 4 \kappa + 5 p = 4100 \\
& 5 \kappa + 4 p = 4000 .
\end{aligned}
$$
Then $9 \kappa + 9 p = 8100$ and the cost of one doll together with one robot will be 900 rubles. Therefore, 4 dolls and 4 robots cost 3600 rub... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. 1.1. An online electric scooter rental service charges a fixed amount for a ride and a certain amount for each minute of rental. Katya paid 78 rubles for a ride lasting 3 minutes. Lena paid 108 rubles for a ride lasting 8 minutes. It takes Kolya 5 minutes to get from home to work. How much will he pay for this ride? | Answer: 90.
## Solution.
1st method. If the rental of a scooter costs $x$ rubles, and a minute of use costs $y$ rubles, then $x+3y=78$, and $x+8y=108$. Solving this system, we get that $y=6, x=60$. Then Kolya will pay $60+30=90$ rubles.
2nd method. Notice that Lena's trip is 30 rubles more expensive than Katya's tri... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. 2.1. Dealer Dima bought a "LADA Kalina" car from the manufacturer and increased its price by $A \%$. There was no demand, and Dima had to sell the car at a sale with a $20 \%$ discount. As a result, his profit was $20 \%$. Find $A$. | Answer: 50.
## Solution.
1st method. Let's assume Dima bought a car from the manufacturer for $X$ rubles. After raising the price, the car cost $X(1+A / 100)$ rubles. After reducing the price by $20 \%$, the car cost $0.8 X(1+A / 100)$. According to the condition, this is the same as $1.2 X$. Therefore, $1+A / 100=1.... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. 3.1. Petya marked 5 points on a face of a cube, turned it and marked 6 points on an adjacent face, then turned it again and marked 7 points, and so on. He marked points on each face this way. What is the maximum number of points that can be on two opposite faces? | Answer: 18.
## Solution.
Since a cube has only 6 faces, the maximum number of points marked by Petya is 10. Then the number 9 will be on an adjacent face to the number 10 and cannot be on the opposite face. Therefore, the maximum sum will not exceed 18. The sum of 18 can be achieved, for example, as follows:
We see that the ... | 60718293 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. Find the smallest positive integer $n$ such that $A_{n}=1+11+111+\ldots+1 \ldots 1$ (the last term contains $n$ ones) is divisible by 45. | Answer: 35.
## Solution.
For the sum $A_{n}$ to be divisible by 45, it must be divisible by 5 and by 9. By the divisibility rule for 5, we get that $n$ must be a multiple of 5. Since any natural number gives the same remainder when divided by 9 as the sum of its digits, $A_{n}$ gives the same remainder when divided b... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. 8.1. The vertices of a regular 22-sided polygon are numbered. In how many ways can four of its vertices be chosen to form a trapezoid? (A trapezoid is a quadrilateral with one pair of parallel sides and the other two sides not parallel). | Answer: 990.
## Solution.
Note that all trapezoids we obtain will be isosceles. We will count the number of inscribed quadrilaterals that have an axis of symmetry that does not pass through the vertices. Thus, we will count all trapezoids once (since each isosceles trapezoid has only one axis of symmetry), and count ... | 990 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.1. On a circular road, there are three cities: $A, B$ and $C$. It is known that the path from $A$ to $C$ is 3 times longer along the arc not containing $B$ than through $B$. The path from $B$ to $C$ is 4 times shorter along the arc not containing $A$ than through $A$. How many times shorter is the path from $A$ to $B... | Answer: 19
Solution. From the condition of the problem, the following relationship follows:
$$
\begin{aligned}
& x=3(y+z) \\
& 4 z=x+y \\
& 3 y+3 z=4 z-y \\
& 4 y=z \\
& x=15 y \\
& x+z=19 y
\end{aligned}
$$
$ if for any real $x$ and $y$ the equality $f(x+f(y))=x+y$ holds. | Answer: 2021.
Solution. Let $f(0)=b$, then for $y=0$ we get $f(x+b)=x$, from which $f(x)=x-b$. Thus, $f(x+f(y))=f(x+(y-b))=f(x+y-b)=x+y-b-b=x+y-2b$.
Since $f(x+f(y))=x+y$ for any real $x$ and $y$, then $b=0$, so $f(x)=x$, and $f(2021)=2021$.
Comment. Only the example of the function $f(x)=x-1$ gets a point. | 2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.4. There is a grid board $2015 \times 2015$. Dima places a detector in $k$ cells. Then Kolya places a grid ship in the shape of a square $1500 \times 1500$ on the board. The detector in a cell reports to Dima whether this cell is covered by the ship or not. For what smallest $k$ can Dima place the detectors in such ... | Answer. $k=2(2015-1500)=1030$.
Solution. We will show that 1030 detectors are enough for Dima. Let him place 515 detectors in the 515 leftmost cells of the middle row of the square, and the remaining 515 detectors in the 515 top cells of the middle column. Note that for any position of the ship, its left column lies i... | 1030 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4-1. Katya attached a square with a perimeter of 40 cm to a square with a perimeter of 100 cm as shown in the figure. What is the perimeter of the resulting figure in centimeters?
 | Answer: 120.
Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4-5. Given a figure consisting of 33 circles. You need to choose three circles that are consecutive in one of the directions. In how many ways can this be done? The image shows three of the desired ways.
 | Answer: 500.
Solution. The perimeter of the figure consists of 3 segments of length 100 and 4 segments of length 50. Therefore, the length of the perimeter is
$$
3 \cdot 100 + 4 \cdot 50 = 500
$$ | 500 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5-5. Along a straight alley, 400 lamps are placed at equal intervals, numbered in order from 1 to 400. At the same time, from different ends of the alley, Alla and Boris started walking towards each other at different constant speeds (Alla from the first lamp, Boris from the four hundredth). When Alla was at the 55th l... | Answer. At the 163rd lamppost.
Solution. There are a total of 399 intervals between the lampposts. According to the condition, while Allа walks 54 intervals, Boris walks 79 intervals. Note that $54+79=133$, which is exactly three times less than the length of the alley. Therefore, Allа should walk three times more to ... | 163 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5-6. A rectangular table of size $x$ cm $\times 80$ cm is covered with identical sheets of paper of size 5 cm $\times 8$ cm. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right... | Answer: 77.
Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
Let's call such ... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-1. Two rectangles $8 \times 10$ and $12 \times 9$ are overlaid as shown in the figure. The area of the black part is 37. What is the area of the gray part? If necessary, round the answer to 0.01 or write the answer as a common fraction.
.
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 |
11-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 58.
Solution. Angle $ACD$ is a right angle since it subtends the diameter of the circle.
Therefore, $\angle CAD = 90^{\circ} - \angle CDA = 48^{\circ}$. Also, $AO = BO = CO$ as they... | 58 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.6. A square $100 \times 100$ is divided into squares $2 \times 2$. Then it is divided into dominoes (rectangles $1 \times 2$ and $2 \times 1$). What is the smallest number of dominoes that could end up inside the squares of the division?
(C. Berlov)
# | # Answer. 100.
Solution. Example. We will divide the top and bottom horizontals into horizontal dominoes - they will end up in $2 \times 2$ squares. The remaining rectangle $98 \times 100$ will be divided into vertical dominoes - they will not end up in $2 \times 2$ squares.
Estimate. Consider the squares $A_{1}, A_{... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the smallest 10-digit number, the sum of whose digits is not less than that of any smaller number. | Answer: 1899999999.
Solution. Among 9-digit numbers, the largest sum of digits is for the number 999999999, which is 81. Since the desired 10-digit number is greater than 999999999, we need to find the smallest number with a sum of digits no less than 81. The sum of the last eight digits of the desired number is no mo... | 1899999999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In a chess tournament, everyone played against each other once. The winner won half of the games and drew the other half. It turned out that he scored 13 times fewer points than all the others. (1 point for a win, 0.5 for a draw, 0 for a loss.) How many chess players were there in the tournament? | Answer: 21 chess players.
Solution. If the number of participants is $n$, then each played $n-1$ games. The winner won half of the games and scored $\frac{1}{2}(n-1)$ points. The winner drew half of the games and scored another $\frac{1}{4}(n-1)$ points. In total, the winner scored $\frac{1}{2}(n-1)+\frac{1}{4}(n-1)=\... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. The length of the diagonal of a rectangular parallelepiped is 3. What is the maximum possible value of the surface area of such a parallelepiped? | Answer: 18.
Solution. Let the edges of the rectangular parallelepiped be $-a, b$, and $c$. According to the problem, its diagonal is $\sqrt{a^{2}+b^{2}+c^{2}}=3$, and thus, $a^{2}+b^{2}+c^{2}=9$. The surface area of the parallelepiped is $f=2(a b+b c+c a)$. We will prove the inequality $a b+b c+c a \leqslant a^{2}+b^{... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.6. Ali-baba came to a cave where there was gold and diamonds. Ali-baba had one large bag with him. It is known that a full bag of gold weighs 200 kg, and if the entire bag is filled with diamonds, it will weigh 40 kg (the empty bag weighs nothing). A kilogram of gold costs 20 dinars, and a kilogram of diamonds costs... | # Solution:
Method 1. Let Ali-Baba have placed a set of gold and diamonds that gives him the maximum possible revenue - we will call such a bag optimal. Then either the bag is filled to the brim, or it weighs exactly 100 kg: otherwise, more diamonds could be added to it and the revenue would increase. Consider the fir... | 3000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.5. Let's call a trapezoid with bases 1 and 3 a "boat," which is obtained by gluing two triangular half-cells to the opposite sides of a unit square. In a $100 \times 100$ square, an invisible boat is placed (it can be rotated, it does not go beyond the boundaries of the square, its middle cell is entirely on one of ... | Answer: 4000 shots.
First solution. We will call a boat horizontal or vertical depending on whether its parallel sides are horizontal or vertical.
First, we will show that 4000 shots are enough. We divide the $100 \times 100$ square into 400 squares of size $5 \times 5$, and in each square, we make 10 shots as shown ... | 4000 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 11.2. On the table, there are 30 coins: 23 ten-ruble coins and 7 five-ruble coins, with 20 of these coins lying heads up and the remaining 10 tails up. What is the smallest $k$ such that among any $k$ randomly selected coins, there will definitely be a ten-ruble coin lying heads up? | Answer: 18.
Solution. If you choose 18 coins, then among them there will be no more than 10 lying heads down, so at least 8 coins will be lying heads up. Among these coins, no more than 7 will be five-ruble coins, so at least one will be a ten-ruble coin, which is what we need.
On the other hand, if initially on the ... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.4. A convex pentagon $A B C D E$ is such that $\angle A B C=128^{\circ}, \angle C D E=9^{\circ}$ $104^{\circ}, A B=B C, A E=E D$. How many degrees does the angle $A D B$ measure?
 | Answer: 26.
Fig. 14: to the solution of problem 11.4
Solution. The angles at the base of the isosceles triangle $ABC$ are equal to $\frac{1}{2}\left(180^{\circ}-128^{\circ}\right)=26^{\circ}... | 26 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. For what least natural $n$ can the numbers from 1 to $n$ be arranged in a circle so that each number is either greater than all 40 following it in the clockwise direction, or less than all 30 following it in the clockwise direction? | Answer: 70.
Solution. If $n \leqslant 39$, then for the number $n$ the condition cannot be satisfied: it cannot be greater than the next 40 numbers (since it is not greater than itself), nor can it be less than the next 30 numbers (since it is the largest).
If $40 \leqslant n \leqslant 69$, then for the number 40 the... | 70 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. The polynomial $P(x)$ has all coefficients as non-negative integers. It is known that $P(1)=4$ and $P(5)=152$. What is $P(11) ?$ | Answer: 1454.
Solution. Suppose the degree of the polynomial $P$ is not less than 4, then its leading coefficient is not less than 1. Since all other coefficients of $P(x)$ are non-negative, then $P(5) \geqslant 5^{4}=625$, which contradicts the condition $P(5)=152$.
Therefore, the degree of the polynomial $P$ is not... | 1454 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.8. In a $28 \times 35$ table, some $k$ cells are painted red, another $r-$ in pink, and the remaining $s-$ in blue. It is known that
- $k \geqslant r \geqslant s$
- each boundary cell has at least 2 neighbors of the same color;
- each non-boundary cell has at least 3 neighbors of the same color.
What is th... | Answer: 28.
Solution. From the condition, it is easy to understand that each cell can have no more than one neighbor of a different color.
We will prove that the coloring of the table must be "striped," meaning that either each row or each column is completely colored in one color. To do this, it is sufficient to sho... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. At a round table, 30 people are sitting, each of whom is either a knight, who always tells the truth, or a liar, who always lies. Each person was asked: "How many knights are among your neighbors?" (Two people are called neighbors of each other if there is no one else sitting between them.) 10 people answered "one,"... | Answer: 22 knights.
Solution. From the condition of the problem, it follows that the number of knights who have two knight-neighbors does not exceed 10; the same can be said about the number of knights who have one knight-neighbor. Therefore, the number of pairs of knight-neighbors does not exceed $(2 \cdot 10 + 1 \cd... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In triangle $A B C$, angle $A$ is $75^{\circ}$, and angle $C$ is $60^{\circ}$. On the extension of side $A C$ beyond point $C$, segment $C D$ is laid off, equal to half of side $A C$. Find angle $B D C$. | Answer: $\angle B D C=45^{\circ}$.
Solution. In this triangle, angle $A B C$ is obviously equal to $180^{\circ}-75^{\circ}-60^{\circ}=45^{\circ}$. Draw the altitude $A H$ and note that triangle $A H B$ is isosceles, since $\angle B A H=90^{\circ}-45^{\circ}=\angle A B H$ (Fig. 1). Therefore, $A H=B H$. In the right tr... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 10.5
The field is a $41 \times 41$ grid, in one of the cells of which a tank is hidden. The fighter aircraft shoots one cell per shot. If a hit occurs, the tank moves to an adjacent cell along a side; if not, it remains in place. After each shot, the pilot does not know whether a hit occurred. To destroy the ta... | # Answer:
2521 shots
## Solution
## Example.
Let's color the cells in a checkerboard pattern so that the corner cells are black. Suppose the pilot first shoots at all the white cells, then at all the black ones, and then again at all the white ones. If the tank was on a white cell, the pilot will destroy it in the ... | 2521 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In a company of 8 people, each person is acquainted with exactly 6 others. In how many ways can four people be chosen such that any two of them are acquainted? (We assume that if A is acquainted with B, then B is also acquainted with A, and that a person is not acquainted with themselves, as the concept of acquainta... | Answer: 16.
Solution. Each person in the company does not know $8-1-6=1$ person in this company. This means the company can be divided into four pairs of strangers. From each pair, we can only take one person to form a company of four people who are all acquainted with each other. The selection of such a company consi... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.4. A set of composite numbers from the set $\{1,2,3,4, \ldots, 2016\}$ is called good if any two numbers in this set do not have common divisors (other than 1). What is the maximum number of numbers that a good set can have? | Answer: 14 numbers.
Example. $2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}, 17^{2}, 19^{2}, 23^{2}, 29^{2}, 31^{2}, 37^{2}, 41^{2}, 43^{2}$. All the presented numbers are composite, as they are squares of natural numbers greater than 1. It is also clear that different prime numbers do not have common divisors greater th... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. Write the number 2013 several times in a row so that the resulting number is divisible by 9. Explain your answer. | Answer: for example, 201320132013.
Solution. We will provide several ways to justify this.
First method. The sum of the digits of the written number is $(2+0+1+3) \cdot 3=18$, so it is divisible by 9 (by the divisibility rule for 9).
Second method. The sum of the digits of the number 2013 is divisible by 3, so if we... | 201320132013 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. In the sum $+1+3+9+27+81+243+729$, any addends can be crossed out and the signs before some of the remaining numbers can be changed from “+” to “-”. Masha wants to use this method to first obtain an expression whose value is 1, then, starting over, obtain an expression whose value is 2, then (starting over again) ... | Answer: up to the number 1093 (inclusive).
Solution. The number 1 is obtained by erasing all addends except the first. Then Masha can get the numbers $2=-1+3, 3=+3$, and $4=+1+3$.
We will show that by adding the addend 9, one can obtain any integer from $5=9-4$ to $13=9+4$. Indeed, $5=-1-3+9; 6=-3+9; 7=+1-3+9; 8=-1+9... | 1093 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. On January 1, 2013, a little boy was given a bag of chocolate candies, containing 300 candies. Each day, the little boy ate one candy. On Sundays, Karlson would fly over, and the little boy would treat him to a couple of candies. How many candies did Karlson eat? (January 1, 2013, was a Tuesday).
2. Petya can swap ... | 1. Answer: 66.
Solution. In a full week, Little One eats 7 candies (one each day), and Karlson eats 2 (on Sunday), so together they eat 9 candies. Dividing 300 by 9 with a remainder, we get the quotient 33 and the remainder 3 (33$\cdot$9+3=300). In the last incomplete week, Little One will eat 3 candies: on Wednesday,... | 66 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. 50 students from fifth to ninth grade published a total of 60 photos on Instagram, each not less than one. All students of the same grade (same parallel) published an equal number of photos, while students of different grades (different parallels) published a different number. How many students published only one ph... | Solution. Let each of the students publish one photo first, during which 50 photos out of 60 will be published. There are 10 photos left to be published, which we will call additional photos. We have a total of five different classes (parallels), and these 10 additional photos must be published by different students fr... | 46 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Find the value of the expression
$$
2^{2}+4^{2}+6^{2}+\ldots+2018^{2}+2020^{2}-1^{2}-3^{2}-5^{2}-\ldots-2017^{2}-2019^{2}
$$ | # Solution.
$$
\begin{aligned}
& 2020^{2}-2019^{2}+2018^{2}-2017^{2}+\ldots+6^{2}-5^{2}+4^{2}-3^{2}+2^{2}-1^{2}= \\
& =(2020-2019)(2020+2019)+(2018-2017)(2018+2017)+\ldots+ \\
& +(6-5)(6+5)+(4-3)(4+3)+(2-1)(2+1)= \\
& =2020+2019+2018+2017+\ldots+6+5+4+3+2+1= \\
& =\frac{2020+1}{2} \cdot 2020=2041210
\end{aligned}
$$
... | 2041210 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Given the natural numbers $1,2,3 \ldots, 10,11,12$. Divide them into two groups such that the quotient of the product of all numbers in the first group divided by the product of all numbers in the second group is an integer, and takes the smallest possible value. What is this quotient? | Solution. Let's factorize the given numbers into prime factors and find their product.
$$
1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 10 \cdot 11 \cdot 12=2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7 \cdot 11
$$
The factors 7 and 11 do not have pairs. One of the factors 3 does not have a pair. Therefore, to make the quoti... | 231 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3.1. A number was multiplied by its first digit and the result was 494, by its second digit - 988, by its third digit - 1729. Find this number. | Answer: 247
Solution. It can be noticed that all three obtained numbers are divisible by 247, from which we get the answer. | 247 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. On the counter lie 10 weights with masses $n, n+1, \ldots, n+9$. The seller took one of them, after which the total weight of all the remaining ones was 1457. Which weight did the seller take? | # Answer: 158
Solution. The total weight of all 10 weights is $10 n+45$. Suppose the weight taken has a weight of $n+x$, where $x<10$. Then
$10 n+45-n-x=1457$.
$9 n+45-x=1457$.
We can see that 1457, when divided by 9, gives a remainder of 8, so $x=1$. Then
$9 n+45-1=1457$
$9 n=1413$.
$n=157$.
$n+x=158$. | 158 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. In a circle, 58 balls of two colors - red and blue - are arranged. It is known that the number of triples of consecutive balls, among which there are more red ones, is the same as the number of triples with a majority of blue ones. What is the smallest number of red balls that could be present?
, the elements of each row and each column form arithmetic progressions. What is the number $x$ in the central cell?
 | Answer: 111.
Solution. The middle term of an arithmetic progression is equal to the half-sum of the extreme terms, so in the middle cell of the first row, the number is $(3+143) / 2=73$, and in the middle cell of the last row, the number is $(82+216) / 2=149$.
$ points. According to the problem, the ratio of the number of points scored by girls to the number of points scored by boys is 1:4. Therefore, the gir... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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