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Problem 7.1. In the picture, nine small squares are drawn, with arrows on eight of them. The numbers 1 and 9 are already placed. Replace the letters in the remaining squares with numbers from 2 to 8 so that the arrows from the square with the number 1 point in the direction of the square with the number 2 (the number 2... | Answer: In square $A$ there is the number 6, in $B-2$, in $C-4$, in $D-5$, in $E-3$, in $F-8$, in $G-7$.
Solution. Let's order all the squares by the numbers in them. This "increasing chain" contains all nine squares.
Notice that in this chain, immediately before $C$ can only be $E$ (only the arrows from $E$ point to... | In\\A\there\is\the\\6,\in\B-2,\in\C-4,\in\D-5,\in\E-3,\in\F-8,\in\G-7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,440 |
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?
=9 \cdot(18+S)$. Solving this linear equation, we get $S=13.5$. | 13.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,442 |
Problem 8.4. Given a square $A B C D$. Point $L$ is on side $C D$ and point $K$ is on the extension of side $D A$ beyond point $A$ such that $\angle K B L=90^{\circ}$. Find the length of segment $L D$, if $K D=19$ and $C L=6$.
Fig. 1: to the solution of problem 8.4
Notice that $\angle ABK = \angle CBL$, since they both complement $\angle ABL$ to ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,444 |
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 | 13,445 |
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 | 13,448 |
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 | 13,451 |
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 | 13,452 |
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 | 13,454 |
9.6. Consider all 100-digit numbers divisible by 19. Prove that the number of such numbers that do not contain the digits 4, 5, and 6 is equal to the number of such numbers that do not contain the digits 1, 4, and 7.
(I. Efremov) | Solution. To each remainder $a$ from division by 19, we associate a remainder $b(a)$ such that $b(a) \equiv 3a \pmod{19}$. Note that the remainders $0, 1, 2, 3, 7, 8, 9$ are mapped to the remainders $0, 3, 6, 9, 2, 5, 8$ respectively. Moreover, from the remainder $b$, the remainder $a = a(b) \equiv -6b \pmod{19}$ can b... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,455 |
9.7. Given a trapezoid $A B C D$, where $A D \| B C$, and rays $A B$ and $D C$ intersect at point $G$. The common external tangents to the circumcircles of triangles $A B C$ and $A C D$ intersect at point $E$. The common external tangents to the circumcircles of triangles $A B D$ and $B C D$ intersect at point $F$. Pro... | Solution. Let the line $E C$ intersect the circle $(A B C)$ again at point $X$, and the line $E A$ intersect the circle $(A C D)$ again at point $Y$ (we will consider the arrangement of points as shown in Fig. 2; other cases are treated similarly).
Consider the homothety with center $E$ that maps $(A B C)$ to $(A C D)... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,456 |
1. A rectangle of size $m \times n (m > n)$, consisting of $1 \times 1$ squares, is laid out with matches (Fig. 22.54a). a) How many matches were used for its construction, assuming the length of a match is 1? b) How many possible rectangles, composed of these matches, can be counted in this figure? c) How many possibl... | Solution. a) Let $\mathrm{m}$ be the length of the rectangle, and $\mathrm{n}$ be the height. Then, vertically, there will be $\mathrm{n} \cdot(\mathrm{m}+1)$ matches, and horizontally, there will be $-\mathrm{m} \cdot(\mathrm{n}+1)$ matches. Therefore, to construct such a rectangle, it will require $\mathrm{n}(\mathrm... | )2mn++n;b)\frac{(+1)n(n+1)}{4};)\frac{n(n+1)(3m-n+1)}{6} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,458 |
2. Solve the inequality $2 \cdot 5^{2 x} \cdot \sin 2 x-3^{x} \geq 5^{2 x}-2 \cdot 3^{x} \cdot \sin 2 x$. | Solution. Transform the inequality:
$$
5^{2 x}(2 \sin 2 x-1) \geq-3^{2 x}(2 \sin 2 x-1) \Rightarrow\left(5^{2 x}+3^{2 x}\right)(2 \sin 2 x-1) \geq 0
$$
Note that $5^{2 x}+3^{2 x}>0$ for any $x \in \mathbf{R}$, so the given inequality is equivalent to the inequality $(2 \sin 2 x-1) \geq 0$. Solving this, we find $\fra... | \frac{\pi}{12}+2\pik\leqx\leq\frac{5\pi}{12}+2\pik,k\in{Z} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 13,459 |
4. In an equilateral triangle with side $a$, a circle is inscribed. From the vertex of the triangle, a second circle with radius $0.5 a$ is drawn. Find the area enclosed between these circles. | Solution. Let $ABC$ be the given equilateral triangle with side length $a$. We will make the necessary constructions (see the figure). The required area is the difference between the area of a $60^{\circ}$ sector of the second circle and the difference of one-third of the area of the triangle and the area of the inscri... | \frac{^{2}}{72}(5\pi-6\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,460 |
5. What should $k$ be so that the inequality
$$
\left|\frac{x^{2}-k x+1}{x^{2}+x+1}\right|<3
$$
holds for any value of $x$? | Solution. Domain of definition: $x \in \mathbf{R}$, since $x^{2}+x+1>0$ for any real $x$. Therefore, the given inequality is equivalent to the inequality
$$
-30 \\
4 x^{2}-(k-3) x+4>0
\end{array}\right.
$$
For the inequalities in the system to be satisfied simultaneously for all $x$, it is necessary and sufficient th... | k\in[-5;1] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 13,461 |
1. The quadratic trinomial $x^{2}+p x+q$ has two distinct non-zero integer roots $a$ and $b$. It is known that $a+p$ is divisible by $q-2 b$. What can the root $a$ be equal to? (Provide all answers and prove that there are no others.) | Answer: $a=3$ or $a=1$.
Solution. By Vieta's theorem, $p=-a-b, q=ab$. Then, according to the condition, the number $a+p=a+(-a-b)=-b$ is divisible by $q-2b=ab-2b=(a-2)b$. When $a=2$, the last number equals zero, and it is incorrect to talk about divisibility. When $a=3$ or $a=1$, divisibility holds. For other integer v... | =3or=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,462 |
2. Do there exist natural numbers $x$ and $y$ such that $\operatorname{GCD}(x ; y)+$ $\operatorname{LCM}(x ; y)+x+y=2019$? | Answer: They do not exist.
Solution: Let's analyze the equation in terms of parity. If both numbers $x$ and $y$ are even, then their GCD and LCM are also even, and the sum of four even numbers must also be even. If they are both odd, then their GCD and LCM are odd, and the sum of four odd numbers is even. Finally, if ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,463 |
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 | 13,464 |
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 | 13,465 |
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 | 13,466 |
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 | 13,467 |
6.2. Mark 6 different points on the plane and draw 6 lines such that there are two marked points on each line and two marked points on either side of it. | Solution: For example, as shown in the figure.
Note: The required construction can also be described in words. For example, as follows: Consider some triangle $ABC$ and draw a line through each of its vertices intersecting the opposite side. Mark points $D, E, F$ on these lines such that the rays $AD, BE$, and $CF$ do... | \ | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,468 |
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 | 13,469 |
6.4. When Winnie-the-Pooh came to visit Rabbit, he ate 3 plates of honey, 4 plates of condensed milk, and 2 plates of jam, and after that, he couldn't get out of the hole because he had become too fat from all the food. But it is known that if he had eaten 2 plates of honey, 3 plates of condensed milk, and 4 plates of ... | # Solution:
Method 1. Let one bowl of honey have a caloric value of $a$ units, one bowl of condensed milk $-b$ units, and one bowl of jam $c$ units. (The higher the caloric value of a product, the more it makes one gain weight.) The condition of the problem means that $3 a+4 b+2 c>2 a+3 b+4 c$ and $3 a+4 b+2 c>4 a+2 b... | Condensed\milk\makes\one\\\weight | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,470 |
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 | 13,471 |
1.1. Long ago, Kirill thought about today and realized the following:
- Tomorrow is not September;
- In a week, it will be September;
- The day after tomorrow is not Wednesday.
On which of the days listed below are all three statements true at the same time? Select all correct options:
- Monday, August 30
- Tuesday,... | Answer: No; No; Yes.
Solution. Note that the first option does not work, as the day after tomorrow (two days after Monday) would be Wednesday. The second option does not work either, as tomorrow (the day after August 31) would be September. The third option fits all the conditions (two days after Wednesday would be Fr... | not | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 13,472 |
2.1. The numbers 7, 8, 9, 10, 11 are arranged in a row in some order. It turned out that the sum of the first three of them is 26, and the sum of the last three is 30. Determine the number standing in the middle.
Solution. The first digit must be divisible by 3, be odd, and not less than 5. Only the digit 9 fits. The second digit must be even and divisible by 3. The digits 0 and 6 fit. The last digit is even and does not divide by 3. The digits 2, 4, and 8 fit. Therefore, all sui... | 902,904,908,962,964,968 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,477 |
7.1. Each of the 37 hexagonal cells has 3, 4, or 6 neighbors. Some of the cells were filled with chips. After that, in each free cell that is adjacent to at least two cells containing chips, the number of these neighboring chips was written. Then the chips were removed, and only the numbers remained on the diagram. How... | # Answer: 8
Solution. Let's look at the picture on the left. Consider the triplet and the central pair. We understand that the chips must be placed at the positions of the yellow crosses. Next, we understand the red minuses - the absence of chips. Then we place the chips at the positions of the green crosses. And the ... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,478 |
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 | 13,479 |
# Task 8.1
Can natural numbers be placed in the cells of a $7 \times 7$ table so that the sum of the numbers in any $2 \times 2$ square and any $3 \times 3$ square is odd?
## Number of points 7 | Answer
no
## Solution
Suppose it is possible. Consider a square of side 6 cells in the table. Since it can be divided into four squares of size $3 \times 3$, the sum of the numbers in this square will be even. On the other hand, this same square can be divided into nine squares of size $2 \times 2$, so this same sum... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,480 |
# 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 | 13,481 |
# Task 8.3
Can the numbers $1,4,9,16,25,36,49,64,81$ be inserted into the circles in the figure so that the sum of the numbers on each radial line and in each triangle is the same?
Number of... | Answer:
impossible
Solution
Let $\mathrm{S}$ be this sum, then $6 \mathrm{~S}=2(1+4+9+16+25+36+49+64+81)$.
Since $1+4+9+16+25+36+49+64+81=285$, then $6 \mathrm{~S}=570, \mathrm{~S}=95$.
But to the sum containing the number 81, one needs to add 14 to reach 95, and there are no two numbers among them with such a sum... | impossible | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,482 |
# Task 8.4
Inside an acute angle, there is a convex quadrilateral $\mathrm{ABCD}$. It turns out that for each of the two lines containing the sides of the angle, the following condition is met: the sum of the distances from vertices A and C to this line is equal to the sum of the distances from vertices B and D to the... | # Solution
Let the line $\mathrm{m}$ contain one of the sides of the given angle, and let $a$ and $c$ be the distances from vertices $A$ and $C$ to $\mathrm{m}$, and $P$ be the midpoint of segment $\mathrm{AC}$. Then the distance from $P$ to $\mathrm{m}$ is $(a + c) / 2$ (the midline of the trapezoid, see the figure).... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,483 |
# 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 | 13,484 |
3. The lengths of the sides of a parallelogram are 3 and 5. The bisectors of all its internal angles limit a polygon on the plane. Find the ratio of its area to the area of the parallelogram.
 | Solution. The bisectors of the internal angles of a parallelogram intersect to form a rectangle. Divide the parallelogram into rhombuses with a side length of 1 by lines parallel to the sides of the parallelogram. It is clear that the area of the red rectangle is twice the area of a rhombus with a side length of 1. And... | \frac{2}{15} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,486 |
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 | 13,487 |
7.1. Arrange 6 non-zero digits (not necessarily distinct) in a circle so that each of them equals the last digit of the sum of its neighbors. | Solution. For example, $-4-2-8-6-8-2-$.
Remark. There are other examples.
Comment. Any correct example -7 points. | -4-2-8-6-8-2- | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,489 |
7.2. Petya, Kolya, and Vasya were collecting mushrooms. Petya said that he found 7 fewer mushrooms than Kolya and Vasya combined, and Kolya said that he found 10 fewer mushrooms than Petya and Vasya combined. Prove that one of the boys was mistaken. | First solution. Suppose that none of the boys made a mistake. Since Petya found 7 fewer mushrooms on an odd number than Kolya and Vasya found together, the number of mushrooms collected by Petya and the number of mushrooms collected by Kolya and Vasya together are of different parity. Therefore, the total number of col... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,490 |
7.3. From a $6 \times 5$ grid rectangle, a $2 \times 1$ rectangle was cut out from the center, as shown in the figure. Can the resulting figure be cut into 6 triangles? | Answer. Yes, it is possible.
Solution. One of the examples of cutting is shown in Fig. 1.
Fig. 1
Comment. Any correct example - 7 points. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,491 |
7.4. In the decimal representation of 13 numbers, the same digit $N$ is used and no other digits are used. Can the sum of these numbers be equal to 8900098? | Answer: It cannot.
Solution: Suppose the sum could equal 8900098. Each of the addends has the same last digit \( N \). Therefore, the last digit of the sum is the same as the last digit of the number \( 13N \). This implies that \( N = 6 \). But then each of the addends is divisible by 6, which means they are divisibl... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,492 |
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 | 13,493 |
1. Can the number 8 be represented as the sum of eight integers such that the product of these numbers is also equal to 8? | 1. For example, $4+2+1+1+1+1+(-1)+(-1)=8$
and $4 \cdot 2 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot(-1) \cdot(-1)=8$.
Answer: Yes. | Yes | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,494 |
3. Let AC be the largest side of triangle ABC. Points M and N are chosen on side AC such that AM = AB, CN = CB. Prove that if BM = BN, then triangle ABC is isosceles. | 3. From the condition, it follows that $\angle A B M=\angle A M B=\angle N M B=\angle M N B=\angle C N B=\angle C B N$. Then $\angle A B N=\angle C B M$. Moreover, $\angle A N B=\angle C M B$, since $\angle B N M=\angle B M N$. Therefore, $\triangle A B N=\triangle C B M$ by the second criterion. Hence, it follows that... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,496 |
4. In a football tournament where each team played against each other once, teams A, B, C, D, E, and F participated. Teams received 3 points for a win, 1 point for a draw, and 0 points for a loss. In the end, it turned out that teams A, B, C, D, and E each had 7 points. What is the maximum number of points that team $\... | 4. In a match where one of the teams won, the teams together score 3 points; in a match that ended in a draw, - 2 points. Since 7 is not divisible by 3, a team that scored 7 points must have at least one draw. Since there are five such teams, there were at least three draws in the tournament. In total, as is easy to ch... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,497 |
5. Four numbers are written on the board, none of which are equal to 0. If each of them is multiplied by the sum of the other three, the result is four identical values. Prove that the squares of the numbers written on the board are equal. | 5. Let the numbers $a, b, c$ and $d$ be given. By the condition $a(b+c+d)=b(a+c+d)$, we get $(a-b)(c+d)=0$. Similarly, from the equality $c(a+b+d)=d(a+b+c)$, we obtain $(c-d)(a+b)=0$. Since either $c+d$ or $c-d$ is not equal to 0, then either $a=b$ or $a=-b$. In both cases, the squares of the numbers $a$ and $b$ are eq... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,498 |
1. Let $D$ be the discriminant of the reduced quadratic trinomial $x^{2}+a x+b$. Find the roots of the trinomial, given that they are distinct and one of them is equal to $D$, and the other is equal to $2D$. | 1. By Vieta's theorem $b=D \cdot 2D=2D^{2}, a=-(D+2D)=-3D$, i.e., the quadratic trinomial is equal to $x^{2}-3Dx+2D^{2}$. Its discriminant $D=(-3D)^{2}-4 \cdot 2D^{2}=D^{2}$, from which $D=D^{2}$, i.e., $D=0$ (in this case both roots are the same and equal to 0) or $D=1$ (in this case the roots are 1 and 2). | D=0orD=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,499 |
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
$, points $\mathrm{K}, \mathrm{L}, \mathrm{M}$, and $\mathrm{N}$ are taken such that $\mathrm{AK}+\mathrm{LC}+\mathrm{CM}+\mathrm{NA}=2$ (points $\mathrm{K}, \mathrm{L}, \mathrm{M}$, and $\mathrm{N}$ are not ... | 4. 1 of the possible solutions.
Draw SR $\| \mathrm{MK}$ and $\mathrm{DQ} \| \mathrm{NL}$, then SRKP and DQLN are parallelograms, which means $\mathrm{KP}=\mathrm{CM}$ and $\mathrm{LQ}=\mathrm{DN}$. Then $\mathrm{AK}+\mathrm{LC}+\mathrm{CM}+\mathrm{NA}=\mathrm{AK}+\mathrm{LQ}+\mathrm{CQ}+\mathrm{CM}+\mathrm{NA}=\mathr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,502 |
5. Is it possible to color the cells of an 8 x 8 board in three colors: 21 cells in white, 21 cells in blue, 22 cells in red - so that no diagonal (not only the two main diagonals, but all those parallel to them) contains cells of all three colors simultaneously? | 5. It is possible. We will color all cells in a checkerboard pattern with white and red, so all diagonals will be monochromatic. Then we will repaint 11 any white cells and 10 any red cells to blue. The resulting coloring is such that on one diagonal, white and red cells will not appear simultaneously. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,503 |
4. Let \( x \) and \( y \) be the cost (in million rubles) of the first and second packages of shares, respectively. According to the conditions of the problem, we have the following system:
$$
\begin{aligned}
& \left\{\begin{array}{c}
1.28(x+y)=7.68 \\
1.28(x+y)=1.4 x+1.2 y
\end{array}\right. \\
& \left\{\begin{array... | Answer: 2 million 400 thousand rubles and 3 million 600 thousand rubles. | 2.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,504 |
9.1. Four consecutive natural numbers are given. The sum of the first three is $A$, and the sum of the last three is $B$. Can the product $A B$ be equal to 20192019? | Answer: No, it cannot.
Solution 1: Let the numbers be $a, a+1, a+2$, and $a+3$. By the condition, $A=a+a+1+a+2=3a+3$, and $B=a+1+a+2+a+3=3a+6$, i.e., both $A$ and $B$ are divisible by 3 (as sums of three consecutive numbers). Then the product $AB$ must be divisible by 9, but 20192019 is not divisible by 9 (the sum of ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,505 |
9.2. The graphs of the functions $y=a x^{2}+b x+1$ and $y=x^{2}+c x+d$ intersect at the point with coordinates ( $2 ; 4$ ). What is the value of the expression $4 a+d$, if $b+c=1$ ? | Answer: 1.
Solution: Since the graphs pass through the point with coordinates (2; 4), then $4=4a+2b+1$ and $4=4+2c+d$. Therefore, $4a+2b=3$, and $2c+d=0$, or $4a=3-2b, d=-2c$. Summing up the obtained expressions: $4a+d=3-2b-2c=3-2(b+c)=3-2=1$. Note: The condition is satisfied by the functions $y=ax^2+bx+1$ and $y=x^2+... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,506 |
9.3. Given a trapezoid $A B C D$, where $A D \| B C, B C=A C=5$, and $A D=6$. The angle $A C B$ is twice the angle $A D B$. Find the area of the trapezoid. | # Answer: 22.
Solution: Drop a perpendicular CH from vertex C to the base AD. Denote the intersection of BD and CH as E.
Let $\angle \mathrm{ADB}=\alpha=\angle \mathrm{CBD}$ (alternate interior angles), 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 | 13,507 |
9.4. Dima and Tsiren are playing a game. At the beginning of the game, the number 0 is written on the board. On a turn, a player adds any natural number not greater than 10 to the written number and writes the result on the board instead of the original number. The player who first gets a three-digit number wins. Dima ... | Answer: Masha.
Solution: The player who can get the numbers: $1,12,23,34,45,56,67,78,89$ will win. Indeed, starting from one, the next player can only land on the numbers $2, \ldots, 11$. From there, the other player will move to 12 and so on. From the number 89, the player will land on the numbers $90, \ldots, 99$ an... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,508 |
9.5. A square field 5 by 5 meters is divided into 25 equal plots of 1 m by 1 m. In each plot, there sits a grasshopper. At a certain moment, each grasshopper jumps to an adjacent plot by either horizontally or vertically. Will there be a plot left without any grasshopper? | Answer: Yes, it will remain.
Solution: Let's color the field in a checkerboard pattern. Since the total number of cells in the $5 \times 5$ field is odd (=25), the number of black and white cells cannot be equal. Let's assume, for definiteness, that there are more black cells. Then the number of grasshoppers sitting o... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,509 |
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 | 13,510 |
11.2. Prove that the equation $x^{2}+y^{2}=2018^{2021}$ is solvable in natural numbers. | Solution. For example, $\left(2018^{1010} \cdot 13 ; 2018^{1010} \cdot 43\right)$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,511 |
11.3. On one main diagonal and all edges of a cube, directions are chosen. What is the smallest length that the sum of the resulting 13 vectors can have, if the edge length is 1, and the length of the main diagonal is $\sqrt{3}$. | Answer: $\sqrt{3}$
Solution: Let's choose a basis of three vectors along the edges of the cube, such that the vector of the diagonal equals $\overrightarrow{e_{1}}+\overrightarrow{e_{2}}+\overrightarrow{e_{3}}$. Then, four vectors will be equal to $\pm \overrightarrow{e_{1}}$, another four to $\pm \overrightarrow{e_{2... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,512 |
11.4. For each integer value of $x$, the polynomial $P(x)$ takes integer values. Can one of its coefficients be equal to $\frac{1}{2021}$? | Answer. Yes, it can.
Solution. Consider, for example, the polynomial $P(x)=\frac{1}{2021}(x+1)(x+2) \cdots(x+2021)$, whose leading coefficient is $\frac{1}{2021}$. For any integer value of $x$, the product $(x+1)(x+2) \cdots(x+2021)$ is divisible by 2021, because one of the 2021 consecutive integers is divisible by 20... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,513 |
11.5. A chess tournament was held in a round-robin system, where each participant played against every other participant once. Let's call a game incorrect if the winner of the game scored fewer points in total than the loser. (A win gives 1 point, a draw - $1 / 2$, a loss - 0). Can incorrect games make up more than $75... | Answer: No.
Solution: Let $N$ be the number of players, $M=[N / 2]$. Players who took the first $M$ places are called strong, and the rest are called weak (among participants with the same total score, places are distributed arbitrarily). Let $X$ be the number of correct matches between strong and weak players. The to... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,514 |
# 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 | 13,515 |
# 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.
A
Now, we will prove that figure V cannot be cut. Highlight three squares in it and note that no two of them can rem... | A,B,G | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 13,523 |
1. Do there exist 3 consecutive numbers, for each of which the sum of the digits is a perfect square? | 1. Answer: No. Among three consecutive integers, at least two have digit sums that are consecutive integers (in this case, there can only be one transition to the next place value). However, two consecutive integers cannot both be squares.
Grading recommendations: answer only - 0 points. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,524 |
3. In a volleyball tournament, 6 teams participated. Prove that after one round of games (each team played with each other exactly once), there will be a team that won at least $3$ games (there are no draws in volleyball). | 3. Six teams played a total of $6 * 5 / 2 = 15$ games. The number of games matches the sum of the number of wins by each team. But if each team won fewer than three games, then the total number of games does not exceed $2 * 6 = 12$. Contradiction. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,526 |
4. The numbers 20 and 21 are written on the board. Every minute, another number is written on the board, equal to the sum of any two of the already written numbers. Is it possible to eventually obtain the number $2020$ this way? | 4. Answer: Yes, it is possible. Since the numbers 20 and 21 are already present on the board, we can obtain any number of the form 20a + 21b. Note that $2020=$ $101 * 20=(101-21) * 20+21 * 20=20 * 80+21 * 20$.
Grading recommendations: only answer - 0 points. | 2020 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,527 |
5. In an acute-angled triangle $\mathrm{ABC}$, a line $s$ passing through vertex A and the median from vertex B divide the height dropped from vertex C into three equal parts. In what ratio does the line $s$ divide the side BC? | 5. Answer: 1:4. Since the line l and the median divide the height into three equal parts, the height is divided by the intersection point with the median in the ratio $2: 1$, counting from its vertex, meaning the height from vertex C is also the median of the isosceles triangle $\mathrm{ABC}(\mathrm{AC}=\mathrm{BC})$.
... | 1:4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,528 |
11.1. The price of a product was first increased by $x$ percent, and then decreased by $y$ percent. As a result, the price remained the same. Find all values that the difference $\frac{1}{x}-\frac{1}{y}$ can take. | Solution: Let the item cost 1 monetary unit. Then after the price increase, its cost became $a=1+\frac{x}{100}$, and after the decrease, it became
$$
a-\frac{a y}{100}=a\left(1-\frac{y}{100}\right)=\left(1+\frac{x}{100}\right)\left(1-\frac{y}{100}\right)
$$
The condition of the problem, therefore, is described by the... | -\frac{1}{100} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,529 |
11.3. On the sides $AB$, $BC$, and $CA$ of triangle $ABC$, points $M$, $N$, and $P$ were marked, respectively. It turned out that the areas of all four triangles $BMN$, $CNP$, $APM$, and $MNP$ are equal. Is it true that points $M$, $N$, and $P$ must be the midpoints of the corresponding sides of the triangle? Justify y... | Solution: Let the sides of the triangle be denoted as $a = BC$, $b = AC$, and $c = AB$. Then, for some numbers $x, y, z$ from the interval $[0; 1]$, the following relationships hold: $BM = xc$, $MA = (1-x)c$, $AP = zb$, $PC = (1-z)b$, $CN = ya$, $NB = (1-y)a$ - see the figure.
Given that $S_{MBN} = \frac{1}{4} S_{ABC}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,531 |
11.4. Given the function $f(x)=\left(1-x^{3}\right)^{-1 / 3}$. Find $f(f(f \ldots f(2018) \ldots))$ (the function $f$ is applied 2019 times) | Solution: Let's carry out equivalent transformations (for $x \neq 1$ and $x \neq 0$):
$$
\begin{gathered}
f(f(x))=\left(1-f(x)^{3}\right)^{-1 / 3}=\left(1-\left(1-x^{3}\right)^{-1}\right)^{-1 / 3}=\left(1-\frac{1}{1-x^{3}}\right)^{-1 / 3}= \\
=\left(\frac{-x^{3}}{1-x^{3}}\right)^{-1 / 3}=\frac{-x^{-1}}{\left(1-x^{3}\r... | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,532 |
11.5. A rectangular parallelepiped with edge lengths $\sqrt{70}, \sqrt{99}, \sqrt{126}$ is orthogonally projected onto all possible planes. Find the maximum value of the projection area. Justify your answer. | Solution: The projection represents a hexagon with pairwise parallel sides (some angles may be degenerate). It is divided into three parallelograms, so its area is twice the area of the triangle $M N P-$ see the figure.
This area - the area of the projection of the triangle - is maximal when the projection is made per... | 168 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,533 |
11.6. In a round-robin chess tournament (each chess player plays one game against each other), 20 chess players participated, 6 of whom were from Russia. It is known that Vladimir, who scored more points than anyone else, took first place. Levon from Armenia took second place, also outscoring each of the other 18 chess... | Solution: Let's provide an example showing that Russian chess players could collectively score 96 points. Suppose Vladimir won all his games except the one against Levon, which ended in a draw. Additionally, suppose Levon drew all his games with the other Russians and consistently won against non-Russians. Finally, sup... | 96 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,534 |
Problem 7.1. Inside a large square, there is a small square, the corresponding sides of which are parallel. The distances between some sides of the squares are marked on the diagram. By how much is the perimeter of the large square greater than the perimeter of the small square?
-4x=32
$$ | 32 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,535 |
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 | 13,536 |
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 | 13,537 |
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 | 13,538 |
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 | 13,539 |
Problem 7.6. On a plane, points $A, B, C, D, E$ were marked such that triangles $A B C$ and $A D E$ are equal: $A B=A D, A C=A E, B C=D E$. Then, point $E$ and the labels of points $B, C$, and $D$ were erased from the drawing.
 any of the numbers $14,18,40$. (b) $14,18,40$.
Solution. From the condition $A B=A D$ it follows that $B$ and $D$ are the two rightmost marked points (in unknown order). The remaining unmarked point is $C$.
Case 1. Let point $B$ be at the top, and point $D$ at the bottom (Fig. 1a). It is clear that for po... | 14,18,40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,540 |
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 | 13,542 |
Variant 7.6.1. On a plane, points $A, B, C, D, E$ were marked such that triangles $A B C$ and $A D E$ are equal: $A B=A D, A C=A E, B C=D E$. Then, point $E$ and the labels of points $B, C$, and $D$ were erased from the drawing.
 any of the numbers $14,18,40$. (b) $14,18,40$. | 14,18,40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,543 |
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 | 13,544 |
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 | 13,545 |
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 | 13,546 |
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 | 13,547 |
1. Vasya's dad is good at math, but on the way to the garage, he forgot the code for the digital lock on the garage. In his memory, he recalls that all the digits of the code are different and their sum is 28. How many different codes does dad need to try to definitely open the garage, if the opening mechanism of the l... | Solution. Among the digits of the code, there are 9 and 8. Otherwise, since all digits of the code are different, the largest sum of the digits of the code would not exceed $9+7+6+5$. Then the other two digits of the code are no more than 7, and their sum is 11. Only two options are possible: $\{7,4\}$ and $\{6,5\}$. L... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,548 |
2. Two candles of different thickness, each 24 cm long, are lit simultaneously. The thin candle burns out in 4 hours, the thick one in 6 hours. After what time will one candle be half the length of the other? It is assumed that each candle burns uniformly, meaning it decreases by the same height over equal time interva... | Solution. Let $x$ hours $(x<4)$ have passed since the candles were lit. Since the thin candle decreases by $24: 4=6$ cm every hour, it will decrease by a length of $6 x$ in this time, leaving an unburned stub of length $24-6 x$. Similarly, the stub of the second candle will have a length of $24-4 x$. For one candle to ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,549 |
3. Can an infinite sheet of graph paper be divided into dominoes $1 \times 2$ cells in such a way that every line running along the grid lines cuts only a finite number of dominoes in half? | Solution. Yes, it is possible. A possible tiling scheme is shown in the figure. The boundaries of areas with different orientations of dominoes form a diagonal cross, diverging from some center. Any horizontal or vertical line running along the grid lines intersects only a finite part of the cross, where the dominoes m... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,550 |
4. The bisector of angle $B A C$ of square $A B C D$ intersects side $B C$ at point $M$. Prove that $A C=A B+B M$. | Solution. First method. Drop a perpendicular $M H$ from point $M$ to the diagonal $A C$ (see fig.). Then triangles $A B M$ and $A M H$ are equal (right triangles with a common hypotenuse and equal acute angles). We get $A H=A B=B C, M H=B M$. Triangle $H M C$ is isosceles (as a right triangle with an acute angle of $45... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,551 |
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