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9.3. Fill the cells of a $5 \times 5$ table with integers such that the sum of all numbers in the table is positive, while the sum of the numbers in any $3 \times 3$ square is negative. | Answer: For example,
| 1 | 1 | 1 | 1 | 1 |
| :---: | :---: | :---: | :---: | :---: |
| 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | -10 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 |.
## Criterion.
7 points. Any correct example.
Comment. There are several different correct examples. | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,487 |
9.4. Natural numbers from 1 to 100 are written on cards, one number per card. Vasya forms sets of three cards such that in each set, one of the numbers equals the product of the other two. What is the maximum number of such sets Vasya could have formed? (The same card cannot be used in two sets.) | Answer: 8 sets.
## Solution.
Estimation. It is obvious that each set must contain at least one single-digit number, since the product of two different two-digit numbers is greater than 100. The number 1 cannot be included in any set, as all our numbers are distinct. Therefore, no more than 8 sets can be formed.
Exam... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,488 |
9.5. A semicircle with diameter $A B$ and center at point $O$ is divided by points $C$ and $D$ into three parts such that point $C$ lies on the arc $A D$. Perpendiculars $D E$ and $D F$ are dropped from point $D$ to segments $O C$ and $A B$ respectively. It turns out that $D E$ is the angle bisector of triangle $A D C$... | Answer: $20^{\circ}$.
Solution. Triangle $A O D$ is isosceles ($O D=O A$, as radii), hence, $\angle O A D=\angle O D A$. Since $D O$ is the bisector of angle $A D F$, then $\angle O A D=\angle O D F$. Calculation of angles in the right triangle $A F D$ shows that $\angle O A D=30^{\circ}$. Let $G$ be the point of inte... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,489 |
1. All cells of a $2 \times 25$ board are filled with crosses and noughts. Each cross has exactly one neighboring cross, and each nought has exactly two neighboring noughts. Provide an example of such an arrangement (neighbors are considered to be signs in cells sharing a common side). | Solution. Note that if the columns on the board alternate between pairs of crosses and 2x2 squares with zeros, then all conditions are met. Repeat the 2x3 group 8 times: in the left column, a pair of crosses, the rest - zeros. And add to this construction a column of two crosses on the right.
Criteria. Full solution -... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,490 |
2. Seryozha decided to start eating properly and every day he ate one fewer chocolate cookie and one more sugar-free cookie than the previous day. In total, during the time of proper nutrition, he ate 264 chocolate cookies and 187 sugar-free cookies. How many days was Seryozha on a proper diet? | Answer: 11 days.
Solution 1. The total number of cookies eaten each day is the same. In total, Seryozha ate $264+187=451=11 \cdot 41$ cookies. Since the proper diet lasted more than one day and he ate more than one cookie each day, either he ate for 11 days, 41 cookies each day, or for 41 days, 11 cookies each day. Bu... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,491 |
3. Once, 20 people gathered in a castle, 11 of them were sages who always tell the truth, and the other 9 were jesters who always lie. Around a table, $n \geq 3$ people sat, each of whom claimed, "Exactly one of my two neighbors is a jester." For which $n$ is this possible? (It is necessary to find all possible $n$ and... | Answer. $n=3,4,5,6,7,8,9,12,15$.
Solution. First, there is a case where all the people at the table are jesters: in this case, all $n \leq 9$ are suitable.
Now, let there be at least one sage at the table. Let this sage be $A$, and his neighbors be sage $B$ and jester $C$. Then, on the other side of $B$ sits a jester... | 3,4,5,6,7,8,9,12,15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,492 |
4. As punishment for bad grades, Vasya's parents sent him to his grandfather in the village. To educate his grandson Vasya, the grandfather first prohibited him from using his smartphone and offered to play the following game. Vasya had a well with water and three buckets with capacities of 3, 5, and 7 liters. Vasya ne... | Answer: Yes, he will be able to.
Solution. Let's denote Vasya's moves with the letter V, and Grandpa's moves with the letter D. See the diagram below.
Criteria. Full solution - 7 points. | Yes,hewillbeableto | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,493 |
5. There is one card with the digit 5, two cards with the digit 3, and a hundred cards with the digit 2.
a) In how many ways can a ten-digit number be formed from them, such that the product of the digits ends in 0?
b) All such numbers were written in ascending order. What number is in the 455th position? | Answer. a) 460 ways; b) 5322222322.
Solution. a) Among the selected cards, there will definitely be cards with a two, so the product of the digits is even. It ends in 0 if and only if the card with a five is also among the selected ones. The number of cards with a three can be 0, 1, or 2. | 460;5322222322 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,494 |
11.1. The quadratic trinomial $y=x^{2}+a x+b$ has real roots $t$ and $l$, where $-1<t<0$. Prove that if $\kappa$ is added to the coefficients $a$ and $b$ of the root $t$, then the resulting trinomial will also have two distinct real roots. | Solution: By Vieta's theorem $b=t \cdot s, a=-t-s$. Then the obtained quadratic trinomial has the form $g(x)=x^{2}+(a+t) x+(b+t)=x^{2}-s x+s t+t$. Its discriminant is $D=s^{2}-4 s t-4 t=(s-2 t)^{2}-4 t(1+t) \geqslant-4 t(1+t)$. Since $-10$, the discriminant of the new quadratic trinomial is positive, which proves the s... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,498 |
11.2. Let $\alpha, \beta$ and $\gamma$ be the plane angles of a trihedral angle. Prove that the numbers $\sin \frac{\alpha}{2}, \sin \frac{\beta}{2}$ and $\sin \frac{\gamma}{2}$ are the lengths of the sides of some triangle. | Solution: Let the vertex of the angle be point $O$. On all edges of the given trihedral angle, we lay off equal segments $O A=O B=O C=1 / 2$ and consider the tetrahedron $O A B C$. From the triangles forming its lateral faces, it is easy (for example, by the cosine theorem) to find the segments $A B, A C$, and $B C$. T... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,499 |
11.3. All 100 natural numbers from 2 to 101 were divided into two groups of 50 numbers each. The numbers in each group were multiplied, and the two resulting products were added. Prove that this sum is a composite number. | Solution: Note that the result of the described action is always a natural number greater than 101. There are two possible cases.
1) Each group contains an even number. Then the product of the numbers in each group is even, as is the sum of these products. An even number greater than 2 is composite, so the statement i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,500 |
11.4. Solve the equation $2^{\sqrt{x y}}+5^{\sqrt{x+y}}=3^{-x}+4^{-y}$. | Solution: The domain of the equation is defined by two conditions: $x y \geqslant 0$ and $x+y \geqslant 0$. From the first condition, it follows that either one of the numbers $x$ and $y$ is 0, or these two numbers have the same sign. From the second condition, it is clear that the case where both are negative is impos... | (0;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,501 |
11.5. Through a point $M$ inside a circle of radius $r$ with center at point $O$, two perpendicular chords are drawn. It is known that the sum of the squares of the lengths of these chords is $m^{2}$. Find the length of the segment $O M$.
. The quadrilateral $OEMF$ is a rectangle, so the equality $OM = \sqrt{OE^2 + OF^2}$ holds. Point $E$ is the midpoint of chord $AB$, and from triangle $AOE$ by the Pythagorean theorem, $AB = 2AE = 2\... | OM=\sqrt{2r^2-0.25m^2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,502 |
11.6. At the factory, there are exactly 217 women, among whom 17 are brunettes, and the remaining 200 are blondes. Before New Year's, all of them dyed their hair, and each of these women wrote in "VK" the surnames of exactly 200 women from the factory, whom they believed to be definitely blondes. Each of the brunettes ... | Solution: According to the problem, the correct list of all 200 blondes will be on "Vkontakte" exactly for 17 female workers at the plant: brunettes will write exactly this list, and a blonde will never write it, as otherwise she would have to include herself in it. Therefore, if a certain list appears not 17 times but... | 13 | Combinatorics | proof | Yes | Yes | olympiads | false | 14,503 |
2. Three bear cubs were dividing three pieces of cheese weighing 10 g, 12 g, and 15 g. A fox came to help them. She can simultaneously bite and eat 1 g of cheese from any two pieces. Can the fox leave the bear cubs equal pieces of cheese? | Answer: She can.
Solution. Let's provide one of the possible examples of how the fox could do it. For convenience, let's record the results of the fox's "work" in a table.
| 10 | 12 | 15 |
| :---: | :---: | :---: |
| 9 | 12 | 14 |
| 8 | 12 | 13 |
| 7 | 12 | 12 |
| 7 | 11 | 11 |
| 7 | 10 | 10 |
| 7 | 9 | 9 |
| 7 | 8 |... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,505 |
3. In the underwater kingdom, there live octopuses with seven and eight legs. Those with 7 legs always lie, while those with 8 legs always tell the truth. One day, a conversation took place between three octopuses.
Green octopus: "We have 21 legs together." Blue octopus (to the green one): "You're lying!" Red octopus:... | Answer. 1) Could not.
2) The green octopus has 7 legs, the blue one has 8 legs, and the red one has 7 legs.
Solution.
1) If the green octopus had told the truth, then each octopus would have 7 legs. This means that the green octopus, according to the condition, should have lied. We get a contradiction, so the green ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,506 |
4. On a grid paper, a square with a side of 5 cells is drawn. It needs to be divided into 5 parts of equal area, drawing segments inside the square only along the grid lines. Do this so that the sum of the lengths of all the segments drawn is equal to 16 cells. | Solution. One possible example is shown in the figure.
## Grading Criteria.
Correct solution - 7 points.
The square is divided into 5 equal-area parts, but the total length of the drawn s... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,507 |
5. Private Petrov took a bucket of unpeeled potatoes and cleaned them in 1 hour. In the process, $25 \%$ of the potatoes went into peels. How long did it take for him to have half a bucket of peeled potatoes? | Answer. In 40 minutes.
Solution. Since a quarter of the potatoes went into peels, Petrov received three quarters of a bucket of peeled potatoes in 1 hour. This means that a quarter of a bucket of peeled potatoes Petrov received in 20 minutes, and half a bucket - in 40 minutes.
## Grading Criteria.
- Correct solution... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,508 |
Problem 2. Does there exist a fraction equal to $\frac{7}{13}$, the difference between the denominator and the numerator of which is 24? | Answer: Yes, it exists, $\frac{28}{52}$.
Solution. Since $\frac{7}{13}$ is an irreducible fraction, any fraction equal to it has the form $\frac{7 x}{13 x}$, where $x$ is some natural number. In this case, the difference between the denominator and the numerator of such a fraction will be equal to $6 x$. We have $6 x=... | \frac{28}{52} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,509 |
Problem 3. The figure shown on the right is made of matches (the side of a small square is one match). The area of the entire shaded figure is 300 square centimeters. Find the total length of all the matches used.
.
Then Petya will fill his five pools in time $5 x$, and Vasya his in $25 x$.
We get the equation $25 x-5 x=1$, from which $x=3$ min. Then Vasya will fill his pools in $25 x=2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,511 |
Problem 5. The king has 10 wise men. One day he gave one gold coin to the first wise man, two coins to the second, three to the third, ..., and ten to the tenth. Then he said that every minute the wise men could ask him to give nine of them one gold coin each. If at any point all the wise men have the same number of co... | Answer: Yes, they will.
Solution. Each action of the king is represented as two consecutive actions:
- give one coin to each sage;
- take one coin from one of the sages.
Then, we will call an action of the first type the one where a coin is taken from the first sage; an action of the second type where a coin is take... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,512 |
1. Immediately after harvesting, the water content in 1 ton of tomatoes was $99 \%$. By the time of sale, the water share had decreased by $4 \%$ (after drying). What did the total weight of the tomatoes become? | Solution. The weight of the dry matter in tomatoes is $1 \%$ of 1 ton - 10 kg. After drying, the weight of the dry matter does not change and constitutes $5 \%(100 \%-$ $(99 \%-4 \%))$ of the total weight of the tomatoes. In other words, the total weight is 200 kg.
Remark. Perhaps, the student can be given 2 - 3 point... | 200 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,513 |
3. How to find the lightest and heaviest stones among $2 N$ stones, any two of which differ in weight, in $3 N-2$ weighings? All weighings are performed on a two-pan balance without weights. | Solution. In $N$ weighings, we weigh all $2 N$ stones pairwise. The stones that turned out to be heavier are placed in one pile, and those that are lighter are placed in another. In the first pile, the heaviest stone is found as follows: we number the stones from 1 to $N$, take stones 1 and 2, weigh them, choose the st... | 3N-2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,515 |
4. Find two numbers whose sum is 2017, and the sum of the numbers written with the same digits but in reverse order is 8947. | Solution. Let the first desired number be $\overline{a b c d}$, and the second be $\overline{\text { efgh }}$. These numbers are such that $\overline{a b c d} \geq \overline{e f g h}$. Note that $d$ and $h$ can sum to either 7 or 17. If $d+h=7$, then $\overline{a b c d}$ and $\overline{e f g}$ are four-digit numbers, $... | 1408,609 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,516 |
5. The numbers from 0 to 9 are written in a row. In one move, we can either add 1 to two adjacent numbers, or multiply three arbitrary numbers by 3. Is it possible to make all the numbers in the row equal using the allowed actions? | Solution. Initially, the sequence contains an odd number (5) of both even and odd numbers. Let's consider how the number of odd numbers (more precisely, the parity of the number of odd numbers) will change when performing the allowed operations. When performing the first operation, the number of odd numbers: 1) does no... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,517 |
9.1. Five children had ten candies in total. The first said: "I have one candy", the second: "I have two candies", ..., the fifth: "I have five candies". How many of these statements could be false? | Answer. From 1 to 5 statements.
Solution. The total number of "claimed" candies is not equal to ten, so at least one of the statements is false. Examples of candy distribution for the number of false statements given in the answer: 1) $1,2,3,4,0 ; 2) 1,2,3,0,4 ; 3) 1,2,0,0,7 ; 4) 1,0,0,0,9 ; 5) 0,0,0,0,10$.
Comment. ... | From1to5statements | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,520 |
9.2. Baron Munchausen told a story that once he arranged all natural numbers from 1 to 100 in a certain order in a circle, and then calculated all possible sums of any three consecutive numbers and wrote them on the board. It turned out that all the numbers on the board were prime. Is the baron mistaken? | Answer: Incorrect.
Solution: Suppose the numbers can be arranged in the required manner. Note that the sum of any three numbers from 1 to 100 is greater than 2, so if the sum of any three numbers is a prime number, then this prime number must be odd. Let the numbers $a, b, c, d$ be arranged consecutively in a circle. ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,521 |
9.3. Natural numbers $a, b, c$ are such that $\frac{a^{2}}{a^{2}+b^{2}}+\frac{c^{2}}{a^{2}+c^{2}}=\frac{2 c}{b+c}$. Prove that $bc$ is a perfect square. | Solution. Let's move everything to one side and transform:
$$
\begin{gathered}
\frac{a^{2}}{a^{2}+b^{2}}+\frac{c^{2}}{a^{2}+c^{2}}-\frac{2 c}{b+c}=\left(\frac{a^{2}}{a^{2}+b^{2}}-\frac{c}{b+c}\right)+\left(\frac{c^{2}}{a^{2}+c^{2}}-\frac{c}{b+c}\right)= \\
=\frac{b\left(a^{2}-b c\right)}{\left(a^{2}+b^{2}\right)(b+c)}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,522 |
9.4. In an acute-angled triangle $A B C$ ($A B>A C$), the length of the altitude $A H$ is equal to the radius of the circumcircle of triangle $A B C$. Point $O$ is the center of the circumcircle of triangle $A B C$. The bisector of triangle $A B C$, drawn from vertex $A$, intersects side $B C$ at point $D$, and line $D... | Solution. First, note that $\angle B A O=\angle C A H$. Indeed, $\angle A O B$ is a central angle, so $\angle A O B=2 \angle A C B$, and triangle $A O B$ is isosceles, hence $\angle B A O$ equals $90^{\circ}-\angle A C B$. At the same time, $\angle C A H$ is found from the right triangle $A C H$ and equals $\angle C A ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,523 |
9.5. In a $5 \times 5$ table, all numbers from 1 to 25 are written, one number per cell. Consider all pairs of numbers that are in the same row or the same column. For each pair, the ratio of the larger number to the smaller number is calculated. The minimum of these ratios among all pairs is denoted by $A$. What is th... | # Answer: $\frac{6}{5}$.
Solution. Consider the numbers $25,24,23,22,21$. If any two of them are in the same row or the same column, then $A \leq \frac{25}{21}\frac{6}{5}$. In the first column, the numbers differ by a multiple of 4; the smallest ratio is $\frac{21}{17}>\frac{6}{5}$. In the second column, the minimum r... | \frac{6}{5} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,524 |
# 11.1. (7 points)
Check the truth of the inequality $\frac{1}{\log _{\pi} 2}+\frac{1}{\log _{2} \pi}>2$. | Solution: $\log _{2} \pi+\frac{1}{\log _{2} \pi}>2$,
$$
\begin{gathered}
\frac{\log _{2}^{2} \pi+1}{\log _{2} \pi}>2, \log _{2} \pi>0 \\
\log _{2}^{2} \pi+1>2 \log _{2} \pi \\
\log _{2}^{2} \pi-2 \log _{2} \pi+1>0 \\
\left(\log _{2} \pi-1\right)^{2}>0
\end{gathered}
$$
This inequality is true. | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 14,525 |
# 11.2. (7 points)
The master makes a whole number of parts in one hour, more than 5, and the apprentice makes 2 parts less. The master completes the order in a whole number of hours, and two apprentices together - one hour faster. How many parts does the order consist of? | Answer: 24.
Solution: Let $x$ be the number of parts in the order; $y$ (parts per hour) be the master's productivity $(y>5)$; $y-2$ (parts per hour) be the apprentice's productivity; $2 y$ - 4 (parts per hour) be the productivity of two apprentices.
Then $\frac{x}{y}$ is the time for the master to complete the order,... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,526 |
# 11.3. (7 points)
The base of the pyramid with vertex $P$ is a quadrilateral $A B C D$, where the sum of angles $A$ and $D$ is five times smaller than the sum of angles $B$ and $C$. Find the angle between the planes of the faces $P A B$ and $P C D$, if both are perpendicular to the base. | Answer: $60^{\circ}$.
Solution: Let the planes $A P B$ and $C P D$ intersect along the line $P X$ (point $X$ lies in the plane $A B C$, see the figure).
Since $(A P B) \perp(A B C)$ and $(C ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,527 |
11.4. (7 points)
Solve the equation $(\sqrt[5]{7+4 \sqrt{3}})^{x}+(\sqrt[5]{7-4 \sqrt{3}})^{x}=194$. | Answer: -10; 10.
Solution: $\left(\sqrt[5]{\frac{(7+4 \sqrt{3})(7-4 \sqrt{3})}{7-4 \sqrt{3}}}\right)^{x}+(\sqrt[5]{7-4 \sqrt{3}})^{x}=194$,
$\frac{1}{(\sqrt[5]{7-4 \sqrt{3}})^{x}}+(\sqrt[5]{7-4 \sqrt{3}})^{x}=194$,
Let $(\sqrt[5]{7-4 \sqrt{3}})^{x}=\mathrm{t}$, then $\frac{1}{t}+t=194$.
$$
t^{2}-194 t+1=0, t_{1,2}=... | -10;10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,528 |
# 11.5. (7 points)
At a joint conference of the party of liars and the party of truth-tellers, 32 people were elected to the presidium and seated in four rows of eight. During the break, each member of the presidium claimed that among their neighbors there are representatives of both parties. It is known that liars al... | Answer: with eight liars.
Solution: Divide all the seats in the presidium into eight groups as shown in the figure. If there are fewer than eight liars, then in one of these groups, only truth-tellers will be sitting, which is impossible. The contradiction obtained shows that there are no fewer than eight liars. The f... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,529 |
10.1. The lines containing the sides of a given acute triangle $T$ were painted red, green, and blue. Then these lines were rotated clockwise around the center of the circumscribed circle of the given triangle by an angle of $120^{\circ}$ (the line retains its color after the rotation). Prove that the three points of i... | Solution. Let $ABC$ be the given triangle, $O$ be the center of its circumscribed circle, $D, E, F$ be the midpoints of its sides $BC$, $CA$, $AB$ respectively, so that $DEF$ is similar to $ABC$ with a coefficient of $1/2$ and $OD \perp BC$, $OE \perp CA$, $OF \perp AB$.
Comment. Let $ABC$ be the given triangle, $A'B'... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,530 |
10.2. Each of 100 schoolchildren has a stack of 101 cards numbered from 0 to 100. The first schoolchild shuffles the stack, then takes one card from the top of the resulting stack, and each time a card is taken (including the first time), writes on the board the arithmetic mean of the numbers on all the cards taken by ... | Solution. On the 1st step, each of the 100 people was given one of the numbers from the set $A_{1}=\{0,1,2, \ldots, 100\}$.
On the 2nd step - one of the numbers from the set $A_{2}=$ $=\left\{\frac{1}{2}, \frac{2}{2}, \frac{3}{2}, \ldots, \frac{199}{2}\right\}$.
On the 100th step, one of the numbers from the set $A_{... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,531 |
10.3. Given natural numbers $a$ and $b$ such that $a \geqslant 2 b$. Does there exist a polynomial $P(x)$ of degree greater than 0 with coefficients from the set $\{0,1,2, \ldots, b-1\}$ such that $P(a)$ is divisible by $P(b)$?
(T. Korotchenko) | Answer. Exists for $b>1$.
Solution. It is easy to see that if $b=1$, then any polynomial with coefficients from 0 to $b-1$ is zero.
Let $b>1$. Represent $a-b$ in base $b$: $a-b=$ $=c_{n} b^{n}+\ldots+c_{1} b+c_{0}$, where $c_{i} \in\{0,1,2, \ldots, b-1\}$. Since $a-b \geqslant b$, in this representation $n \geqslant ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,532 |
1. It is known that the numbers $a+b$ and $\frac{a b}{a+b}$ are integers. Prove that the number $\frac{a^{2}+b^{2}}{a+b}$ is also an integer. | 1. $\frac{a^{2}+b^{2}}{a+b}=\frac{(a+b)^{2}-2 a b}{a+b}=(a+b)-2 \cdot \frac{a b}{a+b}$ - an integer. | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,534 |
2. It is known that the quadratic equations $x^{2}+a x+b=0$ and $x^{2}+c x+d=0$ do not have real roots. Prove that the equation $x^{2}+\frac{a+c}{2} x+\frac{b+d}{2}=0$ also does not have real roots. | 2. Given $a^{2}<4 b, c^{2}<4 d$. We will show that $\left(\frac{a+c}{2}\right)^{2}<4\left(\frac{b+d}{2}\right)$.
We have
$\left(\frac{a+c}{2}\right)^{2}=\frac{1}{4}\left(a^{2}+2 a c+c^{2}\right) \leq \frac{1}{4}\left(a^{2}+\left(a^{2}+c^{2}\right)+c^{2}\right)=\frac{1}{2}\left(a^{2}+c^{2}\right)<\frac{1}{2}(4 b+4 d)$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,535 |
3. Laid out in a row from left to right are 8 wallets, each containing 13 identical coins. One coin was moved from one wallet to the adjacent wallet on the right. The wallets cannot be opened. How can you find the wallet with the fewest coins using two weighings on a balance scale without weights? | 3. Number the wallets from left to right. Compare the 3rd and 6th wallets. Equality means that the transfer was from the 1st, 4th, or 7th. By comparing any two, we can find the lightest one. If the 3rd is heavier than the 6th, then the transfer was either from the 2nd or the 6th. Compare them. If the 3rd is lighter tha... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,536 |
4. Let $A L, B M, C N$ be the medians of triangle $A B C$, intersecting at point $K$. It is known that a circle can be circumscribed around quadrilateral $C L K M$, and $A B=2$. Find the length of the median $C N$. | 4. Answer. $\sqrt{3}$.
Since $M L$ is the midline of triangle $A B C$, we get that $M L \| A B$ and $M L=\frac{1}{2} A B=1$. If $P$ is the intersection point of the median $C N$ and the midli... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,537 |
5. Solve the system of equations
$$
\left\{\begin{array}{l}
x+y=z^{2} \\
y+z=x^{2} \\
z+x=y^{2}
\end{array}\right.
$$ | 5. Answer. $(0,0,0)$ and $(2,2,2)$.
Subtract the second equation from the first, we get
$$
x-z=z^{2}-x^{2} \text {, that is } x-z=(z-x)(z+x) \text {. }
$$
Considering that $z+x=y^{2}$, we arrive at the equation $x-z=(z-x) y^{2}$.
Subtracting the third equation from the second, by similar reasoning we get the equati... | (0,0,0)(2,2,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,538 |
1. It is known that the quadratic trinomial $p(x)=a x^{2}+b x+c$ takes integer values for all integer values of $x$. Can it happen that $a=\frac{1}{2}$? | 1. Answer: It can.
Note that for all integer values of $x$, the value of the expression $x(x+1)$ is divisible by 2, as it is the product of two consecutive integers (thus, one of the numbers is necessarily even). Therefore, the desired quadratic trinomial can have the form $p(x)=\frac{1}{2} x^{2}+\frac{1}{2} x$. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,539 |
2. Petya came up with a four-digit number, in which all the digits are different. It is known that the sum of the first three digits of this number is divisible by 9 and the sum of the last three digits of this number is divisible by 9. What values can the sum of all the digits of this number take? Find all possible va... | 2. Answer. The sum of the digits is 18.
According to the condition, the sum of the first three digits is divisible by 9, and the sum of the last three is also divisible by 9. Therefore, the difference between the first and the last digit is divisible by 9. Since all digits are different, the first digit is 9, and the ... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,540 |
4. In triangle $ABC$, $AB = 3 \cdot BC$, point $M$ is the midpoint of side $AB$, and $BD$ is the angle bisector. Find the angle $MDB$.
---
Here is the translation of the given text, maintaining the original formatting and structure. | 4. Answer. $90^{\circ}$.
Let $X$ be the midpoint of segment $M B$, and draw segment $D X$. By the property of the angle bisector, we have
$$
A D: D C = A B: B C = 3: 1
$$
On the other hand,... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,542 |
5. There are 4 identical jars, each containing 4 different paints, and each jar is filled to $\frac{3}{4}$. It is allowed to pour any part of the liquid from one jar to another (it is assumed that any amount of liquid can be poured from one jar to another if it fits). Is it possible to make the same mixture in all jars... | # 5. Answer. Yes.
Solution. First, pour all the contents from the first jar into the others until they are full. Then pour half of the second jar into the first, half of the third jar into the first, and half of the third jar into the second. The mixtures in the first and second jars are now the same. Fill them into t... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,543 |
2. Cheburashka took several natural numbers and multiplied them. The product turned out to be 1792. What numbers did Cheburashka multiply, if the smallest of these numbers was half the largest? Justify your answer.
Answer: $4,7,8,8$ or $8,14,16$. | Solution. Let's factorize the number 1792 into prime factors: $1792=2^{8} \times 7$. Seven does not appear in either the smaller or the larger factors, as otherwise they would differ by a factor of seven or more. Therefore, the smaller and larger factors must be consecutive powers of two. The highest power of two is 8 ... | 4,7,8,8or8,14,16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,544 |
4. On an island, there live knights who always tell the truth, and liars who always lie. One day, 9 natives, among whom were both knights and liars, stood in a circle, and each said: "The two people standing opposite me are liars." Opposite a native stand the fourth native from him in a clockwise direction and the four... | Solution. Let's divide all the natives into three groups (see the figure, natives from the same group have the same numbers).
In each such triplet, there cannot be three liars (since then one of the liars would be telling the truth) and there cannot be three knights (since then one of the knights would be lying). Ther... | 3or4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,545 |
# 1. CONDITION
Consider quadratic trinomials of the form $x^{2}+p x+q$ with integer coefficients, where $p+q=30$. How many such trinomials have integer roots? | Solution. Let $x_{1}$ and $x_{2}$ be the integer roots of a quadratic polynomial of the form $x^{2} + px + q$. Then $p = -(x_{1} + x_{2})$, $q = x_{1} x_{2}$. Therefore, $30 = p + q = (x_{1} - 1)(x_{2} - 1) - 1$, i.e., $(x_{1} - 1)(x_{2} - 1) = 31$. Since the number 31 is prime, it can be represented as the product of ... | Two\quadratic\polynomials:\x^{2}-34x+64\\x^{2}+30x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,547 |
# 2. CONDITION
Find some function $f(x)$ for which the equations $f(x)=0$ and $f^{\prime}(x)+x^{2}+1=0$ have the same non-empty set of roots. | Solution. Indeed, the equations $-x=0$ and $(-x)'+x^{2}+1=0$ both have the unique root $\mathrm{x}=0$.
Answer. For example, $f(x)=-x$. | f(x)=-x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 14,548 |
# 3. CONDITION
In what minimum number of colors should natural numbers be painted so that any two numbers, the difference between which is 3, 4, or 6, are of different colors? | Solution. Let the number $n=1$ be color $A$, then the numbers 4, 5, and 7 must be painted in another color. Let $n=4$ be color $B$, then from $7-4=3$, it follows that the number $n=7$ is of the third color $C$. Therefore, at least 3 colors are required. The coloring $A A A B B B C C C A A A B B B \ldots$ is the desired... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,549 |
# 4. CONDITION
On the sheet, coordinate axes and branches of the hyperbola $y = k / x$ are drawn ($k$ is unknown, the scale on the coordinate axes is also not specified). A point is marked on one of the branches. Using a compass and a ruler, construct the tangent to the hyperbola at the marked point. | Solution. Let $A$ and $B$ be the projections of the marked point $K$ onto the coordinate axes. We will show that the tangent $l$ to the hyperbola is parallel to $AB$.
Indeed, let $K\left(x_{0}, k / x_{0}\right), A\left(x_{0}, 0\right), B\left(0, k / x_{0}\right)$ be the coordinates of these points, then $\operatorname... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,550 |
# 6. CONDITION
Do there exist natural numbers $a$ and $b$, greater than 1000, such that for any $c$, which is a perfect square, the three numbers $a, b$, and $c$ cannot be the lengths of the sides of a triangle? | Solution. Example. $a=10000^{2}+10000, b=1001$. Suppose there exists such a $c=d^{2}$ that the numbers $a, b, c$ are the lengths of the sides of some triangle. Then the triangle inequalities $a+b>c$, $b+c>a$, $a+c>b$ must hold. Consider the first two of these: $a+b>c$ and $c>a-b$. Note that $a+b=10000^{2}+10000+1001=10... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,551 |
9.5. After watching the movie, viewers rated it one by one with an integer score from 0 to 10. At any given time, the movie's rating was calculated as the sum of all the given scores divided by their number. At some point in time $T$, the rating became an integer, and then with each new voting viewer, it decreased by o... | Answer: 5.
Solution. Consider a moment when the rating has decreased by 1. Suppose that before this, $n$ people had voted, and the rating was an integer $x$. Thus, the sum of the scores was $n x$. Let the next viewer give $y$ points. Then the sum of the scores becomes $n x + y = (n + 1)(x - 1)$, from which $y = x - n ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,552 |
9.6. Given a right triangle $A B C$ with a right angle at $C$. Let $B K$ be the angle bisector of this triangle. The circumcircle of triangle $A K B$ intersects side $B C$ again at point $L$. Prove that $C B + C L = A B$.
( | The first solution. On the extension of $BC$ beyond point $C$, we mark a segment $CN = LC$ (see Fig. 1). Then $CB + CL = NB$, and we need to prove that $AB = NB$.
Since the quadrilateral $ABLK$ is cyclic, we have $\angle CKB = 180^\circ - \angle AKB = 180^\circ - \angle ALB = \angle ALC$. On the other hand, the right ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,553 |
9.7. The numbers $a, b, c$ and $d$ are such that $a^{2}+b^{2}+c^{2}+d^{2}=4$. Prove that $(2+a)(2+b) \geqslant c d$. | Solution. By the inequality of means, we have
$$
c d \leqslant|c d|=\sqrt{c^{2} d^{2}} \leqslant \frac{c^{2}+d^{2}}{2}=\frac{4-a^{2}-b^{2}}{2}
$$
Therefore, it is sufficient to show that
$$
(2+a)(2+b) \geqslant \frac{4-a^{2}-b^{2}}{2}
$$
Move everything to the left side, multiply by 2, and expand the brackets. We o... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 14,554 |
9.8. Pete wants to list all possible sequences of 100 natural numbers, in each of which a triple appears at least once, and any two adjacent members differ by no more than 1. How many sequences will he have to list?
(I. Bogdanov) | Answer. $3^{100}-2^{100}$.
First solution. Let $n=100$. We call a sequence of $n$ natural numbers interesting if any two adjacent members differ by no more than 1. To each interesting sequence $a_{1}, a_{2}, \ldots, a_{n}$, we associate a difference sequence $b_{i}=a_{i+1}-a_{i}(i=1,2, \ldots, n-1)$. All members of th... | 3^{100}-2^{100} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,555 |
10.5. After watching the movie, viewers rated it one by one with an integer score from 0 to 10. At any given time, the movie's rating was calculated as the sum of all the given scores divided by their number. At some point in time $T$, the rating became an integer, and then with each new voting viewer, it decreased by ... | # Answer: 5.
Solution. Consider a moment when the rating has decreased by 1. Suppose that before this, $n$ people had voted, and the rating was an integer $x$. This means the sum of the scores was $n x$. Let the next viewer give $y$ points. Then the sum of the scores becomes $n x + y = (n + 1)(x - 1)$, from which $y =... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,556 |
10.6. Given a right triangle $ABC$ with a right angle at $C$. Let $BK$ be the angle bisector of this triangle. The circumcircle of triangle $AKB$ intersects side $BC$ again at point $L$. Prove that $CB + CL = AB$.
(D. Barabash, B. Obukhov) | The first solution. On the extension of $BC$ beyond point $C$, we mark a segment $CN = LC$ (see Fig. 3). Then $CB + CL = NB$, and we need to prove that $AB = NB$.
Since the quadrilateral $ABLK$ is cyclic, we have $\angle CKB = 180^\circ - \angle AKB = 180^\circ - \angle ALB = \angle ALC$. On the other hand, the right ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,557 |
10.7. The coefficients $a, b, c$ of the quadratic trinomial $f(x)=a x^{2}+b x+$ $+c$ are natural numbers, the sum of which is 2000. Pasha can change any coefficient by 1, paying 1 ruble. Prove that he can obtain a quadratic trinomial that has at least one integer root, paying no more than 1050 rubles. | Solution. We will show that from the original quadratic polynomial, we can obtain a new one that has at least one integer root by changing the coefficients $a, b, c$ by a total of no more than 1022.
If the coefficient $c$ is no more than 1022, then by setting it to zero, we obtain the desired quadratic polynomial $f_{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,558 |
10.8. Given a natural number $n \geqslant 2$. Consider all colorings of the cells of an $n \times n$ board in $k$ colors such that each cell is painted in exactly one color, and all $k$ colors are used. For what smallest $k$ will any such coloring contain four cells painted in four different colors, located at the inte... | Answer: $k=2 n$.
First solution. First, we will show how to color the cells in $2 n-1$ colors so that there is no set of four cells of different colors lying at the intersection of two rows and two columns.
We number the horizontal rows from top to bottom with even numbers $0,2,4, \ldots, 2 n-2$, and the vertical row... | 2n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,559 |
11.5. A quadratic trinomial $f(x)$ has two distinct roots. It turns out that for any numbers $a$ and $b$, the inequality $f\left(a^{2}+b^{2}\right) \geqslant$ $\geqslant f(2 a b)$ holds. Prove that at least one of the roots of this trinomial is negative.
(A. Khryabrov) | First solution. Substitute $b=0$ into the condition and we get that $f\left(a^{2}\right) \geqslant f(0)$ for all $a$. Therefore, $f(t) \geqslant f(0)$ for any positive $t$. From this and the properties of the graph of a quadratic function, it follows, in particular, that the branches of the parabola are directed upward... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,561 |
11.6. There is a hemispherical vase covered with a flat lid. In the vase, there are four identical oranges touching the vase, and one grapefruit touching all four oranges. Is it true that all four points of contact between the grapefruit and the oranges necessarily lie in the same plane? (All fruits are spheres.)
(L. ... | Answer. Correct.
Solution. Let the centers of the vase, grapefruit, and oranges be denoted by $V, G$, and $A_{1}, A_{2}, A_{3}, A_{4}$, respectively. Note that all these six points are distinct, as all the fruits are under the lid. Let the oranges touch the grapefruit at points $K_{1}, K_{2}, K_{3}, K_{4}$, and the va... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,562 |
11.7. There are 300 positive numbers arranged in a circle. Could it happen that each of these numbers, except one, is equal to the difference of its neighbors?
(S. Berlov) | Answer: It could not.
Solution: Suppose the required arrangement exists. It is clear that the largest of the numbers cannot equal the difference of its neighbors; therefore, each of the other numbers equals the difference of its neighbors. In particular, the largest number appears exactly once; denote it by $m$.
Let ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,563 |
11.8. Petya wants to list all possible sequences of 100 natural numbers, in each of which the number 4 or 5 appears at least once, and any two adjacent members differ by no more than 2. How many sequences will he have to list?
(I. Bogdanov) | Answer. $5^{100}-3^{100}$.
First solution. Let $n=100$. We will call a sequence of $n$ natural numbers interesting if any two adjacent members differ by no more than 2. To each interesting sequence $a_{1}, a_{2}, \ldots, a_{n}$, we associate a difference sequence $b_{i}=a_{i+1}-a_{i}(i=1,2, \ldots, n-1)$. All members ... | 5^{100}-3^{100} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,564 |
1. The discriminants of three given quadratic trinomials are 1, 4, and 9. Prove that it is possible to choose one root from each of these trinomials such that the sum of the three chosen roots equals the sum of the three remaining roots. | 1. The discriminants of three given quadratic trinomials are 1, 4, and 9. Prove that it is possible to choose one root from each of these trinomials such that the sum of the three chosen roots equals the sum of the three remaining roots.
Solution. **I** method. Let $b_{1}, b_{2}, b_{3}$ be the coefficients of $x$ in t... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,565 |
2. For what least natural $n$ do there exist such natural $a$ and $b$ that
$$
\text { GCD }(a, b)=999 \text { and } \text{LCM}(a, b)=n!
$$
(here $n!=1 \cdot 2 \cdot \ldots \cdot n)$ ? | 2. For what smallest natural $n$ do there exist such natural $a$ and $b$ that
$$
\text { GCD }(a, b)=999 \text { and } \text{LCM}(a, b)=n!
$$
(Here $n!=1 \cdot 2 \cdot \ldots \cdot n) ?$
ANSWER. $n=37$.
SOLUTION. Since $\text{LCM}(a, b)$ is divisible by $\text{GCD}(a, b)=999=27 \cdot 37$, and the number 37 is prime... | 37 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,566 |
3. Prove that $5 x^{2}+5 y^{2}+8 x y+2 y-2 x+2 \geqslant 0$ for any real numbers $x, y$. | 3. Prove that $5 x^{2}+5 y^{2}+8 x y+2 y-2 x+2 \geqslant 0$ for any real numbers $x, y$.
Solution. Method I. By completing the square in the left-hand side, we obtain the expression $5(x+y)^{2}-2 x y+2 y-2 x+2$, which is $5(x+y)^{2}-2(y-1)(x+1)$. The substitution $a=x+1, b=y-1$ gives the equivalent inequality $5(a+b)^... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 14,567 |
4. In triangle $ABC$, the median $AF$ is drawn. Point $D$ is the midpoint of segment $AF$, and $E$ is the point of intersection of line $CD$ and side $AB$. It is known that $BD = BF$. Prove that $AE = DE$. | 4. In triangle $ABC$, median $AF$ is drawn. Point $D$ is the midpoint of segment $AF$, and $E$ is the point of intersection of line $CD$ and side $AB$. It is known that $BD = BF$. Prove that $AE = DE$.
Solution. Angles $BDF$ and $BFD$ are equal as the base angles of isosceles triangle $BDF$ ($BD = BF$ by the condition... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,568 |
5. Vova has 19 math grades in his journal, all twos and threes, and the first four grades are twos. It turned out that among the quartets of consecutive grades, all 16 possible combinations of four twos and threes are present. What are Vova's last four grades? | 5. Vova has 19 math grades in his journal, all twos and threes, and the first four grades are twos. It turned out that among the quartets of consecutive grades, all 16 possible combinations of four twos and threes are present. What are Vova's last four grades?
ANSWER. 3222.
SOLUTION. Note that there are exactly 16 qu... | 3222 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,569 |
2. Prove that the inequality $x^{2}-2 x \sqrt{y-5}+y^{2}+y-30 \geq 0$ holds for any valid values of $x$ and $y$. | Solution. Domain of definition: $x$ - any number, $y \geq 5$.
$x^{2}-2 x \sqrt{y-5}+y^{2}+y-30=(x-\sqrt{y-5})^{2}+y^{2}-25 \geq 0$,
since $(x-\sqrt{y-5})^{2} \geq 0$ and $y^{2}-25 \geq 0$ for $y \geq 5$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 14,573 |
3. The lateral sides $K L$ and $M N$ of trapezoid $K L M N$ are equal to 15 and 12, respectively, and the base $L M=3$. The bisector of angle $N K L$ passes through the midpoint of side $M N$. Find the area of the trapezoid. Answer: 80. | # Solution.
Let $Q$ be the midpoint of segment $M N$. Extend the bisector of angle $N K L$ to intersect line $L M$ at point $P$. $\angle N K Q = \angle Q K L$.
$\angle N K Q = \angle K P L$ ... | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,574 |
5. In the cells of a table with 10 rows and 10 columns, the numbers +1 and -1 are placed. The products of the numbers in each column and each row are taken. Prove that the sum of these 20 products cannot be equal to two.
# | # Solution.
## 1st method.
Fill the table with ones, then the sum of the products is 20. We will replace 1 with -1. The sum either decreases by 4, remains unchanged, or increases by 4, since one cell participates in two products, so in the sum, two terms will change to opposite values. For example: it was \( S_{1} = ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,575 |
1. The working day at the enterprise lasted 8 hours. During this time, the labor productivity was as planned for the first six hours, and then it decreased by $25 \%$. The director (in agreement with the labor collective) extended the shift by one hour. As a result, it turned out that again the first six hours were wor... | 1. Answer: by 8 percent. Solution. Let's take 1 for the planned labor productivity (the volume of work performed per hour). Then before the shift extension, workers completed $6+1.5=7.5$ units of work per shift. And after the extension, $8+2.1=8.1$ units. Thus, the overall productivity per shift became $8.1: 7.5 \times... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,576 |
2. Each cell of a $5 \times 5$ square contains a number. The numbers in any row and any column form an arithmetic progression. The numbers in the corners of the square are $1, 25, 17, 81$. What number is in the central cell? Don't forget to justify your answer.
| 1 | | | | 25 |
| :--- | :--- | :--- | :--- | :--- |
... | 2. Answer: 31. Solution. In any arithmetic progression $a_{5}=a_{1}+4 d, a_{3}=a_{1}+2 d \Rightarrow$ $a_{3}=\left(a_{1}+a_{5}\right) / 2$. Therefore, in the third cell of the first column stands the number 41, in the third cell of the last column stands the number 21, and in the middle cell of the second row (i.e., in... | 31 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,577 |
3. We have a sequence of numbers $1,2,-3,-4,5,6,-7,-8, \ldots$ From this, another sequence is constructed as $1,1+2=3,1+2-3=0, \ldots$. Prove that the second sequence contains infinitely many zeros. | 3. Solution. The first three numbers in the sum give zero. We will divide the remaining sequence into consecutive quadruples. In each such quadruple, the sum is zero. By listing the first three numbers and an arbitrary number of quadruples, we get a set with a zero sum, and there are infinitely many such sets. Grading ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,578 |
4. Triangle $A B C$ is similar to the triangle formed by its altitudes. Two sides of triangle $A B C$ are 4 cm and 9 cm. Find the third side. | 4. Answer. 6 cm. Solution. Let $A B=9, B C=4, A C=x, S$ - the area of triangle $A B C$. Then the heights of the triangle are $2 S / 4, 2 S / 9, 2 S / x$. By the triangle inequality, side $A C$ can be either the largest or the middle in length. Applying the triangle inequality to the triangle formed by the heights, we s... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,579 |
5. In a right triangle $ABC$ with a right angle at $A$, angles $B$ and $C$ are not equal. A circle with center at some point $O$, passing through $B$ and $C$, intersects the legs $AB$ and $AC$ at points $P$ and $Q$ respectively. The altitude $AS$ intersects the segment $PQ$ at point $K$. $M$ is the midpoint of the hypo... | 5. Solution. By the condition, segments $A S$ and $O M$ are perpendicular to the hypotenuse and therefore parallel to each other. It remains to prove that segments $O K$ and $A M$ are parallel. Let $\angle A C B=\alpha$. Then $\angle A B C=90^{\circ}-\alpha$, $\angle B P Q=180^{\circ}-\alpha$, $\angle A P K=\alpha=\ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,580 |
6. Tenth-grader Dima came up with two natural numbers (not necessarily different). Then he found the sum and the product of these numbers. It turned out that one of these numbers coincides with the arithmetic mean of the other three. What numbers did Dima come up with? Find all possible answers and prove that others ar... | 6. Answer. There are three solutions $(1,2),(4,4)$ and $(3,6)$. Solution. Let the numbers Dim invented be $a, b$. If each of these numbers is different from 1, then their product $a b$ is greater than the other three and cannot be their arithmetic mean. Therefore, there are two fundamentally different cases: $3 a=b+(a+... | (1,2),(4,4),(3,6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,581 |
1. Each half-liar alternates between lying and telling the truth every other day (if they lie today, they will tell the truth tomorrow, and vice versa). Exactly a week ago, one half-liar made two statements: "Yesterday was Wednesday"; "Tomorrow will be Thursday." Today he made two other statements: "Yesterday was Frida... | Solution. The statements "Yesterday was Wednesday" and "Tomorrow will be Thursday" cannot both be true at the same time, so the half-liar was lying a week ago. Since there is an odd number of days in a week, today he is telling the truth. Therefore, yesterday was Friday, which means today is Saturday. Answer: yes, Satu... | Saturday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,582 |
2. Find the smallest natural number that has exactly 12 different natural divisors, the largest prime divisor of which is the number 101, and the last digit is zero. | Solution. Let $n$ be the desired number. According to the condition, $n: 101, n: 2, n: 5$. Consider the number $m=2 \cdot 5 \cdot 101$, note that it has exactly 8 different natural divisors $(1$, $2,5,101,2 \cdot 5, \ldots, 2 \cdot 5 \cdot 101$ ), so $n>m$. Since $n$ is the smallest natural number satisfying the condit... | 2020 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,583 |
3. How many four-digit numbers exist where the digit in the thousands place is greater than the digit in the hundreds place? | Solution. The digit in the thousands place can take one of 9 possible values: $1,2,3, \ldots, 9$ (we cannot take 0, since the number is four-digit). For each of these options, we can specify the corresponding number of options for the hundreds digit: $1,2,3$, ..., 9. That is, in total $1+2+\ldots+9=45$ options. The oth... | 4500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,584 |
4. In triangle $ABC$, the bisector of angle $C$ intersects side $AB$ at point $M$, and the bisector of angle $A$ intersects segment $CM$ at point $T$. It turns out that segments $CM$ and $AT$ divide triangle $ABC$ into three isosceles triangles. Find the angles of triangle $ABC$. | Solution. Since the sum of angles $A$ and $C$ of triangle $ABC$ is less than $180^{\circ}$, then $\angle TAC + \angle TCA < 90^{\circ}$, so angle $\angle ATC$ is obtuse (see the figure).
This... | 36,72,72 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,585 |
5. What is the minimum number of chips that need to be taken so that in any arrangement of them on the cells of a chessboard, there would necessarily be 4 chips standing in a row horizontally? | Solution. Consider one row. To ensure that there are 4 chips in it, there must be no less than 7 chips (6 chips can be placed in two groups of 3 chips). If there are a total of $6 \cdot 8+1=49$ chips, then by the pigeonhole principle, there will definitely be a row with no fewer than 7 chips. On the other hand, it is i... | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,586 |
1. In triangle $ABC$, the bisector $AL$ is drawn. Prove that $BL < AB$. | 1. Solution: Angle $A L B$ is an exterior angle in triangle $L B C$, so it is greater than the interior angle $L B C$. According to the condition, angles $L B C$ and $L B A$ are equal, which means in triangle $A B L$, angle $A L B$ is greater than angle $A B L$. But the side opposite the larger angle is the longer side... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,587 |
2. A five-digit number is called humped if its first three digits are in strictly increasing order, and its last three digits are in strictly decreasing order, and it is called ravine if its first three digits are in strictly decreasing order, and its last three digits are in strictly increasing order. Which numbers st... | 2. Answer: ravine.
Solution: We will only consider numbers starting with 5. Between the ravine numbers, the second digit of which is different from 4, and all humpback numbers, there is a one-to-one correspondence according to the formula: $x \leftrightarrow 109999$ - x. Ravine numbers starting with 54 are extra.
Ins... | ravine | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,588 |
3. A hexagon is formed from 3 yellow and 3 blue sticks, such that the colors of the sticks alternate along its perimeter. From any three consecutive sticks, a triangle can be formed. Prove that a triangle can also be formed from the sticks of one of the colors. | 3. Proof: Let the sticks lying in a circle be: $a, b, c, d, e, f$ (where $a, c, e$ are blue).
For simplicity, let their lengths be denoted by the same letters.
Suppose, for definiteness, that $a$ is the largest blue stick, and the largest yellow stick lies next to $a$. Let this be $b$.
We have $d+c > b$ (by conditio... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,589 |
4. Among 25 coins, two are counterfeit. There is a device into which two coins can be placed, and it will show how many of them are counterfeit. Can both counterfeit coins be identified in 13 tests. | 4. Answer: Yes.
Solution: The first 11 trials, we place new coins in the device each time.
If we get an answer of "2" at least once, the goal is achieved.
If we get an answer of "1" twice, we place one coin from the suspicious pair and one genuine coin (such coins are identified by this point) in the device. This tr... | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,590 |
5. We consider all possible $100-$-digit numbers, where each digit is equal to 1 or 2. For each, the remainder when divided by 1024 is calculated. How many different remainders are there among these? | 5. Solution: Subtract the number $11 \ldots 1$ (one hundred ones) from all the numbers. The numbers under consideration have turned into numbers consisting of zeros and ones from 0 to $11 \ldots 1$. We will show that if two of these numbers have differences in the last 10 digits, then they have different remainders whe... | 1024 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,591 |
11.1 In triangle $\mathrm{ABC}$, angle $\mathrm{A}$ is the largest, and angle $\mathrm{C}$ is the smallest. Which of the vertices A, B, C is closest to the center of the inscribed circle? | Solution: See Fig.
$$
\begin{aligned}
& \frac{r}{A O}=\sin \frac{A}{2} \\
& A O=\frac{r}{\sin \frac{A}{2}}
\end{aligned}
$$
Similarly, $B O=\frac{r}{\sin \frac{B}{2}}, \quad C O=\frac{r}{\sin \frac{C}{2}}$.
(y+z)(z+x)}{x y z}$. Justify that there can be no other values. | Solution: Let $S=x+y+z$. Then $\frac{x+y}{z}=\frac{S-z}{z}=\frac{S}{z}-1, \frac{y+z}{x}=\frac{S}{x}-1$, $\frac{z+x}{y}=\frac{S}{y}-1$, and the condition reduces to the equality: $\frac{S}{z}=\frac{\frac{S}{S}}{y}=\frac{S}{x}$. From this, either $x=y=z$, and then the desired expression equals 8, or $S=0$, and then it eq... | 8or-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,598 |
9.3. In a right triangle $A B C$ on the hypotenuse $A B$, a point $E$ is chosen such that $A C=C E$. The angle bisectors $C L$ and $E K$ of triangle $B C E$ intersect at point $I$. It is known that triangle $I K C$ is isosceles. Find the ratio $C L: A B$. | Solution: Angle $B$ must be less than angle $A$, otherwise there will be no point $E$ on the hypotenuse $AB$ with the property $AC = CE$. Let $\angle B = \beta$. Let, moreover, $\angle ECB = \gamma$, and $\angle CEB = \delta$. Clearly, $\beta + \gamma + \delta = 180^{\circ}$. The equalities $\angle CAE = 90^{\circ} - \... | 0.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,599 |
9.4. Six different natural numbers from 6 to 11 are placed on the faces of a cube. The cube was rolled twice. The first time, the sum of the numbers on the four side faces was 36, and the second time it was 33. What number is written on the face opposite the one with the number 10? Justify your answer. | Solution: The sum of the numbers on all 6 faces of the die is 51. Therefore, the sum of the numbers on the top and bottom faces during the first roll is 15, and during the second roll, it is 18. The number 15 can be obtained (as the sum of two different integers from the interval [6; 11]) in two ways: 15 = 9 + 6 = 8 + ... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,600 |
9.5. Prove that
$$
\frac{2}{1} \cdot \frac{4}{3} \cdot \frac{6}{5} \cdot \ldots \cdot \frac{98}{97} \cdot \frac{100}{99}>10
$$ | Solution: Let
$$
A=\frac{2}{1} \cdot \frac{4}{3} \cdot \cdots \cdot \frac{98}{97} \cdot \frac{100}{99}, \quad B=\frac{3}{2} \cdot \frac{5}{4} \cdot \cdots \cdot \frac{97}{96} \cdot \frac{99}{98}
$$
Then
$$
A B=\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \ldots \cdot \frac{99}{98} \cdot \frac{100}{99}=100
$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 14,601 |
9.1. On a line, several points were marked. Then, between each pair of neighboring points, one more point was marked, and this operation was repeated once more. As a result, 101 points were obtained. How many points were marked initially? | 9.1. Answer: 26. Solution. Let there initially be $k$ points marked. Then $k-1$ more points were added to them (one between the first and second, second and third, $\ldots, k-1$-th and $k$-th marked points), and then another $(k+(k-1))-1=$ $2 k-2$ points. In total, the number of points became $4 k-3$. Solving the equat... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,603 |
9.2. There is a fraction $\frac{1}{3}$. Every second, 1 is added to its numerator, and 7 is added to its denominator. An Eastern belief states: the end of the world will come at the moment when the resulting fraction is divisible by 11. Prove that there is no need to fear the end of the world. | 9.2. Solution. After $n$ seconds, the fraction will have the form $\frac{1+n}{3+7 n}$. Suppose this fraction is reducible by 11, i.e., the numbers $a=1+n$ and $b=3+7 n$ are divisible by 11. But then the number $7 a-b$ should also be divisible by 11, which is not true, since $7 a-b=4$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,604 |
9.3. Perpendiculars $B E$ and $D F$, dropped from vertices $B$ and $D$ of parallelogram $A B C D$ to sides $A D$ and $B C$ respectively, divide the parallelogram into three parts of equal area. On the extension of diagonal $B D$ beyond vertex $D$, segment $D G$ is laid off, equal to segment $B D$. Line $B E$ intersects... | 9.3. Answer: $1: 1$. Solution. By the condition $(A E \cdot B E): 2=E D \cdot B E$, hence $A E=2 E D$. Note that $A D$ is the median of triangle $A B G$. Therefore, the segment $B H$, dividing the median $A D$ in the ratio $A E: E D=2$, is also a median of triangle $A B G$. | 1:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,605 |
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