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3. At present, the exchange rates of the US dollar and the euro are as follows: $D=6$ yuan and $E=7$ yuan. The People's Bank of China determines the yuan exchange rate independently of market conditions and adheres to a policy of approximate equality of currencies. One bank employee proposed the following scheme for ch... | # Solution.
Note that if $D$ and $E$ have different parity, their sum is odd, and if they have the same parity, their sum is even. The second number in the pair is always odd. Initially, we have a pair $О Н$ (Odd, Odd).
After one year
After two years $\quad \mathrm{H}+\mathrm{H}=$ О, Н. And so on.
That is, after an... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,749 |
4. Given an acute-angled triangle $A B C$. The feet of the altitudes $B M$ and $C N$ have perpendiculars $M L$ to $N C$ and $N K$ to $B M$. Find the angle at vertex $A$, if the ratio $K L: B C=3: 4$. | # Solution.
We will prove that triangles $N A M$ and $C A B$ are similar with a similarity coefficient of $\cos A$. The ratio of the leg $A M$ to the hypotenuse $A B$ is $\cos A$. Similarly, $A M / A C=\cos A$.
Since angle $A$ is common to both triangles
=x^{2}+a_{i} x+b_{i}, i=1, \ldots, 2016$ only one quadratic polynomial $\mathrm{P}_... | Solution: If $a_{1}=3, a_{2}=1, \ldots, a_{2016}=\frac{1}{3^{2016}}, b_{1}=1, b_{2}=2, \ldots, b_{2016}=2016$, then $k=1$. If $a_{1}=\frac{1}{3^{2016}}, \quad a_{2}=\frac{1}{3^{2015}}, \ldots, \quad a_{2015}=1, \quad a_{2016}=3, \quad b_{1}=2016, b_{2}=2015, \ldots, \quad b_{2015}=2, \quad b_{2016}=1$, then $k=2016$.
... | 1;2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,751 |
1. Does there exist a natural number $n$, greater than 1, such that the value
of the expression $\sqrt{n \sqrt{n \sqrt{n}}}$ is a natural number? | Answer: It exists.
Solution. For example, $\mathrm{n}=2^{8}$. | 256 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,752 |
2. What is the minimum number of factors that need to be crossed out from the number 99! (99! is the product of all numbers from 1 to 99) so that the product of the remaining factors ends in 2? | Answer: 20.
Solution. From the number 99! all factors that are multiples of 5 must be removed, otherwise the product will end in 0. There are 19 such factors in total.
The product of the remaining factors ends in 6. Indeed, the product $1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9$ ends in 6, and similar... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,753 |
3. Sasha marked several cells in an $8 \times 13$ table such that in any $2 \times 2$ square, there was an odd number of marked cells. Then he marked a few more cells, as a result of which in each $2 \times 2$ square, there became an even number of marked cells. What is the smallest total number of cells that Sasha cou... | Answer: 48.
Solution. See example in the figure (the digit 1 is in the cells marked the first time, the digit 2 - the second time)
Estimate. In the table, 24 independent 2x2 squares can be placed. In the first round, at least one cell in each of them was marked. Since each of them ended up with an odd number of marke... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,754 |
4. For what values of $x$ and $y$ is the equality $x^{2}+(1-y)^{2}+(x-y)^{2}=1 / 3$ true? | Answer: $\mathrm{x}=1 / 3, \mathrm{y}=2 / 3$.
Solution. After expanding the brackets and combining like terms, we get $0=2 x^{2}+2(1-y)^{2}+2(x-y)^{2}-2 / 3=4 x^{2}+4 y^{2}-4 x y-4 y+4 / 3=(2 x-y)^{2}+3(y-2 / 3)^{2}$. From here $2 x$ $=y=2 / 3$. | 1/3,2/3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,755 |
5. Given an increasing sequence of natural numbers, in which any three consecutive numbers form a progression - arithmetic or geometric. It is known that the first two numbers in the sequence are divisible by 4. Prove that there are no prime numbers in the sequence. | Solution. Consider three consecutive numbers a, b, c. Let the numbers a and b be divisible by a prime number p.
If they form a geometric progression, then \( b^2 = a \cdot c \). Then \( b^2 \) is divisible by p, which means b is divisible by p. We get that b and c are divisible by p, so they are not coprime.
If they ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,756 |
9.1. It is known that $a^{2}+b=b^{2}+c=c^{2}+a$. What values can the expression
$$
a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right) ?
$$
take? | Answer: 0.
Solution. From the condition, it follows that $a^{2}-b^{2}=c-b$, $b^{2}-c^{2}=a-c$, and $c^{2}-a^{2}=b-a$. Therefore, $a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right)=a(c-b)+b(a-c)+c(b-a)=0$.
## Grading Criteria:
+ a complete and justified solution is provided
the correct an... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,757 |
9.2. Can the products of all non-zero digits of two consecutive natural numbers differ by exactly 54 times? | Answer: Yes.
Solution. For example, the numbers 299 and 300 have this property. Indeed, $2 \cdot 9 \cdot 9: 3=54$.
These two numbers are the smallest possible. Other examples can be obtained by choosing any two consecutive numbers ending in 299 and 300.
Evaluation criteria:
+ correct example provided
- only the ans... | Yes | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,758 |
9.3. A circle with center $O$ is inscribed in triangle $A B C$. A point $P$ is chosen on side $A B$, and a point $Q$ is chosen on the extension of side $A C$ beyond point $C$ such that segment $P Q$ is tangent to the circle. Prove that $\angle B O P = \angle C O Q$. | Solution. First method. The center of the circle inscribed in an angle lies on its bisector. Applying the theorem about the external angle to triangle $B O P$ (see Fig. 9.3), we get: $\angle B O P=\angle A P O-\angle A B O=\frac{1}{2} \angle A P Q-\frac{1}{2} \angle A B C$. Similarly, for triangle $C O Q: \angle C O Q=... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,759 |
9.4. From Zlatoust to Miass, a "GAZ", a "MAZ", and a "KAMAZ" set off simultaneously. The "KAMAZ", having reached Miass, immediately turned back and met the "MAZ" 18 km from Miass, and the "GAZ" - 25 km from Miass. The "MAZ", having reached Miass, also immediately turned back and met the "GAZ" 8 km from Miass. What is t... | Answer: 60 km.
Solution. Let the distance between the cities be $x$ km, and the speeds of the trucks: "GAZ" $-g$ km/h, "MAZ" - $m$ km/h, "KAMAZ" $-k$ km/h. For each pair of vehicles, we equate their travel time until they meet. We get $\frac{x+18}{k}=\frac{x-18}{m}, \frac{x+25}{k}=\frac{x-25}{g}$ and $\frac{x+8}{m}=\f... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,760 |
9.5. Square $A B C D$ and isosceles right triangle $A E F$ $\left(\angle A E F=90^{\circ}\right)$ are positioned such that point $E$ lies on segment $B C$ (see figure). Find the angle $D C F$.
---
The square $A B C D$ and the isosceles right triangle $A E F$ with $\angle A E F = 90^{\circ}$ are arranged so that point... | Answer: $45^{\circ}$.
Solution. First method. Let $P$ be the foot of the perpendicular dropped from point $F$ to line $B C$ (see Fig. 9.5a). Since $\angle F E P=90^{\circ}-\angle B E A=\angl... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,761 |
9.6. Waiting for customers, a watermelon seller sequentially weighed 20 watermelons (weighing 1 kg, 2 kg, 3 kg, ..., 20 kg), balancing the watermelon on one scale pan with one or two weights on the other pan (possibly identical). In the process, the seller wrote down on a piece of paper the weights he used. What is the... | Answer: 6 numbers.
Solution. Let's check that with weights of 1 kg, 3 kg, 5 kg, 7 kg, 9 kg, and 10 kg, one can weigh any of the given watermelons. Indeed, $2=1+1 ; 4=3+1 ; 6=5+1 ; 8=7+1 ; 11=10+1$; $12=9+3 ; 13=10+3 ; 14=9+5 ; 15=10+5 ; 16=9+7 ; 17=10+7 ; 18=9+9 ; 19=10+9 ;$ $20=10+10$. Thus, 6 different numbers could... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,762 |
1. Find at least one root of the equation
$$
\frac{x-36}{x-1}+\frac{x-18}{x-2}+\frac{x-12}{x-3}+\frac{x-9}{x-4}+\frac{x+9}{x+4}+\frac{x+12}{x+3}+\frac{x+18}{x+2}+\frac{x+36}{x+1}=0
$$ | 1. Solution. Let's regroup the terms as follows:
$\frac{x-36}{x-1}+\frac{x+36}{x+1}+\frac{x-18}{x-2}+\frac{x+18}{x+2}+\frac{x-12}{x-3}+\frac{x+12}{x+3}+\frac{x-9}{x-4}+\frac{x+9}{x+4}=$
$\frac{2 x^{2}-72}{x^{2}-1}+\frac{2 x^{2}-72}{x^{2}-1}+\frac{2 x^{2}-72}{x^{2}-9}+\frac{2 x^{2}-72}{x^{2}-16}$
Now it is clear that... | \6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,763 |
2. 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... | 2. Answer: by 8 percent. Solution. Let's take 1 as 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 | 12,764 |
3. Place the numbers 2, 3, 4, 5, 6, 8 in the free cells of the square so that the sum of the three-digit number in the first row and the three-digit number in the second row is equal to the three-digit number in the third row, and the sum of the three-digit number in the first column and the three-digit number in the s... | 3. Answer.
| $\mathbf{1}$ | 5 | $\mathbf{7}$ |
| :--- | :--- | :--- |
| 4 | 8 | 2 |
| 6 | 3 | $\mathbf{9}$ |
Grading Criteria. Correct answer (even without any explanations or checks) - 7 points. In all other cases - 0 points. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,765 |
4. On the base $AB$ of the isosceles trapezoid $ABCD$, a point $P$ is chosen, dividing the base in the ratio $AP: BP = 4: 1$. It turns out that $\angle CPD = \angle PAD$. Find the ratio $PD / PC$. | 4. Answer. 2:1. Solution. Draw segments $P C$ and $P D$. By the property of parallel lines, $\angle P C D = \angle C P B$ and $\angle C D P = \angle A P D$. Then, taking into account the condition, we obtain the similarity of triangles $C P D$ and $A P D$ and the proportion $P D / P C = A P / A D$. Similarly, due to th... | 2:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,766 |
5. A natural number is called good if it can be represented as the product of two consecutive natural numbers. Prove that any good number greater than 6 can be represented as the sum of a good number and a number that is three times a good number. | 5. Solution. The statement of the problem means that for any natural number $a>2$, one can find natural numbers $x, y$ such that the equality $a(a+1)=x(x+1)+3 y(y+1) \Leftrightarrow (a-x)(a+x+1)=3 y(y+1)$ holds. To satisfy the obtained equality, it is sufficient that at least one of the two systems is satisfied: $\left... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,767 |
6. In pentagon $A B C D E \quad A B=A E=C D=1, \quad B C+D E=1$, $\angle A B C=\angle A E D=90^{\circ}$. Find the area of the pentagon. | 6. Answer: 1. Solution. Extend the segment $C B$ beyond point $B$ to segment $B F=D E$, and then draw segments $C A, C E$ and $A F$. It is easy to see that triangles $C D E$ and $A B F$ are equal. Therefore, $C E=A F$ and the areas of the pentagon and quadrilateral $A E C F$ coincide. But triangles $A E C$ and $A F C$ ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,768 |
10.5. Does there exist a natural number $n$ such that for any non-zero digits $a$ and $b$, the number $\overline{a n b}$ is divisible by $\overline{a b}$? (Here, $\overline{x \ldots y}$ denotes the number obtained by concatenating the decimal representations of numbers $x, \ldots, y$.)
(V. Senderov) | Answer. No, it does not exist.
Solution. Suppose such a number $n=\overline{n_{k} n_{k-1} \ldots n_{1}}$ exists. Then $\overline{1 n 2} \vdots 12 \vdots 4$. But a number is divisible by 4 if and only if the number formed by its last two digits is divisible by 4; hence, $\overline{n_{1} 2} \vdots 4$. Similarly, from $\... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,769 |
10.6. Petya and Vasya came up with ten polynomials of the fifth degree. Then Vasya, in turn, called out consecutive natural numbers (starting from some number), and Petya substituted each called number into one of the polynomials of his choice and wrote down the obtained values on the board from left to right. It turne... | Answer: 50 numbers.
Solution. We will show that Petya could substitute no more than five numbers into each polynomial $P(x)$. Indeed, let the $n$-th term of the resulting arithmetic progression be $a n+b$, and the $n$-th of Vasya's consecutive numbers be $k+n$. Then Petya could substitute this number into $P(x)$ if $P... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,770 |
10.7. The circle with center $I$, inscribed in triangle $A B C$, touches the sides $B C, C A, A B$ at points $A_{1}, B_{1}, C_{1}$ respectively. Let $I_{a}, I_{b}, I_{c}$ be the centers of the excircles of triangle $A B C$, touching the sides $B C, C A$, $A B$ respectively. The segments $I_{a} B_{1}$ and $I_{b} A_{1}$ ... | Solution. The lines $B_{1} C_{1}$ and $I_{b} I_{c}$ are parallel because both of these lines are perpendicular to the bisector $A I$ of angle $B A C$. Similarly, $C_{1} A_{1} \| I_{c} I_{a}$ and $A_{1} B_{1} \| I_{a} I_{b}$; hence, triangles $A_{1} B_{1} C_{1}$ and $I_{a} I_{b} I_{c}$ are homothetic.
 polygons, each of which has an area (measured in cm ${ }^{2}$ ) numerically equal to the perimeter (measured in cm)?
(D. Khranchov) | Answer: No.
Solution. A polygon, the area of which (measured in cm $^{2}$) numerically equals its perimeter (measured in cm), is called good.
Notice that the original triangle is good: it is a right triangle with legs of 5 cm and 12 cm, so its area is 30 cm $^{2}$ and numerically coincides with its perimeter, which i... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,774 |
10.3. In city $\mathrm{N}$, 50 city olympiads in different subjects were held, with exactly 30 schoolchildren participating in each of these olympiads, but there were no two olympiads with the same set of participants. It is known that for any 30 olympiads, there is a schoolchild who participated in all these 30 olympi... | Solution. Suppose the opposite, and let there be distinct 30-element subsets $A_{1}, A_{2}, \ldots, A_{50}$ (sets of participants of each Olympiad) in the set of all schoolchildren such that the intersection of any 30 of them is non-empty, while the intersection of all of them is empty.
Suppose among the sets $A_{1}, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,775 |
10.4. Given natural numbers $a, b, c$ such that $a>1, b>c>1$, and the number $a b c+1$ is divisible by $a b-b+1$. Prove that $b$ is divisible by $a$.
(M. Antipov) | Solution. From the condition, it follows that $(a b c+1)-(a b-b+1)=$ $=a b c-a b+b=b(a c-a+1)$ is divisible by $a b-b+1$. Note that $b$ and $a b-b+1=(a-1) b+1$ are coprime, hence we get that $a c-a+1$ is divisible by $a b-b+1$.
Next, we observe that $0(2 a-2) c+2=a c+(a-2) c+$ $+2 \geqslant a c+2>a c-a+1$. Therefore, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,776 |
10.5. In an acute-angled triangle $A B C$, the altitude $B D$ is drawn and the orthocenter $H$ is marked. The perpendicular bisector of the segment $H D$ intersects the circumcircle of triangle $B C D$ at points $P$ and $Q$. Prove that $\angle A P B + \angle A Q B = 180^{\circ}$.
(M. Turevsky, M. Diden) | Solution. Note that $P Q \| C D$, so $P Q$ is the midline of the right triangle $A H D$. Therefore, $P Q$ intersects the hypotenuse $A H$ at its midpoint $M$, so $M A = M D = M H$ (see Fig. 6).
(1-b)=1
$$
Since $a, b, c$ are integers, there are two possible cases:
1) $a-c=1, 1-b=1$, i.e., $b=0$. Substituting into the system, we get $c=94, a=95$.
2) $a-c=-1, 1-b=-1$, i.e., $c=a+1, b=2$.... | =95,b=0,=94or=31,b=2,=32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,780 |
1. (7 points) In a building, on all floors in all entrances there is an equal number of apartments (more than one). Also, in all entrances there are an equal number of floors. The number of floors is greater than the number of apartments per floor, but less than the number of entrances. How many floors are there in the... | Answer: 11.
Solution. Let the number of apartments per floor be K, the number of floors be F, and the number of entrances be E. According to the condition, $1<K<F<E$. The number 715 can be factored into numbers greater than one in only one way: $715=5 \cdot 11 \cdot 13$. Therefore, $K=5, F=11, E=13$.
Criterion. Corre... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,781 |
2. (7 points) Among four people, there are no three with the same first name, or middle name, or last name, but any two have either the same first name, or middle name, or last name. Is this possible? | Answer: It is possible.
Solution. For example, the 1st and 2nd have the name Andrey, the 3rd and 4th have the name Boris. The 1st and 3rd have the surname Ivanov, the 2nd and 4th have the surname Zhukov. The 1st and 4th have the patronymic Petrovich, the 2nd and 3rd have the patronymic Semenovich.
Criteria. Any corre... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,782 |
4. (7 points) Provide an example of an expression consisting of ones, parentheses, and the signs «+» and «X», such that
- its value is 11
- if in this expression all signs «+» are replaced with signs «X», and signs «X» with signs «+», the result is still 11. | Solution. Here is one of the possible examples:
$$
1+1+1+1+1+1+1+1+1+1+1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1
$$
Criteria. Any correct example: 7 points.
Conceptually correct example, not working due to a technical error (something like an extra or missing one in ... | 1+1+1+1+1+1+1+1+1+1+1\times1\times1\times1\times1\times1\times1\times1\times1\times1\times1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,784 |
11.1. Anya left the house, and after some time Vanya left from the same place and soon caught up with Anya. If Vanya had walked twice as fast, he would have caught up with Anya three times faster. How many times faster would Vanya have caught up with Anya (compared to the actual time) if, in addition, Anya walked twice... | Answer: 7.
Solution: Let Ane's speed be $v$, and Vanya's speed be $V$. The distance Ane has walked is proportional to $v$, while Vanya catches up with her at a speed of $u=V-v$. When Vanya doubles his speed, this difference triples, i.e., $u+V=2V-v=3u$. From this, we get $V=2u=2v$. If Vanya's speed doubles and Ane's s... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,786 |
11.2. In a box, there are white and blue balls, with the number of white balls being 8 times the number of blue balls. It is known that if you pull out 100 balls, there will definitely be at least one blue ball among them. How many balls are there in the box? | Answer: 108.
Solution: Since the number of white balls in the box is 8 times the number of blue balls, the total number of balls in the box is divisible by 9. Since 100 balls are drawn from it, the smallest possible number of balls in the box $n=108$, of which 12 are blue and 96 are white. Since there are fewer than a... | 108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,787 |
11.3. Solve the system of equations $\left\{\begin{array}{l}\sqrt{1-3 x}-1=\sqrt{5 y-3 x} \\ \sqrt{5-5 y}+\sqrt{5 y-3 x}=5 .\end{array}\right.$ | Answer. $\left(-\frac{8}{3},-\frac{4}{5}\right)$.
Solution. Introduce new variables $u=\sqrt{1-3 x}, v=\sqrt{5 y-3 x}, w=$ $\sqrt{5-5 y}$.
The system can be rewritten as $\left\{\begin{array}{l}u-1=v, \\ v+w=5\end{array}\right.$ where $u \geq 0, v \geq 0, w \geq 0$.
Notice that $u^{2}-v^{2}-w^{2}=1-3 x-5 y+3 x-5+5 y... | (-\frac{8}{3},-\frac{4}{5}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,788 |
11.4. Given 15 different pairwise coprime natural numbers from the interval $[2,2022]$. Prove that one of these numbers is prime. | Solution. Suppose this is not the case. Let these numbers be denoted by $a_{1}, a_{2}, \ldots, a_{15}$, and their smallest prime divisors by $p_{1}, p_{2}, \ldots, p_{15}$, respectively. Since all the numbers are pairwise coprime, these prime divisors are distinct. The first 15 prime numbers are 2, 3, 5, 7, 11, 13, 17,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,789 |
11.5. Find all values that angle $B$ of triangle $ABC$ can take if it is known that the distance between the bases of the altitudes dropped from vertices $A$ and $C$ is half the radius of the circumscribed circle around this triangle. | Answer: $15^{\circ}, 75^{\circ}, 105^{\circ}, 165^{\circ}$.
Solution. Let $A_{1}$ and $C_{1}$ be the feet of the altitudes from vertices $A$ and $C$ of triangle $A B C$, respectively. Let $\angle B=\beta$, and $R$ be the radius of the circumcircle of $A B C$. Regardless of the type of triangle $A B C$, triangle $A_{1}... | 15,75,105,165 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,790 |
11.1. Prove that for any polynomial $f(x)$ with integer coefficients and integers $a$ and $b$, the number $f(a-\sqrt{b})+f(a+\sqrt{b})$ is an integer. | Solution. The numbers $a-\sqrt{b}$ and $a+\sqrt{b}$ are roots of the polynomial $g(x)$ $=x^{2}-2 a x+\left(a^{2}-b\right)$ with integer coefficients.
Write $f(x)=g(x) \cdot q(x)+c x+d$, where $c$ and $d$ are integers, then
$f(a-\sqrt{b})=c(a-\sqrt{b})+d, f(a+\sqrt{b})=c(a+\sqrt{b})+d$.
Thus, $f(a-\sqrt{b})+f(a+\sqrt... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,791 |
11.2. Prove that for integers $a, b, c, d$ the product $A=(b-a)(c-a)(d-a)(b-c)(d-c)(d-b)$ is divisible by 12. | Solution. Let's divide the set of integers into four classes $\{4 t\}$, $\{4 t+1\},\{4 t+2\},\{4 t+3\}$.
If among the numbers $a, b, c, d$ there are two that belong to the same class, then their difference, and therefore the number $A$, is divisible by 4.
If no two of the numbers $a, b, c, d$ belong to the same class... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,792 |
11.3. Solve the equation $\sqrt{2 x-1}+\sqrt[3]{x}=\sqrt[4]{17-x}$. | Answer: 1.
Solution. It is easy to verify that $x=1$ is a root of the equation.
Since the left side of the equation is an increasing function (as the sum of two increasing functions), and the right side is a decreasing function, there are no other roots.
Comment. The correct answer without proof of its uniqueness - ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,793 |
11.4. The lateral surface of a right circular cone is a circle of radius $R$, from which a sector corresponding to the central angle $(2-\sqrt{3}) \pi$ has been removed. Find the maximum value of the area of a plane section of the cone passing through its vertex. | Answer: $\frac{1}{2} R^{2}$
Solution. The arc of the circle of development is $2 \pi-(2-\sqrt{3}) \pi=\pi \sqrt{3}$. From the ratio $2 \pi r=2 \pi R \cdot \frac{\pi \sqrt{3}}{2 \pi}$, where $r$ is the radius of the base, we get $r=\frac{R \sqrt{3}}{2}$. If $\alpha$ is the angle at the vertex of the axial section, then... | \frac{1}{2}R^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,794 |
11.5. Varya and Miron are playing the following game. There is a pile of $n$ stones on the table. The players take turns, with Varya starting first. On their turn, a player divides any pile that has more than one stone into several equal piles. The player who cannot make a move (when there is exactly one stone in each ... | Answer: Miron wins when $n$ is prime or $n=2^{k}$. For other values of $n$, Varya wins.
Solution. We will call a pile with one stone "unitary," and a pile with a prime number of stones "prime." A prime pile of $p$ stones can only be divided into $p$ unitary piles.
If $n$ is prime, then Varya's only move is to divide ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,795 |
8.1. Prove that for any natural numbers $a$ and $b$ the number $4 a^{2}+4 a b+4 a+2 b+1$ is composite
| Solution. Note that $4 a^{2}+4 a b+4 a+2 b+1=\left(4 a^{2}+4 a+1\right)+2 b+4 a b=$ $=(2 a+1)^{2}+2 b(2 a+1)=(2 a+1)(2 a+2 b+1)$. Since the numbers $a$ and $b$ are natural, each of the brackets is greater than 1. Therefore, the number is composite.
Comment. The factorization is obtained, but it is not verified that bo... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,796 |
8.2. Five friends played several games (not necessarily the same number) of table tennis with each other (there were no draws). After the games, the first noticed that he had 4 more wins than losses; the second and third noticed that each of them had 5 more losses than wins. The fourth noticed that he had as many wins ... | Answer: Could not.
Solution: Suppose the fifth player won all the games. In each game, one player wins, the other loses. Therefore, the number of wins should equal the number of losses. From the statements of the first four players, it follows that they have 6 fewer wins than losses. Thus, the number of wins of the fi... | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 12,797 |
8.3. In triangle $A B C$, the bisector $A L$ is drawn. A point $P$ is taken on side $A C$ such that $L A$ is the bisector of angle $B L P$. Prove that if $B L = C P$, then angle $A B C$ is twice the angle $B C A$. | Solution. From the condition, it follows that triangles $A P L$ and $A B L$ are equal by the second criterion (see Fig. 1). Then $P L=B L$. But by the condition $B L=C P$. Therefore, $C P=P L$. Then $\angle P L C=\angle P C L$, and the external angle $A P L$ of triangle $C P L$ is twice the angle $P C L . \mathrm{C}$ O... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,798 |
8.4. In a chess tournament, each of the 10 players played one game against each other, and Petya came in last place (scored fewer points than any other participant). Then two players were disqualified, and all points earned in matches against them were annulled, and these two players were removed from the table. It tur... | Answer: 4 points.
Solution. In a tournament with 10 players, played in a single round, $\frac{10 \cdot 9}{2}=45$ points are distributed. Therefore, there will be a player who scores no more than $45: 10=4.5$ points. This means that the player who finished in absolute last place scored no more than 4 points. Similarly,... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,799 |
8.5. Given positive numbers $a, b, c, d$. It is known that any two of them differ by no more than 3 times. Prove that $a^{2}+b^{2}+c^{2}+d^{2}<2(a b+a c+a d+b c+b d+c d)$. | Solution. Since the numbers differ by no more than three times, each of them is not greater than the sum of the others. Moreover, if $a \leq b \leq c \leq d$, then $a<b+c+d, b<a+c+d$, $c<a+b+d, d \leq a+b+c$. Multiplying the first inequality by $a$, the second by $b$, the third by $c$, and the fourth by $d$, we get: $a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,800 |
8.1. Fill in the cells of a $3 \times 3$ table with integers such that the sum of all numbers in the table is positive, while the sum of the numbers in any $2 \times 2$ square is negative. | Answer: For example,
| 1 | 1 | 1 |
| :---: | :---: | :---: |
| 1 | -6 | 1 |
| 1 | 1 | 1 |.
## Criterion.
7 points. Any correct example.
Comment. There are several different correct examples. | \begin{pmatrix}\hline1&1&1\\\hline1&-6&1\\\hline1&1&1\\\hline\end{pmatrix} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,801 |
8.2. In the notebook, all irreducible fractions with the numerator 15 are written, which are greater than $\frac{1}{16}$ and less than $\frac{1}{15} \cdot$ How many such fractions are written in the notebook? | Answer: 8 fractions.
Solution. We look for all suitable irreducible fractions of the form $\frac{n}{15}$. Since $\frac{1}{16} < \frac{15}{n}$, then $\frac{15}{225} > \frac{15}{n}$, and $n > 225$. Therefore, $225 < n < 240$. The fraction $\frac{n}{15}$ is irreducible, meaning $n$ is not divisible by 3 or 5. It is not d... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,802 |
8.3. In the Banana Republic, parliamentary elections were held in which all residents participated. All those who voted for the "Mandarin" party love mandarins. Among those who voted for other parties, $90\%$ do not like mandarins (the rest do like them). What percentage of the votes did the "Mandarin" party receive in... | Answer: $40 \%$.
Solution. Let $x \%$ be the percentage of votes for the party "Mandarin". Then the percentage of votes for the other parties is $(100-x) \%$. One tenth of $(100-x) \%$ love mandarins, so we get the equation $x+(100-x) / 10=46$. Solving this, we find that $x=40$.
## Criteria.
5 points. A correct equa... | 40 | Other | math-word-problem | Yes | Yes | olympiads | false | 12,803 |
8.4. A three-digit number, which does not contain any zeros, is written on the board. Petya wrote down in his notebook all different numbers that can be obtained from the original number by rearranging its digits (including the number written on the board). The sum of all numbers written by Petya in his notebook turned... | Answer: $112,121,211,444$.
Solution. If all the digits in the original three-digit number were different, Petya would have written down 6 different three-digit numbers, and their sum would be no less than 600. Therefore, there are two possible cases. 1) If all the digits were the same, Petya wrote down only one origin... | 112,121,211,444 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,804 |
8.5. In an isosceles triangle \(ABC\), the angle \(A\) at the base is \(75^\circ\). The bisector of angle \(A\) intersects side \(BC\) at point \(K\). Find the distance from point \(K\) to the base \(AC\), if \(BK=10\). | Answer: 5.
Solution. In triangle $ABC$, $\angle B = (180^0 - 75^0 - 75^0) = 30^0$. Drop perpendiculars from point $K$: $KH$ to side $AB$, and $KN$ to side $AC$. In the right triangle $HBK$, the hypotenuse $BK$ is 10. Angle $B$ is 30 degrees, and the side opposite to it is half the hypotenuse, so $HK = 5$. Triangles $A... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,805 |
1. Before leaving for the store, Mom asked Anya, Varya, and Svetlana to do some work. The girls work with the same productivity. At first, they worked together, but after 20 minutes, Anya had to leave, and Varya and Svetlana continued working alone. After another 15 minutes, Varya also had to leave, and Svetlana finish... | # Solution
If the entire work is divided into equal parts, each of which one girl completes in 5 minutes, then Anya completed 4 parts (worked for 20 minutes), Varya - 7 parts (worked for 35 minutes), and Svetlana - 9 parts (worked for 45 minutes). In total, there are 20 such parts.
Therefore, the apples should be dis... | Anya:2,Varya:3.5,Svetlana:4.5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,806 |
3. The residents of the Country of Fairy Tales invited Baba Yaga, Koschei, and Zmey Gorynych as honored guests to their celebration. The residents found out the following. If Baba Yaga does not come to the celebration, then Koschei will come. If Koschei does not come, then Zmey Gorynych will come. If Zmey Gorynych does... | # Solution
The condition "If Baba Yaga does not come to the party, then Koschei will come" is equivalent to the statement "It is impossible for neither Baba Yaga nor Koschei to come to the party."
By similar conditions, it is impossible for both Koschei and Zmey Gorynych, or Zmey Gorynych and Baba Yaga, to be absent ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 12,808 |
5. Each point on the plane is painted in one of four colors (each color is used at least once). Are there lines on this plane whose points are painted in at least three colors? Justify your answer. | # Solution
Let's choose any four points of different colors and draw a pair of lines such that exactly two of the chosen points belong to these lines. If the constructed lines intersect, a point colored in one of the chosen colors appears. Thus, on one of the lines, there are three points of different colors. If the c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,810 |
9.1. In a round-robin chess tournament, six people participated: two boys and four girls. Could the boys have scored twice as many points as the girls by the end of the tournament? (In a round-robin chess tournament, each player plays one game against each of the others. A win earns 1 point, a draw 0.5, and a loss 0 po... | Answer: No, they could not.
Solution. In a round-robin tournament with six participants, $\frac{6 \cdot 5}{2}=15$ points are played, and each participant can score no more than five points. If the boys had scored twice as many points as the girls, they would have scored 10 points in total, meaning 5 points each. Howev... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,811 |
9.2. For the coefficients $a, b, c$ and $d$ of the two quadratic trinomials $x^{2}+b x+c$ and $x^{2}+a x+d$, it is known that $0<a<b<c<d$. Can these trinomials have a common root? | Answer: No, they cannot.
Solution. Since the coefficients of both trinomials are positive, their roots cannot be positive (this can be derived from Vieta's theorem or by direct substitution).
Let $x_{0}$ be a common root of these trinomials. Then $x_{0}^{2}+b x_{0}+c=0$ and $x_{0}^{2}+a x_{0}+d=0$, hence $x_{0}^{2}+b... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,812 |
9.3. Given a triangle $A B C$. A line parallel to $A C$ intersects sides $A B$ and $B C$ at points $P$ and $T$ respectively, and the median $A M$ at point $Q$. It is known that $P Q=3$, and $Q T=5$. Find the length of $A C$. | Answer: $AC=11$.
Solution. First method. Draw a line through point $Q$ parallel to $BC$ (where $N$ and $L$ are the points of intersection of this line with sides $AB$ and $AC$ respectively, see Fig. 9.3a). Since $AM$ is the median of triangle $ABC$, then $LQ=NQ$. Moreover, $PT \| AC$, so $PQ$ is the midline of triangl... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,813 |
9.4. The sum of ten natural numbers is 1001. What is the greatest value that the GCD (greatest common divisor) of these numbers can take? | Answer: 91.
Solution. Example. Consider nine numbers equal to 91, and the number 182. Their sum is 1001.
Estimate. We will prove that the GCD cannot take a value greater than 91. Note that $1001=7 \times 11 \times 13$. Since each term in this sum is divisible by the GCD, the GCD is a divisor of the number 1001. On th... | 91 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,814 |
9.5. Quadrilateral $ABCD$ is inscribed. Points $K$ and $L$ are marked on its diagonals $AC$ and $BD$ respectively, such that $AK = AB$ and $DL = DC$. Prove that lines $KL$ and $AD$ are parallel. | Solution. First method. Since quadrilateral $ABCD$ is inscribed, then $\angle BAC = \angle BDC$ (see Fig. 9.5). Then, in isosceles triangles $ABK$ and $DLC$, the angles at the bases are equal, hence $\angle BLC = \angle BKC$, which means that quadrilateral $BCKL$ is inscribed. Thus, $\angle KLO = \angle BCO = \angle BD... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,815 |
9.6. From a chessboard of size $8 \times 8$, a square of size $2 \times 2$ was cut out such that the remaining board could be divided into rectangles of size $1 \times 3$. Determine which square could have been cut out. (List all possible options and prove that there are no others.) | Answer: any of the nine squares marked on figure 9.6v could have been cut out.
Solution. Let's color the chessboard in three colors diagonally, starting from the lower left corner of the board (see fig. 9.6a). Then, when cutting parts of the board into $1 \times 3$ rectangles, each rectangle will contain cells of all ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,816 |
4. Find the smallest four-digit number, the product of all digits of which is equal to 512.
Answer. 1888 | Solution. $512=2^{9}$, which means all the digits of the number are powers of two, i.e., they can be $1, 2, 4$ or 8. Clearly, the first digit of the number should be as small as possible. Let's take the smallest of the possible ones, which is 1. The rest should be equal to 8, otherwise the product will be less than 512... | 1888 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,818 |
5. Andrey thought of 4 integers and told Vanya all their pairwise sums. Vanya remembered only 4 of them accurately: $0, 3, 6$ and 18. About the two remaining sums, Vanya only remembered that they were two-digit natural numbers. What numbers did Andrey think of? | Answer: $-6 ; 6 ; 9 ; 12$.
Solution. Let the numbers (in ascending order) be a; b; c; d. Considering that the two remaining sums are two-digit, the minimum sums are 0, 3, and 6. Therefore, $\mathrm{a}+\mathrm{b}=0 ; \mathrm{a}+\mathrm{c}=3$, and either $\mathrm{a}+\mathrm{d}=6$ or $\mathrm{b}+\mathrm{c}=6$. From the f... | -6;6;9;12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,819 |
11.1 In a $3 \times 3$ table, numbers from 1 to 9 (each used once) were placed, and then 6 products of numbers in rows and columns were calculated. Could it happen that three of them coincided | Solution: Yes, it could, for example, like this:
| 5 | 7 | 2 |
| :--- | :--- | :--- |
| 6 | 3 | 4 |
| 1 | 8 | 9 |
## Criteria:
- The correct example is provided and it is shown which products are equal, -7 points;
- The example is correct, but it is not shown which products are equal, -6 points;
- Attempts to explai... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,820 |
# 11.2 Solve the equation
$$
\frac{1}{\sqrt{x+20}+\sqrt{x+21}}+\frac{1}{\sqrt{x+21}+\sqrt{x+22}}=1
$$ | Solution 1: By multiplying both fractions by their conjugates, we get $\sqrt{x+22}-\sqrt{x+20}=1$. Multiplying this by its conjugate and transforming, we get $\sqrt{x+22}+\sqrt{x+20}=2$. Adding the two equations, we get $\sqrt{x+22}=\frac{3}{2}, x=-\frac{79}{4}$. Substituting this number into the original equation, we ... | -\frac{79}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,821 |
11.3 Prove that $1!\cdot 2!\cdot 3!\cdot 4!\cdot 5!\cdot \ldots \cdot 2021!\cdot 2022!$ is not a power of a natural number higher than the first. (Reminder: $n!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot n$) | Solution: A number is a $k$-th power if and only if each prime factor appears in its factorization in a multiple of $k$ power. Note that 1009 and 1013 are prime numbers. The factor 1009 appears once in each of 1009!, 1010!, ..., 2017!, and twice from 2018! to 2022!. In total, the degree of inclusion is $1009 + 5 \cdot ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,822 |
11.4 Let $P_{n}(x)=1+x+x^{2}+\cdots+x^{n}$. Prove that for all positive $x$ the inequality
$$
P_{20}(x) \cdot P_{21}\left(x^{2}\right) \leqslant P_{20}\left(x^{2}\right) \cdot P_{22}(x)
$$
holds. | Solution: When $x=1$, the inequality is true. Let $x \neq 1$, then $P_{n}(x)=\frac{x^{n+1}-1}{x-1}$. Then our inequality is equivalent to
$$
\frac{x^{21}-1}{x-1} \cdot \frac{x^{44}-1}{x^{2}-1} \leqslant \frac{x^{42}-1}{x^{2}-1} \cdot \frac{x^{23}-1}{x-1}
$$
The denominator of the product is positive, so by eliminatin... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,823 |
7.1. Replace each asterisk in the example of adding decimal fractions
$$
0, * *+0, * *+0, * *+0, * *=1
$$
with the digit 2 or the digit 3 so that the equation is correct. | Solution: The sum of the four digits in the hundredths place must be a multiple of 10. Since each of the digits is either 2 or 3, there must be exactly two 2s and two 3s among these digits. Then, the sum of the digits in the tens place should end in the digit 9, so among the tens digits, there are three 2s and one 3. I... | 0.32+0.22+0.23+0.23=1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,824 |
7.2. There are nuts in three boxes. In the first box, there are six nuts fewer than in the other two boxes combined, and in the second box, there are ten fewer nuts than in the other two boxes combined. How many nuts are in the third box? Justify your answer. | Solution: Let there be $x$ nuts in the first box, $y$ and $z$ in the second and third boxes, respectively. Then the condition of the problem is defined by the equations $x+6=y+z$ and $x+z=y+10$. From the first equation, $x-y=z-6$, and from the second, $x-y=10-z$. Therefore, $z-6=10-z$, from which $z=8$.
Answer: 8 nuts... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,825 |
7.3. Between the houses of four dwarfs, straight roads are laid, and no three houses are located on the same straight line. The distance between the houses of Doc and Sony is 5 km, Doc and Grumpy is 4 km, Sony and Simple is 10 km, Grumpy and Simple is 17 km. What can the distance between the houses of Sony and Grumpy b... | Solution: Let's denote the houses of Doc, Sony, Grumpy, and Simple by $D, S, V$, and $P$ - see the figure. Since no three houses are collinear, the points $D, S, V$, and $P$ are the vertices of a quadrilateral (not necessarily convex). From triangle $D C V$, by the triangle inequality, we have $S V V P$, from which $S ... | 12or13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,826 |
7.4. Three students were solving the same test, which consisted of several problems. For each correct answer, a student was given two points, for each incorrect answer, one point was deducted, and if the answer to a problem was not written, zero points were given. Together, these three students scored exactly 100 point... | # Solution:
Method 1. Assume the opposite. Then, if all three solved a problem correctly, the total score for it is 6; if all three solved it incorrectly, the total score is -3; if two solved it correctly, the total score is 3; and if one solved it correctly, the total score is 0. All these numbers $(6, -3, 3, 0)$ are... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 12,827 |
7.5. Misha and Masha had the same multi-digit integer written in their notebooks, ending in 9876. Masha placed a plus sign between the third and fourth digits from the right, while Misha placed a plus sign between the fourth and fifth digits from the right. To the surprise of the schoolchildren, both resulting sums tur... | Solution: Let the written number have the form $\overline{x 9876}$, where $x$ is also some natural number. Then Misha got the sum $x+9876$, and Masha got the sum $10 x+9+876$. From the equality $x+9876=10 x+9+876$, we find that $x=999$.
Answer: 9999876 and there is no other number.
Recommendations for checking:
| is... | 9999876 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,828 |
4. Answer. 190. Solution 1. Among the lines intersecting $y=20-x$, there is the line $y=x$. Moreover, the picture is symmetric with respect to the line $y=x$, so the sum of the abscissas is equal to the sum of the ordinates. Through the point $(19 ; 19)$, $180: 9=20$ lines are drawn, of which 19 intersect the line $y=2... | Solution 2. Through the point $(19 ; 19)$, $180: 9=20$ lines are drawn, of which 19 intersect the line $y=20-x$. Let the line $y=x$ intersect the line $y=20-x$ at point $A$, then $x_{A}=10$. The remaining 18 lines are divided into pairs, intersecting the line $y=20-x$ at symmetric points $B$ and $C$ (triangles $M A B$ ... | 190 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,830 |
Problem 1. The sum of the digits of a natural number $a$ is to the sum of the digits of the number $2a$ as 19 to 9. Prove that the decimal representation of the number $a$ has at least 29 digits. | Solution. From the condition, it follows that the sum of the digits of the number $2a$ is divisible by 9. Therefore, the number $2a$ itself, and consequently the number $a$, and the sum of its digits, are divisible by 9. In addition, the sum of the digits of the number $a$ is divisible by 19. Thus, it is divisible by 1... | 29 | Number Theory | proof | Yes | Yes | olympiads | false | 12,832 |
Problem 2. Prove that the cube of any natural number greater than 1 can be represented as the difference of two squares of natural numbers. | Solution. Let $n^{3}=a^{2}-b^{2}=(a+b)(a-b)$. Try to solve the equation $a+b=$ $n^{2}, a-b=n$. Adding these two equations, we get that $2 a=n^{2}+n$, and subtracting them, we get that $2 b=n^{2}-n$. Therefore, $n^{3}={\frac{n^{2}+n}{2}}^{2}-{\frac{n^{2}-n}{2}}^{2}$. Both terms in the parentheses are natural numbers, si... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,833 |
Problem 2. On a plane, 100 points are marked. It turns out that on two different lines a and b, there are 40 marked points each. What is the maximum number of marked points that can lie on a line that does not coincide with a and b? | Answer: 23. Solution: Lines $a$ and $b$ have no more than one common point. If such a point exists and is marked, then together on lines $a$ and $b$ there are 79 marked points; otherwise, there are 80. Therefore, outside lines $a$ and $b$, there are no more than 21 marked points.
Take an arbitrary line $c$ that does n... | 23 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,834 |
Problem 3. A row of 101 numbers is written (the numbers are not necessarily integers). The arithmetic mean of all the numbers without the first one is 2022, the arithmetic mean of all the numbers without the last one is 2023, and the arithmetic mean of the first and last numbers is 51. What is the sum of all the writte... | Answer. 202301. Solution. Let the sum of all numbers be $S$, the first number be $a$, and the last number be $b$. According to the conditions, $(S-a) / 100=2022, \quad(S-b) / 100=2023$, $(a+b) / 2=51$, from which we get $S-a=2022 \cdot 100, S-b=2023 \cdot 100, a+b=51 \cdot 2$. Adding these three equations, we get $2 S=... | 202301 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,835 |
Problem 5. Do there exist three quadratic trinomials, each of which has two roots, the sum of any two of them is one root, and the sum of all three has no roots? | Answer. They exist. Solution. For example, $x^{2}-x, x^{2}+x,-x^{2}+2 x+3$.
- How was this example devised? The fact that the coefficient of $x^{2}$ is 1 for two of the trinomials and -1 for one ensures the uniqueness of the root for the two sums that define linear functions. The uniqueness of the root for the third s... | notfound | Algebra | proof | Yes | Yes | olympiads | false | 12,836 |
1. In an online store, two types of New Year's gifts are sold. The first type of gift contains a toy, 3 chocolate, and 15 caramel candies and costs 350 rubles. The second type of gift contains 20 chocolate and 5 caramel candies and costs 500 rubles. Eugene wants to buy an equal number of caramel and chocolate candies, ... | Answer: 3750 rubles.
Solution. Consider integer values $m$ and $n$ - the quantities of purchased gift sets of candies of the 1st and 2nd types, respectively. These quantities must satisfy the conditions of the problem: the total number of caramel and chocolate candies in them is the same, and this number is the smalle... | 3750 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,837 |
2. Do there exist 2023 integers such that their product is 2023 and their sum is zero? | Answer: No, it does not exist.
Solution. We will prove that this is impossible. Consider a set of 2023 integers. Since the number of integers is odd, their sum is even if and only if the number of odd integers in the sum is even. Consider the possible divisors of the number 2023 - from them, the necessary product can ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,838 |
3. It is known that the quadratic trinomial $x^{2}+b x+c$ has two distinct roots. If we add the coefficients $b$ and $c$ and the two roots (four numbers), we get the number -3, and if we multiply these same four numbers, we get the number 36. Find all such quadratic trinomials. | Answer: $x^{2}+4 x-3$, the only trinomial.
Solution. Consider the roots of the quadratic trinomial $x^{2}+b x+c$. Let's denote them as $x_{1}$ and $x_{2}$. Then, according to the problem, we have the following pair of relations:
$$
\left\{\begin{array}{l}
b+c+x_{1}+x_{2}=-3 \\
b \cdot c \cdot x_{1} \cdot x_{2}=36
\en... | x^{2}+4x-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,839 |
4. There is a checkered board of size $22 \times 23$. Each cell of the board has one checker. In one move, you can choose any two checkers and move each of them to an adjacent cell. If at least two checkers end up on the same cell, you can remove exactly two checkers from that cell. Is it possible to remove all the che... | Answer: No, it is impossible.
Solution. We will prove by contradiction that it is impossible to remove all checkers from the board. Suppose it is possible to remove all checkers from the board after several moves. We will color all the cells of the board in black and white in a checkerboard pattern. For definiteness, ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,840 |
6. The school dance studio calculated that this year they have already performed the dance "Circle Dance" 40 times, and in each performance, exactly 10 people participated, and any two dancers performed together no more than once. Prove that the studio has at least 60 dancers. | Solution. First solution. According to the problem, any two dancers could meet at no more than one performance. We will consider such two dancers as a pair. Consider any performance: from 10 participants of the performance, no more than $\frac{10 \cdot 9}{2}=45$ pairs can be formed (the number of "handshakes," $C_{10}^... | 60 | Combinatorics | proof | Yes | Yes | olympiads | false | 12,842 |
1. In the school, there are boys and girls. The average age of the boys differs from the average age of the girls, but the average of these two numbers coincides with the average age of all students. Who is more in the school - boys or girls? | Answer: equally.
Solution. Let $m$ be the number of boys and $d$ be the number of girls in the school, and let the sum of the ages of all boys be $M$, and the sum of the ages of all girls be $D$. Then the average age of all boys is $\frac{M}{m}$, the average age of all girls is $\frac{D}{d}$, and the average age of al... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,843 | |
2. Prove that for any non-negative numbers $x$ and $y$ the following inequality holds:
$$
2^{x} x+2^{y} y \geqslant 2^{y} x+2^{x} y
$$ | Solution. The difference between the left and right sides of the inequality can be easily factored, it is equal to $(x-y)\left(2^{x}-2^{y}\right)$. The function $f(x)=2^{x}$ is increasing for $x \geqslant 0$, so if $x \geqslant y$, then $2^{x} \geqslant 2^{y}$, and conversely, if $y \geqslant x$, then $2^{y} \geqslant ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,844 |
3. Points $O$ and $I$ are the centers of the circumcircle and incircle of triangle $ABC$, and $M$ is the midpoint of the arc $AC$ of the circumcircle (not containing $B$). It is known that $AB=15, BC=7$, and $MI=MO$. Find $AC$. | Answer: $A C=13$.
Solution. (Fig. 5). First, we will prove that $M I=M A$ (trident lemma).
Indeed, the external angle $A I M$ of triangle $A I B$ is equal to the sum of angles $B A I$ and $A B I$, and since $A I$ and $B I$ are angle bisectors, $\angle A I M=\frac{1}{2} \angle A+\frac{1}{2} \angle B$. Angle $I A M$ is... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,845 |
4. Find all pairs of natural numbers $a$ and $b$ such that $3^{a}+4^{b}$ is a perfect square of an integer. | Answer: $a=b=2$.
Solution. From the equality $3^{a}+4^{b}=n^{2}$, it is clear that the number $n$ is odd, and therefore $n=2 x+1$. Then the original equality can be written as: $3^{a}+4^{b}=(2 x+1)^{2}=4 x^{2}+4 x+1$, from which it follows that the number $3^{a}$ has a remainder of 1 when divided by 4. The latter is o... | =b=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,846 |
5. Given $n$ distinct positive numbers. Sums are formed from them with any number of addends from 1 to $n$.
a) What is the smallest number of different sums that can be obtained
b) What is the largest number of different sums that can be obtained | Answer: a) $\frac{1}{2} n(n+1)$; b) $2^{n}-1$.
Solution. We can assume that the original positive numbers are arranged in ascending order: $a_{1}<a_{2}<\ldots<a_{n}$. Consider the numbers
\... | \frac{1}{2}n(n+1) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,847 |
3. Solve the system of equations
$$
\left\{\begin{array}{l}
x^{3}+2 x=y^{3}+2 y \\
x^{2014}+y^{2014}=2^{2015}
\end{array}\right.
$$ | Solution. The function $f(t)=t^{3}+2 t$ is monotonically increasing on the entire number line as the sum of two monotonically increasing functions $f_{1}(t)=t^{3}$ and $f_{2}(t)=2 t$. Therefore, from the first equation we have $f(x)=f(y) \Leftrightarrow x=y$. The equality $X=Y$ can also be obtained by factoring the exp... | \2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,849 |
4. For what least natural $\mathrm{k}$ does the quadratic trinomial
$$
y=k x^{2}-p x+q \text { with natural coefficients } p \text{ and } q \text{ have two }
$$
distinct positive roots less than 1? | Solution. First, let's estimate the coefficient $\mathrm{k}_{\text{from below. Let }} 0k^{2} x_{1}\left(1-x_{1}\right) x_{2}\left(1-x_{2}\right) \geq 1 \Rightarrow k>4 \text{. Thus, we have obtained the }$
$$
$$
lower bound. To show the accuracy of the estimate, it is sufficient to construct an example with $k=5: y=5 x... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,850 |
5. Point $Q$ lies outside the circle $\omega_{1} \cdot Q A$ and $Q B$ are tangents to the circle ( $A$ and $B$ belong to $\omega_{1}$ ). Through points $A$ and $B$, a second circle $\omega_{2}$ is drawn with its center at point $Q$. On the arc $A B$ of circle $\omega_{2}$, which lies inside circle $\omega_{1}$, an arbi... | Let $\angle A Q B = A \breve{K K} = \alpha$. Then
$$
\angle A K B = \frac{360^{\circ} - \alpha}{2} = 180^{\circ} - \frac{\alpha}{2}
$$
From $\triangle \mathrm{AQ}$, we have $\angle \mathrm{Q} A B = \angle \mathrm{QBA} = 90^{\circ} - \frac{\alpha}{2}$.
On the other hand, $\angle Q A B$ is the angle between the tangen... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,851 |
1. A string of 5 consecutive natural numbers is written. Is it possible that the sum of the digits of the first number is 52, and that of the fifth number is 20? (If it is possible, provide an example of such numbers; if it is impossible, explain why it is impossible) | Answer: possibly.
Solution. Example $889999, \ldots, 890003$. | 889999,\ldots,890003 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,853 |
2. The magpie-crow was cooking porridge, feeding her chicks. The third chick received as much porridge as the first two combined. The fourth chick received as much as the second and third. The fifth chick received as much as the third and fourth. The sixth chick received as much as the fourth and fifth. The seventh chi... | Answer: 40 g.
Solution. Let the first chick receive a grams of porridge, the second - b grams, Then the others received $\mathrm{a}+\mathrm{b}, \mathrm{a}+2 \mathrm{~b}, 2 \mathrm{a}+3 \mathrm{~b}, 3 \mathrm{a}+5 \mathrm{~b}, 0$ grams respectively. In total, they received $8 \mathrm{a}+12 \mathrm{~b}$ grams of porridg... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,854 |
3. Petya and Vasya are playing the following game. Petya has 100 cards, each with a unique number from 1 to 100. On each turn, Petya lays out two cards, after which Vasya immediately takes one of them. After 50 turns, Petya will run out of cards, and 50 cards will remain on the table. Petya's goal is to make the sum of... | Answer: can.
Solution. Let Petya lay out x pairs of even-even type, y pairs of odd-odd type, then he laid out $50-\mathrm{x}-\mathrm{y}$ pairs of even-odd type. The number of even numbers is $2 \mathrm{x}+(50-\mathrm{x}-\mathrm{y})=50+\mathrm{x}-\mathrm{y}$, and the number of odd numbers is $50-x+y$. Since there are 5... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,855 |
5. Among the numbers from 1 to $10^{23}$, which are more numerous - those with a two-digit sum of digits or those with a three-digit sum? Answer: those with a three-digit sum. | Solution. The number $10^{23}$ has a digit sum of 1, so we will not consider it. We will write numbers as 23-digit numbers, adding leading zeros. We will call digits "complementary" if their sum is 9. To each number with a two-digit digit sum, we will correspond a number by replacing each digit with its "complementary"... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,856 |
Problem 4.2. Petya has 25 coins, each with a denomination of $1, 2, 5$, or 10 rubles. Among these coins, 19 are not two-ruble coins, 20 are not ten-ruble coins, and 16 are not one-ruble coins. How many five-ruble coins does Petya have? | Answer: 5.
Solution. Since among Petya's coins, 19 are not two-ruble coins, Petya has a total of 6 two-ruble coins. Similarly, it follows that Petya has 5 ten-ruble coins and 9 one-ruble coins.
Thus, Petya has $25-6-5-9=5$ five-ruble coins. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,858 |
Problem 4.4. In the cafeteria queue, there are five schoolchildren: Anya, Borya, Vera, Gena, and Denis.
- Borya stands at the beginning of the queue.
- Vera stands next to Anya, but not next to Gena.
- Among Anya, Borya, and Gena, no two stand next to each other.
Who stands next to Denis? | Answer: Anya and Gena.
Solution. Let's number the places in the queue from 1 to 5: place №1 - the place at the beginning of the queue, place №5 - at the end of the queue.
From the first condition, we understand that Borya is standing at place №1, and from the third condition, it becomes clear that Anya and Gena are s... | AnyaGena | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,860 |
Problem 4.5. Anton guessed a three-digit number, and Lesha is trying to guess it. Lesha sequentially named the numbers 109, 704, and 124. Anton noticed that each of these numbers matches the guessed number in exactly one digit place. What number did Anton guess? | Answer: 729.
Solution. Note that the first and third numbers have a common hundreds digit 1, the first and second have a common tens digit 0, and the second and third have a common units digit 4.
Suppose the first digit of the guessed number is 1. Then the first and third numbers no longer have common digits with the... | 729 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,861 |
Problem 4.6. Fill in the letters $A, B, C, D, E$ with the digits $1,2,3,4,5$ so that the sum of the digits in all $1 \times 3$ rectangles (both horizontal and vertical) equals 13. Each of the digits from 1 to 5 should appear in the table exactly once.
| $\mathbf{7}$ | $A$ | |
| :---: | :---: | :---: |
| $B$ | $\mathb... | Answer: a3 b5 c2 d1 e4.
Solution. The sum of all numbers in the table is $1+2+3+4+5+6+7+8=36$, and in each of the first two columns it is 13. Therefore, the total sum of the numbers in the table is $36=13+13+(8+C)$, from which $C=2$.
In the second row, the sum is $13=B+6+C=B+6+2$, from which $B=5$.
In the first colu... | a3b5c2d1e4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,862 |
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