problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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|---|---|---|---|---|---|---|---|---|---|
13. In a trapezoid, the lengths of the diagonals are 6 and 8, and the length of the midline is 5. Find the height of the trapezoid. | Answer: $4.8$.
Solution: Parallel to the bases AD and $BC$, translate one of the diagonals BD by the length of BC. We obtain a triangle $ACE$ with sides $6, 8, 10$, which, by the converse of the Pythagorean theorem, is a right triangle. Writing the area in two ways: $S=\frac{1}{2} AC \times CE = \frac{1}{2} AE \times ... | 4.8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,322 |
2. Let's call an integer "extraordinary" if it has exactly one even divisor other than 2. How many extraordinary numbers exist in the interval $[1 ; 75]$? | Answer: 12
Solution: This number should be equal to a prime multiplied by 2. There are 12 such numbers:
$\begin{array}{llllllllllll}2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & 23 & 29 & 31 & 37\end{array}$ | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,324 |
4. A pirate schooner has taken a merchant ship by boarding. Ten pirates did not participate in the fight, while the rest lost either a hand, a leg, or both a hand and a leg in the battle. $54\%$ of the participants in the battle lost a hand, and $34\%$ lost both a hand and a leg. It is known that $2/3$ of all the pirat... | Answer: 60 pirates.
Solution - similar to option v3a. | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,326 |
5. In a regular 1000-gon, all diagonals are drawn. What is the maximum number of diagonals that can be selected such that among any three of the selected diagonals, at least two have the same length? | Answer: 2000
Solution: For the condition of the problem to be met, it is necessary that the lengths of the diagonals take no more than two different values. The diagonals connecting diametrically opposite vertices are 500. Any other diagonal can be rotated to coincide with a diagonal of the corresponding length, i.e.,... | 2000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,327 |
6. In how many ways can the number 1024 be factored into three natural factors such that the first factor is divisible by the second, and the second is divisible by the third? | Answer: 14
Solution: Note that the factors have the form $2^{a} \times 2^{b} \times 2^{c}$, where $\mathrm{a}+\mathrm{b}+\mathrm{c}=10$ and $a \geq b \geq c$. Obviously, c is less than 4, otherwise the sum would be greater than 10. Let's consider the cases:
$c=0)$ Then $b=0, \ldots, 5, a=10-b-6$ options
c=1) Then $b... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,328 |
7. In a trapezoid, the diagonals of which intersect at a right angle, it is known that the midline is 6.5 and one of the diagonals is 12. Find the second diagonal
# | # Answer 5
Solution: Let's parallel translate one of the diagonals so that it forms a right-angled triangle with the other. Then, in this triangle, one leg is 12, and the hypotenuse is 13, so the remaining leg is 5.
## Option $2-$ - | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,329 |
1. Find two positive irreducible fractions with denominators not exceeding 100, the sum of which is equal to 86/111. | Answer: $2 / 3+4 / 37$
Solution: $111=37 * 3$, i.e., one fraction should have a denominator of 37, and the other should have a denominator of 3. It is clear that the numerator of the second fraction can be 1 or 2. 1/3 does not work, but for $2 / 3$ we get $4 / 37$ | 2/3+4/37 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,330 |
3. A rectangle $A D E C$ is described around a right triangle $A B C$ with legs $A B=5$ and $B C=6$, as shown in the figure. What is the area of $A D E C$? | # Answer 30.
Solution: The area of the rectangle is twice the area of the triangle. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,332 |
4. Vasya is coming up with a 4-digit password for a code lock. He doesn't like the digit 2, so he doesn't use it. Moreover, he doesn't like when two identical digits are next to each other. Additionally, he wants the first digit to match the last digit. How many options need to be tried to guarantee guessing Vasya's pa... | Answer: 504
Solution: The password must have the form ABCA, where A, B, C are different digits (not equal to 2). They can be chosen in $9 * 8 * 7=504$ ways.
Comment for graders: Half a point can be given to those who consider that the first digit cannot be zero. | 504 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,333 |
5. In the Empire of Westeros, there were 1000 cities and 2017 roads (each road connected some two cities). From any city, you could travel to any other. One day, an evil wizard enchanted $N$ roads, making it impossible to travel on them. As a result, 7 kingdoms formed, such that within each kingdom, you could travel fr... | Answer: 1024.
Solution: Suppose the evil wizard enchanted all 2017 roads. This would result in 1000 kingdoms (each consisting of one city). Now, imagine that the good wizard disenchants the roads so that there are 7 kingdoms. He must disenchant at least 993 roads, as each road can reduce the number of kingdoms by no m... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,334 |
6. Jack Sparrow needed to distribute 150 piastres into 10 purses. After placing a certain number of piastres in the first purse, he put more in each subsequent purse than in the previous one. As a result, it turned out that the number of piastres in the first purse was not less than half the number of piastres in the l... | # Answer: 16
Solution: Let there be $x$ piastres in the first purse. Then in the second there are no less than $\mathrm{x}+1$, in the third - no less than $\mathrm{x}+2$... in the 10th - no less than $\mathrm{x}+9$. Thus, on one side $x+$ $x+1+\cdots+x+9=10 x+45 \leq 150$, From which $x \leq 10$. On the other side $x ... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,335 |
7. Find the smallest natural number $N$, such that the decimal representation of the number $N \times 999$ consists entirely of sevens. | Answer 778556334111889667445223
Solution: $N \times 999=77 \ldots 7$, then $N$ is a multiple of 7, denote $n=N / 7$. We get $999 n=$ $1000 n-n=11 \ldots 1$, so $1000 n-111 \ldots 1=n$. Write it as a column subtraction and repeat the found digits of $\mathrm{N}$ with a shift of 3 to the left $* * * * * * * 000$
111111... | 778556334111889667445223 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,336 |
1. How many natural numbers from 1 to 2017 have exactly three distinct natural divisors? | Answer: 14.
Solution: Only squares of prime numbers have exactly three divisors. Note that $47^{2}>2017$, so it is sufficient to consider the squares of prime numbers from 2 to 43. There are 14 of them. | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,337 |
2. Vovochka approached a slot machine, on the screen of which the number 0 was glowing. The game rules stated: «The screen shows the number of points. If you insert a 1-ruble coin, the number of points will increase by 1. If you insert a 2-ruble coin, the number of points will double. If you score 50 points, the machin... | Solution: Let's try to solve it from the end - how to get the number 1 from 50 with the least amount of rubles, if you can only divide by 2 and subtract 1. We get: 50>25->24->12->6->3->2->1. That is, it will require 4 two-ruble and 3 one-ruble coins. It is obvious that if you use 3 two-ruble coins and fewer than 5 one-... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,338 |
3. Petya is coming up with a password for his smartphone. The password consists of 4 decimal digits. Petya wants the password not to contain the digit 7, and at the same time, the password should have at least two (or more) identical digits. In how many ways can Petya do this? | # Answer 3537.
Solution: The total number of passwords not containing the digit 7 will be $9^{4}=6561$. Among these, 9x8x7x6=3024 consist of different digits. Therefore, the number of passwords containing identical digits is 6561-3024=3537 passwords. | 3537 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,339 |
4. In the computer center, there are 200 computers, some of which (in pairs) are connected by cables, a total of 345 cables are used. We will call a "cluster" a set of computers such that a signal from any computer in this set can reach all other computers via cables (possibly through intermediate computers). Initially... | Answer: 153.
Solution: Let's try to imagine the problem this way: an evil hacker has cut all the wires. What is the minimum number of wires the admin needs to restore to end up with 8 clusters? Obviously, by adding a wire, the admin can reduce the number of clusters by one. This means that from 200 clusters, 8 can be ... | 153 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,340 |
5. In trapezoid $A B C D$ with bases $A D / / B C$, the diagonals intersect at point $E$. It is known that the areas $S(\triangle A D E)=12$ and $S(\triangle B C E)=3$. Find the area of the trapezoid. | Answer: 27
Solution: Triangle ADE and CBE are similar, their areas are in the ratio of the square of the similarity coefficient. Therefore, this coefficient is
 | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,341 |
6. 2017 numbers are written. It is known that the sum of the squares of any 7 of them is 7, the sum of any 11 of them is positive, and the sum of all 2017 numbers is divisible by 9. Find these numbers. | Answer: Five numbers are equal to -1, the rest are equal to 1.
Solutions: The sum of the squares of any 7 numbers is equal to 7. Therefore, all these squares are equal to 1. All these numbers are equal to +/-1. The sum of 11 is positive, which means the number of -1 does not exceed 5. If all are equal to 1, then the s... | Five\\\equal\to\-1,\the\rest\\equal\to\1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,342 |
7. Find the smallest natural number ending in the digit 2 that doubles when this digit is moved to the beginning. | Answer: 105263157894736842
Solution: Let's write the number in the form ***...** 2 and gradually restore the "asterisks" by multiplying by 2:
```
\(* * \ldots * * 2 \times 2=* * * . . . * * 4\)
\(* * * . . * 42 \times 2=* * * \ldots * 84\)
\(* * * . . .842 \times 2=* * * \ldots * 684\)
***... \(* 6842 \times 2=* * * ... | 105263157894736842 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,343 |
1. The Mad Hatter's clock is fast by 15 minutes per hour, while the March Hare's clock is slow by 10 minutes per hour. One day, they set their clocks by the Dormouse's clock (which is stopped and always shows 12:00) and agreed to meet at 5 o'clock in the evening for their traditional
five-o'clock tea. How long will the... | # Answer: 2 hours.
Solution: The Mad Hatter's clock runs at a speed of 5/4 of normal, so it will take 4 hours of normal time to pass 5 hours. Similarly, the March Hare's clock runs at a speed of 5/6 of normal, so it will take 6 hours of normal time to pass 5 hours. | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,344 |
2. At the international StarCraft championship, 100 participants gathered. The game is played in a knockout format, meaning in each match, two players compete, the loser is eliminated from the tournament, and the winner remains. Find the maximum possible number of participants who won exactly two games? | # Answer: 49
Solution: Each participant (except the winner) lost one game to someone. There are 99 such participants, so no more than 49 participants could have won 2 games (someone must lose 2 games to them).
We can show that there could be 49 such participants. Let's say №3 won against №1 and №2, №5 won against №3 ... | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,345 |
3. The robot moves along straight segments, making turns of 90 degrees to the right or left every minute (neglect the time for turning). The robot travels 10 meters per minute. What is the minimum distance from the starting position that the robot can be after 9 minutes, if it did not turn during the first minute?
# | # Answer: $10 \text{~m}$
Solution: We can consider a coordinate grid with nodes spaced 10m apart. Clearly, the robot travels along the nodes of this grid. It cannot return to the initial position in 9 minutes. However, it can be at a distance of 10 m - it is easy to construct an example. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,346 |
4. A box of sugar has the shape of a rectangular parallelepiped. It contains 280 sugar cubes, each of which is a $1 \times 1 \times 1$ cm cube. Find the surface area of the box, given that the length of each of its sides is less than 10 cm. | # Answer: 262
Solution: $\mathbf{2 8 0}=\mathbf{2}^{3} \mathbf{x} \mathbf{x} \mathbf{x}$. From the condition that the lengths of the sides are integers less than 10, it follows that its edges are equal to 5, 7, and 8. The surface area is $2 *\left(5^{*} 7+5 * 8+7 * 8\right)=262$ | 262 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,347 |
5. We will call a number "remarkable" if it has exactly 4 distinct natural divisors, and among them, there are two such that neither is a multiple of the other. How many "remarkable" two-digit numbers exist? | Answer 36.
Solution: Such numbers must have the form $p_{1} \cdot p_{2}$, where $p_{1}, p_{2}$ are prime numbers. Note that the smaller of these prime numbers cannot be greater than 7, otherwise the product will be no less than 121. It is sufficient to check $p_{1}=2,3,5,7$. For 2, we get the second factor: 3, 5, 7, 1... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,348 |
7. Find all three-digit numbers $\overline{\Pi B \Gamma}$, consisting of different digits $\Pi, B$, and $\Gamma$, for which the equality $\overline{\Pi B \Gamma}=(\Pi+B+\Gamma) \times(\Pi+B+\Gamma+1)$ holds. | Answer: 156.
Solution: Note that П+В $\Gamma \geq 3$ and $\leq 24$ (since the digits are different). Moreover, the numbers
$\overline{\Pi В \Gamma}$ and $(П+B+\Gamma)$ must give the same remainder when divided by 9. This is only possible when П+В $\Gamma) \times(П+B+\Gamma+1)=90$-a two-digit number. By trying $12,15,... | 156 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,350 |
1. Petya's watch gains 5 minutes per hour, while Masha's watch loses 8 minutes per hour. At 12:00, they set their watches to the school clock (which is accurate) and agreed to go to the rink together at half past six. How long will Petya wait for Masha if each arrives at the rink exactly at 18:30 by their own watches? | Answer: 1.5 hours.
Solution: Petya's clock runs at a speed of 13/12 of the normal speed, so 6.5 hours on his clock pass in 6 hours of real time. Masha's clock runs at a speed of $13 / 15$ of the normal speed, so 6.5 hours on her clock pass in 7.5 hours of real time. | 1.5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,351 |
2. At the international table tennis championship, 200 participants gathered. The game is played in a knockout format, i.e., in each match, two players participate, the loser is eliminated from the championship, and the winner remains. Find the maximum possible number of participants who won at least three matches. | Answer: 66.
Solution: Each participant (except the winner) lost one game to someone. There are 199 such participants, so no more than 66 participants could have won 3 games (someone must lose 3 games to them).
We will show that there could be 66 such participants. Let №4 win against №1,2,3; №7 - against №4,5,6,... №1... | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,352 |
3. The turtle crawled out of its house and moved in a straight line at a constant speed of 5 m/hour. After an hour, it turned $90^{\circ}$ (right or left) and continued moving, then crawled for another hour, then turned $90^{\circ}$ (right or left) again... and so on. It crawled for 11 hours, turning $90^{\circ}$ at th... | Answer 5 m.
Solution: We can consider a coordinate grid with nodes spaced 5 m apart. It is clear that the turtle crawls along the nodes of this grid. In 11 hours, it cannot return to the initial position. However, it can be 5 m away - it is easy to construct an example. | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,353 |
4. Kolya has 440 identical cubes with a side length of 1 cm. Kolya assembled a rectangular parallelepiped from them, with all edges having a length of at least 5 cm. Find the total length of all the edges of this parallelepiped | Answer: 96.
Solution: $440=2^{3} \times 5 \times 11$. Since all edges have a length of at least 5, their lengths are 8, 5, and 11. Each edge is included 4 times, so the total length is ( $8+5+11)^{*} 4=96$. | 96 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,354 |
5. Let's call a number "marvelous" if it has exactly 3 different odd natural divisors (and an arbitrary number of even divisors). How many "marvelous" two-digit numbers exist? | Answer: 7.
Solution: Such numbers have the form $2^{k} \times p^{2}$, where $p-$ is an odd prime number. Clearly, $p$ does not exceed 7, since the result must be a two-digit number. If $p=3$, then $k=0,1,2,3$; If $p=5$, then $k=0,1$; if $p=7$, then $k=0$. In total, there are 7 options.
$. | Solution. Let $t=2^{1+\sqrt{x-1}}$, we get
$$
\frac{2 t-24}{t-8} \geqslant 1 \Longleftrightarrow \frac{t-16}{t-8} \geqslant 0 \Longleftrightarrow t \in(-\infty ; 8) \cup[16 ;+\infty)
$$
From this, either $2^{1+\sqrt{x-1}}<2^{3}, \sqrt{x-1}<2, x \in[1 ; 5)$, or $2^{1+\sqrt{x-1}} \geqslant 2^{4}, \sqrt{x-1} \geqslant 3... | 526 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,358 |
2. Solve the equation
$$
\sin ^{4} x+5(x-2 \pi)^{2} \cos x+5 x^{2}+20 \pi^{2}=20 \pi x
$$
Find the sum of its roots that belong to the interval $[-\pi ; 6 \pi]$, and provide it in your answer, rounding to two decimal places if necessary. | Solution. The original equation is equivalent to the equation
$$
\sin ^{4} x+5(x-2 \pi)^{2}(\cos x+1)=0
$$
the left side of which is non-negative. Therefore, $\sin x=0$ and $(x-2 \pi)^{2}(\cos x+1)=0$. Therefore, the solution to the equation is: $x=2 \pi$ and $x=\pi+2 \pi n, n \in \mathbb{Z}$. On the interval $[-\pi ... | 31.42 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,359 |
3. Inside a right triangle $ABC$ with hypotenuse $AC$, a point $M$ is taken such that the areas of triangles $ABM$ and $BCM$ are one-third and one-fourth of the area of triangle $ABC$, respectively. Find $BM$, if $AM=60$ and $CM=70$. If the answer is not an integer, round it to the nearest integer. | Solution. Denoting $A B=c, B C=a$, we get
$$
\left\{\begin{array}{l}
\left(c-\frac{c}{4}\right)^{2}+\left(\frac{a}{3}\right)^{2}=60^{2} \\
\left(\frac{c}{4}\right)^{2}+\left(a-\frac{a}{3}\right)^{2}=70^{2}
\end{array}\right.
$$
We solve the system and find $B M=\left(\frac{a}{3}\right)^{2}+\left(\frac{c}{4}\right)^{2... | 38 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,360 |
5. Find all $a$ for which the system
$$
\left\{\begin{array}{l}
x^{2}+4 y^{2}=1 \\
x+2 y=a
\end{array}\right.
$$
has a unique solution. If necessary, round it to two decimal places. If there are no solutions, put 0 in the answer. | Solution. Since $2 y=a-x$, from the first equation we get:
$$
x^{2}+(a-x)^{2}=1 \Longleftrightarrow 2 x^{2}-2 a x+a^{2}-1=0
$$
This equation has a unique solution when $\frac{D}{4}=2-a^{2}=0$. Therefore, $a= \pm \sqrt{2}$, and the smallest value is $-\sqrt{2} \approx-1.414214 \ldots$
Answer: $-1.41$.
I-A. The sum o... | 1894 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,362 |
9. The sum of 1265 natural numbers is $2016+33$, and their product is $-\left(2016^{2}+33\right)$. Find all possible sets of such numbers. In your answer, indicate the sum of the largest and smallest numbers from all the found sets. If such numbers do not exist, then in your answer, indicate the number 0.
Hint: $2016^... | Answer 574.
II-A. Solve the equation
$$
(1+\cos x+\cos 2 x+\cos 3 x+\cos 4 x)^{4}+(\sin x+\sin 2 x+\sin 3 x+\sin 4 x)^{4}=\frac{3}{4}+\frac{\cos 8 x}{4}
$$
Find the sum of all roots on the interval $A$, rounding to the nearest integer if necessary. If there are no roots or infinitely many roots on this interval, wri... | 574 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,363 |
12. It is known that $x=2026$, find $y=BC$. The answer is 90.02.
IV-A. Define $f(a)$ as the function equal to the number of distinct solutions to the equation
$$
\sin \frac{a \pi x}{x^{2}+1} + \cos \frac{\pi (x^{2} + 4 a x + 1)}{4 x^{2} + 4} = \sqrt{2 - \sqrt{2}}
$$
For example, $f(a_1) = 3$ means that when $a = a_1... | Solution. Let $t=\frac{a \pi x}{x^{2}+1}$. Then the equation takes the form $\sin t+\cos \left(\frac{\pi}{4}+t\right)=\sqrt{2-\sqrt{2}}$ or
\[
\begin{aligned}
& \sin t+\sin \left(\frac{\pi}{4}-t\right)=\sqrt{2-\sqrt{2}} \Longleftrightarrow \\
& 2 \sin \frac{\pi}{8} \cos \left(\frac{\pi}{8}-t\right)=\sqrt{2-\sqrt{2}} \... | 8.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,365 |
8. Equation:
$$
\cos \frac{a \pi x}{x^{2}+4}-\cos \frac{\pi\left(x^{2}-4 a x+4\right)}{4 x^{2}+16}=-\sqrt{2-\sqrt{2}}
$$
inequality $f(a) \leqslant 5$. Answer: $a \in\left[-\frac{21}{2} ; \frac{21}{2}\right]$, the length of the interval is 21.
V-A. Let $a_{n}$ be the number of permutations $\left(k_{1}, k_{2}, \ldot... | Solution. We will prove in several steps:
Step 1. Prove that there is a recurrence relation: $a_{n}=a_{n-1}+a_{n-3}+1$. The following are the possible beginnings of permutations:
- In the sequence $(1,2, \ldots, n)$. Discard the first one, and decrease the remaining numbers by 1. What remains satisfies the conditions... | 6585451 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,366 |
1. A coin was tossed 2021 times. What is the probability that an even number of "heads" will appear? | Answer: $1 / 2$
Solution: Let $p$ be the probability that an even number of "heads" appeared in the first 2020 tosses, and $q=1-p$ be the probability that an odd number appeared. Then, on the 2021st toss, heads will appear with a probability of $1 / 2$, and the probability will be $\frac{1}{2} q$ or tails, then the pr... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,367 |
2. Ivan Semenovich leaves for work at the same time every day, drives at the same speed, and arrives exactly at 9:00. One day he overslept and left 40 minutes later than usual. To avoid being late, Ivan Semenovich drove at a speed 60% faster than usual and arrived at 8:35. By what percentage should he have increased hi... | Answer: By $30 \%$.
Solution: By increasing the speed by $60 \%$, i.e., by 1.6 times, Ivan Semenovich reduced the time by 1.6 times and gained 40+25=65 minutes. Denoting the usual travel time as $T$, we get $\frac{T}{1.6}=T-65$, from which $T=\frac{520}{3}$. To arrive in $T-40=\frac{400}{3}$, the speed needed to be in... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,368 |
3. Compare the numbers $\frac{100}{101} \times \frac{102}{103} \times \ldots \times \frac{1020}{1021} \times \frac{1022}{1023}$ and $\frac{5}{16}$ | Answer: $\frac{100}{101} \times \frac{102}{103} \times \ldots \times \frac{2020}{2021} \times \frac{2022}{2023}<\frac{5}{16}$
Solution: Notice that $\frac{n}{n+1}<\frac{n+1}{n+2}$. Therefore, $A=\frac{100}{101} \times \frac{102}{103} \times \ldots \times \frac{2020}{2021} \times \frac{2022}{2023}<\frac{101}{102} \time... | A<\frac{5}{16} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,369 |
4. The English club is attended by 20 gentlemen. Some of them are acquainted (acquaintances are mutual, i.e., if A knows B, then B knows A). It is known that there are no three gentlemen in the club who are pairwise acquainted.
One day, the gentlemen came to the club, and each pair of acquaintances shook hands with ea... | Answer: 100 handshakes.
Solution: Choose a gentleman with the maximum number of acquaintances (if there are several, choose any one). Suppose he has $n$ acquaintances. These acquaintances cannot be pairwise acquainted with each other. Consider the remaining $(20-n-1)$ gentlemen, each of whom has no more than $n$ acqua... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,370 |
5. Ms. Olga Ivanovna, the class teacher of 5B, is staging a "Mathematical Ballet." She wants to arrange the boys and girls so that there are exactly 2 boys 5 meters away from each girl. What is the maximum number of girls that can participate in the ballet, given that 5 boys are participating? | Answer: 20 girls.
Solution: Let's select and fix two arbitrary boys - $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$. Suppose they are 5 m away from some girl - G. Then $\mathrm{M}_{1}, \mathrm{M}_{2}$, and G form an isosceles triangle with the legs being 5 m. Given the fixed positions of $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$,... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,371 |
6. A space probe, moving in a straight line at a constant speed, flies past Mars and measures the distance to this planet every day at exactly 12:00. It is known that on February 1st, the distance was 5 million km, on February 10th - 2 million km, and on February 13th - 3 million km.
Determine when the probe will pass... | Answer: February 9.
Let $d(t)$ be the distance to Mars at time $t$. Suppose Mars is at the origin, the probe starts from the point $\left(x_{0}, y_{0}, z_{0}\right)$, and its velocity vector is $\left(v_{x}, v_{y}, v_{z}\right)$.
Then at time $t$, the probe will be at the point $\left(x_{0}+t \cdot v_{x}, y_{0}+t \cd... | February9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,372 |
7. Given a polynomial $P(x)$, not identically zero. It is known that for all $x$ the identity $(x-2020) \cdot P(x+1)=(x+2021) \cdot P(x)$ holds. How many roots does the equation $P(x)=0$ have? Answer: 4042 | Solution:
First, we prove two auxiliary lemmas:
Lemma 1. If $x_{0} \neq 2020$ is a root of $P(x)$, then $x_{0}+1$ is also a root.
Proof: Substitute $\left(x_{0}-2020\right) \cdot P\left(x_{0}+1\right)=\left(x_{0}+2021\right) \cdot P\left(x_{0}\right)=0$ but $\left(x_{0}-2020\right) \neq 0$.
Lemma 2. If $x_{0} \neq ... | 4042 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,373 |
Problem 2. 2-1. Find the minimum value of the function
$$
f(x)=x^{2}+(x-2)^{2}+(x-4)^{2}+\ldots+(x-100)^{2}
$$
If the result is a non-integer, round it to the nearest integer and write it as the answer. | Solution. Knowledge of arithmetic progression is required. It turns out to be a quadratic function
$f(x)=51 x^{2}-2(2+4+6+\ldots+100) x+(2^{2}+4^{2}+6^{2}+\ldots+100^{2})=51 x^{2}-2 \cdot 50 \cdot 51 x+4 \cdot(1^{2}+2^{2}+3^{2}+\ldots+50^{2})$.
The minimum is achieved at the point $x_{0}=50$. In this case,
$f(50)=50... | 44200 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,374 |
2-2. Find the minimum value of the function
$$
f(x)=x^{2}+(x-2)^{2}+(x-4)^{2}+\ldots+(x-102)^{2}
$$
If the result is a non-integer, round it to the nearest integer and write it as the answer. | Solution. $f(x)=52 x^{2}-2(2+4+\ldots+102) x+2^{2}+4^{2}+\ldots+102^{2}=52 x^{2}-2 \cdot 51 \cdot 52 x+4\left(1^{2}+2^{2}+3^{2}+\ldots+51^{2}\right)$.
The minimum of the function $f$ is achieved at the point $x_{0}=51 . f(51)=51^{2}+49^{2}+\ldots 1^{2}+1^{2}+3^{2}+\ldots+49^{2}+51^{2}=$ $2\left(1^{2}+3^{2}+\ldots 51^{... | 46852 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,375 |
2-3. Find the minimum value of the function
$$
f(x)=(x-1)^{2}+(x-3)^{2}+\ldots+(x-101)^{2}
$$
If the result is a non-integer, round it to the nearest integer and write it in the answer. | Solution. $f(x)=51 x^{2}-2(1+3+\ldots+101) x+\left(1^{2}+3^{2}+\ldots+101^{2}\right)=51 x^{2}-2 \cdot 51^{2} x+\left(1^{2}+3^{2}+\ldots+101^{2}\right)$.
The minimum of the function $f$ is achieved at the point $x_{0}=51$. Since $1^{2}+3^{2}+\ldots+(2 n-1)^{2}=\frac{n\left(4 n^{2}-1\right)}{3}$, then $f(51)=51^{3}-2 \c... | 44200 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,376 |
2-4. Find the minimum value of the function
$$
f(x)=x^{2}+(x-2)^{2}+(x-4)^{2}+\ldots+(x-104)^{2}
$$
If the result is a non-integer, round it to the nearest integer and write it in the answer. | Solution. $f(x)=53 x^{2}-2(2+4+\ldots 104) x+4\left(1^{2}+2^{2}+\ldots+52^{2}\right)=53 x^{2}-2 \cdot 52 \cdot 53 x+4 \cdot \frac{52 \cdot 53 \cdot 105}{6}$ (since $\left.1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}\right)$.
The minimum of the function $f$ is achieved at the point $x_{0}=52 . f(52)=53 \cdot ... | 49608 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,377 |
Problem 3. Determine how many roots of the equation
$$
4 \sin 2 x + 3 \cos 2 x - 2 \sin x - 4 \cos x + 1 = 0
$$
are located on the interval $\left[10^{2014!} \pi ; 10^{2014!+2015} \pi\right]$. In your answer, write the sum of all digits of the found number. | Solution. Let $t=\sin x+2 \cos x$. Then $t^{2}=\sin ^{2} x+2 \sin 2 x+4 \cos ^{2} x=2 \sin 2 x+\frac{3}{2} \cos 2 x+\frac{5}{2}$. The original equation is equivalent to $t^{2}-t-2=0$. From which
$$
\left[\begin{array}{c}
\operatorname{sin} x + 2 \operatorname{cos} x = -1, \\
\operatorname{sin} x + 2 \operatorname{cos}... | 18135 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,378 |
Task 4. Find the smallest natural $m$, for which there exists such a natural $n$ that the sets of the last 2014 digits in the decimal representation of the numbers $a=2015^{3 m+1}$ and $b=2015^{6 n+2}$ are the same, and $a<b$. | Solution. The coincidence of the last 2014 digits of the two specified powers means divisibility by \(10^{2014}\) of the difference
\[
2015^{6 n+2}-2015^{3 m+1}=2015^{3 m+1} \cdot\left(2015^{6 n-3 m+1}-1\right)
\]
since the first factor in the obtained decomposition is not divisible by 2, and the second is not divisi... | 671 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,381 |
Problem 5. From vertex $B$ of triangle $A B C$, a line is drawn intersecting side $A C$ at point $E$. Find the height $B F$ of triangle $A B C$, given that the center of the circumscribed circle around triangle $A B C$ lies on ray $B E, A F \cdot F E=5$ and $\operatorname{ctg} \angle E B C: \operatorname{ctg} \angle B ... | Solution. Drop a perpendicular $CD$ from point $C$ to the line $BE$. Then the ratio of the cotangents gives the proportion $BD: DE = 3: 4$. Since $DE > BD$, we have $EC > BC$. As in a right triangle the hypotenuse is greater than the leg, $BC > CF$, and therefore, point $F$ lies between points $E$ and $C$ (see Fig. 1).... | 1.94 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,382 |
Problem 6. Find the value of $a$ for which the sum of all real roots of the equation
$$
\frac{f(a) \cdot x^{2}+1}{x^{2}+g(a)}=\sqrt{\frac{x g(a)-1}{f(a)-x}}
$$
is minimized, where $f(a)=a^{2}-\sqrt{21} a+26, g(a)=\frac{3}{2} a^{2}-\sqrt{21} a+27$. Provide the found value, rounding it to two decimal places if necessar... | Solution. Method 1. Note that for all values of $a: g(a)>f(a) \geqslant 20$, and therefore, the domain of the equation does not contain negative numbers. On the set $x \geqslant 0$, the functions $F(x)=\frac{f(a) \cdot x^{2}+1}{x^{2}+g(a)}$ and $F^{-1}(x)=\sqrt{\frac{x g(a)-1}{f(a)-x}}$ are mutually inverse. It is not ... | \sqrt{21}/2\approx2.29 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,383 |
5. At the vertices of a regular 9-gon
(see fig.) place the numbers $2016, 2017, \ldots, 2024$ in such a way that for any three vertices forming an equilateral triangle, one of the numbers is equal to the arithmetic mean of the other two.
![](https://cdn.mathpix.com/cropped/2024_05_06_eb02360d9d89ae1b4e45g-04.jpg?heig... | Solution: notice that any three numbers, spaced at equal intervals, work
Lomonosov Moscow State University
## School Olympiad "Conquer Sparrow Hills" in Mathematics
Final Round Tasks for 2015/2016 Academic Year for Grades 5-6 | 2016,2017,2018,2019,2020,2021,2022, | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,386 |
4. Can the numbers 2016, 2017, ..., 2024 be placed in the specified positions (see fig.), so that the sum of the numbers on each side of the triangle is the same?
ANSWER: Yes, it is possible... | Solution: First, place the average value 2020 at all points. Then arrange $0, \pm1, \pm2, \pm3, \pm4$ in such a way that the sum on each side is equal to 0 (see the figure). Other arrangements are also possible.
+110(b+$ $c)=4983$. Notice that $a+d$ must end in 3, so $a+d=3$ or $a+d=13$. The second case is impossible, since then $1001(a+d)+110(b+c)>13000$. Therefore, $a+d=3$, substituting into the original equation, w... | 19922991 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,388 |
4. Points $A, B, C$ are located sequentially, with the distance $A B$ being $a$ km and the distance $B C$ being $b$ km. A cyclist left point $A$ and headed for point $C$. At the same time, a pedestrian left point $B$ and headed for point $A$. It is known that the pedestrian and the cyclist arrived at points $A$ and $C$... | Answer: $\frac{a(a+b)}{2 a+b}$ km.
Solution: Let $D$ be the meeting point. Since the speeds of the cyclist and the pedestrian are in the ratio $(a+b): a$, point $D$ divides the segment $A B$ in the same ratio. Therefore, $A D=k(a+b), B D=k a$ and $k(a+b)+k a=a$, from which $k=\frac{a}{2 a+b}$. In the end, we get $A D=... | \frac{(+b)}{2+b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,391 |
6. In quadrilateral $A B C D$, it is known that $A B=B C, \angle A B C=$ $\angle A D C=90^{\circ}$. A perpendicular $B H$ is dropped from vertex $B$ to side $A D$. Find the area of quadrilateral $A B C D$, if it is known that $B H=h$. | Answer: $h^{2}$.
Solution:
Cut off the triangle $A B H$, attach it to the top - we get a square with side $h$. | ^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,393 |
7. It is known that for some natural numbers $a, b$, the number $N=\frac{a^{2}+b^{2}}{a b-1}$ is also natural. Find all possible values of $N$.
---
The provided text has been translated into English while preserving the original formatting and structure. | Answer: 5.
Solution: It is not difficult to find the solution $a=2, b=1, N=5$. Let's prove that no other $N$ is possible.
Let $\left(a_{0}, b_{0}\right)$ be a solution corresponding to the smallest value of $a^{2}+b^{2}$. Without loss of generality, we can assume that $a_{0}>b_{0}$. If $b_{0}=1$, then $N=\frac{a^{2}+... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,394 |
8. In a trapezoid, the diagonals intersect at a right angle, and one of them is equal to the midline. Determine the angle this diagonal forms with the bases of the trapezoid. | Answer: $60^{\circ}$.
Solution: We perform a parallel translation of the diagonal - we will get a right triangle, in which the leg is half the hypotenuse. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,395 |
9. The numbers $1,2, \ldots, 2016$ were divided into pairs, such that the product of the numbers in each pair does not exceed some natural number $N$. What is the smallest $N$ for which this is possible? | Answer: $1017072=1008 \times 1009$.
Solution: If we consider the numbers $1008,1009, \ldots, 2016$, then some two must fall into one pair. Therefore, $N$ cannot be less than $1008 \cdot 1009$. We will show that such a partition is possible when $N=1008 \cdot 1009$. We can partition into pairs: $(1,2016),(2,2016), \ldo... | 1017072=1008\times1009 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,396 |
7. Given a sequence of positive numbers $a_{1}, a_{2}, \ldots, a_{10}$, satisfying the relation $a_{n}\left(a_{n-1}+a_{n+1}\right)=2 a_{n-1} a_{n+1}\left(a_{n}+1\right)$ for $n=2,3, \ldots, 9$. Find $a_{5}$ if it is known that $a_{1}=1$ and $a_{10}=0.01$ | Solution: Divide both sides by $a_{n-1} a_{n} a_{n+1} \neq 0$ (since the numbers are positive) and make the substitution $\frac{1}{a_{n}}=b_{n}$, we get the relation $b_{n-1}+b_{n+1}=2 b_{n}+2 \Leftrightarrow b_{n+1}-b_{n}=b_{n}-b_{n-1}+2$. It is clear that the distances between adjacent terms of the sequence $b_{n}$ i... | 0.04 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,397 |
1. The trousers are cheaper than the jacket, the jacket is cheaper than the coat, the coat is cheaper than the fur, and the fur is cheaper than the diamond necklace by the same percentage. By what percentage is the fur more expensive than the trousers, if the diamond necklace is 6.25 times more expensive than the coat? | Answer: by $1462.5\%$.
Solution. According to the condition $\frac{B K}{Sh}=\frac{\Pi}{\Pi}=\frac{\Pi}{K}=\frac{K}{B}=\alpha$. In this case, the cost of the diamond necklace $(B K)$ is related to the cost of the coat (P) by the ratio $B K=\Pi \cdot \alpha^{2}$. Then $\alpha=\sqrt{6.25}=\sqrt{\frac{25}{4}}=\frac{5}{2}$... | 1462.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,398 |
2. In the wallet of the merchant Hans, there are 20 silver coins worth 2 crowns each, 15 silver coins worth 3 crowns each, and 3 gold ducats (1 ducat equals 5 crowns). In how many ways can Hans pay a sum of 10 ducats? Coins of the same denomination are indistinguishable. | Answer: 26
Solution. If the merchant paid $x$ coins at 2 crowns each, $y$ coins at 3 crowns each, and $z$ ducats (i.e., $z$ times 5 crowns), we get the system: $2 x+3 y+5 z=50, x \in[0 ; 20], y \in[0 ; 15], z \in[0 ; 3]$.
a) When $z=0$, we get the equation $2 x+3 y=50$, which has suitable solutions under the conditio... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,399 |
3. Determine for which values of $n$ and $k$ the equation $\sin x + \sin y = \frac{\pi k}{2017}$ is a consequence of the equation $x + y = \frac{\pi n}{48}$. | Answer: $k=0, n=96 m, m \in Z$.
Solution. The pair $\left(0, \frac{\pi n}{48}\right)$ is a solution to the second equation, hence $\sin \frac{\pi n}{48}=\frac{\pi k}{2017}$. The pair $\left(\pi ; \frac{\pi n}{48}-\pi\right)$ is also a solution to the second equation, hence $-\sin \frac{\pi n}{48}=\frac{\pi k}{2017}$. ... | k=0,n=96,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,400 |
5. Solve the inequality $x+\sqrt{x^{2}+4} \geq\left(2 x+1+\sqrt{4 x^{2}+4 x+2}\right) \cdot 8^{x+1}$. | Answer: $x \in\left(-\infty ;-\frac{2}{3}\right]$.
Solution. The original inequality is equivalent to the inequality:
$$
2\left(\frac{x}{2}+\sqrt{\left(\frac{x}{2}\right)^{2}+1}\right) \geq\left(2 x+1+\sqrt{(2 x+1)^{2}+1}\right) \cdot 8^{x+1} \Leftrightarrow \frac{x}{2}+\sqrt{\left(\frac{x}{2}\right)^{2}+1} \geq\left... | x\in(-\infty;-\frac{2}{3}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,402 |
1. According to the condition, these numbers are written only with the digits $0,1,2,6,8$. Then, three-digit numbers that are multiples of 4 can end with exactly 10 variants: $00,08,12,16,20,28,60,68,80,88$. In each of these 10 variants, the first place can be occupied by any of the 4 digits $1,2,6,8$. Additionally, th... | # Answer: 49.
Answer to option: 1-2: 53.
## Solutions for option 1-2.
According to the condition, these numbers are written only with the digits $0,2,3,4,5,7$. Then, three-digit numbers that are multiples of 4 can end with exactly 9 variants: $00,04,20,24,32,40,44,52,72$. In this case, the first place in each of the... | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,403 |
5. 1 way to solve.
$$
\begin{aligned}
& \left(\operatorname{tg} 9^{\circ}-\operatorname{tg} 63^{\circ}\right)+\left(\operatorname{tg} 81^{\circ}-\operatorname{tg} 27^{\circ}\right)=-\frac{\sin 54^{\circ}}{\cos 9^{\circ} \cos 63^{\circ}}+\frac{\sin 54^{\circ}}{\cos 81^{\circ} \cos 27^{\circ}}= \\
& =\frac{\sin 54^{\cir... | Answer: The second number is greater.
Answer to option: 1-2: The first number is greater. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,407 |
1. Let $t=\pi / 4-x / 3$. Then the original inequality is equivalent to:
$$
\begin{aligned}
3 \cos 2 t+2 \cos t-5 \geqslant 0 \Longleftrightarrow 3\left(2 \cos ^{2} t-1\right)+2 \cos t-5 \geqslant 0 \Longleftrightarrow \\
\Longleftrightarrow 6(\cos t-1)\left(\cos t+\frac{4}{3}\right) \geqslant 0
\end{aligned}
$$
We o... | Answer: $x=\frac{3 \pi}{4}+6 \pi k, k \in \mathbb{Z}$.
Answer to variant: $3-2: x=-\frac{3 \pi}{4}+6 \pi k, k \in \mathbb{Z}$. | \frac{3\pi}{4}+6\pik,k\in\mathbb{Z} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,408 |
4. The condition means that the polynomial $x^{4}+5 x^{3}+a x^{2}+b x+c$ is divisible by the polynomial of the third degree $x^{3}+(8+b) x^{2}+(b+4) x+(c+3)$ without a remainder. This condition is equivalent to the system
$$
\left\{\begin{array}{l}
a-b-4=(8+b)(-3-b) \\
b-(c+3)=(4+b)(-3-b) \\
c=(c+3)(-3-b)
\end{array}\... | Answer: $a=5, b=-5, c=-6$.
Answer to variant: 3-2: $a=5, b=-6, c=-5$. | =5,b=-5,=-6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,411 |
5. After substituting the variables $X=\sqrt{|x-1|}, \sqrt{3|y|}$, we obtain the system:
$$
\left\{\begin{array}{l}
X+Y=1 \\
X^{4}+Y^{4}=-a \\
X \geqslant 0, Y \geqslant 0
\end{array}\right.
$$
Each solution $\left(X_{0}, Y_{0}\right)$ of this system such that $0<X_{0}<1,0<Y_{0}<1, X_{0} \neq Y_{0}$ generates exactly... | Answer: $a=-1, a=-1 / 8$. Answer to the option: $4-2: a=1 / 2, a=1 / 16$.
Answer and solution to option $5-1$ (2) 1. Answer: $x= \pm \frac{\pi}{12}+\frac{\pi n}{2}$. Since $\operatorname{ctg}^{2} x-\operatorname{tg}^{2} x=\frac{4 \cos 2 x}{\sin ^{2} 2 x}$, the original equation is equivalent to the equation
$$
\frac{... | =-1,=-\frac{1}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,417 |
3. A parallelepiped is inscribed in a sphere of radius $\sqrt{3}$, and the volume of the parallelepiped is 8. Find the surface area of the parallelepiped. | 3. $S_{\text {full }}=24$
The translation is provided while maintaining the original text's formatting and structure. | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,420 |
4. A circle with center at point $O$ touches side $B C$ of triangle $A B C$ at point $C$ and touches median $A D$ of triangle $A B C$ at point $A$. $O B$ is the bisector of angle $A O D$. Find the angles of triangle $A B C$. | 4. $\frac{\pi}{2}, \arccos \frac{\sqrt{17}-1}{4}, \arcsin \frac{\sqrt{17}-1}{4}$. | \frac{\pi}{2},\arccos\frac{\sqrt{17}-1}{4},\arcsin\frac{\sqrt{17}-1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,421 |
1. Determine whether the number $N=7 \times 9 \times 13+2020 \times 2018 \times 2014$ is prime or composite. | Answer: Composite.
Solution: Note that 7+2020=9+2018=13+2014=2027. Therefore, $N \equiv 7 \times 9 \times 13 + (-7) \times (-9) \times (-13) = 0 \pmod{2027}$ | Composite | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,423 |
3. Prove that the sum of 6-digit numbers, not containing the digits 0 and 9 in their decimal representation, is divisible by 37. | Solution: Let's break it down into pairs, each complementing the other to make 9 in each digit (for example, 123456+876543). We get a sum of 999999 in each pair, which is equal to $9 \times 1001 \times 111$. It is easy to verify that 111 is divisible by 37. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 5,424 |
6. Ms. Olga Ivanovna, the class teacher of 5B, is staging a "Mathematical Ballet." She wants to arrange the boys and girls so that exactly 2 boys are 5 meters away from each girl. What is the maximum number of girls that can participate in the ballet if it is known that 5 boys are participating? | Answer: 20 girls.
Solution: Let's select and fix two arbitrary boys - $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$. Suppose they are 5 m away from some girl - G. Then $\mathrm{M}_{1}, \mathrm{M}_{2}$, and G form an isosceles triangle with the legs being 5 m. Given the fixed positions of $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$,... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,426 |
7. Given an equilateral triangle. Is it possible to cut it into two figures, one of which is a 2020-gon, and the other is a 2021-gon (polygons do not have to be convex)? | Answer: Yes, it is possible.
Solution: Draw a 2019-segment broken line inside the triangle, connecting two vertices of the triangle. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,427 |
1.1 Misha noticed that a tram passed by him in 3 seconds, and a tunnel 100 meters long in 13 seconds. Find the speed of the tram (in meters per second), assuming that it remains the same throughout the entire observation period. | Solution. Let the speed of the tram (in meters per second) be $v$, and the length of the tram (in meters) be $l$. If $t_{1}$ and $t_{2}$ are the times it takes for the tram to pass by Mishka and through a tunnel of length $a$ respectively, then
$$
\left\{\begin{array} { l }
{ l = v \cdot t _ { 1 } } \\
{ a + l = v \c... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,428 |
1.2 Mishа noticed that the tram passed by him in 4 seconds, and a tunnel 64 meters long in 12 seconds. Find the length of the tram (in meters), assuming that its speed remains constant throughout the entire observation period. | Answer: 32. (E)
 | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,429 |
1.3 Mishа noticed that the tram passed by him in 2 seconds, and a tunnel 96 meters long - in 10 seconds. Find the speed of the tram (in meters per second), assuming that it remains the same throughout the entire observation period. | Answer: 12. (B)
 | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,430 |
2.1 Find $f(2013)$, if for any real $x$ and $y$ the equality holds
$$
f(x-y)=f(x)+f(y)-2xy
$$ | Solution. Substitute $x=y=0$. We get $f(0)=2 f(0)+0$, from which we obtain that $f(0)=0$. Substitute $x=y$. We get $0=f(0)=f(x)+f(x)-2 x^{2}$. Hence, $f(x)=x^{2}$.
Answer: 4052169. (C)
 | 4052169 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,431 |
3-1. Calculate the sum
$$
S=\frac{2014}{2 \cdot 5}+\frac{2014}{5 \cdot 8}+\frac{2014}{8 \cdot 11}+\ldots+\frac{2014}{2012 \cdot 2015}
$$
In your answer, specify the remainder from dividing by 5 the even number closest to the obtained value of $S$. | Solution. Since
$$
\begin{aligned}
& \frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots+\frac{1}{2012 \cdot 2015}= \\
& =\frac{1}{3} \cdot\left(\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\ldots\left(\frac{1}{2012}-\frac{1}{2015}\righ... | 336 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,432 |
3-2. Calculate the sum
$$
S=\frac{2013}{2 \cdot 6}+\frac{2013}{6 \cdot 10}+\frac{2013}{10 \cdot 14}+\ldots+\frac{2013}{2010 \cdot 2014}
$$
In your answer, specify the remainder when the nearest even number to the obtained value of $S$ is divided by 5. | Answer: 2. (C)
Options.
$$
\begin{array}{|l|l|l|l|l|l|l|l|l|}
\hline \mathbf{A} & 0 & \mathbf{B} & 1 & \mathbf{C} & 2 & \mathbf{D} & 3 & \mathbf{E} \\
4 & \mathbf{F} \\
\hline
\end{array}
$$ | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,433 |
3-3. Calculate the sum
$$
S=\frac{2014}{3 \cdot 7}+\frac{2014}{7 \cdot 11}+\frac{2014}{11 \cdot 15}+\ldots+\frac{2014}{2011 \cdot 2015}
$$
In your answer, specify the remainder when dividing by 5 the natural number closest to the obtained value of $S$ | Answer: 3. ( $\mathbf{D}$)
Options.
$$
\begin{array}{|l|l|l|l|l|l|l|l|l|}
\hline \mathbf{A} & 0 & \mathbf{B} & 1 & \mathbf{C} & 2 & \mathbf{D} & 3 & \mathbf{E} \\
\hline
\end{array}
$$ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,434 | |
3-4. Calculate the sum
$$
S=\frac{2015}{3 \cdot 8}+\frac{2015}{8 \cdot 13}+\frac{2015}{13 \cdot 18}+\ldots+\frac{2015}{2008 \cdot 2013}
$$
In your answer, specify the remainder when dividing by 5 the natural number closest to the obtained value of $S$. | Answer: 4. ( ( $\mathbf{2}$ )
Options.
 | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,435 |
4-2. A circle touches the sides of an angle at points $A$ and $B$. The distance from a point $C$ lying on the circle to the line $A B$ is 6. Find the sum of the distances from point $C$ to the sides of the angle, given that one of these distances is 5 more than the other. | Answer: 13. (C)
Options.
 | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,437 |
4-3. A circle touches the sides of an angle at points $A$ and $B$. The distance from a point $C$ lying on the circle to the line $A B$ is 6. Find the sum of the distances from point $C$ to the sides of the angle, given that one of these distances is nine times smaller than the other. | Answer: 20. (E)
Answer options.
 | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,438 |
4-4. A circle touches the sides of an angle at points $A$ and $B$. The distance from a point $C$ lying on the circle to the line $A B$ is 8. Find the sum of the distances from point $C$ to the sides of the angle, given that one of these distances is 30 less than the other. | Answer: 34. (B)
Answer options.
 | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,439 |
5-1. Solve the inequality
$$
\sqrt{3 x-7}-\sqrt{3 x^{2}-13 x+13} \geqslant 3 x^{2}-16 x+20
$$
In your answer, specify the sum of all integer values of $x$ that satisfy the inequality. | Solution. As a result of the substitution $v=\sqrt{3 x-7}, u=\sqrt{3 x^{2}-13 x+13}$, we obtain the equivalent inequality $u \leqslant v$. Therefore, $x$ satisfies the inequality $3 x^{2}-13 x+13 \leqslant 3 x-7 \Longleftrightarrow 2 \leqslant x \leqslant 10 / 3$. Out of the two integer values $x=2$ and $x=3$, only $x=... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,440 |
5-2. Solve the inequality
$$
\sqrt{6 x-13}-\sqrt{3 x^{2}-13 x+13} \geqslant 3 x^{2}-19 x+26
$$
In your answer, specify the sum of all integer values of $x$ that satisfy the inequality. | Solution. The inequality is satisfied only by the following integer values: $x=3, x=4$.
Answer: (D) 7.
Answer choices.
$$
\begin{array}{|l|l|l|l|l|l|l|l|l|}
\hline \mathbf{A} & 2 & \mathbf{B} & 3 & \mathbf{C} & 5 & \mathbf{D} & 7 & \mathbf{E} \\
\hline
\end{array}
$$ | 7 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,441 |
5-3. Solve the inequality
$$
\sqrt{5 x-11}-\sqrt{5 x^{2}-21 x+21} \geqslant 5 x^{2}-26 x+32
$$
In your answer, specify the sum of all integer values of $x$ that satisfy the inequality. | Solution. The inequality is satisfied by only one integer value: $x=3$.
Answer: (B) 3.
Answer choices.
$$
\begin{array}{|l|l|l|l|l|l|l|l|l|}
\hline \mathbf{A} & 2 & \mathbf{B} & 3 & \mathbf{C} & 5 & \mathbf{D} & 7 & \mathbf{E} \\
\hline
\end{array}
$$ | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,442 |
5-4. Solve the inequality
$$
\sqrt{10 x-21}-\sqrt{5 x^{2}-21 x+21} \geqslant 5 x^{2}-31 x+42
$$
In your answer, indicate the sum of all integer values of $x$ that satisfy the inequality. | Solution. The inequality is satisfied only by the following integer values: $x=3, x=4$.
Answer: (D) 7.
Answer choices.
 | 7 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,443 |
6-2. From three mathematicians and ten economists, a committee of seven people needs to be formed. At the same time, the committee must include at least one mathematician. In how many ways can the committee be formed? | Solution. Direct calculation gives: $C_{13}^{7}-C_{10}^{7}=1596$.
Answer: 1596. | 1596 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,444 |
6-3. From twelve students and three teachers, a school committee consisting of nine people needs to be formed. At the same time, at least one teacher must be included in it. In how many ways can the committee be formed? | Solution. Direct calculation gives: $C_{15}^{9}-C_{12}^{9}=4785$.
Answer: 4785. | 4785 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,445 |
6-4. From eleven students and three teachers, a school committee consisting of eight people needs to be formed. At the same time, at least one teacher must be included in it. In how many ways can the committee be formed? | Solution. Direct calculation gives: $C_{14}^{8}-C_{11}^{8}=2838$.
Answer: 2838. | 2838 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,446 |
8-1. Find all common points of the graphs
$$
y=8 \cos \pi x \cdot \cos ^{2} 2 \pi x \cdot \cos 4 \pi x \quad \text { and } \quad y=\cos 9 \pi x
$$
with abscissas belonging to the segment $x \in[0 ; 1]$. In your answer, specify the sum of the abscissas of the found points. | Solution. Multiply by $\sin \pi x$. All $x=n, n \in \mathbb{Z}$ need to be removed from the answer, as they are not roots of the original equation. Next, using the equality $8 \sin \pi x \cos \pi x \cos 2 \pi x \cos 4 \pi x=\sin 8 \pi x$, we find
$$
\begin{aligned}
\sin 8 \pi x \cdot \cos 2 \pi x & =\sin \pi x \cdot \... | 3.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,451 |
8-2. Find all common points of the graphs
$$
y=8 \cos ^{2} \pi x \cdot \cos 2 \pi x \cdot \cos 4 \pi x \quad \text { and } \quad y=\cos 6 \pi x
$$
with abscissas belonging to the interval $[-1 ; 0]$. In your answer, specify the sum of the abscissas of the found points. | Solution. This is equivalent to $\sin 9 \pi x=\sin (-5 \pi x), x \neq n, n \in \mathbb{Z}$. From this, $x=k / 7$ or $x=1 / 4+l / 2, k, l \in \mathbb{Z}$.
The roots of the equation in the interval $[-1 ; 0]$ are: $-1 / 7,-2 / 7,-3 / 7,-4 / 7,-5 / 7,-6 / 7$ and $-1 / 4,-3 / 4$.
Answer: -4 . | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,452 |
8-3. Find all common points of the graphs
$$
y=8 \cos \pi x \cdot \cos 2 \pi x \cdot \cos ^{2} 4 \pi x \quad \text { and } \quad y=\cos 11 \pi x
$$
with abscissas belonging to the interval $[0 ; 1]$. In the answer, specify the sum of the abscissas of the found points. | Solution. This is equivalent to $\sin 4 \pi x=\sin (-10 \pi x), x \neq n, n \in \mathbb{Z}$. From this, $x=k / 7$ or $x=1 / 6+l / 3, k, l \in \mathbb{Z}$.
The roots of the equation in the interval $[0 ; 1]$ are: $1 / 7, 2 / 7, 3 / 7, 4 / 7, 5 / 7, 6 / 7$ and $1 / 6, 1 / 2, 5 / 6$.
Answer: 4,5 . | 4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,453 |
8-4. Find all common points of the graphs
$$
y=8 \cos \pi x \cdot \cos ^{2} 2 \pi x \cdot \cos 4 \pi x \quad \text { and } \quad y=\cos 5 \pi x
$$
with abscissas belonging to the interval $[-1 ; 0]$. In your answer, specify the sum of the abscissas of the found points. | Solution. This is equivalent to $\sin 10 \pi x=\sin (-4 \pi x), x \neq n, n \in \mathbb{Z}$. From this, $x=k / 7$ or $x=1 / 6+l / 3, k, l \in \mathbb{Z}$.
The roots of the equation in the interval $[-1 ; 0]$ are: $-1 / 7, -2 / 7, -3 / 7, -4 / 7, -5 / 7, -6 / 7$ and $-1 / 6, -1 / 2$, $-5 / 6$.
Answer: $-4.5$. | -4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,454 |
9-1. Find all positive $a$ for which the equation
$$
\frac{2 \pi a + \arcsin (\sin x) + 2 \arccos (\cos x) - a x}{\tan^2 x + 1} = 0
$$
has exactly three distinct solutions in the set $(-\infty; 7\pi]$. In your answer, provide the sum of all such $a$ (if no such $a$ exists, indicate 0; if the number $a$ is not an inte... | Solution. The domain of permissible values - $x \in \mathbb{R} \backslash\{\pi / 2+\pi n\}_{n=1}^{+\infty}$. On the domain of permissible values, we solve the equation
$$
\arcsin (\sin x)+2 \arccos (\cos x)=a x-2 \pi a
$$
The function
$$
f(x)=\arcsin (\sin x)+2 \arccos (\cos x)
$$
is periodic with a period of $2 \p... | 1.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,455 |
9-2. Find all positive $a$ for which the equation
$$
\frac{4 \pi a + \arcsin (\sin x) + 3 \arccos (\cos x) - a x}{2 + \tan^2 x} = 0
$$
has exactly three solutions. In your answer, specify the sum of all found $a$ (if such $a$ do not exist, then specify 0; if the number $a$ is not an integer, then round it to the hund... | Solution. For $a=1, a=2/3, a=4/5$. The sum is $2.4(6)$.
Answer: 2.47. | 2.47 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,456 |
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