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9-3. Find all negative $a$ for which the equation
$$
\frac{6 \pi a - \arcsin (\sin x) + 2 \arccos (\cos x) - a x}{\tan^2 x + 4} = 0
$$
has exactly three solutions in the set $[\pi, +\infty)$. In your answer, provide the sum of all such $a$ (if no such $a$ exists, indicate 0; if the number $a$ is not an integer, round... | Solution. For $a=-1/3, a=-2/3, a=-3/5$. The sum is $-1.6$.
Answer: $-1.6$. | -1.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,457 |
9-4. Find all negative $a$ for which the equation
$$
\frac{8 \pi a-\arcsin (\sin x)+3 \arccos (\cos x)-a x}{3+\operatorname{tg}^{2} x}=0
$$
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... | 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 | 5,458 |
10-1. At the base of the pyramid $S A B C D$ lies a trapezoid $A B C D$ with bases $B C$ and $A D$. Points $P_{1}$, $P_{2}, P_{3}$ belong to the side $B C$, such that $B P_{1}<B P_{2}<B P_{3}<B C$. Points $Q_{1}, Q_{2}, Q_{3}$ belong to the side $A D$, such that $A Q_{1}<A Q_{2}<A Q_{3}<A D$. Denote the points of inter... | Solution. From the properties of the trapezoid, it follows that the triangles (see Fig. 2), shaded with the same color, have the same area. From this, the equality of the sums of the areas marked
![](https://cdn.mathpix.com/cropped/2024_05_06_997c53c06135e2de9c16g-10.jpg?height=317&width=934&top_left_y=1980&top_left_x... | 2028 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,459 |
1. The villages "Upper Vasyuki" and "Lower Vasyuki" are located on the bank of a river. A steamboat takes one hour to travel from Upper Vasyuki to Lower Vasyuki, while a motorboat takes 45 minutes. It is known that the speed of the motorboat in still water is twice the speed of the steamboat (also in still water). Dete... | Answer: 90 minutes.
Solution: Let's take the distance between the villages as a unit of length (unit). Then the speed of the steamboat downstream is 1 unit/h, and the speed of the motorboat is $4 / 3$ units/h. Therefore, the own speed of the steamboat is $1 / 3$ units/h, from which the speed of the current is $2 / 3$ ... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,460 |
5. Let $\Sigma(n)$ denote the sum of the digits of the number $n$. Find the smallest three-digit $n$ such that $\Sigma(n)=\Sigma(2 n)=\Sigma(3 n)=\ldots=\Sigma\left(n^{2}\right)$ | Answer: 999.
Solution: Let the desired number be $\overline{a b c}$. Note that this number is not less than 101 (since 100 does not work). Therefore, $101 \cdot \overline{a b c}=\overline{a b c 00}+\overline{a b c}$ also has the same sum of digits. But the last digits of this number are obviously $b$ and $c$, so the s... | 999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,462 |
6. Compare the numbers
$$
\left(1+\frac{1}{1755}\right)\left(1+\frac{1}{1756}\right) \ldots\left(1+\frac{1}{2015}\right) \text { and } \sqrt{\frac{8}{7}}
$$
Indicate in the answer «1» if the first number is greater; «2», if the second number is greater; «0», if the numbers are equal. | Answer: The first one is greater.
Solution: Convert each parenthesis on the left to a common denominator, then use the inequality $\frac{n \cdot n}{(n-1)(n+1)}>1$. Then we get that the square of the first number is greater than $\frac{2015}{1755}=\frac{31}{27}>\frac{8}{7}$. | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,463 |
8. Several buses (more than three) at the beginning of the workday sequentially leave from one point to another at constant and equal speeds. Upon arrival at the final point, each of them, without delay, turns around and heads back in the opposite direction. All buses make the same number of trips back and forth, and t... | Answer: 52 or 40.
Solution: If we denote $n-$ as the number of buses and $k-$ as the number of trips, then we can calculate the number of encounters (for example, by a schedule illustrating the movement of buses). It turns out that every two buses meet $2 k-1$ times. In total, we get $n(n-1)(2 k-1)=$ 300, i.e., we nee... | 52or40 | Other | math-word-problem | Yes | Yes | olympiads | false | 5,465 |
1. Eight numbers stand in a row such that the sum of any three consecutive numbers equals 50. The first and last numbers of these eight are known. Fill in the six empty places:
11 12.
ANSWER: $11,12,27,11,12,27,11,12$ | Solution: It follows from the condition that the sequence of numbers is periodic with a period of 3. | 11,12,27,11,12,27,11,12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,466 |
3. In the test, there are 4 sections, each containing the same number of questions. Andrey answered 20 questions correctly. The percentage of his correct answers was more than 60 but less than 70. How many questions were in the test?
OTBET: 32. | Solution. According to the condition $\frac{60}{100}<\frac{20}{x}<\frac{70}{100}$, from this $28 \frac{4}{7}=\frac{200}{7}<x<\frac{100}{3}=33 \frac{1}{3}$, that is $29 \leq x \leq 33$. From the first condition of the problem, it follows that the number of questions must be divisible by 4. | 32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,467 |
4. Find all 4-digit numbers that are 7182 less than the number written with the same digits in reverse order.
ANSWER: 1909 | Solution. Let's write the desired number as $\overline{a b c d}$
Then $\overline{a b c d}=\overline{d c b a}-7182$, from which $111(d-a)+10(c-b)=798$.
Obviously, $(d-a)$ can only be 7 or 8.
In the case $d-a=7$, we get $10(c-b)=21-$ which does not fit.
In the case $d-a=8$, we get $10(c-b)=-90$, hence $b-c=9$, from w... | 1909 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,468 |
6. How many different right-angled triangles exist, one of the legs of which is equal to $\sqrt{2016}$, and the other leg and the hypotenuse are expressed as natural numbers?
ANSWER: 12. | Solution. According to the condition $c^{2}-b^{2}=a^{2}=2016$, that is, $(c-b)(c+b)=2^{5} \cdot 3^{2} \cdot 7$. The system $\left\{\begin{array}{l}c-b=n, \\ c+b=k\end{array}\right.$ (here $n-$ is one of the divisors of the number 2016, and $k=\frac{2016}{n}$) has natural solutions $c=\frac{n+k}{2}, b=\frac{k-n}{2}$, if... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,470 |
1. In a row, there are 8 numbers such that the sum of any three consecutive numbers equals 100. The first and last numbers of these eight are known. Fill in the six empty places:
$$
\text { 20, _, , , , , , } 16 \text {, }
$$
ANSWER: $20,16,64,20,16,64,20,16$ | Solution: It follows from the condition that the sequence of numbers is periodic with a period of 3. | 20,16,64,20,16,64,20,16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,471 |
2. Insert arithmetic operation signs and parentheses into the expression consisting of three numbers, $\frac{1}{8} \ldots \frac{1}{9} \ldots \frac{1}{28^{\prime}}$ so that the result of the calculations is equal to $\frac{1}{2016}$.
ANSWER: $\frac{1}{8} \times \frac{1}{9} \times \frac{1}{28}$ or $\left(\frac{1}{8}-\fr... | Solution: factorize 2016 into prime factors and notice that 2016=8x9x28. | \frac{1}{8}\times\frac{1}{9}\times\frac{1}{28} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,472 |
3. The test consists of 5 sections, each containing the same number of questions. Pavel answered 32 questions correctly. The percentage of his correct answers turned out to be more than 70 but less than 77. How many questions were in the test?
ANSWER: 45. | Solution: from the condition $0.7<32 / x<0.77$ it follows that $41<x<46$, but $x$ is a multiple of 5, so $x=45$. | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,473 |
4. Find all 4-digit numbers that are 8802 less than the number written with the same digits in reverse order.
ANSWER: 1099 | Solution: Note that $x+8802<10000$, therefore, $x<1198$, so the first digit is 1, and the second digit is 0 or 1. The sum of x+8802 ends in 1, so the last digit is 9. We get a number of the form 10a9 or 11a9, we get the equation of the form $10 \mathrm{a} 9+8802=9 \mathrm{a} 01$ or $11 \mathrm{a} 9+8802=9 \mathrm{a} 11... | 1099 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,474 |
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. ANSWER: place the numbers in sequence (other options are possible).
(b+a)=1001=7 \times 11 \times 13$. We can represent 1001 as the product of two factors $1 \times 1001=7 \times 143=11 \times 91=13 \times 77$ - the first factor must be smaller - there are only 4 options
Lomonosov Moscow St... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,476 |
1. An integer was increased by 2, as a result, its square decreased by 2016. What was the number initially (before the increase)?
ANSWER: -505. | Solution: Solve the equation $(x+2)^{2}=x^{2}-2016$. | -505 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,477 |
2. Znayka told Neznayka that to convert kilolunes to kilograms, one needs to divide the mass in kilolunes by 4 and then decrease the obtained number by $4 \%$. Neznayka decided that to convert from kilograms to kilolunes, one should multiply the mass in kilograms by 4 and then increase the obtained number by $4 \%$. By... | Solution: One kilolun constitutes $0.25 * 0.96=0.24$ kg. Therefore, in one kilogram there are $25 / 6$ kiloluns. If Nезнайка translates 1 kg, he will get $4 * 1.04=4.16$, which is $99.84\%$ of $25 / 6$. | 0.16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,478 |
4. Find the largest natural number that cannot be represented as the sum of two composite numbers.
OTBET: 11 | Solution: Even numbers greater than 8 can be represented as the sum of two even numbers greater than 2. And odd numbers greater than 12 can be represented as the sum of 9 and an even composite number. By direct verification, we are convinced that 11 cannot be represented in this way. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,480 |
5. Solve the equation $\frac{\sqrt{(-x)^{2}}+(\sqrt{-x})^{2}}{x^{2}+(-x)^{2}}=\frac{1}{2016}$ ANSWER: $x=-2016$. | Solution: Note that $\mathrm{x}<0$ from the domain, we can represent the equation as $\frac{-2 x}{2 x^{2}}=\frac{1}{2016}$. | -2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,481 |
6. How many five-digit numbers of the form $\overline{a b 16 c}$ are divisible by 16? ( $a, b, c-$ are arbitrary digits, not necessarily different).
ANSWER: 90. | Solution: Note that the first digit does not affect divisibility, hence $a=1, \ldots, 9$. On the other hand, divisibility by 8 implies that $c=0$ or 8. If $c=0$, then $\mathrm{b}$ must be even, and if $\mathrm{c}=8$ - odd. In both cases, we get 5 options, from which the total number is $9 *(5+5)=90$.
Lomonosov Moscow ... | 90 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,482 |
4. Philatelist Andrey decided to distribute all his stamps equally into 3 envelopes, but it turned out that one stamp was extra. When he distributed them equally into 5 envelopes, 3 stamps were extra; finally, when he distributed them equally into 7 envelopes, 5 stamps remained. How many stamps does Andrey have in tota... | Solution. If the desired number is $x$, then the number $x+2$ must be divisible by 3, 5, and 7, i.e., it has the form $3 \cdot 5 \cdot 7 \cdot p$. Therefore, $x=105 p-2$. Since by the condition $150<x \leq 300$, then $p=2$. Therefore, $x=208$. | 208 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,483 |
5. Solve the equation $(\sqrt{x})^{2016}+(\sqrt{1-x})^{2016}=1$
ANSWER: $x=0$ or 1. | Solution: From the domain of definition $0<=x<=1$. It is clear that 0 and 1 are valid. If $0<x<1$, then $(\sqrt{x})^{2016}+(\sqrt{1-x})^{2016}<x+(1-x)=1$, and cannot be a solution. | 0or1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,484 |
6. Find all natural numbers that are 36 times the sum of their digits. ANSWER: $324 ; 648$. | Solution. Let $S(x)$ denote the sum of the digits of the number $x$. Then the equation $x=36 \cdot S(x)$ has no solutions for $n \geq 5$, since $x \geq 10^{n-1}$, while $36 \cdot S(x) \leq 36 \cdot 9 \cdot n = 324 n$. For $36(a+b+c+d) \Leftrightarrow 964 a + 64 b > 26 \cdot 9 + 35 \cdot 9 \geq 26 c + 35 d$. For $n=3: a... | 324;648 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,485 |
2. Five runners ran a relay. If the first runner ran twice as fast, they would have spent $5 \%$ less time. If the second runner ran twice as fast, they would have spent $10 \%$ less time. If the third runner ran twice as fast, they would have spent $12 \%$ less time. If the fourth runner ran twice as fast, they would ... | Solution: If each ran twice as fast, they would run 50% faster. This means that if the 5th runner ran faster, the time would decrease by $50-5-10-12-15=8 \%$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,486 |
3. 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 possibl... | 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.
.
Solution. Let's compare the numbers $\log _{k-1} k$ and $\log _{k}(k+1)$ for all $k \geqslant 3$. Note that by the Cauchy inequality: $\sqrt{\log _{k}(k-1) \cdot \log _{k}(k+1)}<0.5 \log _{k}\left(k^{2}-1\right)<1$. Therefore, $\log _{k}(k+1)<\log _{k-1... | \log_{2010}2011>\log_{2011}2012 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,490 |
2. Several numbers form an arithmetic progression, where their sum is 63, and the first term is one and a half times the common difference of the progression. If all terms of the progression are decreased by the same amount so that the first term of the progression is equal to the common difference of the progression, ... | # Problem 2.
Answer: the difference is $21 / 8$ or 2.
Solution. Let the numbers $a_{1}, a_{2}, \ldots, a_{n}$ form an arithmetic progression with a common difference $d$. Then $a_{1}+a_{2}+\cdots+a_{n}=63, a_{1}=3 d / 2 \Rightarrow 0.5 d(n+2) n=63$.
After the described change in the condition, these numbers will aga... | 21/8or2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,491 |
3. Find the area of the figure defined on the coordinate plane by the inequality
$$
\sqrt{\arcsin y} \leqslant \sqrt{\arccos x}
$$ | # Problem 3.
Answer: $1+(\pi / 4)$ (in all versions).
Solution (for versions I-1 and I-2). The domain of admissible values of the inequality is determined by the system $\left\{\begin{array}{l}\arcsin x \geqslant 0, \\ \arccos y \geqslant 0\end{array} \Longleftrightarrow\left\{\begin{array}{l}0 \leqslant x \leqslant ... | 1+\frac{\pi}{4} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,492 |
4. The shortest distance from vertex $B$ of triangle $A B C$ to the points of the opposite side is 12. Find the sides $A B$ and $B C$ of this triangle, if $\sin \angle C=\sqrt{3} / 2$ and $A C=5$. | # Problem 4.
Answer: one side is 12, and the other is either $(5+\sqrt{501}) / 2$ or $\sqrt{229}$.
Solution. Consider three possible cases.
1) Angles $A$ and $C$ are acute. Then $\angle C=60^{\circ}$ and the height $B H$ is 12. But in this case $C H=6$ and the base $H$ of the height cannot lie on the side $A C$.
2) ... | AB=12,BC=\frac{5+\sqrt{501}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,493 |
5. Find all values of $a$, for each of which the system
$$
\left\{\begin{array}{l}
|y|+|y-x| \leqslant a-|x-1| \\
(y-4)(y+3) \geqslant(4-x)(3+x)
\end{array}\right.
$$
has exactly two solutions. | # Problem 5.
Answer: when $a=7$.
Solution. By the triangle inequality:
$$
|y|+|y-x|+|x-1| \geqslant|y-(y-x)-(x-1)|=1
$$
therefore, the first inequality can have solutions only when $a \geqslant 1$. In this case, it is equivalent to the system of inequalities
$$
-\frac{a-1}{2} \leqslant y \leqslant \frac{a+1}{2}, \... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,494 |
12. (9) On a plane, there are 9 points arranged in a $3 \times 3$ grid, as shown in the figure.
a) Lines are drawn through all possible pairs of points. How many different lines are obtaine... | Answer: a) 20 ; b) 76.
Solution: a) The number of pairs of points will be $C_{9}^{2}=36$, but lines coincide when three points lie on the same line. There are $8=3$ horizontals +3 verticals +2 diagonals such cases. Therefore, subtract $36-8 \times 2=20$.
b) The number of triplets of points is $C_{9}^{3}=84$, but not ... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,495 |
12. (9) On a plane, there are 9 points arranged in a $3 \times 3$ grid, as shown in the figure.
a) Through all possible pairs of points, lines are drawn. How many different lines are obtain... | Answer: a) 20 ; b) 76.
Solution: a) The number of pairs of points will be $C_{9}^{2}=36$, but lines coincide when three points lie on the same line. There are $8=3$ horizontals +3 verticals +2 diagonals such cases. Therefore, subtract $36-8 \times 2=20$.
b) The number of triplets of points is $C_{9}^{3}=84$, but not ... | )20;b)76 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,496 |
1. The decagon $A B C D P Q R S T U V W$ has two parallel bases: square $A B C D$ and octagon $P Q R S T U V W$, all angles of which are equal, and eight lateral faces: triangles $A P Q, B R S, C T U, D V W$ and rectangles $D A P W, A B R Q, B C T S$ and $C D V U$. It is known that the area of the section of this decag... | Answer: 3.61 . (Exact value $\sqrt{13}(m=1, k=1)$.) | 3.61 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,497 |
1. On the Island of Knights and Liars, knights always tell the truth, while liars always lie. In a school on this island, both knights and liars study in the same class. One day, the teacher asked four children: Anu, Banu, Vanu, and Danu, who among them had completed their homework. They answered:
- Anu: Banu, Vanu, a... | Answer: 1.
Solution: If Vanu is a knight, then all the others are liars.
Let Vanu be a liar. Then Danu is also a liar (since he says Vanu is a knight). And at least one of Anu and Banu must be a knight. Both of them cannot be knights, as they contradict each other. In any case, only one of the children is a knight. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,498 |
2. In triangle $\triangle A B C$, the sides $A B=5$ and $A C=6$ are known. What should the side $B C$ be so that the angle $\angle A C B$ is as large as possible? Provide the length of side $B C$, rounded to the nearest integer. | Answer: 3
Solution: Construct $AC=6$. Then the geometric locus of points $B$ will be a circle of radius 5 centered at point $A$. The angle $\angle ACB$ will be the largest when $CB$ is tangent to the circle (see figure). Then $CB \perp AB$ and by the Pythagorean theorem we get $BC=\sqrt{AC^{2}-AB^{2}}=\sqrt{11} \appro... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,499 |
1. What time between $14:10$ and $15:10$ do the clocks show when the angle between the minute and hour hands is $90^{\circ}$? | Answer: 15:10, 14 hours $27 \frac{3}{11}$ minutes $=14 \frac{5}{11}$ hours
1) If the clock shows $15: 00$, then obviously the angle between the minute and hour hands is $90^{\circ}$.
2) In $n$ minutes, the minute hand passes through an angle of $(6 n)^{\circ}$ degrees, and the hour hand through $\left(\frac{n}{2}\righ... | 14:27\frac{3}{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,504 |
3. Petya sequentially writes down integers, starting from 21, such that each subsequent number is 4 less than the previous one, and Vasya, looking at the current number, calculates the sum of all the numbers written down by this point. Which of the sums found by Vasya will be closest to 55? | Answer: 56.
First solution. Petya sequentially writes down the members of the arithmetic progression $a_{n}=$ $21-4(n-1)$, while Vasya writes down the sum of these members $S_{n}=\frac{2 \cdot 21-4(n-1)}{2} \cdot n=(21-2(n-1)) n=23 n-2 n^{2}$. The equation $23 x-2 x^{2}=55$ has the following solutions: $\frac{23 \pm \... | 56 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,506 |
4. Natural numbers $m, n$ are such that the fraction $\frac{m}{n}$ is irreducible, while the fraction $\frac{4 m+3 n}{5 m+2 n}$ is reducible. By which natural numbers can it be reduced? | # Answer: 7
Let the numbers $4 m+3 n$ and $5 m+2 n$ have a common divisor $d>1$. Then there exist natural numbers $a$ and $b$ such that the system holds:
$$
\left\{\begin{array}{l}
4 m+3 n=a d \\
5 m+2 n=b d
\end{array}\right.
$$
From this system, we can express $m$ and $n$: $7 n=d(5 a-4 b), 7 m=d(3 b-2 a)$. If 7 do... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,507 |
5. Solve the equation $\sqrt[3]{15 x+1-x^{2}}+\sqrt[3]{x^{2}-15 x+27}=4$. | Answer: $x=0, x=2, x=13, x=15$.
Let's make a substitution: $a=\sqrt[3]{15 x+1-x^{2}}, b=\sqrt[3]{x^{2}-15 x+27}$. The equation is equivalent to the system
$$
\left\{\begin{array} { l }
{ a + b = 4 , } \\
{ a ^ { 3 } + b ^ { 3 } = 2 8 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ a + b = 4 , } \\
{ ( a ... | 0,2,13,15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,508 |
7. Find the maximum value of the expression
$$
\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\ldots+\left(x_{2010}-x_{2011}\right)^{2}+\left(x_{2011}-x_{1}\right)^{2}
$$
for $x_{1}, \ldots, x_{2011} \in[0 ; 1]$.
# | # Answer: 2010.
Let's prove this using mathematical induction for $2n+1$ numbers $x_{1}, \ldots, x_{2n+1} \in [0; 1]$. Specifically, we will show that the maximum value of the expression
$$
\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\ldots+\left(x_{2n}-x_{2n+1}\right)^{2}+\left(x_{2n+1}-x_{1}\right)^{2... | 2010 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,509 |
8. Solve the system
$$
\left\{\begin{array}{l}
5 x^{2}+3 y^{2}+3 x y+2 x z-y z-10 y+5=0 \\
49 x^{2}+65 y^{2}+49 z^{2}-14 x y-98 x z+14 y z-182 x-102 y+182 z+233=0
\end{array}\right.
$$ | Answer: $(0,1,-2),\left(\frac{2}{7}, 1,-\frac{12}{7}\right)$.
Consider the second equation of the system as a quadratic equation in terms of the difference $(x-z)$:
$$
49(x-z)^{2}-14(x-z)(y+13)+65 y^{2}-102 y+233=0 .
$$
$\frac{D}{4}=49(y+13)^{2}-49\left(65 y^{2}-102 y+233\right)=-(56)^{2}(y-1)^{2} \leqslant 0$.
The... | (0,1,-2),(\frac{2}{7},1,-\frac{12}{7}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,510 |
9. In a tetrahedron, all plane angles at one vertex are right angles. A certain point in space is at a distance of 3 from the specified vertex of the tetrahedron, and at distances of $\sqrt{5}, \sqrt{6}$, and $\sqrt{7}$ from the other vertices. Find the distances from the center of the circumscribed sphere around the t... | Answer: $\frac{\sqrt{3}}{2}, 1, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{\sqrt{13}}$.
Let $ABCD$ be the given tetrahedron, and all plane angles at vertex $D$ are right angles. Let the lengths of segments $DA, DB$, and $DC$ be $a, b$, and $c$ respectively. We associate the rays $DA, DB$, and $DC$ with the rectangular coordi... | \frac{\sqrt{3}}{2},1,\frac{\sqrt{2}}{2},\frac{\sqrt{3}}{\sqrt{13}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,511 |
10. There are 12 pencils of pairwise different lengths. In how many ways can they be placed in a box in 2 layers of 6 pencils each, so that in each layer the pencils are arranged in increasing order of length (from left to right), and each pencil in the upper layer lies strictly above a pencil in the lower layer and is... | Answer: 132
Let $0 \leqslant m \leqslant n \leqslant 6$. Denote by $K(m, n)$ the set of all arrangements of $m+n$ pencils of different lengths with the conditions: 1) in the bottom row, there are $n$ pencils, starting from the right edge without gaps in decreasing order of length; 2) in the top row, there are also $m$... | 132 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,512 |
12. (9) On a plane, there are 9 points arranged in a $3 \times 3$ grid, as shown in the figure.
a) Through all possible pairs of points, lines are drawn. How many different lines are obtain... | Answer: a) 20 ; b) 76.
Solution: a) The number of pairs of points will be $C_{9}^{2}=36$, but lines coincide when three points lie on the same line. There are $8=3$ horizontals +3 verticals +2 diagonals such cases. Therefore, subtract $36-8 \times 2=20$.
b) The number of triplets of points is $C_{9}^{3}=84$, but not ... | )20;b)76 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,513 |
1. Six natural numbers (possibly repeating) are written on the faces of a cube, such that the numbers on adjacent faces differ by more than 1. What is the smallest possible value of the sum of these six numbers? | Answer: 18.
Solution: Consider three faces that share a common vertex. The numbers on them differ pairwise by 2, so the smallest possible sum would be for $1+3+5=9$. The same can be said about the remaining three faces. | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,514 |
2. The Slytherin faculty has 30 students. Some of them are friends (friendship is mutual), but there are no 3 people who are all friends with each other. On New Year's, everyone sent cards to all their friends. What is the maximum number of cards that could have been sent? | Answer: 450.
Solution: Let's find the person with the most friends. Suppose there are no fewer than 15, and denote their number as $15+a$. We will divide the students into two groups: the first group will consist of these $15+a$ students. According to the condition, they cannot be friends with each other, so each of t... | 450 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,515 |
3. Given an isosceles right triangle with legs equal to $\sqrt{2}$. Find the straight-line cut of the smallest length that divides the area of the triangle in half.
In the answer, specify the length of the cut, rounded to 2 decimal places after the decimal point. Answer: $0.91$. | Solution: Let in triangle $ABC$ angle $B$ be a right angle. The cut intersects two sides of the triangle at points $P$ and $Q$.
First, assume that these points are located on the legs. Let $x = BP, y = BQ$.
Then $\frac{1}{2} xy = \frac{1}{2} S(ABC) = \frac{1}{2}$, and $PQ^2 = x^2 + y^2 = x^2 + \frac{1}{x^2} = \left(x... | 0.91 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,516 |
4. Given a sequence of natural numbers $a_{n}$, the terms of which satisfy the relations $a_{n+1}=k \cdot \frac{a_{n}}{a_{n-1}}$ (for $n \geq 2$). All terms of the sequence are integers.
It is known that $a_{1}=1$, and $a_{2018}=2020$. Find the smallest natural $k$ for which this is possible. | Answer: 2020
Solution: Let a2=x. Then all terms of the sequence will have the form $x^{m} k^{n}$.
The powers of $k$ will repeat with a period of 6: $0,0,1,2,2,1,0,0, \ldots$
The powers of $x$ will also repeat with a period of 6: $0,1,1,0,-1,-1,0,1, \ldots$
Since 2018 gives a remainder of 2 when divided by 6, then $... | 2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,517 |
5. Find the maximum possible value of the expression $x y(x-y)$, given that $x, y \in$ $[0 ; 1]$ | Answer: 0.25.
Solution: Consider $x y(x-y)=-x y^{2}+x^{2} y$ as a quadratic function of $y$. Its maximum is achieved at the vertex of the corresponding parabola $y_{0}=\frac{x}{2}$. Clearly, it is located on the given segment. Substituting into the expression, we get $-x y_{0}^{2}+x^{2} y_{0}=\frac{x^{3}}{4}$. Clearly... | 0.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,518 |
6. Points $A_{1}, \ldots, A_{12}$ are the vertices of a regular 12-gon. How many different 11-segment open broken lines without self-intersections with vertices at these points exist? Broken lines that can be transformed into each other by rotation are considered the same. | Answer: 1024.
Solution: The first vertex of the broken line can be chosen in 12 ways. Each subsequent vertex (except the last one) can be chosen in two ways - it must be adjacent to the already marked vertices to avoid self-intersections. The last vertex is chosen uniquely. We get $12^{*} 2^{10}$ ways. Considering 12 ... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,519 |
7. Find the smallest three-digit number with the property that if a number, which is 1 greater, is appended to it on the right, then the result (a six-digit number) will be a perfect square. Answer: 183 | Solution: Let the required number be \(a\), then \(1000a + a + 1 = n^2\). We can write it as: \(1001a = (n - 1)(n + 1)\). Factorize \(1001 = 7 \times 11 \times 13\), so the product \((n - 1)(n + 1)\) must be divisible by 7, 11, and 13. Moreover, for the square to be a six-digit number, \(n\) must be in the interval \([... | 183 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,520 |
8. Solve the equation $\sqrt[3]{x+2}+\sqrt[3]{3 x-1}=\sqrt[3]{16 x+4}$. In the answer, specify the root if there is only one, and the sum of the roots if there are several. | Answer: 1.25
Solution: Let $a=\sqrt[3]{x+2}, b=\sqrt[3]{3 x-1}$.
Then $a+b=\sqrt[3]{4\left(a^{3}+b^{3}\right)}$. Raising both sides to the cube, we get $(a+b)^{3}=4\left(a^{3}+b^{3}\right)$. Moving to one side and factoring, we have $(a+b)(a-b)^{2}=0$.
Therefore, either $a+b=0$, in which case $x+2=1-3 x$, leading to... | 1.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,521 |
14. 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 | 5,522 |
15. A right-angled triangle was cut along a straight line into two parts and these parts were assembled into a square (see fig). What is the length of the shorter leg if the longer leg is 10?
... | # Answer 5.
Solution: The short leg is equal to the side of the square, and the long leg is twice the side of the square. | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,523 |
17. 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 | 5,525 |
18. 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
$c=1)$ The... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,526 |
19. Find the last two digits of the sum
$$
1^{2}+2^{2}+\ldots+50^{2}-51^{2}-\ldots-100^{2}+101^{2}+\ldots 150^{2}-151^{2}-\ldots 200^{2}+\ldots-2000^{2}+2001^{2}+\ldots+2017^{2}
$$
(i.e., 50 numbers with a plus sign, 50 with a minus sign, and so on.) | # Answer: 85
Solution: Note that the numbers $\mathrm{n}^{2}$ and $(\mathrm{n}+50)^{2}$ give the same remainder when divided by 100. Therefore, in each hundred, the sum will end in two zeros. The last digits of the squares from 2001 to 2017 are:
| 01 | 04 | 09 | 16 | 25 | 36 | 49 | 64 | 81 | 00 | 21 | 44 | 69 | 96 | ... | 85 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,527 |
20. Find the sum of the fourth powers of the real roots of the equation
$$
x^{4}-1000 x^{2}+2017=0
$$ | # Answer: 1991932
Solution: The roots of the equation are of the form $\pm \sqrt{t_{1}}, \pm \sqrt{t_{2}}$, where $t_{1,2}$ are the roots of the equation $t^{2}-1000 t+2017=0$. Therefore, the sum of the fourth powers is $2\left(t_{1}^{2}+t_{2}^{2}\right)=2\left(t_{1}+t_{2}\right)^{2}-4 t_{1} t_{2}=2 \cdot 1000^{2}-4 \... | 1991932 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,528 |
1. 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 at least 121. It is sufficient to check $p_{1}=2,3,5,7$. For 2, we get the second factor: 3, 5, 7, 11, 1... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,529 |
2. Arrange the smallest square area using square tiles of sizes $1 \times 1$, $2 \times 2$, and $3 \times 3$, such that the number of tiles of each size is the same. | Answer:
Solution. Let $n$ be the number of squares of each type. Then $n + 4n + 9n = 14n$ must be a perfect square. The smallest $n$ for which this is possible is 14. The figure shows an exa... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,530 |
3. The company conducted a survey among its employees - which social networks they use: VKontakte or Odnoklassniki. Some employees said they use VKontakte, some - Odnoklassniki, some said they use both social networks, and 40 employees said they do not use social networks. Among all those who use social networks, 75% u... | Answer: 540
Solution: Since 75% of social media users use VKontakte, it follows that only 25% use Odnoklassniki. Additionally, 65% use both networks, so in total, Odnoklassniki is used by $65+25=90\%$ of social media users. These $90\%$ constitute $5 / 6$ of the company's employees, so $100\%$ constitutes $10 / 9 * 5 ... | 540 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,531 |
4. In a regular 2017-gon, all diagonals are drawn. Petya randomly selects some $\mathrm{N}$ diagonals. What is the smallest $N$ such that among the selected diagonals, there are guaranteed to be two of the same length? | Answer: 1008.
Solution: Let's choose an arbitrary vertex and consider all the diagonals emanating from it. There are 2014 of them, and by length, they are divided into 1007 pairs. Clearly, by rotating the polygon, any of its diagonals can be aligned with one of these. Therefore, there are only 1007 different sizes of ... | 1008 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,532 |
5. In how many ways can the number 10000 be factored into three natural factors, none of which is divisible by 10? Factorizations that differ only in the order of the factors are considered the same. | Answer: 6.
Solution: Each of the factors should only include powers of 2 and 5 (they cannot be included simultaneously, as it would be a multiple of 10). There can be two factors that are powers of two, in which case the third factor is 625. Or conversely, two factors are powers of five, and the third factor is 16. In... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,533 |
7. How many zeros does the number
$$
(1 \underbrace{000 \ldots 001}_{2017 \text { zeros }})^{2017}-1 ?
$$
end with? | Answer 2018
Solution: Factorize:
$(1 \underbrace{000 \ldots 001}_{2017 \text { zeros }})^{2017}-1=(1 \underbrace{000 \ldots 00}_{2017 \text { zeros }} 1-1) \times(1 \underbrace{000 \ldots 00}_{2017 \text { zeros }} 1^{2016}+1 \underbrace{000 \ldots 00}_{2017 \text { zeros }} 1^{2015}+$
$\ldots .+1 \underbrace{000 \l... | 2018 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,535 |
8. Find $a$ such that the sum of the squares of the real roots of the equation $x^{4}+a x^{2}-$ $2017=0$ is 4. | Answer: $1006.5$.
Solution: Let's make the substitution $t=x^{2}$. The equation $t^{2}+a t-2017=0$ has two roots, the product of which is -2017 according to Vieta's theorem. Therefore, one of them is negative. Let $t_{1}>0$, then the roots of the original equation are $\pm \sqrt{t_{1}}$. The sum of their squares is $2... | 1006.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,536 |
1. Vovochka approached a slot machine, on the screen of which the number 0 was lit. The game rules stated: «The screen shows the number of points. If you throw a 1 ruble coin, the number of points will increase by 1. If you throw a 2 ruble coin, the number of points will double. If you score 50 points, the machine will... | # Answer: 11 rubles.
Solution: Let's try to solve it from the end - how to get from 50 to 1 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 take 4 two-ruble and 3 one-ruble coins. Obviously, if you use 3 two-ruble coins and fewer than 5 ... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,537 |
2. 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 is $9^{4}=6561$. Of these, 9 98x7x6=3024 consist of different digits. Therefore, 6561-3024=3537 passwords contain identical digits. | 3537 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,538 |
3. 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 | 5,539 |
4. In trapezoid $A B C D$ with bases $A D / / B C$
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: Triangles ADE and CBE are similar, their areas are in the ratio of the square of the similarity coefficient. Therefore, this coefficient is 2. This means that point E d... | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,540 |
5. Twenty seventeen 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 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,541 |
6. 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:
$* * * . . . * * 2 \times 2=* * * . . * * 4$
$* * * . . . * 42 \times 2=* * * . . . * 84$
$* * * \ldots * 842 \times 2=* * * \ldots * 684$
$* * * . . * 6842 \times 2=* * * \... | 105263157894736842 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,542 |
7. A point $M$ is chosen on the diameter $A B$.
Points $C$ and $D$, lying on the circle on the same side of $AB$ (see figure), are chosen such that $\angle A M C = \angle B M D = 30^{\circ}$. Find the diameter of the circle if it is known that
^{11} \cdot \sqrt{3 x-x^{2}}=0$. | Answer: $x=0, x=2$
Solution: The domain of this equation is the interval [0;2]. On this interval, both terms are non-negative. Their sum is zero only when both are zero. The first term is zero at $x=-1,0,2$, and the second term is zero at $x=0,2,3$. Only 0 and 2 coincide.
## Variant 2-b | 0,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,544 |
1. Find the smallest $n \geq 2017$, such that $1^{\mathrm{n}}+2^{\mathrm{n}}+3^{\mathrm{n}}+4^{\mathrm{n}}$ is not divisible by 10. | Answer: 2020.
Solution: Powers of 1 always end in 1. The last digit of powers of 2 changes with a period of 4:
$2,4,8,6$. Powers of 3 also follow the pattern $3,9,7,1$. Powers of 4 change... | 2020 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,545 |
3. 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. Additionally, he doesn't like when two identical digits are next to each other. He also wants the first digit to match the last digit. How many options need to be tried to guarantee guessing Vasya's passwor... | Answer: 504
Solution: The password should 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 | 5,547 |
4. 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 | 5,548 |
5. 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 $\mathrm{x}$ 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 the one hand, $x+$ $x+1+\cdots+x+9=10 x+45 \leq 150$, from which $x \leq 10$. On the ot... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,549 |
7. Solve the equation $\sqrt{x}\left(x^{2}-3 x-4\right)^{3}+\sqrt{4-x}\left(x^{2}-8 x\right)^{5}=0$. | Answer: $x=0 ; x=4$.
Solution The domain of definition (ODZ) of the given equation is the interval [0;4]. On this interval, the quadratic trinomials $x^{2}-3 x-4 n x^{2}-8 x$ do not exceed 0. The sum of two non-positive terms equals 0 only if each of them is 0. The first one is 0 at $x=-1,0,4$. The second one is 0 at ... | 0;4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,551 |
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 | 5,552 |
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, which means no more than 49 participants could have won 2 games (someone must lose 2 games to them).
We will show that there could have been 49. Let's say №3 won against №1 and №2, №5 - against №3 and №... | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,553 |
3. 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, because otherwise the product will be at least 121. It is sufficient
. Then all the grid lines inside the triangle were outlined. What is the maximum number of triangles that can be found in this drawing?
# | # Answer: 28 triangles
Solution: One of the sides of the triangle must be inclined, i.e., lie on the segment BC. If we fix some diagonal segment, the remaining vertex is uniquely determined. That is, we need to choose 2 points out of 8, which can be done in $7 \times 8 \backslash 2=28$ ways. | 28 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,555 |
5. In trapezoid $A B C D$ with bases $A D$ and $B C$, the angle $B C D$ is known to be $120^{\circ}$. A circle with radius 1 is inscribed in this angle, passing through points $A, B$ and $D$. Find the area of triangle $A B D$. Answer: $\frac{\sqrt{3}}{4}$. | Solution: Triangle BCD is isosceles, the angles at the base are $(180-120)/2=30^{\circ}$. Therefore, $\angle B D A=$ $\angle C B D=30^{\circ}$. On the other hand, $\angle B O D=$ $360-90-90-\angle B C D=60^{\circ}$. Then the angle
 \times(П+B+\Gamma+1)$ holds. | Answer: 156.
Solution: Note that $P+B+\Gamma \geq 3$ and $\leq 24$ (since the digits are different). Moreover, the numbers $\overline{P B \Gamma}$ and $(P + B + \Gamma)$ should give the same remainder when divided by 9. This is only possible when $P+B+\Gamma$ is a multiple of 3. Note that $P+B+\Gamma=9$ does not work,... | 156 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,557 |
7. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$. We will call a point "equidistant" if there exist two vertices of the cube for which this point is the midpoint of the segment. How many "equidistant" points are there in the cube | Answer: 19.
Solution: Midpoints of 12 edges, centers of 6 faces, and the center of the cube. | 19 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,558 |
8. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-x-3=0$. Find $\left(x_{1}^{5}-20\right) \cdot\left(3 x_{2}^{4}-2 x_{2}-35\right)$. | Answer: -1063.
Solution: If $x$ is one of the roots of the equation, then $x^{2}=x+3$. Squaring, we get $x^{4}=$ $x^{2}+6 x+9=7 x+12$. Multiplying by $x$, we get $x^{5}=7 x^{2}+12 x=19 x+21$. Substituting the roots of the equation for $x$, we get: $\left(x_{1}^{5}-20\right) \cdot\left(3 x_{2}^{4}-2 x_{2}-35\right)=\le... | -1063 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,559 |
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 | 5,560 |
3. We will 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 $\mathrm{k}=0,1$; if $\mathrm{p}=7$, then $k=0$. In total, there are 7 options.
 was drawn. Then all the grid lines inside the triangle were outlined. What is the maximum number of rectangles that can be found in this drawing? | Answer: 126
Solution: For each cell, find the number of rectangles in which this cell is the top right corner. This is not difficult to do, you can simply multiply the cell number horizontally and vertically (if starting from the lower left corner and numbering from one).
| | | | | | | |
| :--- | :--- | :--- |... | 126 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,562 |
8. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-x-4=0$. Find $\left(x_{1}^{5}-20 x_{1}\right) \cdot\left(x_{2}^{4}+16\right)$. | Answer 76
Solution: If $x$ is one of the roots of the equation, then $x^{2}=x+4$. Squaring, we get $x^{4}=$ $x^{2}+8 x+16=9 x+20$. Multiplying by $x$, we get $x^{5}=9 x^{2}+20 x=29 x+36$. Substituting $x$ with the roots of the equation, we get: $\left(x_{1}^{5}-20 x_{1}\right) \cdot\left(x_{2}^{4}+16\right) .=\left(29... | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,566 |
8. Find $\sqrt{\frac{x}{63}-32} \times \sqrt{\frac{y}{63}-32}$, given that $\frac{1}{x}+\frac{1}{y}=\frac{1}{2016}$.
ANSWER: 32. | Solution: Let's make the substitution: $a=x / 63, b=y / 63$. Then we can rewrite the condition as: find $\sqrt{(a-32) \cdot(b-32)}$, given that $\frac{1}{a}+\frac{1}{b}=\frac{1}{32}$, i.e., $a b=32(a+b)$. Transforming: $\sqrt{(a-32) \cdot(b-32)}=\sqrt{a b-32(a+b)+1024}=32$. | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,573 |
1. In the village of Big Vasyuki, there is a riverside promenade 50 meters long, running along the river. A boat 10 meters long passes by it in 5 seconds when traveling downstream and in 4 seconds when traveling upstream. How many seconds will it take for a paper boat to float from one end of the promenade to the other... | Answer: $33 \frac{1}{3}$ sec.
Solution: The speed of the boat downstream is $15 \mathrm{~m} / \mathrm{c}$, and upstream it is $12 \mathrm{~m} /$ s, so the speed of the current is $1.5 \mathrm{~m} / \mathrm{c}$. Therefore, the boat will pass by her in 50/1.5 $=33 \frac{1}{3}$ sec.
Comment for graders: The times for tr... | 33\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,574 |
2. 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 obvious 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 | 5,575 |
3. Find the smallest $n>2016$, such that $1^{n}+2^{n}+3^{n}+4^{n}$ is not divisible by 10. | Answer: 2020.
Solution: Powers of 1 always end in 1. The last digit of powers of 2 changes with a period of 4: 2,4,8,6. Powers of 3 also change with a period of 4: 3,9,7,1. Powers of 4 change with a period of 2: 4,6,4,6. That is, the last digit will repeat with a period of 4. Checking for $n=1,2,3$ we get that the las... | 2020 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,576 |
4. Masha has 2 kg of "Swallow" candies, 3 kg of "Truffle" candies, 4 kg of "Bird's Milk" candies, and 5 kg of "Citron" candies. What is the maximum number of New Year's gifts she can make if each gift must contain 3 different types of candies, 100 grams of each? | Answer: 45
Solution: Even if Masha puts "Citron" in all the gifts, she will still have 2+3+4 = 9 kg of candies left, and in each gift, she must put at least 200g (and if she doesn't put Citron in all of them, then more than 200g). This means there can be no more than 45 gifts. 45 gifts can be made if she creates 5 gif... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,577 |
5. 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 | 5,578 |
6. 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 ... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,579 |
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