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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. 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 |
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. 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 |
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 |
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 |
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 |
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. 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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
$$
\left\{\begin{array} { l }
{ l = v \cdot t _ { 1 } } \\
{ a + l = v \c... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.2 Mishа noticed that the tram passed by him in 4 seconds, and a tunnel 64 meters long in 12 seconds. Find the length of the tram (in meters), assuming that its speed remains constant throughout the entire observation period. | Answer: 32. (E)
 | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.3 Mishа noticed that the tram passed by him in 2 seconds, and a tunnel 96 meters long - in 10 seconds. Find the speed of the tram (in meters per second), assuming that it remains the same throughout the entire observation period. | Answer: 12. (B)
 | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.1 Find $f(2013)$, if for any real $x$ and $y$ the equality holds
$$
f(x-y)=f(x)+f(y)-2xy
$$ | Solution. Substitute $x=y=0$. We get $f(0)=2 f(0)+0$, from which we obtain that $f(0)=0$. Substitute $x=y$. We get $0=f(0)=f(x)+f(x)-2 x^{2}$. Hence, $f(x)=x^{2}$.
Answer: 4052169. (C)
 | 4052169 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3-1. Calculate the sum
$$
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 |
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 |
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 |
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 |
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 |
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 |
5-1. 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 |
5-2. 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 |
5-3. 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 |
5-4. 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 |
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 |
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 |
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 |
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 |
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
$ 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 |
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 |
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 |
6. How many different right-angled triangles exist, one of the legs of which is equal to $\sqrt{1001}$, and the other leg and the hypotenuse are expressed as natural numbers?
ANSWER: 4. | Solution: Let's write down the Pythagorean theorem: $a^{2}+1001=b^{2}$. From this, we get $(b-a)(b+a)=1001=7 \times 11 \times 13$. 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 |
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 |
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 |
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 |
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 |
6. Find the smallest natural number $N$ such that the number $99 N$ consists only of threes.
ANSWER: 3367. | Solution. The number 33N must consist of all ones. A number is divisible by 33 if it is divisible by 3 and by 11. A number consisting of all ones is divisible by 3 if the number of ones is a multiple of 3, and it is divisible by 11 if the number of ones is a multiple of 2. The smallest such number is 111111, so 33N = 1... | 3367 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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 |
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 |
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.
$$
From this system, we can express $m$ and $n$: $7 n=d(5 a-4 b), 7 m=d(3 b-2 a)$. If 7 do... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. Find the maximum value of the expression
$$
\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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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. 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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. 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 |
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 |
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 |
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 |
6. Find all three-digit numbers $\overline{\Pi B \Gamma}$, consisting of distinct digits П, B, and Г, for which the equality $\overline{\Pi B \Gamma}=(П+B+\Gamma) \times(П+B+\Gamma+1)$ 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
2. Six numbers are given, the arithmetic mean of which is equal to some number A. Petrov calculated the arithmetic mean of the first four numbers - it turned out to be A+10. Vasechkin calculated the arithmetic mean of the last four numbers - it is A - 7. In which direction and by how much does the arithmetic mean of th... | Answer: 3 less.
Solution - similar to option v3a. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 |
1. If you open the cold water tap, the bathtub will fill up in 10 minutes, if you open the hot water tap, it will take 15 minutes. If you pull the plug, the bathtub will completely drain in 12 minutes. How long will it take to fill the bathtub if you open both taps and pull the plug?
# | # Answer: 12 min.
Solution: Let's take one bathtub as a unit of volume. The cold water tap flows at a rate of $1 / 10$ of a bathtub per minute, and the hot water tap flows at a rate of $1 / 15$ of a bathtub per minute. Water drains from the drain hole at a rate of $1 / 12$ of a bathtub per minute. Therefore, the total... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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