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2.1. Find the integer part of the number $a+\frac{9}{b}$, where $a$ and $b-$ are respectively the integer and fractional part of the number $\sqrt{76-42 \sqrt{3}}$. | The given number is $\sqrt{76-42 \sqrt{3}}=\sqrt{(7-3 \sqrt{3})^{2}}=7-3 \sqrt{3}=1+(6-3 \sqrt{3})$, where $6-3 \sqrt{3} \in(0 ; 1)$. Therefore, $a=1, b=6-3 \sqrt{3}$. Thus, $a+\frac{9}{b}=1+\frac{9}{6-3 \sqrt{3}}=1+\frac{3}{2-\sqrt{3}}=1+3(2+\sqrt{3})=7+3 \sqrt{3}$. Since $12<7+3 \sqrt{3}<13$, the integer part of the number $a+\frac{9}{b}$ is 12. | 12 | Algebra | olympiads | null | null | \(\boxed{12}\) | 2.1. Find the integer part of the number $a+\frac{9}{b}$, where $a$ and $b-$ are respectively the integer and fractional part of the number $\sqrt{76-42 \sqrt{3}}$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The given number is $\sqrt{76-42 \sqrt{3}}=\sqrt{(7-3 \sqrt{3})^{2}}=7-3 \sqrt{3}=1+(6-3 \sqrt{3})$, where $6-3 \sqrt{3} \in(0 ; 1)$. Therefore, $a=1, b=6-3 \sqrt{3}$. Thus, |
13.1. [7-8.7 (20 points), 9.8 (15 points), 10.8 (20 points)] There is a rotating round table with 16 sectors, on which numbers $0,1,2, \ldots, 7,8,7,6, \ldots, 2,1$ are written in a circle. 16 players are sitting around the table, numbered in order. After each rotation of the table, each player receives as many points as the number written on the sector they end up in after the table stops. It turned out that after 13 rotations of the table, player number 5 scored a total of 72 points, and player number 9 scored a total of 84 points. How many points did player number 1 score? | Players No. 5 and No. 9 together scored $72+84=156=12 \cdot 13$ points. In one spin, they can together score no more than 12 points. Therefore, in each of the 13 spins, they together scored 12 points. Note that the 12 points they score can be one of the sums $8+4, 7+5$, $6+6, 5+7$ or $4+8$, when the sector 8 is in front of one of them or between them, i.e., in front of one of the players with numbers $6, 7, 8$. In each of these five cases, player No. 1 scores respectively $4, 3, 2, 1$ or 0 points. This can be expressed by the formula $x_{1}=\frac{x_{5}-x_{9}}{2}+2$, where $x_{n}$ is the number of points scored by the player with number $n$ in this spin. Then the number of points scored by player No. 1 after 13 spins is $\frac{72-84}{2}+2 \cdot 13=20$. | 20 | Logic and Puzzles | olympiads | null | null | \(\boxed{20}\) | 13.1. [7-8.7 (20 points), 9.8 (15 points), 10.8 (20 points)] There is a rotating round table with 16 sectors, on which numbers $0,1,2, \ldots, 7,8,7,6, \ldots, 2,1$ are written in a circle. 16 players are sitting around the table, numbered in order. After each rotation of the table, each player receives as many points as the number written on the sector they end up in after the table stops. It turned out that after 13 rotations of the table, player number 5 scored a total of 72 points, and player number 9 scored a total of 84 points. How many points did player number 1 score?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Players No. 5 and No. 9 together scored $72+84=156=12 \cdot 13$ points. In one spin, they can together score no more than 12 points. Therefore, in each of the 13 spins, they together scored 12 points. Note that the 12 points they score can be one of the sums $8+4, 7+5$, $6+6, 5+7$ or $4+8$, when the sector 8 is in front of one of them or between them, i.e., in front of one of t |
20.1. [10.7 (15 points)] The function $f$, defined on the set of integers, satisfies the following conditions:
1) $f(1)+1>0$
2) $f(x+y)-x f(y)-y f(x)=f(x) f(y)-x-y+x y$ for any $x, y \in \mathbb{Z}$;
3) $2 f(x)=f(x+1)-x+1$ for any $x \in \mathbb{Z}$.
Find $f(10)$. | If $h(x)=f(x)+x$, then from condition 2) we get $h(x+y)=h(x) h(y)$. Then, for $x=y=0$, this equality takes the form $h(0)^{2}=h(0)$, i.e., $h(0)=0$ or $h(0)=1$. In the first case, $h(x) \equiv 0$, which is impossible due to condition 1). If $a=h(1)$, then $h(x)=a^{x}$ for any $x \in \mathbb{Z}$. From condition 3), we find $a=2$, then $h(10)=2^{10}=1024, f(10)=h(10)-10=1014$. | 1014 | Algebra | olympiads | null | null | \(\boxed{1014}\) | 20.1. [10.7 (15 points)] The function $f$, defined on the set of integers, satisfies the following conditions:
1) $f(1)+1>0$
2) $f(x+y)-x f(y)-y f(x)=f(x) f(y)-x-y+x y$ for any $x, y \in \mathbb{Z}$;
3) $2 f(x)=f(x+1)-x+1$ for any $x \in \mathbb{Z}$.
Find $f(10)$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If $h(x)=f(x)+x$, then from condition 2) we get $h(x+y)=h(x) h(y)$. Then, for $x=y=0$, this equality takes the form $h(0)^{2}=h(0)$, i.e., $h(0)=0$ or $h(0)=1$. In the first case, $h(x) \e |
20.2. The function $g$, defined on the set of integers, satisfies the conditions:
1) $g(1)>1$
2) $g(x+y)+x g(y)+y g(x)=g(x) g(y)+x+y+x y$ for any $x, y \in \mathbb{Z}$;
3) $3 g(x)=g(x+1)+2 x-1$ for any $x \in \mathbb{Z}$.
Find $g(5)$. | Let $g(1)=a$, then from condition 3) we sequentially find $g(2)=3 a-1, g(3)=$ $9 a-6, g(4)=27 a-23, g(5)=81 a-76$. From condition 2), substituting $x=4$ and $y=1$, we obtain after simplifications the equation $a^{2}-5 a+4=0$, from which $a=1$ or $a=4$. The case $a=1$ contradicts condition 1). Therefore, $a=4$, then $g(5)=81 a-76=81 \cdot 4-76=248$. | 248 | Algebra | olympiads | null | null | \(\boxed{248}\) | 20.2. The function $g$, defined on the set of integers, satisfies the conditions:
1) $g(1)>1$
2) $g(x+y)+x g(y)+y g(x)=g(x) g(y)+x+y+x y$ for any $x, y \in \mathbb{Z}$;
3) $3 g(x)=g(x+1)+2 x-1$ for any $x \in \mathbb{Z}$.
Find $g(5)$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $g(1)=a$, then from condition 3) we sequentially find $g(2)=3 a-1, g(3)=$ $9 a-6, g(4)=27 a-23, g(5)=81 a-76$. From condition 2), substituting $x=4$ and $y=1$, we obtain a |
3.1. (13 points) Ani has blue, green, and red paints. She wants to paint a wooden cube so that after painting, the cube has two faces of each color. In how many different ways can she do this? Ways of painting that can be obtained by rotating the cube are considered the same. | Red opposite red, blue opposite blue, green opposite green is one way. Red opposite red, blue opposite green is one way and there are two similar ways. Opposite each color, there is a face of one of the two remaining colors - two ways. In total, there are 6 ways to paint. | 6 | Combinatorics | olympiads | null | null | \(\boxed{6}\) | 3.1. (13 points) Ani has blue, green, and red paints. She wants to paint a wooden cube so that after painting, the cube has two faces of each color. In how many different ways can she do this? Ways of painting that can be obtained by rotating the cube are considered the same.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Red opposite red, blue opposite blue, green opposite green is one way. Red opposite red, blue opposite green is one way and there are tw |
4.1. (13 points) A guard has detained a stranger and wants to drive him away. But the person caught said that he had made a bet with his friends for 100 coins that the guard would not drive him away (if the guard does, he has to pay his friends 100 coins, otherwise they pay him), and, deciding to bribe the guard, offered him to name a sum. What is the largest number of coins the guard can ask for so that the stranger, guided solely by his own profit, would definitely pay the guard? | If the guard asks for 199 coins, then by agreeing to give him this amount, the outsider will win the dispute and receive 100 coins. In total, he will lose 99 coins. If the outsider refuses, he will lose the dispute and lose 100 coins, which is less favorable (by 1 coin) for the one caught. If the guard demands 200, the outsider might refuse, as there is no difference in benefit. If the guard demands more, it is more advantageous for the outsider to refuse. The guard can ask for less, but the problem requires finding the largest amount. Thus, the answer is 199 coins. | 199 | Logic and Puzzles | olympiads | null | null | \(\boxed{199}\) | 4.1. (13 points) A guard has detained a stranger and wants to drive him away. But the person caught said that he had made a bet with his friends for 100 coins that the guard would not drive him away (if the guard does, he has to pay his friends 100 coins, otherwise they pay him), and, deciding to bribe the guard, offered him to name a sum. What is the largest number of coins the guard can ask for so that the stranger, guided solely by his own profit, would definitely pay the guard?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If the guard asks for 199 coins, then by agreeing to give him this amount, the outsider will win the dispute and receive 100 coins. In total, he will lose 99 coins. If the outsider refuses, he will lose the dispute and lose 100 coins, which is less favorable (by 1 coin) for the one cau |
9.1. (13 points) The surface of a round table is divided into 9 identical sectors, in which the numbers from 1 to 9 are written sequentially clockwise. Around the table sit 9 players with numbers \(1, 2, \ldots, 9\), going clockwise. The table can rotate around its axis in both directions, while the players remain in place. The players are seated at equal distances from each other, so when the table stops rotating, exactly one player is opposite each sector and receives the number of coins written on that sector. It is known that after 11 rotations of the table, player ㄱo4 received 90 coins, and player № 8 received 35 coins. How many coins did player № 1 receive? | If, as a result of spinning the table, one coin went to someone with numbers from 5 to 8, then player No. 8 received 5 fewer coins than player No. 9, and if one coin went to someone else, then No. 8 received 4 more coins than player No. 4. Let the number of spins where one coin went to someone with numbers from 5 to 8 be exactly $k$. Then we get the equation $-5k + 4(11 - k) = 35 - 90$, from which $k = 11$. This means that in all 11 spins, each player from 9 to 4 (if moving clockwise) received exactly 11 more coins than the previous one. Therefore, the first player has $35 + 11 + 11 = 57$ coins. | 57 | Combinatorics | olympiads | null | null | \(\boxed{57}\) | 9.1. (13 points) The surface of a round table is divided into 9 identical sectors, in which the numbers from 1 to 9 are written sequentially clockwise. Around the table sit 9 players with numbers \(1, 2, \ldots, 9\), going clockwise. The table can rotate around its axis in both directions, while the players remain in place. The players are seated at equal distances from each other, so when the table stops rotating, exactly one player is opposite each sector and receives the number of coins written on that sector. It is known that after 11 rotations of the table, player ㄱo4 received 90 coins, and player № 8 received 35 coins. How many coins did player № 1 receive?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If, as a result of spinning the table, one coin went to someone with numbers from 5 to 8, then player No. 8 received 5 fewer coins than player No. 9, and if one coin went to someone else, then No. 8 received 4 more coins than player No. 4. Let the number of spins where one coin went to someone with n |
4. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$ | Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace{4 \ldots 4}_{2012}=\underbrace{4 \ldots 4}_{2011} 3 \underbrace{5 \ldots 5}_{2011} 6$. The sum of the digits is $4 \cdot 2011+3+5 \cdot 2011+6=18108$. | 18108 | Number Theory | olympiads | null | null | \(\boxed{18108}\) | 4. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace |
5.1. (13 points) Find the smallest natural number that is divisible by 11 and whose representation contains 5 zeros and 7 ones. (You can use the divisibility rule for 11: a number is divisible by 11 if the difference between the sum of the digits in the even positions and the sum of the digits in the odd positions is divisible by 11.) | The number 7 is not divisible by 2, so the number cannot have exactly $5+7=12$ digits. Let's consider the case when the number has 13 digits. If the "new" digit is zero, we again get a contradiction. Therefore, the "new" digit must be at least 1. To find the smallest number, we should place this digit as far to the right as possible in the number. Let's try to find the desired number in the form $100000111111 x$. By equating the sums of the digits in the even and odd positions, we get the equation $4+x=3$, from which $x=1$, which is impossible. Let's check the number in the form $a=10000011111 x 1$. By calculating the sums of the digits in the even and odd positions, we get $5=2+x \Rightarrow x=3$. Therefore, the desired number is 1000001111131. | 1000001111131 | Number Theory | olympiads | null | null | \(\boxed{1000001111131}\) | 5.1. (13 points) Find the smallest natural number that is divisible by 11 and whose representation contains 5 zeros and 7 ones. (You can use the divisibility rule for 11: a number is divisible by 11 if the difference between the sum of the digits in the even positions and the sum of the digits in the odd positions is divisible by 11.)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The number 7 is not divisible by 2, so the number cannot have exactly $5+7=12$ digits. Let's consider the case when the number has 13 digits. If the "new" digit is zero, we again get a contradiction. Therefore, the "new" digit must be at least 1. To find the smallest number, we should place this digit as far to the right as possible in the number. Let's try to find the desire |
7.1. (13 points) Yura has unusual clocks with several minute hands moving in different directions. Yura calculated that the minute hands coincided in pairs exactly 54 times in one hour. What is the maximum number of minute hands that can be on Yura's clocks? | Let some two arrows coincide, then after 30 seconds they will coincide again. Therefore, the arrows in each such pair will coincide exactly 2 times per minute. Thus, if $n$ arrows move in one direction, and $m$ arrows move in the other, then $2 m n=54, m n=27$. Therefore, $n$ can be $1,3,9$ or 27. The largest sum $m+n$ is obtained when $n=1, m=27$ (or vice versa) and is equal to 28. | 28 | Combinatorics | olympiads | null | null | \(\boxed{28}\) | 7.1. (13 points) Yura has unusual clocks with several minute hands moving in different directions. Yura calculated that the minute hands coincided in pairs exactly 54 times in one hour. What is the maximum number of minute hands that can be on Yura's clocks?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let some two arrows coincide, then after 30 seconds they will coincide again. Therefore, the arrows in each such pair will coincide exactly 2 times per minute. Thus, if $n$ arrows move in one |
1.1. Find all pairs of two-digit natural numbers for which the arithmetic mean is $25 / 24$ times greater than the geometric mean. In your answer, specify the largest of the arithmetic means for all such pairs. | Let $a, b$ be the required numbers (without loss of generality, we can assume that $a > b$) and let $\frac{a+b}{2}=25x$ and $\sqrt{ab}=24x$. Then $(\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}=98x$ and $(\sqrt{a}-\sqrt{b})^2=$ $a+b-2\sqrt{ab}=2x$. From this, it follows that $\sqrt{a}+\sqrt{b}=7(\sqrt{a}-\sqrt{b})$. Therefore, $3\sqrt{a}=4\sqrt{b}$. Thus, $a:b=16:9$. Considering that $a$ and $b$ are two-digit numbers, the sum will be the largest for $a=16 \cdot 6=96, b=9 \cdot 6=54$. Their arithmetic mean is $\frac{96+54}{2}=75$. | 75 | Algebra | olympiads | null | null | \(\boxed{75}\) | 1.1. Find all pairs of two-digit natural numbers for which the arithmetic mean is $25 / 24$ times greater than the geometric mean. In your answer, specify the largest of the arithmetic means for all such pairs.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $a, b$ be the required numbers (without loss of generality, we can assume that $a > b$) and let $\frac{a+b}{2}=25x$ and $\sqrt{ab}=24x$. Then $(\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}=98x$ and $(\sqrt{a}-\sqrt{b})^2=$ $a+b-2\sqrt{ab}=2x$. From this, it follows t |
4.1. How many numbers from 1 to 1000 (inclusive) cannot be represented as the difference of two squares of integers? | Note that any odd number $2 n+1$ can be represented as $(n+1)^{2}-n^{2}$. Moreover, an even number that is a multiple of 4 can be represented as $4 n=(n+1)^{2}-(n-1)^{2}$. The numbers of the form $4 n+2$ remain. Note that a square can give remainders of 0 or 1 when divided by 4, so numbers of the form $4 n+2$ cannot be obtained as the difference of squares. There is exactly one such number (of the form $4 n+2$) in every set of four consecutive numbers, so the total number of such numbers from 1 to 1000 will be $1000 / 4=250$. | 250 | Number Theory | olympiads | null | null | \(\boxed{250}\) | 4.1. How many numbers from 1 to 1000 (inclusive) cannot be represented as the difference of two squares of integers?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that any odd number $2 n+1$ can be represented as $(n+1)^{2}-n^{2}$. Moreover, an even number that is a multiple of 4 can be represented as $4 n=(n+1)^{2}-(n-1)^{2}$. The numbers of the form $4 n+2$ remain. Note that a square can give remainders of 0 or 1 when |
5.1. The sequence is defined by the relations $a_{1}=1$,
$$
a_{2 n}=\left\{\begin{array}{ll}
a_{n}, & \text { if } n \text { is even, } \\
2 a_{n}, & \text { if } n \text { is odd; }
\end{array} \quad a_{2 n+1}= \begin{cases}2 a_{n}+1, & \text { if } n \text { is even, } \\
a_{n}, & \text { if } n \text { is odd. }\end{cases}\right.
$$
Find the smallest natural $n$ for which $a_{n}=a_{2017}$. | The given rules can be easily interpreted in terms of the binary system: if $n$ ends in 0 and 1 is appended to the right, then 1 is appended to the right of $a_{n}$. If $n$ ends in 1 and 0 is appended, then 0 is appended to the right of $a_{n}$. In all other cases, $a_{n}$ does not change (when 0 is appended to 0 or 1 is appended to 1). Let's write the number 2017 in binary: $2017=11111100001_{2}$. It is easy to see that $a_{2017}=101_{2}=5_{10}$. By checking the first few values, we find $a_{5}=5$. | 5 | Algebra | olympiads | null | null | \(\boxed{5}\) | 5.1. The sequence is defined by the relations $a_{1}=1$,
$$
a_{2 n}=\left\{\begin{array}{ll}
a_{n}, & \text { if } n \text { is even, } \\
2 a_{n}, & \text { if } n \text { is odd; }
\end{array} \quad a_{2 n+1}= \begin{cases}2 a_{n}+1, & \text { if } n \text { is even, } \\
a_{n}, & \text { if } n \text { is odd. }\end{cases}\right.
$$
Find the smallest natural $n$ for which $a_{n}=a_{2017}$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The given rules can be easily interpreted in terms of the binary system: if $n$ ends in 0 and 1 is appended to the right, then 1 is appended to the right of $a_{n}$. If $n$ ends in 1 and 0 is appended, then 0 is appended to the right of $a_{n}$. In all |
7.1. Natural numbers from 1 to some $n$ are written in a row. When one of the numbers was removed, it turned out that the arithmetic mean of the remaining numbers is $40 \frac{3}{4}$. Find the number that was removed. | The sum of numbers from 1 to $n$ is $S_{n}=\frac{n(n+1)}{2}$. Let the number removed be $m$, where $1 \leqslant m \leqslant n$. Then the condition of the problem can be written as
$$
\frac{S_{n}-m}{n-1}=40 \frac{3}{4}
$$
Transforming: $\frac{n+2}{2}-\frac{m-1}{n-1}=40 \frac{3}{4}$. Note that $\frac{m-1}{n-1} \leqslant 1$, so if $n$ is even, then $\frac{m-1}{n-1}=\frac{1}{4}$, and if odd, then $\frac{m-1}{n-1}=\frac{3}{4}$. In the first case, we get $n=80$, from which $m-1=79 \cdot \frac{1}{4}-$ not an integer. In the second case, $n=81$, i.e., $m-1=80 \cdot \frac{3}{4}=60$. | 61 | Number Theory | olympiads | null | null | \(\boxed{61}\) | 7.1. Natural numbers from 1 to some $n$ are written in a row. When one of the numbers was removed, it turned out that the arithmetic mean of the remaining numbers is $40 \frac{3}{4}$. Find the number that was removed.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The sum of numbers from 1 to $n$ is $S_{n}=\frac{n(n+1)}{2}$. Let the number removed be $m$, where $1 \leqslant m \leqslant n$. Then the condition of the problem can be written as
$$
\frac{S_{n}-m}{n-1}=40 \frac{3}{4}
$$
Transforming: $\frac{n+2}{2}-\frac{m-1}{n-1}=40 \frac{3}{4}$. Note t |
2.1. Once, in a company, the following conversation took place:
- We must call Misha immediately! - exclaimed Vanya.
However, no one remembered Misha's phone number.
- I remember for sure that the last three digits of the phone number are consecutive natural numbers, - said Nastya.
- And I recall that the first five digits formed a palindrome, - noted Anton.
- Seven-digit numbers are not memorized as a whole; they are broken down into three groups: first three digits, and then two groups of two digits each. I think the three-digit number obtained in this way was divisible by 9 - remarked Nikita.
- That's right, - supported Mitya, - and there were three consecutive ones in the phone number.
- Only one of the two-digit numbers obtained by Nikita's method was prime, - added Sasha.
Help the guys restore Misha's phone number. | Note that three consecutive ones cannot stand at the beginning, as 111 is not divisible by 9. Therefore, among the first three digits, there is one that is different from one. If this is the second or third digit, then since the first five form a palindrome and the last three are consecutive natural numbers, there will be no three consecutive ones in the number. Therefore, the second, third, and fourth digits are ones, then the first and fifth are sevens. This means that the first two-digit number in the Nikitin division is 17 (a prime number). The last three digits will be consecutive natural numbers if the second two-digit number is 89 or 65. But since 89 is a prime number, the only option left is 65. Therefore, the desired number is 7111765. | 7111765 | Number Theory | olympiads | null | null | \(\boxed{7111765}\) | 2.1. Once, in a company, the following conversation took place:
- We must call Misha immediately! - exclaimed Vanya.
However, no one remembered Misha's phone number.
- I remember for sure that the last three digits of the phone number are consecutive natural numbers, - said Nastya.
- And I recall that the first five digits formed a palindrome, - noted Anton.
- Seven-digit numbers are not memorized as a whole; they are broken down into three groups: first three digits, and then two groups of two digits each. I think the three-digit number obtained in this way was divisible by 9 - remarked Nikita.
- That's right, - supported Mitya, - and there were three consecutive ones in the phone number.
- Only one of the two-digit numbers obtained by Nikita's method was prime, - added Sasha.
Help the guys restore Misha's phone number.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that three consecutive ones cannot stand at the beginning, as 111 is not divisible by 9. Therefore, among the first three digits, there is one that is different from one. If this is the second or third digit, then since the first five form a palindrome and the last three are consecutive natural numbers, there will be no three consecutive ones in the number. Therefore, t |
3.1. From one point on a circular track, a pedestrian and a cyclist started simultaneously in the same direction. The cyclist's speed is $55\%$ greater than the pedestrian's speed, and therefore the cyclist overtakes the pedestrian from time to time. At how many different points on the track will the overtakes occur? | Let's assume the length of the path is 55 (in some units), and the speeds of the pedestrian and the cyclist are $100 x$ and $155 x$. Then, the overtakes will occur every $1 / x$ units of time. During this time, the pedestrian covers 100 units, which means he will be 10 units away in the opposite direction from the start. This will happen at each overtake. At the 11th overtake, the pedestrian will have covered 1100 units and will be back at the starting point, after which the overtake points will start repeating. | 11 | Combinatorics | olympiads | null | null | \(\boxed{11}\) | 3.1. From one point on a circular track, a pedestrian and a cyclist started simultaneously in the same direction. The cyclist's speed is $55\%$ greater than the pedestrian's speed, and therefore the cyclist overtakes the pedestrian from time to time. At how many different points on the track will the overtakes occur?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's assume the length of the path is 55 (in some units), and the speeds of the pedestrian and the cyclist are $100 x$ and $155 x$. Then, the overtakes will occur every $1 / x$ units of time. During this time, the pedestrian covers 100 units, which means he |
4.1. We will call a natural number interesting if all its digits, except the first and last, are less than the arithmetic mean of the two adjacent digits. Find the largest interesting number. | Let $a_{n}$ be the n-th digit of the desired number. By the condition, $a_{n}9$. Similarly, there cannot be four negative differences. Therefore, the maximum number of such differences can be 7, and the desired number is an eight-digit number. To make it the largest, the first digit should be 9, and the first differences between the digits should be as small as possible. If the differences are sequentially $3,2,1,0,-1,-2,-3$, then we get the answer: 96433469. | 96433469 | Number Theory | olympiads | null | null | \(\boxed{96433469}\) | 4.1. We will call a natural number interesting if all its digits, except the first and last, are less than the arithmetic mean of the two adjacent digits. Find the largest interesting number.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $a_{n}$ be the n-th digit of the desired number. By the condition, $a_{n}9$. Similarly, there cannot be four negative differences. Therefore, the maximum number of such differences can be 7, and the desired number is an eight-di |
5.1. Find the smallest natural solution of the inequality $2^{x}+2^{x+1}+\ldots+2^{x+2000}>2^{2017}$. Answer. 17. | The value $x=17$ is clearly a solution. If $x \leqslant 16$, then we have
$$
2^{x}+2^{x+1}+\ldots+2^{x+2000} \leqslant 2^{16}+2^{17}+\ldots+2^{2015}+2^{2016}<2^{2017}
$$
since this inequality reduces sequentially to the following: $2^{16}+2^{17}+\ldots+2^{2015}<$ $2^{2017}-2^{2016}=2^{2016}, 2^{16}+2^{17}+\ldots+2^{2014}<2^{2016}-2^{2015}=2^{2015}, \ldots, 2^{16}+2^{17}<2^{19}-2^{18}=2^{18}$, $2^{16}<2^{18}-2^{17}=2^{17}$, which is true. | 17 | Inequalities | olympiads | null | null | \(\boxed{17}\) | 5.1. Find the smallest natural solution of the inequality $2^{x}+2^{x+1}+\ldots+2^{x+2000}>2^{2017}$. Answer. 17.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The value $x=17$ is clearly a solution. If $x \leqslant 16$, then we have
$$
2^{x}+2^{x+1}+\ldots+2^{x+2000} \leqslant 2^{16}+2^{17}+\ldots+2^{2015}+2^{2016}<2^{2017}
$$
since this inequality reduces sequentially to the |
6.1. How many triangles with integer sides have a perimeter equal to 27? (Triangles that differ only in the order of the sides - for example, $7,10,10$ and $10,10,7$ - are considered the same triangle.) | Arrange the sides in ascending order: $a \leqslant b \leqslant c$. Then the smaller side $a$ does not exceed 9. If $a=1$ or $a=2$, there is one such triangle. If $a=3$ or $a=4$, there are two such triangles. If $a=5$ or $a=6$, there are three such triangles. If $a=7$, there are four such triangles. If $a=8$, there are two such triangles. If $a=9$, there is one such triangle. In total, we get 19 triangles. | 19 | Number Theory | olympiads | null | null | \(\boxed{19}\) | 6.1. How many triangles with integer sides have a perimeter equal to 27? (Triangles that differ only in the order of the sides - for example, $7,10,10$ and $10,10,7$ - are considered the same triangle.)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Arrange the sides in ascending order: $a \leqslant b \leqslant c$. Then the smaller side $a$ does not exceed 9. If $a=1$ or $a=2$, there is one such triangle. If $a=3$ or $a=4$, there are two such triangl |
2. First-grader Petya was laying out a contour of an equilateral triangle with the chips he had, so that each of its sides, including the vertices, contained the same number of chips. Then, with the same chips, he managed to lay out the contour of a square in the same way. How many chips does Petya have, if each side of the square contains 2 chips less than each side of the triangle? | Let the side of the triangle contain $x$ chips, and the side of the square - $y$ chips. The total number of chips, counted in two ways, is $3 x-3=4 y-4$ (we account for the corner chips being counted twice). From the problem statement, it follows that $y=x-2$. Therefore, we get the equation $3(x-1)=4(x-3)$, from which $x=9$. Thus, the total number of chips is $3 \cdot 9-3=24$. | 24 | Number Theory | olympiads | null | null | \(\boxed{24}\) | 2. First-grader Petya was laying out a contour of an equilateral triangle with the chips he had, so that each of its sides, including the vertices, contained the same number of chips. Then, with the same chips, he managed to lay out the contour of a square in the same way. How many chips does Petya have, if each side of the square contains 2 chips less than each side of the triangle?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let the side of the triangle contain $x$ chips, and the side of the square - $y$ chips. The total number of chips, counted in two ways, is $3 x-3=4 y-4$ (we account for the corner chips bein |
4. Kolya started playing $W o W$ at the moment when the hour and minute hands were opposite. He finished playing after an integer number of minutes, and at the end, the minute hand coincided with the hour hand. How long did he play (if it is known that he played for less than 12 hours) | The minute hand catches up with the hour hand at a speed of $\frac{11^{\circ}}{2}$ /min. For them to coincide, the difference in their angles of rotation should be $180+360 k$. This value is a multiple of 11 when $k=5,16, \ldots$. According to the problem, only $k=5$ fits, which gives us 6 o'clock. | 6 | Logic and Puzzles | olympiads | null | null | \(\boxed{6}\) | 4. Kolya started playing $W o W$ at the moment when the hour and minute hands were opposite. He finished playing after an integer number of minutes, and at the end, the minute hand coincided with the hour hand. How long did he play (if it is known that he played for less than 12 hours)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The minute hand catches up with the hour hand at a speed of $\frac{11^{\circ}}{2}$ /min. For them to coincide, the difference in their angles of rotat |
2.1. (16 points) A team consisting of juniors and masters from the "Vimpel" sports society went to a shooting tournament. The average number of points scored by the juniors turned out to be 22, by the masters - 47, and the average number of points for the entire team - 41. What is the proportion (in percent) of masters in this team | Let there be $x$ juniors and $y$ masters in the team. Then the total number of points scored by the team is $22x + 47y = 41(x + y)$, from which we find $19x = 6y$. Therefore, the proportion of masters is $\frac{y}{x+y} = \frac{19y}{19x + 19y} = \frac{19y}{25y} = 0.76$, i.e., $76\%$. | 76 | Algebra | olympiads | null | null | \(\boxed{76}\) | 2.1. (16 points) A team consisting of juniors and masters from the "Vimpel" sports society went to a shooting tournament. The average number of points scored by the juniors turned out to be 22, by the masters - 47, and the average number of points for the entire team - 41. What is the proportion (in percent) of masters in this team
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let there be $x$ juniors and $y$ masters in the team. Then the total number of points scored by the team is $22x + 47y = 41(x + y)$, from whic |
4.1. (16 points) An eraser, 3 pens, and 2 markers cost 240 rubles, while 2 erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers? | From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will cost $2 \cdot 240=480$ rubles. Therefore, one pen costs $480-440=40$ rubles. Then, 3 erasers, 4 pens, and 6 markers cost $680-4 \cdot 40=520$ rubles. | 520 | Algebra | olympiads | null | null | \(\boxed{520}\) | 4.1. (16 points) An eraser, 3 pens, and 2 markers cost 240 rubles, while 2 erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will co |
6.1. (16 points) In a hut, several inhabitants of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The inhabitants of the Ah tribe always tell the truth, while the inhabitants of the Ukh tribe always lie. One of the inhabitants said: "There are no more than 16 of us in the hut," and then added: "We are all from the Ukh tribe." Another said: "There are no more than 17 of us in the hut," and then noted: "Some of us are from the Ah tribe." The third said: "There are five of us in the hut," and, looking around, added: "There are no fewer than three inhabitants of the Ukh tribe among us." How many inhabitants of the Ah tribe are in the hut? | A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 people in the hut. The third one is from tribe Ukh, as he said there are five of them in total. But since he said there are no fewer than three residents from tribe Ukh, there are no more than two. Two have already been found (the third and the first), so there are exactly two. Then the number of residents from tribe Ah is 15. | 15 | Logic and Puzzles | olympiads | null | null | \(\boxed{15}\) | 6.1. (16 points) In a hut, several inhabitants of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The inhabitants of the Ah tribe always tell the truth, while the inhabitants of the Ukh tribe always lie. One of the inhabitants said: "There are no more than 16 of us in the hut," and then added: "We are all from the Ukh tribe." Another said: "There are no more than 17 of us in the hut," and then noted: "Some of us are from the Ah tribe." The third said: "There are five of us in the hut," and, looking around, added: "There are no fewer than three inhabitants of the Ukh tribe among us." How many inhabitants of the Ah tribe are in the hut?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 people in the hut |
7.1. (16 points) Vasya's parents allowed him to buy himself two toys. In the store, there are 7 different remote-controlled cars and 5 different construction sets. In how many ways can he choose his gift? | Vasya can choose one car and one constructor in $5 \cdot 7=35$ ways, two different cars in $-\frac{7 \cdot 6}{2}=21$ ways (one car can be chosen in 7 ways, one of the remaining in 6 ways, but in this way each pair of cars is counted exactly twice), and two constructors in $-\frac{5 \cdot 4}{2}=10$ ways. In total, we get $35+21+10=66$ ways.
The problem can also be solved differently: note that Vasya can buy $7+5=12$ different toys in the store, so a pair of them can be chosen $\frac{12 \cdot 11}{2}=66$ different ways. | 66 | Combinatorics | olympiads | null | null | \(\boxed{66}\) | 7.1. (16 points) Vasya's parents allowed him to buy himself two toys. In the store, there are 7 different remote-controlled cars and 5 different construction sets. In how many ways can he choose his gift?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Vasya can choose one car and one constructor in $5 \cdot 7=35$ ways, two different cars in $-\frac{7 \cdot 6}{2}=21$ ways (one car can be chosen in 7 ways, one of the remaining in 6 ways, but in this way each pair of cars is counted exactly twice), and two const |
1.1. Once, in a company, the following conversation took place:
- We must call Misha immediately! - exclaimed Vanya.
However, no one remembered Misha's phone number.
- I remember for sure that the last three digits of the phone number are consecutive natural numbers, - said Nastya.
- And I recall that the first five digits formed a palindrome, - noted Anton.
- Seven-digit numbers are not memorized as a whole; they are broken down into three groups: first three digits, and then two groups of two digits each. I think the three-digit number obtained in this way was divisible by 9 - remarked Nikita.
- That's right, - supported Mitya, - and there were three consecutive ones in the phone number.
- Only one of the two-digit numbers obtained by Nikita's method was prime, - added Sasha.
Help the guys restore Misha's phone number. | Note that three consecutive ones cannot stand at the beginning, as 111 is not divisible by 9. Therefore, among the first three digits, there is one that is different from one. If this is the second or third digit, then since the first five form a palindrome and the last three are consecutive natural numbers, there will be no three consecutive ones in the number. Therefore, the second, third, and fourth digits are ones, then the first and fifth are sevens. This means that the first two-digit number in the Nikitin partition is 17 (a prime number). The last three digits will be consecutive natural numbers if the second two-digit number is 89 or 65. But since 89 is a prime number, the only option left is 65. Therefore, the desired number is 7111765. | 7111765 | Logic and Puzzles | olympiads | null | null | \(\boxed{7111765}\) | 1.1. Once, in a company, the following conversation took place:
- We must call Misha immediately! - exclaimed Vanya.
However, no one remembered Misha's phone number.
- I remember for sure that the last three digits of the phone number are consecutive natural numbers, - said Nastya.
- And I recall that the first five digits formed a palindrome, - noted Anton.
- Seven-digit numbers are not memorized as a whole; they are broken down into three groups: first three digits, and then two groups of two digits each. I think the three-digit number obtained in this way was divisible by 9 - remarked Nikita.
- That's right, - supported Mitya, - and there were three consecutive ones in the phone number.
- Only one of the two-digit numbers obtained by Nikita's method was prime, - added Sasha.
Help the guys restore Misha's phone number.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that three consecutive ones cannot stand at the beginning, as 111 is not divisible by 9. Therefore, among the first three digits, there is one that is different from one. If this is the second or third digit, then since the first five form a palindrome and the last three are consecutive natural numbers, there will be no three consecutive ones in the number. Therefore, th |
Problem 1. Three friends, weightlifters A, B, and C, came to a competition. They all competed in the same weight category, and one of them became the winner. If the weight lifted by weightlifter A is added to the weight lifted by weightlifter B, the total is 220 kg. If the weights lifted by weightlifters A and C are added together, the total is 240 kg, and if the weights lifted by weightlifters B and C are added together, the total is 250 kg. What weight did the winner of the competition lift? | The sum of three weights is $(220+240+250) / 2=355$. Therefore, V lifted $355-220=135$, B lifted $355-240=115$, A lifted $355-250=105$. The winner is $\mathrm{B}=135$. | 135 | Algebra | olympiads | null | null | \(\boxed{135}\) | Problem 1. Three friends, weightlifters A, B, and C, came to a competition. They all competed in the same weight category, and one of them became the winner. If the weight lifted by weightlifter A is added to the weight lifted by weightlifter B, the total is 220 kg. If the weights lifted by weightlifters A and C are added together, the total is 240 kg, and if the weights lifted by weightlifters B and C are added together, the total is 250 kg. What weight did the winner of the competition lift?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The sum of three weights is $(220+240+250) / 2=355$. Therefore, V lifted $355-220=13 |
Problem 2. How many solutions in integers does the equation
$$
\frac{1}{2022}=\frac{1}{x}+\frac{1}{y} ?
$$ | By eliminating the denominators, we obtain the equation
\[
(x-2022)(y-2022)=2022^{2}.
\]
Since \(2022^{2}=2^{2} \cdot 3^{2} \cdot 337^{2}\), the number \(x-2022\) can have the factors \(2^{0}, 2^{1}, 2^{2}\) - a total of three options. Similarly for the other factors. The number \(x-2022\) thus has \(3 \cdot 3 \cdot 3=27\) possible natural values, which means 54 integer values. However, one of these corresponds to \(x=0\), which does not satisfy the original equation. Therefore, there are 53 roots (for each suitable value of \(x\), we have \(y=\frac{2022^{2}}{x-2022}+2022\) - an integer). | 53 | Number Theory | olympiads | null | null | \(\boxed{53}\) | Problem 2. How many solutions in integers does the equation
$$
\frac{1}{2022}=\frac{1}{x}+\frac{1}{y} ?
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
By eliminating the denominators, we obtain the equation
\[
(x-2022)(y-2022)=2022^{2}.
\]
Since \(2022^{2}=2^{2} \cdot 3^{2} \cdot 337^{2}\), the number \(x-2022\) can have the factors \(2^{0}, 2^{1}, 2^{2}\) - a total of three options. Similarly for the other factors. The number \(x-2022\) thus |
. Find the smallest natural number that has the following property: the remainder of its division by 20 is one less than the remainder of its division by 21, and the remainder of its division by 22 is 2. | The desired number is $20k + a = 21l + a + 1 = 22m + 2$, where $0 \leqslant a \leqslant 19$ and $l, k, m \geqslant 0$. From the first equation and the congruence modulo 20, we get that $l + 1 \equiv 0 \pmod{20}$. Since we are looking for the smallest number, let's try $l = 19$, if this $l$ does not work, then consider $l = 19 + 20 = 39$, and so on. We seek the desired number in the form $21 \cdot 19 + a + 1 = 22m + 2$, compare modulo 22: $3 + a + 1 \equiv 2 \pmod{22}$, hence, $a \equiv 20 \pmod{22}$, which is impossible. Now we seek the desired number in the form $21 \cdot 39 + a + 1 = 22m + 2$, so $5 + a + 1 \equiv 2 \pmod{22}$, that is, $a \equiv 18 \pmod{22}$. The number $21 \cdot 39 + 18 + 1 = 838$ satisfies the condition and is the smallest by construction. | 838 | Number Theory | olympiads | null | null | \(\boxed{838}\) | . Find the smallest natural number that has the following property: the remainder of its division by 20 is one less than the remainder of its division by 21, and the remainder of its division by 22 is 2.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The desired number is $20k + a = 21l + a + 1 = 22m + 2$, where $0 \leqslant a \leqslant 19$ and $l, k, m \geqslant 0$. From the first equation and the congruence modulo 20, we get that $l + 1 \equiv 0 \pmod{20}$. Since we are looking for the smallest number, let's try $l = 19$, if this $l$ does not work, then consider $l = 19 + 20 = 39$, and so on. We seek the desired number in the f |
Problem 3. Find the three last digits of the number $10^{2022}-9^{2022}$. | Since $A=10^{2022}-(10-1)^{2022}=10^{2022}-10^{2022}+2022 \cdot 10^{2021}-C_{2022}^{2} \cdot 10^{2022}+\ldots+$ $C_{2022}^{3} \cdot 10^{3}-C_{2022}^{2} \cdot 10^{2}+C_{2022}^{1} \cdot 10-1$, then $A(\bmod 1000) \equiv-C_{2022}^{2} \cdot 100+C_{2022}^{1} \cdot 10-1(\bmod 1000) \equiv$ $-\frac{2022 \cdot 2021 \cdot 100}{2}+20220-1(\bmod 1000) \equiv-100+220-1 \equiv 119$. | 119 | Number Theory | olympiads | null | null | \(\boxed{119}\) | Problem 3. Find the three last digits of the number $10^{2022}-9^{2022}$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since $A=10^{2022}-(10-1)^{2022}=10^{2022}-10^{2022}+2022 \cdot 10^{2021}-C_{2022}^{2} \cdot 10^{2022}+\ldots+$ $C_{2022}^{3} \cdot 10^{3}-C_{2022}^{2} \cdot 10^{2}+C_{2022}^{1} \cdot 10 |
3. And our cat gave birth to kittens yesterday! It is known that the two lightest kittens weigh a total of 80 g, the four heaviest weigh 200 g, and the total weight of all the kittens is 500 g. How many kittens did the cat give birth to? | The two lightest weigh 80 g, so the others weigh no less than 40 g each. Similarly, we find that all except the 4 heaviest weigh no more than 50 g. Consider the kittens that are not among the 2 lightest and the 4 heaviest. Their total weight is 500-200-80=220g. There must be 5 of them, because if there were 4, the weight could not be more than $4 \times 50=200$ g, and if there were 6, their weight would be no less than $6 \times 40=240$ g. Therefore, there are 11 kittens in total. | 11 | Logic and Puzzles | olympiads | null | null | \(\boxed{11}\) | 3. And our cat gave birth to kittens yesterday! It is known that the two lightest kittens weigh a total of 80 g, the four heaviest weigh 200 g, and the total weight of all the kittens is 500 g. How many kittens did the cat give birth to?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The two lightest weigh 80 g, so the others weigh no less than 40 g each. Similarly, we find that all except the 4 heaviest weigh no more than 50 g. Consider the kittens that are not among the 2 lightest and the 4 heaviest. Their total weight |
4. Find a two-digit number, the digits of which are different and the square of which is equal to the cube of the sum of its digits. | Let's write the condition as $\overline{A B}^{2}=(A+B)^{3}$. Note that the sum of the digits ( $A+B$ ) does not exceed 17, i.e., $\overline{A B}^{2} \leq 17^{3}$. Moreover, this number $\overline{A B}^{2}=n^{6}$, where $n-$
is some natural number that does not exceed 4. However, 1 and 2 do not work, since their cubes are single-digit numbers. Only 3 and 4 remain, and direct verification shows that $27^{2}=(2+7)^{2}=729$. | 27 | Number Theory | olympiads | null | null | \(\boxed{27}\) | 4. Find a two-digit number, the digits of which are different and the square of which is equal to the cube of the sum of its digits.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's write the condition as $\overline{A B}^{2}=(A+B)^{3}$. Note that the sum of the digits ( $A+B$ ) does not exceed 17, i.e., $\overline{A B}^{2} \leq 17^{3}$. Moreover, this number $\overline{A B}^{2}=n^{6}$, |
5. On a circle, 25 points are marked, painted either red or blue. Some of the points are connected by segments, with one end of each segment being blue and the other end red. It is known that there do not exist two red points that belong to the same number of segments. What is the maximum possible number of red points? | Let's take 13 red and 12 blue points. The first red point is not connected to any other, the second is connected to one blue, ..., the 13th is connected to 12 blue points. Obviously, there cannot be more red points, because if there are more than 13, the number of connection options is less than 13, i.e., (by the pigeonhole principle) some two red points will belong to the same number of segments. | 13 | Combinatorics | olympiads | null | null | \(\boxed{13}\) | 5. On a circle, 25 points are marked, painted either red or blue. Some of the points are connected by segments, with one end of each segment being blue and the other end red. It is known that there do not exist two red points that belong to the same number of segments. What is the maximum possible number of red points?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's take 13 red and 12 blue points. The first red point is not connected to any other, the second is connected to one blue, ..., the 13th is connected to 12 blue points. Obviously, there cannot be m |
5. Find the number of 9-digit numbers in which each digit from 1 to 9 appears exactly once, the digits 1, 2, 3, 4, 5 are arranged in ascending order, and the digit 6 appears before the digit 1 (for example, 916238457). | Note that after arranging the digits $7,8,9$, the remaining digits are uniquely determined. Therefore, the number of such numbers is $7 \cdot 8 \cdot 9=504$. | 504 | Combinatorics | olympiads | null | null | \(\boxed{504}\) | 5. Find the number of 9-digit numbers in which each digit from 1 to 9 appears exactly once, the digits 1, 2, 3, 4, 5 are arranged in ascending order, and the digit 6 appears before the digit 1 (for example, 916238457).
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that after arranging the digits $7,8,9$, the remaining digits are uniquel |
3.1. The decreasing sequence $a, b, c$ is a geometric progression, and the sequence $577 a, \frac{2020 b}{7}, \frac{c}{7}$ is an arithmetic progression. Find the common ratio of the geometric progression. | Let $b=a q, c=a q^{2}$. The properties of an arithmetic progression and the conditions of the problem lead to the equation $2 \cdot \frac{2020 a q}{7}=577 a+\frac{a q^{2}}{7} \Leftrightarrow q^{2}-4040 q+4039=0$, from which $q=1$ or $q=4039$. A decreasing geometric progression can only occur when $q=4039$ (for example, if $a=-1$). | 4039 | Algebra | olympiads | null | null | \(\boxed{4039}\) | 3.1. The decreasing sequence $a, b, c$ is a geometric progression, and the sequence $577 a, \frac{2020 b}{7}, \frac{c}{7}$ is an arithmetic progression. Find the common ratio of the geometric progression.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $b=a q, c=a q^{2}$. The properties of an arithmetic progression and the conditions of the problem lead to the equation $2 \cdot \frac{2020 a q}{7}=577 a+\frac{a q |
6.1. On 19 cards, the numbers $15,16,17, \ldots, 33$ are written respectively (one number per card). Members of the math club Vasya, Petya, and Misha decided to divide all these cards among themselves so that each of them gets at least one card and no one ends up with a pair of cards where the difference between the numbers is odd. How many ways are there to distribute the cards? | According to the condition, each participant will have either only even-numbered cards or only odd-numbered cards. We choose a participant (3 ways), and give them all 9 cards with even numbers. The remaining 10 cards with odd numbers are distributed between the two others, which can be done in $2^{10}$ ways. However, there will be 2 ways in which one of these two participants ends up without any cards. Therefore, we have $3 \cdot\left(2^{10}-2\right)$ ways.
Similarly, we give all the cards with odd numbers to one participant, and distribute the remaining 9 cards between the other two. Here, we have $3 \cdot\left(2^{9}-2\right)$ ways.
In total, the number of ways is: $3 \cdot\left(2^{10}+2^{9}-4\right)=3 \cdot 4 \cdot\left(2^{8}+2^{7}-1\right)=4596$. | 4596 | Combinatorics | olympiads | null | null | \(\boxed{4596}\) | 6.1. On 19 cards, the numbers $15,16,17, \ldots, 33$ are written respectively (one number per card). Members of the math club Vasya, Petya, and Misha decided to divide all these cards among themselves so that each of them gets at least one card and no one ends up with a pair of cards where the difference between the numbers is odd. How many ways are there to distribute the cards?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
According to the condition, each participant will have either only even-numbered cards or only odd-numbered cards. We choose a participant (3 ways), and give them all 9 cards with even numbers. The remaining 10 cards with odd numbers are distributed between the two others, which can be done in $2^{10}$ ways. However, there will be 2 ways in which one of these two participants e |
6. How many solutions in integers does the equation $x^{2}+y^{2}=6 x+2 y+15$ have?
ANSWER: 12. | By completing the square, we obtain the equation of the circle $(x-3)^{2}+$ $(y-1)^{2}=25$. This is possible when one of the terms equals 25 and the other equals 0 (4 cases), or when one equals 16 and the other equals 9 (8 cases). | 12 | Algebra | olympiads | null | null | \(\boxed{12}\) | 6. How many solutions in integers does the equation $x^{2}+y^{2}=6 x+2 y+15$ have?
ANSWER: 12.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
By completing the square, we obtain the equation of the circle $(x-3)^{2}+$ $(y-1)^{2}=25$. This is possible when o |
7. Find all natural numbers N such that the remainder of dividing 2017 by $N$ is 17. In your answer, specify the number of such $N$.
ANSWER 13. | As N, all positive divisors of the number 2000 greater than 17 will work. The divisors of 2000 have the form $2^{a} \cdot 5^{b}$, where $a=0,1,2,3,4$ and $b=0,1,2,3$. There will be 20 of them in total, but we need to exclude $1,2,4,5,8,10$ and 16. | 13 | Number Theory | olympiads | null | null | \(\boxed{13}\) | 7. Find all natural numbers N such that the remainder of dividing 2017 by $N$ is 17. In your answer, specify the number of such $N$.
ANSWER 13.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
As N, all positive divisors of the number 2000 greater than 17 will work. The divisors of 2000 have the form $2^{a} \cdot 5^ |
8. On a circle, 2017 different points $A_{1}, \ldots, A_{2017}$ are marked, and all possible chords connecting these points pairwise are drawn. A line is drawn through the point $A_{1}$, not passing through any of the points $A_{2}, \ldots A_{2017}$. Find the maximum possible number of chords that can have at least one common point with this line.
ANSWER: 1018080. | Let $k$ points be located on one side of the given line, then $2016-k$ points are on the other side. Thus, $k(2016-k)$ chords intersect the line and another 2016 pass through point $A_{1}$. Note that $k(2016-k)=-(k-1008)^{2}+1008^{2}$. The maximum of this expression is achieved when $k=1008$. | 1018080 | Combinatorics | olympiads | null | null | \(\boxed{1018080}\) | 8. On a circle, 2017 different points $A_{1}, \ldots, A_{2017}$ are marked, and all possible chords connecting these points pairwise are drawn. A line is drawn through the point $A_{1}$, not passing through any of the points $A_{2}, \ldots A_{2017}$. Find the maximum possible number of chords that can have at least one common point with this line.
ANSWER: 1018080.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $k$ points be located on one side of the given line, then $2016-k$ points are on the other side. Thus, $k(2016-k)$ chords intersect the line a |
6.1. (16 points) In a hut, several inhabitants of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The inhabitants of the Ah tribe always tell the truth, while the inhabitants of the Ukh tribe always lie. One of the inhabitants said: "There are no more than 16 of us in the hut," and then added: "We are all from the Ukh tribe." Another said: "There are no more than 17 of us in the hut," and then noted: "Some of us are from the Ah tribe." The third said: "There are five of us in the hut," and, looking around, added: "There are no fewer than three inhabitants from the Ukh tribe among us." How many inhabitants from the Ah tribe are in the hut? | A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 people in the hut. The third one is from tribe Ukh, as he said there are five of them in total. But since he said there are no fewer than three residents from tribe Ukh, there are no more than two. Two have already been found (the third and the first), so there are exactly two. Then the number of residents from tribe Ah is 15. | 15 | Logic and Puzzles | olympiads | null | null | \(\boxed{15}\) | 6.1. (16 points) In a hut, several inhabitants of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The inhabitants of the Ah tribe always tell the truth, while the inhabitants of the Ukh tribe always lie. One of the inhabitants said: "There are no more than 16 of us in the hut," and then added: "We are all from the Ukh tribe." Another said: "There are no more than 17 of us in the hut," and then noted: "Some of us are from the Ah tribe." The third said: "There are five of us in the hut," and, looking around, added: "There are no fewer than three inhabitants from the Ukh tribe among us." How many inhabitants from the Ah tribe are in the hut?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 people in the hut |
3. Having walked $2 / 5$ of the length of a narrow bridge, a pedestrian noticed that a car was approaching the bridge from behind. Then he walked back and met the car at the beginning of the bridge. If the pedestrian had continued walking forward, the car would have caught up with him at the end of the bridge. Find the ratio of the car's speed to the pedestrian's speed. | In the time $t$ that the pedestrian walked towards the car until they met at the beginning of the bridge, he covered $2 / 5$ of the bridge's length. Therefore, if the pedestrian continued walking forward, in time $t$ he would have covered another $2 / 5$ of the bridge's length, and he would have $1 / 5$ of the bridge's length left to walk. According to the problem, in time $t$, the car would have reached the beginning of the bridge and would have the entire bridge left to travel before meeting the pedestrian. Thus, the ratio of the car's speed to the pedestrian's speed is 5. | 5 | Algebra | olympiads | null | null | \(\boxed{5}\) | 3. Having walked $2 / 5$ of the length of a narrow bridge, a pedestrian noticed that a car was approaching the bridge from behind. Then he walked back and met the car at the beginning of the bridge. If the pedestrian had continued walking forward, the car would have caught up with him at the end of the bridge. Find the ratio of the car's speed to the pedestrian's speed.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
In the time $t$ that the pedestrian walked towards the car until they met at the beginning of the bridge, he covered $2 / 5$ of the bridge's length. Therefore, if the pedestrian continued walking forward, in time $t$ he would have covered another $2 / 5$ of the bridge's length, and he woul |
8. In a commercial football tournament, five teams participated. Each was supposed to play exactly one match against each other. Due to financial difficulties, the organizers canceled some games. In the end, it turned out that all teams had scored a different number of points, and no team had a zero in the points scored column. What is the minimum number of games that could have been played in the tournament, if three points were awarded for a win, one for a draw, and zero for a loss? | The minimum possible total score: $1+2+3+4+5=15$. In one game, a maximum of 3 points (in total) can be scored. Therefore, there were at least 5 games. However, if there were exactly 5 games, then all games would have to end with one of the teams winning, and then no team would have scored exactly 1 point.
Thus, there were at least 6 games. An example of the tournament results can be given by the table.
| $*$ | 1 | 2 | 3 | 4 | 5 | $\sum$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | $*$ | $*$ | $*$ | 3 | 3 | 6 |
| 2 | $*$ | $*$ | $*$ | $*$ | 3 | 3 |
| 3 | $*$ | $*$ | $*$ | 1 | 3 | 4 |
| 4 | 0 | $*$ | 1 | $*$ | 1 | 2 |
| 5 | 0 | 0 | 0 | 1 | $*$ | 1 | | 6 | Combinatorics | olympiads | null | null | \(\boxed{6}\) | 8. In a commercial football tournament, five teams participated. Each was supposed to play exactly one match against each other. Due to financial difficulties, the organizers canceled some games. In the end, it turned out that all teams had scored a different number of points, and no team had a zero in the points scored column. What is the minimum number of games that could have been played in the tournament, if three points were awarded for a win, one for a draw, and zero for a loss?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The minimum possible total score: $1+2+3+4+5=15$. In one game, a maximum of 3 points (in total) can be scored. Therefore, there were at least 5 games. However, if there were exactly 5 games, then all games would have to end with one of the teams winning, and then no team would have scored exactly 1 point.
Thus, there were at least 6 gam |
1. There were 21 consecutive natural numbers written on the board. When one of the numbers was erased, the sum of the remaining numbers became 2017. Which number was erased? | Let the numbers on the board be N-10, N-9,..,N, ..., N+10. Their sum is $21 \mathrm{~N}$. When one of these numbers - x - was erased, the sum became 2017, $21 \mathrm{~N}-$ $x=2017$. Therefore, $x=21 N-2017$, since this is one of these numbers, we get $N-10 \leq 21 N-2017 \leq N+10$. Solving the inequalities $\frac{2007}{20} \leq N \leq \frac{2027}{20}$, we get $N$ $=101$, therefore, $x=21 * 101-2017=104$. | 104 | Number Theory | olympiads | null | null | \(\boxed{104}\) | 1. There were 21 consecutive natural numbers written on the board. When one of the numbers was erased, the sum of the remaining numbers became 2017. Which number was erased?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let the numbers on the board be N-10, N-9,..,N, ..., N+10. Their sum is $21 \mathrm{~N}$. When one of these numbers - x - was erased, the sum became 2017, $21 \mathrm{~N}-$ $x=2017$. Therefore, $x=21 N-20 |
4. For natural numbers $m$ and $n$, it is known that $3 n^{3}=5 m^{2}$. Find the smallest possible value of $m+n$. | Obviously, if $m, n$ contain prime factors other than 3 or 5, then they can be canceled out (and reduce $m+n$).
Let $n=3^{a} \cdot 5^{b}, m=3^{c} \cdot 5^{d}$. Then the condition implies that $3 a+1=2 c, 3 b=2 d+1$.
The smallest possible values are: $a=1, b=1, c=2, d=1$, from which $n=15, m=45$. | 60 | Number Theory | olympiads | null | null | \(\boxed{60}\) | 4. For natural numbers $m$ and $n$, it is known that $3 n^{3}=5 m^{2}$. Find the smallest possible value of $m+n$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Obviously, if $m, n$ contain prime factors other than 3 or 5, then they can be canceled out (and reduce $m+n$).
Let $n=3^{a} \cdot 5^{b}, m=3^{c} \c |
5. Calculate $\sqrt{n}+\sqrt{n+524}$, given that this number is rational and that $n$ is a natural number. | Let the desired number be $a$. We have $\sqrt{n+524}=a-\sqrt{n}, n+524=a^{2}-2 a \sqrt{n}+n$. By the condition, $a$ is rational, so $\sqrt{n}$ is also rational. Therefore, $n=k^{2}, k \in \mathbb{N}$. Then the number $\sqrt{n+524}$ is also rational, so $n+524=m^{2}, m \in \mathbb{N}$. Thus, $m^{2}-k^{2}=524,(m-k)(m+k)=4 \cdot 131$. Note that the numbers $m-k$ and $m+k$ have the same parity, and the number 131 is prime. Therefore, $m-k=2$ and $m+k=2 \cdot 131$. Both equations are satisfied when $m=132, k=130$. Therefore, $a=m+k=262$. | 262 | Algebra | olympiads | null | null | \(\boxed{262}\) | 5. Calculate $\sqrt{n}+\sqrt{n+524}$, given that this number is rational and that $n$ is a natural number.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let the desired number be $a$. We have $\sqrt{n+524}=a-\sqrt{n}, n+524=a^{2}-2 a \sqrt{n}+n$. By the condition, $a$ is rational, so $\sqrt{n}$ is also rational. Therefore, $n=k^{2}, k \in \mathbb{N}$. Then the number $\sqrt{n+524}$ is also rational, so $n+524=m^{2}, m |
Problem 2. How many divisors of the number $2021^{2021}$ have a cube root that is a natural number? | Since $2021=43 \cdot 47$, all divisors of the number $2021^{2021}$ have the form $43^{\alpha} \cdot 47^{\beta}$, where $\alpha, \beta \in[0 ; 2021]$. In this case, the exact cubes are numbers of the form $43^{3 n} \cdot 47^{3 k}$, where $3 n, 3 k \in[0 ; 2021]$, that is, $n, k \in[0 ; 673]$. There are $674^{2}=454276$ such numbers. | 454276 | Number Theory | olympiads | null | null | \(\boxed{454276}\) | Problem 2. How many divisors of the number $2021^{2021}$ have a cube root that is a natural number?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since $2021=43 \cdot 47$, all divisors of the number $2021^{2021}$ have the form $43^{\alpha} \cdot 47^{\beta}$, where $\alpha, \beta \in[0 ; 2021]$. In this case, th |
Problem 4. How many triples of numbers $a, b, c$ exist, each of which is a root of the corresponding equation $a x^{2}+b x+c=0$? | If $a=0$ or $a=b=c$, then $a=b=c=0$. Otherwise: if $a=b \neq c$, then either $a=b=-1, c=0$, or $a=b=1, c=-2$; if $a \neq b=c$, then either $a=1, b=c=-0.5$; if $a=c \neq b$, then $a=c=c_{0}, b=1 / c_{0}$, where the number $c_{0}<0$ is the unique root of the equation $c^{3}+c=-1$. | 5 | Algebra | olympiads | null | null | \(\boxed{5}\) | Problem 4. How many triples of numbers $a, b, c$ exist, each of which is a root of the corresponding equation $a x^{2}+b x+c=0$?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If $a=0$ or $a=b=c$, then $a=b=c=0$. Otherwise: if $a=b \neq c$, then either $a=b=-1, c=0$, or $a=b=1, c=-2$; if $a \neq b=c$, then either $ |
3.1. (12 points) The number
$$
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{2017}{2018!}
$$
was written as an irreducible fraction with natural numerator and denominator. Find the last two digits of the numerator. | We have
$\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{2017}{2018!}=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\ldots+\left(\frac{1}{2017!}-\frac{1}{2018!}\right)=1-\frac{1}{2018!}=\frac{2018!-1}{2018!}$.
In the end, we obtained an irreducible fraction, and the last two digits of the number 2018! - 1 are two nines. | 99 | Algebra | olympiads | null | null | \(\boxed{99}\) | 3.1. (12 points) The number
$$
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{2017}{2018!}
$$
was written as an irreducible fraction with natural numerator and denominator. Find the last two digits of the numerator.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
We have
$\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{2017}{2018!}=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\ldots+\left(\frac{1}{2017!}-\fr |
5.1. (12 points) The decreasing sequence $a, b, c$ is a geometric progression, and the sequence $19 a, \frac{124 b}{13}, \frac{c}{13}$ is an arithmetic progression. Find the common ratio of the geometric progression. | Let $b=a q, c=a q^{2}$. The properties of an arithmetic progression and the conditions of the problem lead to the equation $2 \cdot \frac{124 a q}{13}=19 a+\frac{a q^{2}}{13} \Leftrightarrow q^{2}-248 q+247=0$, from which $q=1$ or $q=247$. A decreasing geometric progression can only occur when $q=247$ (for example, if $a=-1$). | 247 | Algebra | olympiads | null | null | \(\boxed{247}\) | 5.1. (12 points) The decreasing sequence $a, b, c$ is a geometric progression, and the sequence $19 a, \frac{124 b}{13}, \frac{c}{13}$ is an arithmetic progression. Find the common ratio of the geometric progression.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $b=a q, c=a q^{2}$. The properties of an arithmetic progression and the conditions of the problem lead to the equation $2 \cdot \frac{124 a q}{13}=19 a+\frac{a |
10.1. (12 points) For different natural numbers $k, l, m, n$, it is known that there exist such three natural numbers $a, b, c$ that each of the numbers $k, l, m, n$ is a root of either the equation $a x^{2}-b x+c=0$, or the equation $c x^{2}-16 b x+256 a=0$. Find $k^{2}+l^{2}+m^{2}+n^{2}$. | If $k, l$ are the roots of the first equation, then the roots of the second equation are the numbers $m=\frac{16}{k}$, $n=\frac{16}{l}$. Therefore, the numbers $k, l, m, n$ are divisors of the number 16. The divisors of 16 are the numbers 1, 2, 4, 8, and 16, but the number 4 does not fit, as by the condition all numbers $k, l, m, n$ are distinct. Thus, $k^{2}+l^{2}+m^{2}+n^{2}=1^{2}+2^{2}+8^{2}+16^{2}=325$. | 325 | Algebra | olympiads | null | null | \(\boxed{325}\) | 10.1. (12 points) For different natural numbers $k, l, m, n$, it is known that there exist such three natural numbers $a, b, c$ that each of the numbers $k, l, m, n$ is a root of either the equation $a x^{2}-b x+c=0$, or the equation $c x^{2}-16 b x+256 a=0$. Find $k^{2}+l^{2}+m^{2}+n^{2}$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If $k, l$ are the roots of the first equation, then the roots of the second equation are the numbers $m=\frac{16}{k}$, $n=\frac{16}{l}$. Therefore, the numbers $k, l, m, n$ are divisors of the number 16. T |
Problem 4. Vanya thought of a two-digit number, then swapped its digits and multiplied the resulting number by itself. The result turned out to be four times larger than the number he thought of. What number did Vanya think of? | Let the intended number be $\overline{m n}=10 m+n$. Then $4 \overline{m n}=\overline{n m}^{2}$. Therefore, $\overline{n m}^{2}$ is divisible by 4, and $\overline{n m}$ is divisible by 2, so the digit $m$ is even (and not zero). Moreover, $\overline{m n}=\overline{n m}^{2}: 4=(\overline{n m}: 2)^{2}$, which means $\overline{m n}$ is a square of a natural number starting with an even digit. Therefore, $\overline{m n}$ can be 25, 49, 64, or 81. Checking shows that only the last one satisfies the condition. | 81 | Algebra | olympiads | null | null | \(\boxed{81}\) | Problem 4. Vanya thought of a two-digit number, then swapped its digits and multiplied the resulting number by itself. The result turned out to be four times larger than the number he thought of. What number did Vanya think of?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let the intended number be $\overline{m n}=10 m+n$. Then $4 \overline{m n}=\overline{n m}^{2}$. Therefore, $\overline{n m}^{2}$ is divisible by 4, and $\overline{n m}$ is divisible by 2, so the digit $m$ is even (and not zero). Moreover, $\overline{m n}= |
Problem 5. The Martian traffic light consists of six identical bulbs arranged in two horizontal rows (one above the other) with three bulbs in each. A rover driver in the fog can distinguish the number and relative position of the lit bulbs on the traffic light (for example, if two bulbs are lit, whether they are in the same horizontal row or in different rows, whether they are in the same vertical row, or in adjacent vertical rows, or in the two outer vertical rows). However, he cannot distinguish the unlit bulbs and the body of the traffic light. Therefore, if only one bulb is lit, it is impossible to determine which one of the six it is). How many signals of the Martian traffic light can the rover driver distinguish in the fog? If no bulb on the traffic light is lit, the driver cannot see it. | If two traffic light signals differ only by the shift of the lit bulbs, then the driver cannot distinguish them (and vice versa). Therefore, any signal can either be transformed by a shift to the left and/or up into an indistinguishable signal, which has at least one bulb lit in the top horizontal row and at least one in the left vertical row, or the signal already has this property. There are two cases. | 44 | Combinatorics | olympiads | null | null | \(\boxed{44}\) | Problem 5. The Martian traffic light consists of six identical bulbs arranged in two horizontal rows (one above the other) with three bulbs in each. A rover driver in the fog can distinguish the number and relative position of the lit bulbs on the traffic light (for example, if two bulbs are lit, whether they are in the same horizontal row or in different rows, whether they are in the same vertical row, or in adjacent vertical rows, or in the two outer vertical rows). However, he cannot distinguish the unlit bulbs and the body of the traffic light. Therefore, if only one bulb is lit, it is impossible to determine which one of the six it is). How many signals of the Martian traffic light can the rover driver distinguish in the fog? If no bulb on the traffic light is lit, the driver cannot see it.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If two traffic light signals differ only by the shift of the lit bulbs, then the driver cannot distinguish them (and vice versa). Therefore, any signal can either be transformed by a shift to the left and |
1.1. (2 points) The average age of employees in a company consisting of 13 people is 35 years. After a new employee is hired, the average age of the employees becomes 34 years. Find the age of the new employee. | The sum of the employees' ages before the new hire was $13 \cdot 35$, and after the new employee was hired, the sum of the ages became $14 \cdot 34$. Therefore, the age of the new employee is $14 \cdot 34 - 13 \cdot 35 = 35 - 14 = 21$. | 21 | Algebra | olympiads | null | null | \(\boxed{21}\) | 1.1. (2 points) The average age of employees in a company consisting of 13 people is 35 years. After a new employee is hired, the average age of the employees becomes 34 years. Find the age of the new employee.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The sum of the employees' ages before the new hire was $13 \cdot 35$, and after the new employee was hired, the sum of |
3.1. (12 points) An eraser, 3 pens, and 2 markers cost 240 rubles. Two erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers? | From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will cost $2 \cdot 240=480$ rubles. Therefore, one pen costs $480-440=40$ rubles. Then, 3 erasers, 4 pens, and 6 markers cost $680-4 \cdot 40=520$ rubles. | 520 | Algebra | olympiads | null | null | \(\boxed{520}\) | 3.1. (12 points) An eraser, 3 pens, and 2 markers cost 240 rubles. Two erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will co |
5.1. (12 points) The equation $x^{2}+5 x+1=0$ has roots $x_{1}$ and $x_{2}$. Find the value of the expression
$$
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}
$$ | Since $x_{1}^{2}=-5 x_{1}-1$, then $\left(1+x_{1}\right)^{2}=1+2 x_{1}-5 x_{1}-1=-3 x_{1}$. Therefore,
$$
\begin{gathered}
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}=6\left(\frac{-5 x_{1}-1}{-3 x_{2}}+\frac{-5 x_{2}-1}{-3 x_{1}}\right)=\frac{10\left(x_{1}^{2}+x_{2}^{2}\right)+2\left(x_{1}+x_{2}\right)}{x_{1} x_{2}}= \\
=\frac{10\left(x_{1}+x_{2}\right)^{2}-20 x_{1} x_{2}+2\left(x_{1}+x_{2}\right)}{x_{1} x_{2}}=\frac{10 \cdot 25-20-10}{1}=220
\end{gathered}
$$ | 220 | Algebra | olympiads | null | null | \(\boxed{220}\) | 5.1. (12 points) The equation $x^{2}+5 x+1=0$ has roots $x_{1}$ and $x_{2}$. Find the value of the expression
$$
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since $x_{1}^{2}=-5 x_{1}-1$, then $\left(1+x_{1}\right)^{2}=1+2 x_{1}-5 x_{1}-1=-3 x_{1}$. Therefore,
$$
\begin{gathered}
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}=6\left(\frac{-5 x_{1}-1}{-3 x_{2}}+\fr |
6.1. (12 points) From point $A$ to point $B$, a bus and a cyclist departed simultaneously at 13:00. After arriving at point $B$, the bus, without stopping, headed back and met the cyclist at point $C$ at 13:10. Upon returning to point $A$, the bus again, without stopping, headed to point $B$ and caught up with the cyclist at point $D$, which is located $\frac{2}{3}$ km from point $C$. Find the speed of the bus (in km/h), if the distance between points $A$ and $B$ is 4 km, and the speeds of the bus and the cyclist are constant. | Let $v_{1}$ and $v_{2}$ be the speeds (in km/h) of the cyclist and the bus, respectively. By the time of the first meeting, they have collectively traveled $\frac{1}{6} v_{1}+\frac{1}{6} v_{2}=8$ km. Until the next meeting, the time that has passed is equal to $\frac{s_{0}}{v_{1}}=\frac{2 \cdot \frac{1}{6} v_{1}+s_{0}}{v_{2}}$ hours, where $s_{0}=\frac{1}{3}$ km. From this, we find $\frac{1}{4} v_{1}^{2}+v_{1}-24=0, v_{1}=8$ km/h (the second root is negative), $v_{2}=40 \text{km}/$ h. | 40 | Algebra | olympiads | null | null | \(\boxed{40}\) | 6.1. (12 points) From point $A$ to point $B$, a bus and a cyclist departed simultaneously at 13:00. After arriving at point $B$, the bus, without stopping, headed back and met the cyclist at point $C$ at 13:10. Upon returning to point $A$, the bus again, without stopping, headed to point $B$ and caught up with the cyclist at point $D$, which is located $\frac{2}{3}$ km from point $C$. Find the speed of the bus (in km/h), if the distance between points $A$ and $B$ is 4 km, and the speeds of the bus and the cyclist are constant.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $v_{1}$ and $v_{2}$ be the speeds (in km/h) of the cyclist and the bus, respectively. By the time of the first meeting, they have collectively traveled $\frac{1}{6} v_{1}+\frac{1}{6} v_{2}=8$ km. Until the next meeting, the time that has pa |
3.1. (14 points) Represent the number $\frac{201920192019}{191719171917}$ as an irreducible fraction. In the answer, write down the denominator of the resulting fraction. | We have
$$
\frac{201920192019}{191719171917}=\frac{2019 \cdot 100010001}{1917 \cdot 100010001}=\frac{2019}{1917}=\frac{3 \cdot 673}{3 \cdot 639}=\frac{673}{639}
$$
Since $639=3^{2} \cdot 71$ and 673 is not divisible by 3 and 71, the resulting fraction is irreducible. | 639 | Number Theory | olympiads | null | null | \(\boxed{639}\) | 3.1. (14 points) Represent the number $\frac{201920192019}{191719171917}$ as an irreducible fraction. In the answer, write down the denominator of the resulting fraction.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
We have
$$
\frac{201920192019}{191719171917}=\frac{2019 \cdot 100010001}{1917 \cdot 100010001}=\frac{2019}{1917}=\frac{3 \cdot 673}{3 |
4.1. (14 points) How many six-digit numbers exist whose sum of digits equals 51? | Since the maximum sum of the digits of a six-digit number is $6 \cdot 9=54$, the sought numbers are $699999, 789999, 888999$, as well as all numbers obtained from them by rearranging the digits. The first of these numbers can form 6 six-digit numbers (the digit 6 can be in any of the 6 positions), the second can form $6 \cdot 5=30$ numbers (6 options for placing one of the digits 7, 8, and another 5 for the other), and the third can form $\frac{6 \cdot 5 \cdot 4}{1 \cdot 2 \cdot 3}=20$ numbers (6 options for one digit 8, another 5 options for the second, and 4 options for the third, but since these digits are the same, we have counted each number $1 \cdot 2 \cdot 3=6$ times). In total, $6+30+20=56$ numbers. | 56 | Combinatorics | olympiads | null | null | \(\boxed{56}\) | 4.1. (14 points) How many six-digit numbers exist whose sum of digits equals 51?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since the maximum sum of the digits of a six-digit number is $6 \cdot 9=54$, the sought numbers are $699999, 789999, 888999$, as well as all numbers obtained from them by rearranging the digits. The first of these numbers can form 6 six-digit numbers (the digit 6 can be in any of the 6 positions), the second can form $6 \cdot 5=30$ numbers (6 options for p |
5.1. (14 points) The old man was pulling the turnip, and one by one, the old woman, the granddaughter, the dog, and the cat joined him. They pulled and pulled, but couldn't pull out the turnip! The cat called the mouse. They pulled and pulled, and finally pulled out the turnip! It is known that each subsequent participant pulls with a quarter of the strength of the previous one. How many men from the village, pulling with the same strength as the old man, should have been called to help the old man and the cat pull out the turnip? | The old woman, the granddaughter, Zhuchka, and the mouse pull with a combined force of
$$
\frac{3}{4}+\left(\frac{3}{4}\right)^{2}+\left(\frac{3}{4}\right)^{3}+\left(\frac{3}{4}\right)^{5}=\frac{2019}{1024}=1 \frac{995}{1024}
$$
of the force of the old man. Therefore, two men need to be called. | 2 | Algebra | olympiads | null | null | \(\boxed{2}\) | 5.1. (14 points) The old man was pulling the turnip, and one by one, the old woman, the granddaughter, the dog, and the cat joined him. They pulled and pulled, but couldn't pull out the turnip! The cat called the mouse. They pulled and pulled, and finally pulled out the turnip! It is known that each subsequent participant pulls with a quarter of the strength of the previous one. How many men from the village, pulling with the same strength as the old man, should have been called to help the old man and the cat pull out the turnip?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The old woman, the granddaughter, Zhuchka, and the mouse pull with a combined force of
$$
\frac{3}{4}+\left(\frac{3}{4}\right)^{2}+\left(\frac{3}{4 |
6.1. (14 points) Find the smallest natural number whose representation contains 5 zeros and 7 ones, and the sum of the digits in the even positions is equal to the sum of the digits in the odd positions. | The number 7 is not divisible by 2, so the number cannot have exactly $5+7=12$ digits. Consider the case where the number has 13 digits. If the "new" digit is zero, we again get a contradiction. Therefore, the "new" digit must be at least 1. Thus, to find the smallest number, we choose it so that this digit is as far to the right in the number as possible. Let's try to find the desired number in the form 100000111111x. Equating the sums of the digits in the even and odd positions, we get the equation $4+x=3$, from which $x=1$, which is impossible. Let's check the number in the form $a=10000011111 x 1$. Calculating the sums of the digits in the even and odd positions, we get $5=2+x \Rightarrow x=3$. Therefore, the desired number is 1000001111131. | 1000001111131 | Number Theory | olympiads | null | null | \(\boxed{1000001111131}\) | 6.1. (14 points) Find the smallest natural number whose representation contains 5 zeros and 7 ones, and the sum of the digits in the even positions is equal to the sum of the digits in the odd positions.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The number 7 is not divisible by 2, so the number cannot have exactly $5+7=12$ digits. Consider the case where the number has 13 digits. If the "new" digit is zero, we again get a contradiction. Therefore, the "new" digit must be at least 1. Thus, to find the smallest number, we choose it so that this digit is as far to the right in the number as possible. Let's try to find t |
1. On the Island of Knights and Liars, there live knights who always tell the truth and liars who always lie. One day, five residents of this island were asked one by one how many knights were among them.
- One, - answered the first.
- Two, - answered the second.
- Three, - answered the third.
- Don't believe them, they are all liars, - said the fourth.
- You are the liar! - said the fifth to the fourth.
How many knights were there in reality?
Answer: 2. | Among the first three, there is no more than one knight, since they all give different answers. If we consider the 4th and 5th, one of them is a knight and the other is a liar. Therefore, there is one or two knights in total. In any case, one of the first three told the truth. This means there are two knights: the second and the fifth. | 2 | Logic and Puzzles | olympiads | null | null | \(\boxed{2}\) | 1. On the Island of Knights and Liars, there live knights who always tell the truth and liars who always lie. One day, five residents of this island were asked one by one how many knights were among them.
- One, - answered the first.
- Two, - answered the second.
- Three, - answered the third.
- Don't believe them, they are all liars, - said the fourth.
- You are the liar! - said the fifth to the fourth.
How many knights were there in reality?
Answer: 2.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Among the first three, there is no more than one knight, since they all give different answers. If we consider the 4th and 5th, one of them is a knight and the other is |
1.1. (2 points) In a large family, one of the children has 3 brothers and 6 sisters, while another has 4 brothers and 5 sisters. How many boys are there in this family | In the family, boys have fewer brothers than girls have, and girls have fewer sisters than boys have. Therefore, there are 4 boys and 6 girls in the family. | 4 | Logic and Puzzles | olympiads | null | null | \(\boxed{4}\) | 1.1. (2 points) In a large family, one of the children has 3 brothers and 6 sisters, while another has 4 brothers and 5 sisters. How many boys are there in this family
The following text is the beginning part of the answer, which you can refer to for solving the problem:
In the family, boys have fewer brothers than girls have, and girls have fewer |
3.1. (14 points) Svetlana, Katya, Olya, Masha, and Tanya attend a math club, in which more than $60 \%$ of the students are boys. What is the smallest number of schoolchildren that can be in this club? | Let $M$ be the number of boys, $D$ be the number of girls in the club. Then $\frac{M}{M+D}>\frac{3}{5}$, hence $M>\frac{3}{2} D \geqslant \frac{15}{2}$. The minimum possible value of $M$ is 8, and the minimum possible value of $D$ is 5, making a total of 13 children. | 13 | Number Theory | olympiads | null | null | \(\boxed{13}\) | 3.1. (14 points) Svetlana, Katya, Olya, Masha, and Tanya attend a math club, in which more than $60 \%$ of the students are boys. What is the smallest number of schoolchildren that can be in this club?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $M$ be the number of boys, $D$ be the number of girls in the club. Then $\frac{M}{M+D}>\frac{3}{5}$, hence $M>\frac{3}{2} D \geqsl |
4.1. (14 points) Find all integer values that the fraction $\frac{8 n+157}{4 n+7}$ can take for natural $n$. In your answer, write the sum of the found values. | We have $\frac{8 n+157}{4 n+7}=2+\frac{143}{4 n+7}$. Since the divisors of the number 143 are only $1, 11, 13$, and 143, integer values of the fraction are obtained only when $n=1$ and $n=34$, which are 15 and 3, respectively, and their sum is 18. | 18 | Algebra | olympiads | null | null | \(\boxed{18}\) | 4.1. (14 points) Find all integer values that the fraction $\frac{8 n+157}{4 n+7}$ can take for natural $n$. In your answer, write the sum of the found values.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
We have $\frac{8 n+157}{4 n+7}=2+\frac{143}{4 n+7}$. Since the divisors of the number 143 are only $1, 11, 13$, and 143, int |
5.1. (14 points) A guard has detained a stranger and wants to drive him away. But the person caught said that he had made a bet with his friends for 100 coins that the guard would not drive him away (if the guard does, he has to pay his friends 100 coins, otherwise they pay him), and, deciding to bribe the guard, offered him to name a sum. What is the largest number of coins the guard can ask for so that the stranger, guided solely by his own profit, would definitely pay the guard? | If the guard asks for 199 coins, then by agreeing to give him this amount, the outsider will win the dispute and receive 100 coins. In total, he will lose 99 coins. If the outsider refuses, he will lose the dispute and lose 100 coins, which is less favorable (by 1 coin) for the one caught. If the guard demands 200, the outsider might refuse, as there is no difference in profit. If the guard demands more, it is more profitable for the outsider to refuse. The guard can ask for less, but the problem requires finding the largest amount. Thus, the answer is 199 coins. | 199 | Logic and Puzzles | olympiads | null | null | \(\boxed{199}\) | 5.1. (14 points) A guard has detained a stranger and wants to drive him away. But the person caught said that he had made a bet with his friends for 100 coins that the guard would not drive him away (if the guard does, he has to pay his friends 100 coins, otherwise they pay him), and, deciding to bribe the guard, offered him to name a sum. What is the largest number of coins the guard can ask for so that the stranger, guided solely by his own profit, would definitely pay the guard?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If the guard asks for 199 coins, then by agreeing to give him this amount, the outsider will win the dispute and receive 100 coins. In total, he will lose 99 coins. If the outsider refuses, he will lose the dispute and lose 100 coins, which is less favorable (by 1 coin) for the one c |
7.1. (14 points) We will call a natural number a snail if its representation consists of the representations of three consecutive natural numbers, concatenated in some order: for example, 312 or 121413. Snail numbers can sometimes be squares of natural numbers: for example, $324=18^{2}$ or $576=24^{2}$. Find a four-digit snail number that is the square of some natural number. | Note that a snail number can only be a four-digit number if the three numbers from which its record is formed are $8, 9, 10$ (numbers $7, 8, 9$ and smaller still form a three-digit number, while $9, 10, 11$ and larger - already form a number of no less than five digits). It remains to figure out the order of the selected numbers. The square of a natural number cannot end in 10 (there should be an even number of zeros or none at all). Also, a square cannot end in the digit 8. Therefore, the last digit must be 9. The remaining options are 8109 and 1089. Since $8109=90^{2}+9$ and $1089=33^{2}$, the only four-digit snail number that is a square is 1089. | 1089 | Number Theory | olympiads | null | null | \(\boxed{1089}\) | 7.1. (14 points) We will call a natural number a snail if its representation consists of the representations of three consecutive natural numbers, concatenated in some order: for example, 312 or 121413. Snail numbers can sometimes be squares of natural numbers: for example, $324=18^{2}$ or $576=24^{2}$. Find a four-digit snail number that is the square of some natural number.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that a snail number can only be a four-digit number if the three numbers from which its record is formed are $8, 9, 10$ (numbers $7, 8, 9$ and smaller still form a three-digit number, while $9, 10, 11$ and larger - already form a number of no less than five digits). It remains to figure out the order of the selected numbe |
1.1. The sequence $\left\{x_{n}\right\}$ is defined by the conditions $x_{1}=20, x_{2}=17, x_{n+1}=x_{n}-x_{n-1}(n \geqslant 2)$. Find $x_{2018}$. | From the condition, it follows that $x_{n+3}=x_{n+2}-x_{n+1}=x_{n+1}-x_{n}-x_{n+1}=-x_{n}$, therefore $x_{n+6}=$ $-x_{n+3}=x_{n}$, i.e., the sequence is periodic with a period of 6. Since $2018=6 \cdot 336+2$, we get $x_{2018}=x_{2}=17$. | 17 | Algebra | olympiads | null | null | \(\boxed{17}\) | 1.1. The sequence $\left\{x_{n}\right\}$ is defined by the conditions $x_{1}=20, x_{2}=17, x_{n+1}=x_{n}-x_{n-1}(n \geqslant 2)$. Find $x_{2018}$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
From the condition, it follows that $x_{n+3}=x_{n+2}-x_{n+1}=x_{n+1}-x_{n}-x_{n+1}=-x_{n}$, therefore $x_{n+6}=$ $-x_{ |
3.1. Philatelist Andrey decided to distribute all his stamps equally into 2 envelopes, but it turned out that one stamp was left over. When he distributed them equally into 3 envelopes, one stamp was again left over; when he distributed them equally into 5 envelopes, 3 stamps were left over; finally, when he tried to distribute them equally into 9 envelopes, 7 stamps were left over. How many stamps does Andrey have in total, if recently, to accommodate them all, he had to buy a second album for 150 stamps, as one such album was no longer sufficient? | If the desired number is $x$, then from the first sentence it follows that $x$ is odd, and from the rest it follows that the number $x+2$ must be divisible by 3, 5, and 9, i.e., has the form $5 \cdot 9 \cdot p$. Therefore, $x=45(2 k-1)-2=90 k-47$. According to the condition $150<x \leqslant 300$, so $k=3$. Therefore, $x=223$. | 223 | Number Theory | olympiads | null | null | \(\boxed{223}\) | 3.1. Philatelist Andrey decided to distribute all his stamps equally into 2 envelopes, but it turned out that one stamp was left over. When he distributed them equally into 3 envelopes, one stamp was again left over; when he distributed them equally into 5 envelopes, 3 stamps were left over; finally, when he tried to distribute them equally into 9 envelopes, 7 stamps were left over. How many stamps does Andrey have in total, if recently, to accommodate them all, he had to buy a second album for 150 stamps, as one such album was no longer sufficient?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If the desired number is $x$, then from the first sentence it follows that $x$ is odd, and from the rest it follows that the number $x+2$ must be divisible by 3, 5, |
6.1. Find all integer solutions of the equation $x \ln 27 \log _{13} e=27 \log _{13} y$. In your answer, specify the sum $x+y$ for the solution $(x, y)$, where $y$ is the smallest, exceeding 70. | The original equation is equivalent to the equation $\frac{x \ln 27}{\ln 13}=\frac{27 \ln y}{\ln 13} \Leftrightarrow x \ln 27=27 \ln y \Leftrightarrow$ $\ln 27^{x}=\ln y^{27} \Leftrightarrow 27^{x}=y^{27} \Leftrightarrow 3^{x}=y^{9}$, with $y \geqslant 1$, and thus $x \geqslant 0$. Since 3 is a prime number, either $y=1$ (then $x=0$), or $y$ is divisible by 3 and has no other prime divisors. Therefore, $y=3^{n}$, where $n \geqslant 0$. Hence, $x=9n$. Since $y>70$, then $n \geqslant 4$, the solution sought is: $(x, y)=(36,81)$. The answer is $x+y=117$. | 117 | Algebra | olympiads | null | null | \(\boxed{117}\) | 6.1. Find all integer solutions of the equation $x \ln 27 \log _{13} e=27 \log _{13} y$. In your answer, specify the sum $x+y$ for the solution $(x, y)$, where $y$ is the smallest, exceeding 70.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The original equation is equivalent to the equation $\frac{x \ln 27}{\ln 13}=\frac{27 \ln y}{\ln 13} \Leftrightarrow x \ln 27=27 \ln y \Leftrightarrow$ $\ln 27^{x}=\ln y^{27} \Leftrightarrow 27^{x}=y^{27} \Leftrightarrow 3^{x}=y^{9}$, with $y \geqslant 1$, and thus $x \geqslant |
1.1. Find the smallest 12-digit natural number that is divisible by 36 and contains each of the 10 digits at least once. | The number must be divisible by 4 and by 9. Since the sum of ten different digits is 45, the sum of the two remaining digits must be 0, 9, or 18. We need the smallest number, so we add two digits 0 to the digits $0,1, \ldots, 9$ and place the digits 10002345 at the beginning of the desired number (this is the smallest possible "start" of the number). The remaining digits $6,7,8,9$ must ensure divisibility by 4. Therefore, the last two digits can be 76, 96, or 68. The smallest option: 7896. | 100023457896 | Number Theory | olympiads | null | null | \(\boxed{100023457896}\) | 1.1. Find the smallest 12-digit natural number that is divisible by 36 and contains each of the 10 digits at least once.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The number must be divisible by 4 and by 9. Since the sum of ten different digits is 45, the sum of the two remaining digits must be 0, 9, or 18. We need the smallest number, so we add two digits 0 to the digits $0,1, \ldots, 9$ and place the digi |
5.1. Let $S(n)$ be the sum of the digits in the decimal representation of the number $n$. Find $S\left(S\left(S\left(S\left(2017^{2017}\right)\right)\right)\right)$. Answer. 1. | Since $2017^{2017}<10000^{2017}$, the number of digits in the representation of $2017^{2017}$ does not exceed $4 \cdot 2017=$ 8068, and their sum $S\left(2017^{2017}\right)$ does not exceed $9 \cdot 8068=72612$. Then we sequentially obtain $S\left(S\left(2017^{2017}\right)\right) \leqslant 6+9 \cdot 4=42, S\left(S\left(S\left(2017^{2017}\right)\right)\right) \leqslant 3+9=12, S\left(S\left(S\left(S\left(2017^{2017}\right)\right)\right)\right) \leqslant 9$. Note also that the sum of the digits of a number gives the same remainder when divided by 9 as the number itself. Since 2016 is divisible by 9, the number $2017^{2017}=(2016+1)^{2017}$ gives a remainder of 1 when divided by 9. Therefore, $\left.S\left(S\left(S\left(2017^{2017}\right)\right)\right)\right)=1$. | 1 | Number Theory | olympiads | null | null | \(\boxed{1}\) | 5.1. Let $S(n)$ be the sum of the digits in the decimal representation of the number $n$. Find $S\left(S\left(S\left(S\left(2017^{2017}\right)\right)\right)\right)$. Answer. 1.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since $2017^{2017}<10000^{2017}$, the number of digits in the representation of $2017^{2017}$ does not exceed $4 \cdot 2017=$ 8068, and their sum $S\left(2017^{2017}\right)$ does not exceed $9 \cdot 8068=72612$. Then we sequentially obtain $S\left(S\left(2017^{2017}\right)\right) \leqslant 6+9 \cdot 4=42, S\left(S\left(S\left(2017^{2017}\right)\right)\right) \leqslant 3+9=12, S\left |
7. 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. | Let's create a table, in the first row of which we will record the numbers following Petya, and in the second row we will place the results of Vasya's calculations.
| 21 | 17 | 13 | 9 | 5 | 1 | -3 | -7 | -11 | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :--- |
| 21 | 38 | 51 | 60 | 65 | 66 | 63 | 56 | 45 | $\cdots$ |
Since the numbers that Vasya will write down further are negative, the sum calculated by Kolya will decrease, moving away from 55. Therefore, the closest to 55 will be the sum equal to 56. | 56 | Algebra | olympiads | null | null | \(\boxed{56}\) | 7. 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.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's create a table, in the first row of which we will record the numbers following Petya, and in the second row we will place the results of Vasya's calculations.
| 21 | 17 | 13 | 9 | 5 | 1 | -3 | -7 | -11 | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :-- |
1. Two trains, each containing 15 identical cars, were moving towards each other at constant speeds. Exactly 28 seconds after the first cars of the trains met, passenger Sasha, sitting in the third car, passed passenger Valera from the oncoming train, and another 32 seconds later, the last cars of these trains had completely passed each other. In which car, by count, was Valera traveling?
Answer: 12. | Since 60 seconds have passed from the moment the "zero" cars parted ways (which is also the moment the first cars met) to the moment the 15th cars parted ways, the next cars parted ways every $60: 15=4$ seconds. Therefore, after 28 seconds, the 7th cars of the trains had just parted ways, meaning the 7th car of one train had just aligned with the 8th car of the other. At this moment, Sasha's 3rd car aligned with Valery's car, which has the number $8+(7-3)=12$. | 12 | Logic and Puzzles | olympiads | null | null | \(\boxed{12}\) | 1. Two trains, each containing 15 identical cars, were moving towards each other at constant speeds. Exactly 28 seconds after the first cars of the trains met, passenger Sasha, sitting in the third car, passed passenger Valera from the oncoming train, and another 32 seconds later, the last cars of these trains had completely passed each other. In which car, by count, was Valera traveling?
Answer: 12.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since 60 seconds have passed from the moment the "zero" cars parted ways (which is also the moment the first cars met) to the moment the 15th cars parted ways, the next cars parted ways every $60: 15=4$ seconds. Therefore, after 28 |
7. What is the smallest (same) number of pencils that need to be placed in each of 6 boxes so that in any 4 boxes there are pencils of any of 26 predefined colors (there are enough pencils available)? | On the one hand, pencils of each color should appear in at least three out of six boxes, since if pencils of a certain color are in no more than two boxes, then in the remaining boxes, which are at least four, there are no pencils of this color. Therefore, the total number of pencils should be no less than $3 \cdot 26$.
On the other hand, if we take 3 pencils of each of the 26 colors and arrange them so that no pencils of the same color are in the same box, then a pencil of any color will be found in any four boxes (since there are only two other boxes). The required arrangement of pencils can be achieved, for example, by dividing all the pencils into 3 groups of 26 pencils of all colors and distributing each group evenly (13 pencils) into two boxes. | 13 | Combinatorics | olympiads | null | null | \(\boxed{13}\) | 7. What is the smallest (same) number of pencils that need to be placed in each of 6 boxes so that in any 4 boxes there are pencils of any of 26 predefined colors (there are enough pencils available)?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
On the one hand, pencils of each color should appear in at least three out of six boxes, since if pencils of a certain color are in no more than two boxes, then in the remaining boxes, which are at least four, there are no pencils of this color. Therefore, the total number of pencils should be no less than $3 \cdot 26$.
On the other hand, if we take 3 pencils of each of the 26 |
1. Find the sum of the digits of the number $A$, if $A=2^{63} \cdot 4^{25} \cdot 5^{106}-2^{22} \cdot 4^{44} \cdot 5^{105}-1$. Answer: 959. | Since $2^{63} \cdot 4^{25} \cdot 5^{106}=2^{113} \cdot 5^{106}=2^{7} \cdot 10^{106}$ and $2^{22} \cdot 4^{44} \cdot 5^{105}=$ $2^{110} \cdot 5^{105}=2^{5} \cdot 10^{105}$, then $A+1=2^{7} \cdot 10^{106}-2^{5} \cdot 10^{105}=\left(2^{7} \cdot 10-2^{5}\right) \cdot 10^{105}=$ $1248 \cdot 10^{105}$. Therefore, the number $A$ is $1248 \cdot 10^{105}-1=1247 \underbrace{999 \ldots 999}_{105 \text { digits }}$, and the sum of its digits is $1+2+4+7+9 \cdot 105=959$. | 959 | Number Theory | olympiads | null | null | \(\boxed{959}\) | 1. Find the sum of the digits of the number $A$, if $A=2^{63} \cdot 4^{25} \cdot 5^{106}-2^{22} \cdot 4^{44} \cdot 5^{105}-1$. Answer: 959.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since $2^{63} \cdot 4^{25} \cdot 5^{106}=2^{113} \cdot 5^{106}=2^{7} \cdot 10^{106}$ and $2^{22} \cdot 4^{44} \cdot 5^{105}=$ $2^{110} \cdot 5^{105}=2^{5} \cdot 10^{105}$, then $A+1=2^{7} \cdot 10^{106}-2^{5} \cdot 10^{105}=\left(2^ |
1. Each Kinder Surprise contains exactly 3 different gnomes, and there are 12 different types of gnomes in total. In the box, there are enough Kinder Surprises, and in any two of them, the triplets of gnomes are not the same. What is the minimum number of Kinder Surprises that need to be bought to ensure that after they are opened, there is at least one gnome of each of the 12 types? | From 11 different gnomes, a maximum of $C_{11}^{3}=\frac{11 \cdot 10 \cdot 9}{3 \cdot 2 \cdot 1}=165$ different Kinder can be formed, so if you take one more Kinder, there cannot be fewer than 12 different gnomes among them. | 166 | Combinatorics | olympiads | null | null | \(\boxed{166}\) | 1. Each Kinder Surprise contains exactly 3 different gnomes, and there are 12 different types of gnomes in total. In the box, there are enough Kinder Surprises, and in any two of them, the triplets of gnomes are not the same. What is the minimum number of Kinder Surprises that need to be bought to ensure that after they are opened, there is at least one gnome of each of the 12 types?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
From 11 different gnomes, a maximum of $C_{11}^{3}=\frac{11 \cdot 10 \cdot 9}{3 \cdot 2 \cdot 1}=165$ different |
. Find the largest four-digit number in which all digits are different, and moreover, no two of them can be swapped to form a smaller number. | If the digit 0 is not used, then the digits in each such number must be in ascending order, and the largest of such numbers is 6789. Since a number cannot start with zero, using 0, one can obtain additional suitable numbers where 0 is in the hundreds place, and the other digits are in ascending order. The largest of such numbers is 7089. | 7089 | Number Theory | olympiads | null | null | \(\boxed{7089}\) | . Find the largest four-digit number in which all digits are different, and moreover, no two of them can be swapped to form a smaller number.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If the digit 0 is not used, then the digits in each such number must be in ascending order, and the largest of such numbers is 6789. Since a number cannot start with zero |
. On the board, all such natural numbers from 3 to 223 inclusive are written, which when divided by 4 leave a remainder of 3. Every minute, Borya erases any two of the written numbers and instead writes their sum, decreased by 2. In the end, only one number remains on the board. What can it be? | Initially, the total number of written numbers is $224: 4=56$, and their sum is $(3+223) \cdot 56: 2=6328$. Every minute, the number of written numbers decreases by 1, and their sum decreases by 2. Therefore, one number will remain on the board after 55 minutes and will be equal to $6328-55 \cdot 2=6218$. | 6218 | Number Theory | olympiads | null | null | \(\boxed{6218}\) | . On the board, all such natural numbers from 3 to 223 inclusive are written, which when divided by 4 leave a remainder of 3. Every minute, Borya erases any two of the written numbers and instead writes their sum, decreased by 2. In the end, only one number remains on the board. What can it be?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Initially, the total number of written numbers is $224: 4=56$, and their sum is $(3+223) \cdot 56: 2=6328$. Every minute, the number of written numbers d |
Problem 4. Masha has seven different dolls, which she arranges in six different doll houses so that each house has at least one doll. In how many ways can Masha do this? It is important which doll ends up in which house. It does not matter how the dolls are seated in the house where there are two of them. | In each such seating arrangement, in one of the cabins there will be two dolls, and in the other cabins - one each. First, we choose which two dolls will sit in one cabin. This can be done in $7 \cdot 6: 2=21$ ways. Then we choose which cabin to seat these two dolls in. This can be done in 6 ways. Finally, we choose how to seat the remaining 5 dolls in the remaining 5 cabins. This can be done in $5!=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120$ ways. Therefore, the number of seating arrangements is $21 \cdot 6 \cdot 120=15120$. | 15120 | Combinatorics | olympiads | null | null | \(\boxed{15120}\) | Problem 4. Masha has seven different dolls, which she arranges in six different doll houses so that each house has at least one doll. In how many ways can Masha do this? It is important which doll ends up in which house. It does not matter how the dolls are seated in the house where there are two of them.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
In each such seating arrangement, in one of the cabins there will be two dolls, and in the other cabins - one each. First, we choose which two dolls will sit in one cabin. This can be done in $7 \cdot 6: 2=21$ ways. Then we choose which cabin to seat these two do |
4. It is known that $x+y+z=2016$ and $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{1}{32}$. Find $\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}$
ANSWER: 60. | Add 1 to each fraction, we get $\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}+1=\frac{x+y+z}{x+y}+$ $\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}=2016 \cdot\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{2016}{32}=63$. | 63 | Algebra | olympiads | null | null | \(\boxed{63}\) | 4. It is known that $x+y+z=2016$ and $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{1}{32}$. Find $\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}$
ANSWER: 60.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Add 1 to each fraction, we get $\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}+1=\frac{x+y+z}{x+y}+$ $\frac{x+y+ |
5. Find all natural numbers $N$ such that the remainder of dividing 2017 by $N$ is 17. In your answer, specify the number of such $N$.
ANSWER 13. | - As $N$, all positive divisors of the number 2000 greater than 17 will work. The divisors of 2000 have the form $2^{a} \cdot 5^{b}$, where $a=0,1,2,3,4$ and $b=0,1,2,3$. There will be 20 of them in total, but we need to exclude $1,2,4,5,8,10$ and 16. | 13 | Number Theory | olympiads | null | null | \(\boxed{13}\) | 5. Find all natural numbers $N$ such that the remainder of dividing 2017 by $N$ is 17. In your answer, specify the number of such $N$.
ANSWER 13.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
- As $N$, all positive divisors of the number 2000 greater than 17 will work. The divisors of 2000 have the form $2^{a} \cdot |
6. There are 5 identical buckets, each with a maximum capacity of some integer number of liters, and a 30-liter barrel containing an integer number of liters of water. The water from the barrel was distributed among the buckets, with the first bucket being half full, the second one-third full, the third one-quarter full, the fourth one-fifth full, and the fifth one-sixth full. How many liters of water were in the barrel?
ANSWER: 29. | The solution $-\frac{x}{2}+\frac{x}{3}+\frac{x}{4}+\frac{x}{5}+\frac{x}{6}=\frac{87 x}{60}=\frac{29 x}{20}$ will be an integer only when $x$ is a multiple of 20. But $x>20$ cannot be taken, as the sum will exceed 30 liters. | 29 | Algebra | olympiads | null | null | \(\boxed{29}\) | 6. There are 5 identical buckets, each with a maximum capacity of some integer number of liters, and a 30-liter barrel containing an integer number of liters of water. The water from the barrel was distributed among the buckets, with the first bucket being half full, the second one-third full, the third one-quarter full, the fourth one-fifth full, and the fifth one-sixth full. How many liters of water were in the barrel?
ANSWER: 29.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The solution $-\frac{x}{2}+\frac{x}{3}+\frac{x}{4}+\frac{x}{5}+\frac{x}{6}=\frac{87 x}{60}=\frac{29 x}{20}$ will |
6. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$ | Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace{4 \ldots 4}_{2012}=\underbrace{4 \ldots 4}_{2011} 3 \underbrace{5 \ldots 5}_{2011} 6$. The sum of the digits is $4 \cdot 2011+3+5 \cdot 2011+6=18108$. | 18108 | Number Theory | olympiads | null | null | \(\boxed{18108}\) | 6. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace |
1.1. A trip in Moscow using the "Troyka" card in 2016 costs 32 rubles for one subway ride and 31 rubles for one trip on surface transport. What is the minimum total number of trips that can be made under these rates, spending exactly 5000 rubles? | If $x$ is the number of trips by surface transport and $y$ is the number of trips by subway, then we have $31x + 32y = 5000$, from which $32(x + y) = 5000 + x \geqslant 5000$. Therefore, $x + y \geqslant 5000 / 32 = 156 \frac{1}{4}$. The smallest integer value of $x + y$ is 157, which is achieved when $x = 24, y = 133$. | 157 | Algebra | olympiads | null | null | \(\boxed{157}\) | 1.1. A trip in Moscow using the "Troyka" card in 2016 costs 32 rubles for one subway ride and 31 rubles for one trip on surface transport. What is the minimum total number of trips that can be made under these rates, spending exactly 5000 rubles?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If $x$ is the number of trips by surface transport and $y$ is the number of trips by subway, then we have $31x + 32y = 5000$, from which $32(x + y) = 5000 + x \ |
2.1. Determine the number of natural divisors of the number $11!=1 \cdot 2 \cdot \ldots \cdot 10 \cdot 11$ that are multiples of three. Answer. 432. | The prime factorization of the given number is $11!=2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7 \cdot 11$. All multiples of three that are divisors of this number have the form $2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} \cdot 7^{\delta} \cdot 11^{\varphi}$, where $\alpha \in[0 ; 8], \beta \in[1 ; 4], \gamma \in[0 ; 2]$, $\delta \in[0 ; 1], \varphi \in[0 ; 1]$. The total number of such divisors is $(8+1) \cdot 4 \cdot(2+1)(1+1)(1+1)=432$. | 432 | Number Theory | olympiads | null | null | \(\boxed{432}\) | 2.1. Determine the number of natural divisors of the number $11!=1 \cdot 2 \cdot \ldots \cdot 10 \cdot 11$ that are multiples of three. Answer. 432.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The prime factorization of the given number is $11!=2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7 \cdot 11$. All multiples of three that are divisors of this number have the form $2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} \c |
5. How many solutions in integers does the equation $x^{2}+y^{2}=6 x+2 y+15$ have?
ANSWER: 12. | By completing the square, we obtain the equation of the circle $(x-3)^{2}+$ $(y-1)^{2}=25$. This is possible when one of the terms equals 25 and the other equals 0 (4 cases), or when one equals 16 and the other equals 9 (8 cases). | 12 | Algebra | olympiads | null | null | \(\boxed{12}\) | 5. How many solutions in integers does the equation $x^{2}+y^{2}=6 x+2 y+15$ have?
ANSWER: 12.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
By completing the square, we obtain the equation of the circle $(x-3)^{2}+$ $(y-1)^{2}=25$. This is possible when o |
7. On a circle, 2017 different points $A_{1}, \ldots, A_{2017}$ are marked, and all possible chords connecting these points pairwise are drawn. A line is drawn through the point $A_{1}$, not passing through any of the points $A_{2}, \ldots, A_{2017}$. Find the maximum possible number of chords that can have at least one common point with this line.
ANSWER: 1018080. | Let $k$ points be located on one side of the given line, then $2016-k$ points are on the other side. Thus, $k(2016-k)$ chords intersect the line and another 2016 pass through point $A_{1}$. Note that $k(2016-k)=-(k-1008)^{2}+1008^{2}$. The maximum of this expression is achieved when $k=1008$. | 1018080 | Combinatorics | olympiads | null | null | \(\boxed{1018080}\) | 7. On a circle, 2017 different points $A_{1}, \ldots, A_{2017}$ are marked, and all possible chords connecting these points pairwise are drawn. A line is drawn through the point $A_{1}$, not passing through any of the points $A_{2}, \ldots, A_{2017}$. Find the maximum possible number of chords that can have at least one common point with this line.
ANSWER: 1018080.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $k$ points be located on one side of the given line, then $2016-k$ points are on the other side. Thus, $k(2016-k)$ chords intersect the line a |
2.1. (14 points) A team consisting of juniors and masters from the "Vimpel" sports society went to a shooting tournament. The average number of points scored by the juniors turned out to be 22, by the masters - 47, and the average number of points for the entire team - 41. What is the share (in percent) of masters in this team? | Let there be $x$ juniors and $y$ masters in the team. Then the total number of points scored by the team is $22x + 47y = 41(x + y)$, from which we find $19x = 6y$. Therefore, the proportion of masters is $\frac{y}{x+y} = \frac{19y}{19x + 19y} = \frac{19y}{25y} = 0.76$, i.e., $76\%$. | 76 | Algebra | olympiads | null | null | \(\boxed{76}\) | 2.1. (14 points) A team consisting of juniors and masters from the "Vimpel" sports society went to a shooting tournament. The average number of points scored by the juniors turned out to be 22, by the masters - 47, and the average number of points for the entire team - 41. What is the share (in percent) of masters in this team?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let there be $x$ juniors and $y$ masters in the team. Then the total number of points scored by the team is $22x + 47y = 41(x + y)$, from whic |
4.1. (14 points) An eraser, 3 pens, and 2 markers cost 240 rubles, while 2 erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers? | From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will cost $2 \cdot 240=480$ rubles. Therefore, one pen costs $480-440=40$ rubles. Then, 3 erasers, 4 pens, and 6 markers cost $680-4 \cdot 40=520$ rubles. | 520 | Algebra | olympiads | null | null | \(\boxed{520}\) | 4.1. (14 points) An eraser, 3 pens, and 2 markers cost 240 rubles, while 2 erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will co |
6.1. (14 points) In a hut, several residents of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The residents of the Ah tribe always tell the truth, while the residents of the Ukh tribe always lie. One of the residents said: "There are no more than 16 of us in the hut," and then added: "We are all from the Ukh tribe." Another said: "There are no more than 17 of us in the hut," and then noted: "Some of us are from the Ah tribe." The third said: "There are five of us in the hut," and, looking around, added: "There are no fewer than three residents of the Ukh tribe among us." How many residents from the Ah tribe are in the hut? | A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 people in the hut. The third one is from tribe Ukh, as he said there are five of them in total. But since he said there are no fewer than three residents from tribe Ukh, there are no more than two. Two have already been found (the third and the first), so there are exactly two. Then the number of residents from tribe Ah is 15. | 15 | Logic and Puzzles | olympiads | null | null | \(\boxed{15}\) | 6.1. (14 points) In a hut, several residents of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The residents of the Ah tribe always tell the truth, while the residents of the Ukh tribe always lie. One of the residents said: "There are no more than 16 of us in the hut," and then added: "We are all from the Ukh tribe." Another said: "There are no more than 17 of us in the hut," and then noted: "Some of us are from the Ah tribe." The third said: "There are five of us in the hut," and, looking around, added: "There are no fewer than three residents of the Ukh tribe among us." How many residents from the Ah tribe are in the hut?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 people in the hut |
5. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$ | Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace{4 \ldots 4}_{2012}=\underbrace{4 \ldots 4}_{2011} 3 \underbrace{5 \ldots 5}_{2011} 6$. The sum of the digits is $4 \cdot 2011+3+5 \cdot 2011+6=18108$. | 18108 | Number Theory | olympiads | null | null | \(\boxed{18108}\) | 5. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace |
1.1.1. (2 points) Find the sum of the squares of two numbers if it is known that their arithmetic mean is 8, and the geometric mean is $2 \sqrt{5}$. | If $a$ and $b$ are the numbers in question, then $a+b=16, a b=(2 \sqrt{5})^{2}=20$, therefore
$$
a^{2}+b^{2}=(a+b)^{2}-2 a b=256-40=216 .
$$ | 216 | Algebra | olympiads | null | null | \(\boxed{216}\) | 1.1.1. (2 points) Find the sum of the squares of two numbers if it is known that their arithmetic mean is 8, and the geometric mean is $2 \sqrt{5}$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If $a$ and $b$ are the numbers in question, then $a+b=16, a b=(2 \sqrt |
2.1.1. (2 points) Find the greatest value of $x$ that satisfies the inequality
$$
\left(6+5 x+x^{2}\right) \sqrt{2 x^{2}-x^{3}-x} \leqslant 0
$$ | $$
\left[\begin{array} { c }
{ 2 x ^ { 2 } - x ^ { 3 } - x = 0 ; } \\
{ \{ \begin{array} { c }
{ 2 x ^ { 2 } - x ^ { 3 } - x > 0 ; } \\
{ 6 + 5 x + x ^ { 2 } \leqslant 0 ; }
\end{array} }
\end{array} \Longleftrightarrow \left[\begin{array} { l }
{ - x ( x - 1 ) ^ { 2 } = 0 ; } \\
{ \{ \begin{array} { c }
{ - x ( x - 1 ) ^ { 2 } > 0 ; } \\
{ ( x + 2 ) ( x + 3 ) \leqslant 0 ; }
\end{array} }
\end{array} \Longleftrightarrow \left[\begin{array}{l}
x=0 \\
x=1 ; \\
\left\{\begin{array}{l}
x<0 ; \\
-3 \leqslant x \leqslant-2
\end{array}\right.
\end{array}\right.\right.\right.
$$
All solutions of the last system are negative. Therefore, the greatest solution of the original inequality will be 1. | 1 | Inequalities | olympiads | null | null | \(\boxed{1}\) | 2.1.1. (2 points) Find the greatest value of $x$ that satisfies the inequality
$$
\left(6+5 x+x^{2}\right) \sqrt{2 x^{2}-x^{3}-x} \leqslant 0
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
$$
\left[\begin{array} { c }
{ 2 x ^ { 2 } - x ^ { 3 } - x = 0 ; } \\
{ \{ \begin{array} { c }
{ 2 x ^ { 2 } - x ^ { 3 } - x > 0 ; } \\
{ 6 + 5 x + x ^ { 2 } \leqslant 0 ; }
\end{array} }
\end{array} \Longleftrightarrow \left[\begin{array} { l }
{ - x ( x - 1 ) ^ { 2 } = 0 ; } \\
{ \{ \begin{array} { c }
{ - x ( x - 1 ) ^ { 2 } > 0 ; } \\
{ ( x |
3.2.1. (12 points) The equation $x^{2}+5 x+1=0$ has roots $x_{1}$ and $x_{2}$. Find the value of the expression
$$
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}
$$ | Since $x_{1}^{2}=-5 x_{1}-1$, then $\left(1+x_{1}\right)^{2}=1+2 x_{1}-5 x_{1}-1=-3 x_{1}$. Therefore,
$$
\begin{gathered}
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}=6\left(\frac{-5 x_{1}-1}{-3 x_{2}}+\frac{-5 x_{2}-1}{-3 x_{1}}\right)=\frac{10\left(x_{1}^{2}+x_{2}^{2}\right)+2\left(x_{1}+x_{2}\right)}{x_{1} x_{2}}= \\
=\frac{10\left(x_{1}+x_{2}\right)^{2}-20 x_{1} x_{2}+2\left(x_{1}+x_{2}\right)}{x_{1} x_{2}}=\frac{10 \cdot 25-20-10}{1}=220
\end{gathered}
$$ | 220 | Algebra | olympiads | null | null | \(\boxed{220}\) | 3.2.1. (12 points) The equation $x^{2}+5 x+1=0$ has roots $x_{1}$ and $x_{2}$. Find the value of the expression
$$
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since $x_{1}^{2}=-5 x_{1}-1$, then $\left(1+x_{1}\right)^{2}=1+2 x_{1}-5 x_{1}-1=-3 x_{1}$. Therefore,
$$
\begin{gathered}
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}=6\left(\frac{-5 x_{1}-1}{-3 x_{2}}+\fr |
4.1.1. (12 points) From point $A$ to point $B$, a bus and a cyclist departed simultaneously at 13:00. After arriving at point $B$, the bus, without stopping, headed back and met the cyclist at point $C$ at 13:10. Upon returning to point $A$, the bus again, without stopping, headed to point $B$ and caught up with the cyclist at point $D$, which is located $\frac{2}{3}$ km from point $C$. Find the speed of the bus (in km/h), if the distance between points $A$ and $B$ is 4 km, and the speeds of the bus and the cyclist are constant. | Let $v_{1}$ and $v_{2}$ be the speeds (in km/h) of the cyclist and the bus, respectively. By the time of the first meeting, they have collectively traveled $\frac{1}{6} v_{1}+\frac{1}{6} v_{2}=8$ km. Until the next meeting, the time that has passed is equal to $\frac{s_{0}}{v_{1}}=\frac{2 \cdot \frac{1}{6} v_{1}+s_{0}}{v_{2}}$ hours, where $s_{0}=\frac{1}{3}$ km. From this, we find $\frac{1}{4} v_{1}^{2}+v_{1}-24=0, v_{1}=8$ km/h (the second root is negative), $v_{2}=40$ km/h. | 40 | Algebra | olympiads | null | null | \(\boxed{40}\) | 4.1.1. (12 points) From point $A$ to point $B$, a bus and a cyclist departed simultaneously at 13:00. After arriving at point $B$, the bus, without stopping, headed back and met the cyclist at point $C$ at 13:10. Upon returning to point $A$, the bus again, without stopping, headed to point $B$ and caught up with the cyclist at point $D$, which is located $\frac{2}{3}$ km from point $C$. Find the speed of the bus (in km/h), if the distance between points $A$ and $B$ is 4 km, and the speeds of the bus and the cyclist are constant.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $v_{1}$ and $v_{2}$ be the speeds (in km/h) of the cyclist and the bus, respectively. By the time of the first meeting, they have collectively traveled $\frac{1}{6} v_{1}+\frac{1}{6} v_{2}=8$ km. Until the next meeting, the time that ha |
4.2.1. (12 points) Every morning, each member of the Ivanov family drinks an 180-gram cup of coffee with milk. The amount of milk and coffee in their cups varies. Masha Ivanova found out that she drank $\frac{2}{9}$ of all the milk consumed that morning and $\frac{1}{6}$ of all the coffee consumed that morning. How many people are in this family? | Let there be $x$ (for example, grams) of milk and $y$ grams of coffee, and $n$ people in the family. Since each family member drank the same amount of coffee with milk, then $\left(\frac{2 x}{9}+\frac{y}{6}\right) n=$ $=x+y \Leftrightarrow(4 x+3 y) n=18 x+18 y \Leftrightarrow 2 x(2 n-9)=3 y(6-n)$. From this, it follows that $2 n-9$ and $6-n$ have the same sign, that is, $4.5<n<6$. Since $n$ is an integer, then $n=5$. | 5 | Algebra | olympiads | null | null | \(\boxed{5}\) | 4.2.1. (12 points) Every morning, each member of the Ivanov family drinks an 180-gram cup of coffee with milk. The amount of milk and coffee in their cups varies. Masha Ivanova found out that she drank $\frac{2}{9}$ of all the milk consumed that morning and $\frac{1}{6}$ of all the coffee consumed that morning. How many people are in this family?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let there be $x$ (for example, grams) of milk and $y$ grams of coffee, and $n$ people in the family. Since each family member drank the same amount of coffee with milk, then $\left(\frac{2 x}{9}+\frac{y}{6}\rig |
4.3.1. (12 points) On the table, there are 13 weights arranged in a row by mass (the lightest on the left, the heaviest on the right). It is known that the mass of each weight is an integer number of grams, the masses of any two adjacent weights differ by no more than 5 grams, and the total mass of the weights does not exceed 2019 grams. Find the maximum possible mass of the heaviest weight under these conditions. | If the mass of the heaviest weight is $m$, then the masses of the other weights will be no less than $m-5, m-10, \ldots, m-60$ grams, and their total mass will be no less than $13 m-390$ grams. Then $13 m-390 \leq 2019$, from which $m \leq 185$. It remains to verify that the set of weights with masses $185,180,175,170,165,160,155,150,145,140,135,130$ and 129 grams satisfies the condition of the problem. | 185 | Number Theory | olympiads | null | null | \(\boxed{185}\) | 4.3.1. (12 points) On the table, there are 13 weights arranged in a row by mass (the lightest on the left, the heaviest on the right). It is known that the mass of each weight is an integer number of grams, the masses of any two adjacent weights differ by no more than 5 grams, and the total mass of the weights does not exceed 2019 grams. Find the maximum possible mass of the heaviest weight under these conditions.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If the mass of the heaviest weight is $m$, then the masses of the other weights will be no less than $m-5, m-10, \ldots, m-60$ grams, and their total mass will be no less than $13 m-390$ grams. Then $13 |
4.4.1. (12 points) A goat eats 1 hay wagon in 6 weeks, a sheep in 8 weeks, and a cow in 3 weeks. How many weeks will it take for 5 goats, 3 sheep, and 2 cows to eat 30 such hay wagons together? | The goat eats hay at a rate of $1 / 6$ cart per week, the sheep - at a rate of $1 / 8$ cart per week, the cow - at a rate of $1 / 3$ cart per week. Then 5 goats, 3 sheep, and 2 cows together will eat hay at a rate of $\frac{5}{6}+\frac{3}{8}+\frac{2}{3}=\frac{20+9+16}{24}=\frac{45}{24}=\frac{15}{8}$ carts per week. Therefore, 30 carts of hay they will eat in $30: \frac{15}{8}=16$ weeks. | 16 | Algebra | olympiads | null | null | \(\boxed{16}\) | 4.4.1. (12 points) A goat eats 1 hay wagon in 6 weeks, a sheep in 8 weeks, and a cow in 3 weeks. How many weeks will it take for 5 goats, 3 sheep, and 2 cows to eat 30 such hay wagons together?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The goat eats hay at a rate of $1 / 6$ cart per week, the sheep - at a rate of $1 / 8$ cart per week, the cow - at a rate of $1 / 3$ cart per week. Then 5 goats, 3 sheep, and 2 cows together wil |
6.4.1. (12 points) Solve the inequality $\sqrt{x-4}+\sqrt{x+1}+\sqrt{2 x}-\sqrt{33-x}>4$. In your answer, write the sum of all its integer solutions. | Rewrite the inequality as
$$
\sqrt{x-4}+\sqrt{x+1}+\sqrt{2 x}>\sqrt{33-x}+4
$$
Find the domain of the variable $x$:
$$
\left\{\begin{array}{l}
x-4 \geqslant 0 \\
x+1 \geqslant 0 \\
2 x \geqslant 0 \\
33-x \geqslant 0
\end{array}\right.
$$
from which $x \in[4,33]$. We are interested in the integer values from this interval that satisfy the inequality. Note that the left side of the obtained inequality is increasing, while the right side is strictly decreasing. They coincide at $x=8$. For $x \in\{4,5,6,7\}$, the left side is less than the right side, and for $x \in\{9,10, \ldots, 33\}$, the left side is greater than the right side. We calculate the sum. $9+10+\ldots+33=\frac{9+33}{2} \cdot 25=525$. | 525 | Inequalities | olympiads | null | null | \(\boxed{525}\) | 6.4.1. (12 points) Solve the inequality $\sqrt{x-4}+\sqrt{x+1}+\sqrt{2 x}-\sqrt{33-x}>4$. In your answer, write the sum of all its integer solutions.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Rewrite the inequality as
$$
\sqrt{x-4}+\sqrt{x+1}+\sqrt{2 x}>\sqrt{33-x}+4
$$
Find the domain of the variable $x$:
$$
\left\{\begin{array}{l}
x-4 \geqslant 0 \\
x+1 \geqslant 0 \\
2 x \geqslant 0 \\
33-x \geqslant 0
\end{array}\right.
$$
from which $x \in[4,33]$. We are interested in the integer values from this interval that satisfy the inequalit |
8.1.1. (12 points) Among the first hundred elements of the arithmetic progression $3,7,11, \ldots$ find those that are also elements of the arithmetic progression $2,9,16, \ldots$ In your answer, indicate the sum of the found numbers. | $$
\begin{gathered}
a_{n}=a_{1}+(n-1) d_{1}=3+(n-1) \cdot 4 \\
b_{m}=b_{1}+(m-1) d_{2}=2+(m-1) \cdot 7 \\
3+(n-1) \cdot 4=2+(m-1) \cdot 7 \\
4(n+1)=7 m, \quad m=4 k, \quad n=7 k-1
\end{gathered}
$$
Consider the sequence of coinciding terms of the progressions $A_{k}$. For $k=1$, we find $n=6$, $m=4$. For $k=2$, we find $n=13$, $m=8$. The first term $A_{1}=a_{6}=23$, the difference $d=a_{13}-a_{6}=$ $51-23=28$. There are a total of $[100: 7]=14$ terms in the sequence, and their sum is
$$
S=\frac{2 \cdot 23+13 \cdot 28}{2} \cdot 15=2870
$$ | 2870 | Number Theory | olympiads | null | null | \(\boxed{2870}\) | 8.1.1. (12 points) Among the first hundred elements of the arithmetic progression $3,7,11, \ldots$ find those that are also elements of the arithmetic progression $2,9,16, \ldots$ In your answer, indicate the sum of the found numbers.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
$$
\begin{gathered}
a_{n}=a_{1}+(n-1) d_{1}=3+(n-1) \cdot 4 \\
b_{m}=b_{1}+(m-1) d_{2}=2+(m-1) \cdot 7 \\
3+(n-1) \cdot 4=2+(m-1) \cdot 7 \\
4(n+1)=7 m, \quad m=4 k, \quad n=7 k-1
\end{gathered}
$$
Consider the sequence of coinciding terms of the progressions $A_{k}$. Fo |
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