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4.31. Prove that if $2^{n}-2$ is divisible by $n$, then $2^{2^{n}-1}-2$ is divisible by $2^{n}-1$.
保留源文本的换行和格式，直接输出翻译结果。 | 4.31. Let $2^{n}-2=n m$. Then
$$
\frac{2^{2^{n}-1}-2}{2^{n}-1}=2 \frac{2^{2^{n}-2}-1}{2^{n}-1}=2 \frac{2^{n m}-1}{2^{n}-1}=2\left(2^{n(m-1)}+2^{n(m-2)}+\ldots+2^{n}+1\right)
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,195 |
4.32. Let $a, b, m$ and $n$ be natural numbers, and $a$ is coprime with $b$ and $a>1$. Prove that $a^{m}+b^{m}$ is divisible by $a^{n}+b^{n}$ if and only if $m=k n$, where $k$ is an odd number.
$$
\% * \%
$$ | 4.32. If $m=k n$, where $k$ is an odd number, then $a^{m}+b^{m}=\left(a^{n}\right)^{k}+$ $+\left(b^{n}\right)^{k}$ is divisible by $a^{n}+b^{n}$ (Problem 5.1 b).
Let $m=k n+r$, where $k$ is an odd number and $0<r<n$. We will prove that $a^{m}+b^{m}$ is not divisible by $a^{n}+b^{n}$. Indeed,
$$
a^{k n+r}+b^{k n+r}=a^... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,196 |
4.33. a) Prove that the number $1 / 2 + 1 / 3 + \ldots + 1 / n$ cannot be an integer.
b) Prove that the number $\frac{1}{k} + \frac{1}{k+1} + \ldots + \frac{1}{k+n}$, where $k$ and $n$ are natural numbers, cannot be an integer. | 4.33. Let's immediately solve problem b). Suppose the number in question is an integer. Among the numbers $k, k+1, \ldots, k+n$, select those that are divisible by the highest power of two. Let there be only one such number, namely $2^{m} p$, where $p$ is odd. Then the number in question has the form $\frac{1}{2^{m}}\l... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,197 |
4.35. How many zeros does the product of all integers from 1 to 100 inclusive end with? | 4.35. Answer: 24. Among the numbers from 1 to 100, there are 20 numbers divisible by 5, and among the numbers divisible by 5, there are 4 numbers divisible by 25 (there are no numbers divisible by 125 among the given numbers). Therefore, the considered product is divisible by $5^{24}$ and not divisible by $5^{25}$. It ... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,199 |
4.36. Let $p$ be a prime number and $a$ be the largest integer for which $n!$ is divisible by $p^{a}$. Prove that
$$
a=\left[\frac{n}{p}\right]+\left[\frac{n}{p^{2}}\right]+\left[\frac{n}{p^{3}}\right]+\ldots
$$ | 4.36. Among the numbers $1,2, \ldots, n$ there are $[n / p]$ numbers divisible by $p$, $\left[n / p^{2}\right]$ numbers divisible by $p^{2}$, and so on. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,200 |
4.37. Prove that $n$ ! is not divisible by $2^{n}$. | 4.37. Let $a$ be the largest integer for which $n!$ is divisible by $2^{a}$. According to problem 4.36,
$$
a=\left[\frac{n}{2}\right]+\left[\frac{n}{2^{2}}\right]+\left[\frac{n}{2^{3}}\right]+\ldots \leqslant \frac{n}{2}+\frac{n}{2^{2}}+\frac{n}{2^{3}}+\ldots=n
$$
since $[x] \leqslant x$. Moreover, $\left[\frac{n}{2^... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,201 |
4.38. Find the highest power of two that divides the number $(n+1)(n+2) \cdot \ldots \cdot 2 n$.
$$
* * *
$$ | 4.38. Let $(2 n-1)!!=1 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot(2 n-1)$ and $(2 n)!!=2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2 n$. It is clear that $(2 n)!=(2 n-1)!!(2 n)!!$ and $(2 n)!!=2^{n} n!$. Therefore, $(n+1)(n+2) \times \ldots$ $\cdots \times 2 n=\frac{(2 n)!}{n!}=\frac{(2 n-1)!!(2 n)!!}{n!}=2^{n}(2 n-1)!!$, whe... | 2^n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,202 |
4.39. Find all natural $n$ for which $\frac{1}{n}$ and $\frac{1}{n+1}$ are finite decimal fractions. | 4.39. Answer: 1 and 4. For $n>1$, the equalities $2^{k} 5^{l}=n$ and $2^{s} 5^{t}=n+1$ must hold. The numbers $n$ and $n+1$ are coprime, so either the equality $2^{k}+1=5^{t}$ or the equality $5^{l}+1=2^{s}$ must hold. Suppose that $5^{l}+1=2^{s}$. Then the number $2^{s}$ ends in 6, so $s=4 m$. Therefore, $5^{l}=2^{4 m... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,203 |
4.40. a) Prove that for any odd number $a$ and any natural number $m$, there exist infinitely many natural numbers $k$ such that $a^{k}-1$ is divisible by $2^{m}$.
b) Prove that for any odd number $a$, there are only finitely many natural numbers $m$ for which $a^{m}-1$ is divisible by $2^{m}$. | 4.40. a) Let $k=2^{n}$. Then
$$
\begin{aligned}
a^{k}-1=\left(a^{2^{n-1}}-1\right) & \left(a^{2^{n-1}}+1\right)= \\
& =(a-1)(a+1)\left(a^{2}+1\right)\left(a^{4}+1\right) \ldots\left(a^{2^{n-1}}+1\right)
\end{aligned}
$$
The number $a$ is odd, so in this product there are $n+1$ even factors. Therefore, $a^{k}-1$ is di... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,204 |
4.41. Find all natural numbers $m$ for which: a) $3^{m}-1$ is divisible by $2^{m}$; b) $31^{m}-1$ is divisible by $2^{m}$.
## 4.8. Remainders of Division | 4.41. a) Answer: 1, 2, and 4. Using the solution to problem 4.40 b). If \( a=3 \), then \( d(a-1)=1 \) and \( d\left(a^{2}-1\right)=3 \). The inequality \( s \leqslant \frac{2+n}{2^{n}} \) has the following solutions: \((s, n)=(1,1)\) and \((1,2)\).
b) Answer: 1, 2, 4, 6, and 8. If \( a=31 \), then \( d(a-1)=1 \) and ... | 1,2,4,6,8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,205 |
4.42. What remainders can the square of an integer give when divided by: a) 3, b) 4, c) 5, d) $8?$ | 4.42. a) Answer: 0 and 1. Use the equality $(3 k \pm 1)^{2}=$ $=9 k^{2} \pm 6 k+1$.
b) Answer: 0 and 1. Use the equality $(2 k+1)^{2}=4 k^{2}+$ $+4 k+1$.
c) Answer: 0, 1, and 4. Use the equalities $(5 k \pm 1)^{2}=$ $=25 k^{2} \pm 10 k+1$ and $(5 k \pm 2)^{2}=25 k^{2} \pm 20 k+4$.
d) Answer: 0, 1, and 4. Use the equ... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,206 |
4.43. Find the smallest natural number that:
a) When divided by 5 gives a remainder of 4, when divided by 6 gives a remainder of 5, and when divided by 7 gives a remainder of 6.
b) When divided by 5 gives a remainder of 4, when divided by 6 gives a remainder of 5, and when divided by 8 gives a remainder of 7. | 4.43. a) Answer: 209. Let $n$ be the desired number. Then $n+1$ is divisible by $5 \cdot 6 \cdot 7=210$. Therefore, $n=209$.
b) Answer: 119. Let $n$ be the desired number. Then $n+1$ is divisible by $\operatorname{LCM}(5,6,8)=120$. Therefore, $n=119$. | 119 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,207 |
4.44. Find the number that:
a) When divided by 5 gives a remainder of $a$, and when divided by $6$ gives a remainder of $b$.
b) When divided by 5 gives a remainder of $a$, when divided by $6$ gives a remainder of $b$, and when divided by 7 gives a remainder of $c$. | 4.44. a) Answer: $6 a+25 b$. Let $6 n \equiv 1(\bmod 5)$ and $5 m \equiv 1(\bmod 6)$. Then
$$
\begin{aligned}
& 6 n a+5 m b \equiv 6 n a \equiv a(\bmod 5) \\
& 6 n a+5 m b \equiv 5 m b \equiv b(\bmod 6) .
\end{aligned}
$$
In our case $m=5$ and $n=1$.
b) Answer: $126 a+175 b+120 c$. Let $42 k \equiv 1(\bmod 5), 35 n ... | 6a+25b | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,208 |
4.45. a) A number $n$ when divided by 4 gives a remainder of 3. Prove that $n$ cannot be represented as the sum of two squares of integers.
b) A number $n$ when divided by 8 gives a remainder of 7. Prove that $n$ cannot be represented as the sum of three squares of integers. | 4.45. a) According to problem 4.42 b) the sum of the squares of two numbers when divided by 4 can give remainders 0, 1, and 2.
b) According to problem 4.42 g) the sum of the squares of two numbers when divided by 8 can give remainders $0,1,2,4$ and 5, and the sum of the squares of three numbers - remainders $0,1,2,3,4... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,209 |
4.46. Find a four-digit number that, when divided by 131, gives a remainder of 112, and when divided by 132, gives a remainder of 98. | 4.46. Answer: 1946. Let $N$ be the desired number. By the condition, $N=131 k+112=132 l+98$, where $k$ and $l$ are natural numbers. Moreover, $N<10000$, so $l=\frac{N-98}{132}<\frac{10000-98}{132} \leqslant 75$. Further, $131 k+112=132 l+98$, therefore $131(k-l)=l-14$. Consequently, if $k \neq l$, then $|l-14| \geqslan... | 1946 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,210 |
4.47. Append three digits to $523 \ldots$ so that the resulting six-digit number is divisible by 7, 8, and 9. | 4.47. Answer: 523152 or 523656.
The obtained number must be divisible by $7 \cdot 8 \cdot 9=504$. Dividing 523000 by 504 with a remainder: $523000=1037 \cdot 504+352$. Since $504-352=152$, the numbers 523152 and $523152+504=$ $=523656$ are divisible by 504. There are no other numbers divisible by 504 among the numbers... | 523152or523656 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,211 |
4.48. Find the remainder when the number
$$
10^{10}+10^{\left(10^{2}\right)}+10^{\left(10^{3}\right)}+\ldots+10^{\left(10^{10}\right)}
$$
is divided by 7. | 4.48. Answer: 5. Note that $10^{6} \equiv 1(\bmod 7)$, since $10^{3}+1$ is divisible by 7, and $10^{k} \equiv 4(\bmod 6)$ for $k \geqslant 1$, since the number $99 \ldots 96$ is even and divisible by 3. Therefore, $10^{10^{k}} \equiv 10^{4}(\bmod 7)$ for $k \geqslant 1$. Hence, the required remainder is the remainder o... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,212 |
4.49. Let the numbers $m_{1}, \ldots, m_{k}$ be pairwise coprime and $m=m_{1} \ldots m_{k}$. Prove that for any integers $a_{1}, \ldots, \ldots, a_{k}$ the system of congruences $x \equiv a_{i}\left(\bmod m_{i}\right)$, where $i=1, \ldots, \ldots, k$, has a solution, and if $x_{1}$ and $x_{2}$ are two solutions, then $... | 4.49. Let $n_{i}=m / m_{i}$. The number $n_{i}$ is the product of numbers that are coprime with $m_{i}$, so $\gcd \left(n_{i}, m_{i}\right)=1$. In this case, one can choose integers $r_{i}$ and $s_{i}$ such that $r_{i} m_{i}+s_{i} n_{i}=1$ (a proof of this statement using the Euclidean algorithm is given on p. 43). Let... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,213 |
4.50. Find all natural numbers $n$ for which $2^{n}-1$ is divisible by 7. | 4.50. It is clear that $2^{3}=8 \equiv 1(\bmod 7)$. Therefore, $2^{3 k} \equiv 1(\bmod 7)$, $2^{3 k+1} \equiv 2(\bmod 7)$ and $2^{3 k+2} \equiv 4(\bmod 7)$. Consequently, $2^{n}-1$ is divisible by 7 if and only if $n$ is divisible by 3. | n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,214 |
4.51. Let $p$ be a prime number. Suppose we are given $r$ natural numbers $a_{1}, \ldots, a_{r}$, all less than $p$. Prove that if $r<p$, then from these numbers, one can form at least $r+1$ sums that give different remainders when divided by $p$ (the sum of the "empty set" of addends, which is considered to be zero, i... | 4.51. We apply induction on $r$. For $r=1$, there are two sums: 0 and $a_{1}$. Suppose the statement is true for $r<p-1$ numbers and false for $r+1$ numbers. Then the sums $0, s_{1}, \ldots, s_{r}$, formed from the numbers $a_{1}, \ldots, a_{r}$, give different remainders when divided by $p$, but all sums $0+a_{r+1}, s... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,215 |
4.52. Prove that from any $2 n-1$ integers, one can select exactly $n$ numbers, the sum of which is divisible by $n$.
## 4.9. Coprime Numbers
- Natural numbers $m$ and $n$ are called coprime if GCD $(m, n)=1$. | 4.52. First, let's prove that if the required statement is true for \( n=a \) and for \( n=b \), then it is also true for \( n=ab \). Suppose we are given \( 2ab-1 \) integers. The statement is true for \( n=b \), and moreover, \( 2ab-1 \geqslant 2b-1 \). Therefore, from the given \( 2ab-1 \) numbers, we can select \( ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,216 |
4.53. Prove that for no natural number $n$ the number $n(n+1)$ can be a power of a natural number. | 4.53. Suppose that $n(n+1)=m^{k}$, where $m$ and $k$ are natural numbers, $k \geqslant 2$. The numbers $n$ and $n+1$ are coprime, so $n=a^{k}$ and $n+1=b^{k}$, where $a$ and $b$ are natural numbers. Clearly, $b>a$. But $(a+1)^{k}>(a+1) a^{k-1}=a^{k}+a^{k-1} \geqslant n+1$. Therefore, $b^{k} \geqslant(a+1)^{k}>n+1$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,217 |
4.54. a) Prove that for any integer $n$, among the numbers $n, n+1, n+2, n+3, n+4$ there is at least one number that is coprime with the other four numbers.
b) Prove that for any integer $n$, among the numbers $n, n+1, n+2, \ldots, n+9$ there is at least one number that is coprime with the other nine numbers. | 4.54. a) If $|k-l| \leqslant 4$ and $k \neq l$, then the greatest common divisor of the numbers $k$ and $l$ does not exceed 4. Therefore, the greatest common divisor of any pair of selected numbers does not exceed 4. From five consecutive numbers, one can choose a pair of consecutive odd numbers. Among two consecutive ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,218 |
4.55. Let $n$ and $m$ be different natural numbers. Prove that the numbers $F_{n}=2^{2^{n-1}}+1$ and $F_{m}=2^{2^{m-1}}+1$ are coprime.
### 4.10. Prime Numbers | 4.55. Let's assume that $n>m$. Substitute $x=2$ into the identity
$$
(x-1)(x+1)\left(x^{2}+1\right)\left(x^{4}+1\right) \ldots\left(x^{2^{n-2}}+1\right)=x^{2^{n-1}}+1 .
$$
As a result, we get $F_{1} F_{2} \ldots F_{n-1}+2=F_{n}$.
Suppose that the numbers $F_{n}$ and $F_{m}$ are divisible by $d$. Then $2=$ $=F_{n}-F_... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,219 |
4.56. Prove that the square of any prime number $p>3$ when divided by 12 leaves a remainder of 1. | 4.56. Let's see what remainders a prime number \( p > 3 \) can give when divided by 6. It cannot give a remainder of 2 or 4, since otherwise it would be even. It cannot give a remainder of 3, since otherwise it would be divisible by 3. Therefore, a prime number \( p > 3 \) when divided by 6 gives a remainder of 1 or 5,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,220 |
4.57. Prove that there are infinitely many prime numbers (Euclid). | 4.57. Suppose there are only a finite number of distinct prime numbers, namely, $p_{1}, \ldots, p_{r}$. Consider the number $p_{1} \ldots p_{r}+1$. It is not divisible by any of the numbers $p_{1}, \ldots, p_{r}$, so it has a prime divisor different from $p_{1}, \ldots, p_{r}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,221 |
4.58. Prove that there are infinitely many prime numbers of the form $4k-1$. | 4.58. Suppose that $p_{1}, \ldots, p_{r}$ are all the distinct prime numbers of the form $4 k-1$. Consider the number $4 p_{1} \ldots p_{r}-1$. It is odd, so all its prime divisors must be of the form $4 k \pm 1$. It is also clear that all prime divisors cannot be of the form $4 k+1$, since the product of numbers of th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,222 |
4.59. a) Prove that if the number $2^{n}-1$ is prime, then the number $n$ is also prime.
b) Prove that if the number $2^{n}+1$ is prime, then $n=2^{k}$. | 4.59. a) The polynomial $x^{q}-1$ is divisible by $x-1$, therefore $2^{p q}-1=$ $=\left(2^{p}\right)^{q}-1$ is divisible by $2^{q}-1$.
b) If $q$ is odd, then the polynomial $x^{q}+1$ is divisible by $x+1$. Therefore, if the number $n$ has an odd divisor $q>1$, then $2^{n}+1$ is divisible by $2^{q}+1$.
64 Chapter 4. D... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,223 |
4.60. Prove that for any natural $n$ the number $2^{2^{n}}+2^{2^{n-1}}+1$ has at least $n$ distinct prime divisors.
### 4.11. Arithmetic of Residues | 4.60. Let's use the identity $x^{4}+x^{2}+1=\left(x^{2}+1-x\right) \times$ $\times\left(x^{2}+1+x\right)$. For $x=2^{2^{n-2}}$, we get that the number in question is the product of the numbers $2^{2^{n-1}}+2^{2^{n-2}}+1$ and $2^{2^{n-1}}-2^{2^{n-2}}+1$. These numbers are coprime, since they are odd, and their differenc... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,224 |
4.61. Prove that the remainders of the division of $a \pm b$ and $a b$ by $n$ are uniquely determined by the remainders of the division of $a$ and $b$ by $n$. | 4.61. Let $a=a_{1}+a_{2} n$ and $b=b_{1}+b_{2} n$. Then $a \pm b=a_{1} \pm b_{1}+$ $+\left(a_{2} \pm b_{2}\right) n$ and $a b=a_{1} b_{1}+\left(a_{2} b_{1}+a_{1} b_{2}+a_{2} b_{2} n\right) n$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,225 |
4.62. Let $p$ be a prime number, and let the number $a$ not be divisible by $p$. Prove that all remainders of the division by $p$ of the numbers $a$, $2a$, $3a, \ldots, (p-1)a$ are pairwise distinct, i.e., each number from 1 to $p-1$ appears among these remainders exactly once. | 4.62. If $x a \equiv y a(\bmod p)$, then $(x-y) a$ is divisible by $p$. Since $a$ and $p$ are coprime, $x-y$ is divisible by $p$. Therefore, if $1 \leqslant x, y \leqslant p-1$, then $x=y$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,226 |
4.63. Prove that if $p$ is a prime number, and the numbers $a$ and $b$ are not divisible by $p$, then the remainder of the division of $a$ by $p$ is uniquely determined by the remainders of the division of the numbers $b$ and $a b$ by $p$. | 4.63. According to problem 4.62, exactly one of the remainders of dividing by $p$ of the numbers $b, 2b, \ldots, (p-1)b$ is equal to 1. Therefore, there exists a unique integer $\bar{b}, 1 \leqslant \bar{b} \leqslant p-1$, for which $\bar{b} b \equiv 1(\bmod p)$. In this case, $a \equiv \bar{b}(a b)(\bmod p)$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,227 |
5.1. a) $x^{n}-y^{n}$; b) $x^{2 n+1}+y^{2 n+1}$. | 5.1. a) $(x-y)\left(x^{n-1}+x^{n-2} y+\ldots+y^{n-1}\right)$.
b) $(x+y)\left(x^{2 n}-x^{2 n-1} y+x^{2 n-2} y^{2}-\ldots+y^{2 n}\right)$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,228 | |
5.2. $x^{4}+4$. | 5.2. $\left(x^{2}-2 x+2\right)\left(x^{2}+2 x+2\right)$. | (x^{2}-2x+2)(x^{2}+2x+2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,229 |
5.3. $(x+y+z)^{3}-x^{3}-y^{3}-z^{3}$. | 5.3. $3(x+y)(y+z)(z+x)$. | 3(x+y)(y+z)(z+x) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,230 |
5.4. $x^{3}+y^{3}+z^{3}-3 x y z$. | 5.4. $(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$. | (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,231 |
5.5. $(x-y)^{3}+(y-z)^{3}+(z-x)^{3}$. | 5.5. $3(x-y)(y-z)(z-x)$. | 3(x-y)(y-z)(z-x) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,232 |
5.6. $a^{10}+a^{5}+1$. | 5.6. $\left(a^{2}+a+1\right)\left(a^{8}-a^{7}+a^{5}-a^{4}+a^{3}-a+1\right)$. | (^{2}++1)(^{8}-^{7}+^{5}-^{4}+^{3}-+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,233 |
5.7. $a^{4}(b-c)+b^{4}(c-a)+c^{4}(a-b)$. | 5.7. $\left(a^{2}+b^{2}+c^{2}+a b+b c+c a\right)(a-b)(b-c)(a-c)$. | (^{2}+b^{2}+^{2}+++)(-b)(b-)(-) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,234 |
5.8. $x^{4}+x^{3}+x^{2}+x+12$. | 5.8. $\left(x^{2}-2 x+3\right)\left(x^{2}+3 x+4\right)$. | (x^{2}-2x+3)(x^{2}+3x+4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,235 |
5.9. a) Factorize $x^{8}+x^{4}+1$ into two factors.
b) Factorize $x^{8}+x^{4}+1$ into four factors, allowing square roots of natural numbers as coefficients.
## 5.2. Proof of identities | 5.9. a) $\left(x^{4}+x^{2}+1\right)\left(x^{4}-x^{2}+1\right)$.
b) Each of the resulting fourth-degree polynomials can be factored using the identity
$$
\left(x^{2}+a x+1\right)\left(x^{2}-a x+1\right)=x^{4}+\left(2-a^{2}\right) x^{2}+1
$$
by setting $a=1$ and $a=\sqrt{3}$. | (x^{2}+x+1)(x^{2}-x+1)(x^{2}+\sqrt{3}x+1)(x^{2}-\sqrt{3}x+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,236 |
5.10. Prove that for any natural $n$ the following relation holds
$$
\frac{(2 n)!}{n!}=2^{n} \cdot(2 n-1)!!
$$
## 5.3. Sums of Squares | 5.10. It is clear that $n!2^{n}=2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2 n$. Therefore, $n!2^{n}(2 n-1)!!=$ $=(2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2 n) \cdot 1 \cdot 3 \cdot 5 \cdot \ldots \cdot(2 n-1)=(2 n)!$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,237 |
5.11. Prove that if each of the numbers $m$ and $n$ can be represented as the sum of two squares of integers, then
their product $m n$ can also be represented as the sum of two squares of integers. | 5.11. Use the identity $\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=(a c+b d)^{2}+$ $+(a d-b c)^{2}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,238 |
5.12. a) Represent as a sum of squares
\[
\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)-\left(a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\right)^{2}
\]
b) Represent as a sum of squares
\[
\left(a_{1}^{2}+\ldots+a_{n}^{2}\right)\left(b_{1}^{2}+\ldots+b_{n}^{2}\right)-\left(a_{1} b_{1}+\... | 5.12. a) $\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2}+\left(a_{2} b_{3}-a_{3} b_{2}\right)^{2}+\left(a_{1} b_{3}-a_{3} b_{1}\right)^{2}$.
b) $\sum\left(a_{i} b_{j}-a_{j} b_{i}\right)^{2}$, where the summation is over all pairs $i<j$. | \sum(a_{i}b_{j}-a_{j}b_{i})^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,239 |
5.13. It is known that the number $a+1 / a$ is an integer. a) Prove that the number $a^{2}+1 / a^{2}$ is also an integer. b) Prove that the number $a^{n}+1 / a^{n}$ is an integer for any natural $n$. | 5.13. a) Use the identity $a^{2}+1 / a^{2}=(a+1 / a)^{2}-2$.
b) Use the identity
$$
a^{n+1}+\frac{1}{a^{n+1}}=\left(a^{n}+\frac{1}{a^{n}}\right)\left(a+\frac{1}{a}\right)-\left(a^{n-1}+\frac{1}{a^{n-1}}\right)
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,240 |
5.14. Prove that any odd number is the difference of two squares of integers. | 5.14. Use the identity $2 n+1=(n+1)^{2}-n^{2}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The instruction you provided is a meta-instruction about the translation task. The actual translation is as follows:
5.14. ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,241 |
5.15. Prove that the product of four consecutive integers plus one results in a perfect square. | 5.15. It is sufficient to note that
$$
n(n+1)(n+2)(n+3)+1=(n(n+3)+1)^{2} \text {. }
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,242 |
5.16. Prove that if $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ then $\frac{1}{a^{n}}+\frac{1}{b^{n}}+$ $+\frac{1}{c^{n}}=\frac{1}{a^{n}+b^{n}+c^{n}}$ for any odd $n$. | 5.16. The equality $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ is equivalent to the equality $(b c+c a+a b)(a+b+c)=a b c$, i.e., $(a+b)(b+c)(c+a)=0$ (assuming that $a b c \neq 0$ and $a+b+c \neq 0$). Therefore, the triplet of numbers $a$, $b, c$ has the form $x, -x, y$, where $y \neq \pm x$. However, then the... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,243 |
5.17. Let $x, y, z$ be pairwise distinct integers. Prove that the number $(x-y)^{5}+(y-z)^{5}+(z-x)^{5}$ is divisible by $5(y-z)(z-x)(x-y)$. | 5.17. It is not difficult to verify that the quotient is $x^{2}+y^{2}+z^{2}-x y- y z-x z$. Indeed, let $x-y=u$ and $y-z=v$. Then $(x-y)^{5}+ (y-z)^{5}+(z-x)^{5}=u^{5}+v^{5}-(u+v)^{5}=-5\left(u^{4} v+2 u^{3} v^{2}+2 u^{2} v^{3}+v u^{4}\right)=$ $=-5 u v(u+v)\left(u^{2}+u v+v^{2}\right)$ and $5(y-z)(z-x)(x-y)=-5 u v(u+v)... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,244 |
5.18. Prove that for any pairwise distinct rational numbers $a, b$ and $c$, the number $\frac{1}{(a-b)^{2}}+\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}}$ is the square of some rational number. | 5.18. Let $x=\frac{1}{a-b}, y=\frac{1}{b-c}$ and $z=\frac{1}{c-a}$. Then
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=(a-b)+(b-c)+(c-a)=0
$$
therefore $x y+y z+x z=0$. Consequently, $x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-$ $-2(x y+y z+x z)=(x+y+z)^{2}$ - a square of a rational number, as required. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,245 |
5.19. Prove that if $\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0$, then $\frac{a}{(b-c)^{2}}+\frac{b}{(c-a)^{2}}+\frac{c}{(a-b)^{2}}=0$. | 5.19. Let's prove that
$$
\begin{aligned}
& \frac{a}{(b-c)^{2}}+\frac{b}{(c-a)^{2}}+\frac{c}{(a-b)^{2}}= \\
& \quad=\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)
\end{aligned}
$$
Indeed, the sum of three fractions
$$
\begin{aligned}
\frac{a}{b-c}\left(\f... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,246 |
5.20. Prove that $n^{2}+3 n+5$ is not divisible by 121 for any integer $n$. | 5.20. Note that $n^{2}+3 n+5=(n+7)(n-4)+33$. If the number $n^{2}+3 n+5$ is divisible by 121, then the number $(n+7)(n-4)$ is divisible by 11. But $(n+7)-(n-4)=11$, so both factors are divisible or not divisible by 11 simultaneously. Therefore, if the number $(n+7)(n-4)$ is divisible by 11, then it is divisible by 121.... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,247 |
5.21. Prove that the expression
$$
x^{5}+3 x^{4} y-5 x^{3} y^{2}-15 x^{2} y^{3}+4 x y^{4}+12 y^{5}
$$
is never equal to 33 for any integer values of $x$ and $y$. | 5.21. Let's represent the given expression as
$$
(x+2 y)(x-y)(x+y)(x-2 y)(x+3 y)
$$
For $y \neq 0$, all five factors of this product are pairwise distinct, and the number 33 cannot be represented as the product of five pairwise distinct integers (although it can be represented as the product of four pairwise distinct... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,248 |
5.22. Prove that for any natural $n \geqslant 2$ the number $2^{4 n+2}+1$ is not the product of two prime numbers. | 5.22. The number $2^{4 n+2}+1$ can be represented both as the product $\left(2^{2}+1\right)\left(2^{4 n}-2^{4 n-2}+\ldots-2^{2}+1\right)$, and as the product $\left(2^{2 n+1}+2^{n+1}+1\right)\left(2^{2 n+1}-2^{n+1}+1\right)$. For $n \geqslant 2$, these factorizations are distinct. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,249 |
5.23. Prove that any rational number can be represented as the sum of three cubes of rational numbers.
## 5.5. Decompositions of Rational Functions | 5.23. Let's use the identity
$$
(a-b)^{3}+(b-c)^{3}+c^{3}=3 b^{2}(a-c)+\left(a^{3}-3 b\left(a^{2}-c^{2}\right)\right)
$$
Take a rational number $t$ and set $a=12 t(t+1), b=(t+1)^{3}$ and $c=12 t(t-1)$. Then we get
$$
72 t(t+1)^{6}=(a-b)^{3}+(b-c)^{3}+c^{3}
$$
If $t \neq-1$, then the number $w=72 t$ is the sum of th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,250 |
5.24. Represent the fractions $\frac{2}{x^{2}-1}$ and $\frac{2 x}{x^{2}-1}$ as the sum of fractions $\frac{a}{x-1}+\frac{b}{x+1}$, where $a$ and $b$ are real numbers. | 5.24. It is clear that $\frac{a}{x-1}+\frac{b}{x+1}=\frac{(a+b) x+(a-b)}{x^{2}-1}$. Therefore, $\frac{2}{x^{2}-1}=$ $=\frac{1}{x-1}-\frac{1}{x+1}$ and $\frac{2 x}{x^{2}-1}=\frac{1}{x-1}+\frac{1}{x+1}$. | \frac{2}{x^{2}-1}=\frac{1}{x-1}-\frac{1}{x+1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,251 |
5.25. Let the numbers $a_{1}, \ldots, a_{n}$ be pairwise distinct. Prove that one can choose numbers $A_{1}, \ldots, A_{n}$ such that
$$
\frac{1}{\left(x+a_{1}\right) \ldots\left(x+a_{n}\right)}=\frac{A_{1}}{x+a_{1}}+\ldots+\frac{A_{n}}{x+a_{n}}
$$ | 5.25. We will prove the required statement by induction on $n$. For $n=2$, we can set $A_{1}=\frac{1}{a_{2}-a_{1}}$ and $A_{2}=\frac{1}{a_{1}-a_{2}}$. Now suppose that
$$
\frac{1}{\left(x+a_{1}\right) \ldots\left(x+a_{n-1}\right)}=\frac{B_{1}}{x+a_{1}}+\frac{B_{n-1}}{x+a_{n-1}}
$$
Let us represent each fraction $\fra... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,252 |
5.27. Prove that if for all $x, y, z$ the equality
$$
\begin{gathered}
a_{11} x^{2}+a_{22} y^{2}+a_{33} z^{2}+2 a_{12} x y+2 a_{13} x z+2 a_{23} y z= \\
=\left(b_{1} x+b_{2} y+b_{3} z\right)\left(c_{1} x+c_{2} y+c_{3} z\right)
\end{gathered}
$$
holds, then $a_{11} a_{22} a_{33}+2 a_{13} a_{12} a_{23}=a_{23}^{2} a_{11... | 5.27. If the given equality holds for all $x, y, z$, then
$$
\begin{array}{ll}
a_{11}=b_{1} c_{1}, & 2 a_{12}=b_{1} c_{2}+b_{2} c_{1} \\
a_{22}=b_{2} c_{2}, & 2 a_{13}=b_{1} c_{3}+b_{3} c_{1} \\
a_{33}=b_{3} c_{3}, & 2 a_{23}=b_{2} c_{3}+b_{3} c_{2}
\end{array}
$$
Indeed, this statement is equivalent to the assertion... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,254 |
5.28. Prove that
$$
[x]+\left[x+\frac{1}{n}\right]+\left[x+\frac{2}{n}\right]+\ldots+\left[x+\frac{n-1}{n}\right]=[n x]
$$ | 5.28. Consider the function
$$
f(x)=[n x]-[x]-\left[x+\frac{1}{n}\right]-\ldots-\left[x+\frac{n-1}{n}\right]
$$
It is clear that
$$
\begin{array}{r}
f\left(x+\frac{1}{n}\right)=[n x+1]-\left[x+\frac{1}{n}\right]-\left[x+\frac{2}{n}\right]-\ldots-\left[x+\frac{n-1}{n}\right]-[x+1]= \\
=[n x]-[x]-\left[x+\frac{1}{n}\r... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,255 |
5.29. Let $p$ and $q$ be coprime natural numbers. Prove that
$$
\left[\frac{q}{p}\right]+\left[\frac{2 q}{p}\right]+\ldots+\left[\frac{(p-1) q}{p}\right]=\frac{(p-1)(q-1)}{2}
$$ | 5.29. Consider the rectangle $0 \leqslant x \leqslant p, 0 \leqslant y \leqslant q$ and draw its diagonal $y=q x / p$. The sum of interest is the number of points with integer coordinates that lie strictly inside (not on the boundary) of this rectangle and below its diagonal. Symmetry relative to the center of the rect... | \frac{(p-1)(q-1)}{2} | Number Theory | proof | Yes | Yes | olympiads | false | 38,256 |
5.30. Let $p$ and $q$ be coprime odd natural numbers. Prove that
$$
\begin{aligned}
& {\left[\frac{q}{p}\right]+\left[\frac{2 q}{p}\right]+\ldots+\left[\frac{\frac{p-1}{2} \cdot q}{p}\right]+\left[\frac{p}{q}\right]+} \\
&+ {\left[\frac{2 p}{q}\right]+\ldots+\left[\frac{\frac{q-1}{2} \cdot p}{q}\right]=\frac{(p-1)(q-1... | 5.30. Consider the rectangle $1 \leqslant x \leqslant \frac{p-1}{2}, 1 \leqslant y \leqslant \frac{q-1}{2}$ and draw the line $y=q x / p$. The sum of interest consists of two parts. The first part is equal to the number of integer points in this rectangle lying below this line, and the second part - above. There are no... | \frac{(p-1)(q-1)}{4} | Number Theory | proof | Yes | Yes | olympiads | false | 38,257 |
5.31. Prove that $[\sqrt{4 n+1}]=[\sqrt{4 n+2}]=[\sqrt{4 n+3}]$ for any natural $n$. | 5.31. Let $k=[\sqrt{4 n+1}]$. Then $k^{2} \leqslant 4 n+1<(k+1)^{2}$. A perfect square when divided by 4 cannot leave a remainder of 2 or 3, therefore the numbers $4 n+2$ and $4 n+3$ cannot coincide with $(k+1)^{2}$. Consequently, both these numbers are less than $(k+1)^{2}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,258 |
5.32. Prove that $[\sqrt{n}+\sqrt{n+1}]=[\sqrt{4 n+2}]$ for any natural number $n$. | 5.32. First, let's compare the numbers $\sqrt{n}+\sqrt{n+1}$ and $\sqrt{4 n+2}$. Instead of the inequality sign, we will use the sign $\vee$ and consider only transformations that preserve the inequality sign: $\sqrt{n}+\sqrt{n+1} \vee \vee \sqrt{4 n+2} \Leftrightarrow n+2 \sqrt{n(n+1)}+n+1 \vee 4 n+2 \Leftrightarrow 2... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,259 |
5.33. Prove that $[\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}]=[\sqrt{9 n+8}]$ for any natural number $n$. | 5.33. Let $x=\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}$. Then $x^{2}=3 n+3+$ $+2(\sqrt{n(n+1)}+\sqrt{n(n+2)}+\sqrt{(n+1)(n+2)})$. For $n \geqslant 1$, the following inequalities hold:
$$
\begin{gathered}
\left(n+\frac{2}{5}\right)^{2}<n(n+1)<\left(n+\frac{1}{2}\right)^{2} \\
\left(n+\frac{7}{10}\right)^{2}<n(n+2)<(n+1)^{2} \\
\l... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,260 |
5.34. Prove that $[\sqrt[3]{n}+\sqrt[3]{n+1}]=[\sqrt[3]{8 n+3}]$ for any natural number $n$. | 5.34. According to problem 8.42, for any numbers $a>b>0$, the inequality $\frac{a+b}{2}\sqrt[3]{n}+\sqrt[3]{n+1}>\sqrt[6]{64 n^{2}+64 n}>\sqrt[3]{8 n+3}$ holds. Therefore, if $\sqrt[3]{8 n+3}>m$, then $\sqrt[3]{n}+\sqrt[3]{n+1}>m$. Moreover, there does not exist an integer $m$ such that $\sqrt[3]{n}+\sqrt[3]{n+1} \geqs... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,261 |
6.1. Compare which of the numbers is greater: a) $\sqrt{2}+\sqrt{3}$ or $\sqrt{11}$; b) $\sqrt{6}+2 \sqrt{7}$ or $\sqrt{10}+\sqrt{21}$; c) $\sqrt{11}$ or $5-\sqrt[3]{5}$. | 6.1. Instead of the inequality sign, we will use the sign $\vee$ and consider only transformations that preserve the inequality sign.
a) $\sqrt{2}+\sqrt{3} \vee \sqrt{11} \Leftrightarrow 2+2 \sqrt{6}+3 \vee 11 \Leftrightarrow 2 \sqrt{6} \vee 6 \Leftrightarrow 6 \vee 9$. Therefore, $\sqrt{2}+\sqrt{3}5+\sqrt[3]{5}$. $<2... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 38,262 |
6.3. a) $\frac{1}{2+\sqrt{3}}$; b) $\frac{1}{\sqrt{2}+\sqrt{3}}$; c) $\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$. | 6.3. a) $\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=2-\sqrt{3}$.
b) $\frac{1}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}=\sqrt{3}-\sqrt{2}$.
c) We need to use the fact that
$$
\begin{gathered}
(\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}+\sqrt{3}-\sqrt{5})(\sqrt... | 2-\sqrt{3},\sqrt{3}-\sqrt{2},-\frac{1}{24} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,264 |
6.4. a) $\frac{1}{\sqrt[3]{x}-\sqrt[3]{y}} ;$ b) $\frac{1}{\sqrt[3]{x}+\sqrt[3]{y}}$. | 6.4. a) Multiply the denominator by $\sqrt[3]{x^{2}}+\sqrt[3]{x y}+\sqrt[3]{y^{2}}$.
b) Multiply the denominator by $\sqrt[3]{x^{2}}-\sqrt[3]{x y}+\sqrt[3]{y^{2}}$. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,265 |
6.5. a) $\frac{1}{\sqrt[n]{x}-\sqrt[n]{y}}$; b) $\frac{1}{\sqrt[2 n+1]{x}+\sqrt[2 n+1]{y}}$. | 6.5. a) Multiply the denominator by $\sqrt[n]{x^{n-1}}+\sqrt[n]{x^{n-2} y}+\ldots+\sqrt[n]{y^{n-1}}$.
b) Multiply the denominator by
$$
\sqrt[2 n+1]{x^{2 n}}-\sqrt[2 n+1]{x^{2 n-1} y}+\sqrt[2 n+1]{x^{2 n-2} y^{2}}-\ldots+\sqrt[n]{y^{2 n}}
$$ | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,266 |
6.6. $\frac{1}{\sqrt{x}-\sqrt[3]{y}}$.
6.6. $\frac{1}{\sqrt{x}-\sqrt[3]{y}}$. | 6.6. Multiply the denominator by $(\sqrt{x}+\sqrt[3]{y})\left(x^{2}+x \sqrt[3]{y^{2}}+\sqrt[3]{y^{4}}\right)$. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,267 |
6.9. Prove that if $a>\sqrt{b}$, then
$$
\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}} \pm \sqrt{\frac{a-\sqrt{a^{2}-b}}{2}}=\sqrt{a \pm \sqrt{b}}
$$ | 6.9. The left side is a positive number, so it is enough to check that after squaring both sides we get an identity. This is easily verified since $\left(a+\sqrt{a^{2}-b}\right) \times$ $\times\left(a-\sqrt{a^{2}-b}\right)=b$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,270 |
6.10. Represent the following numbers in the form $\sqrt{a} \pm \sqrt{b}$, where $a$ and $b$ are natural numbers:
a) $\sqrt{3+2 \sqrt{2}}$
b) $\sqrt{9+4 \sqrt{5}}$
c) $\sqrt{7-2 \sqrt{10}}$. | 6.10. Using the formula from problem 6.9, we get the following answers: a) $1+\sqrt{2} ;$ b) $2+\sqrt{5}$; c) $\sqrt{5}-\sqrt{2}$. | 1+\sqrt{2},2+\sqrt{5},\sqrt{5}-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,271 |
6.11. Prove that $\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{2}$. | 6.11. The number $\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}$ is positive, and its square is equal to $4+\sqrt{7}-2 \sqrt{16-7}+4-\sqrt{7}=8-2 \sqrt{9}=2$. | \sqrt{2} | Algebra | proof | Yes | Yes | olympiads | false | 38,272 |
6.12. Prove that the number $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ is rational. | 6.12. The identity $(2 \pm \sqrt{2})^{3}=20 \pm 14 \sqrt{2}$ shows that $\sqrt[3]{20 \pm 14 \sqrt{2}}=2 \pm \sqrt{2}$. Therefore, the considered number is 4. | 4 | Algebra | proof | Yes | Yes | olympiads | false | 38,273 |
6.13. Prove that the following numbers are rational:
a) $\sqrt[3]{3+\sqrt{\frac{242}{27}}}+\sqrt[3]{3-\sqrt{\frac{242}{27}}}$
b) $\sqrt[3]{6+\sqrt{\frac{847}{27}}}+\sqrt[3]{6-\sqrt{\frac{847}{27}}}$ | 6.13. Let $x=\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}$. Then $x^{3}=2 a+3 x \sqrt[3]{a^{2}-b}$. In case a) we get $x^{3}=6+x$, and in case b) we get $x^{3}=12+5 x$. The first equation has a root $x=2$, and the second equation has a root $x=3$. Therefore, it is sufficient to check that these equations have no other (re... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,274 |
6.14. Prove that $\sqrt{\sqrt[3]{4}-1}+\sqrt{\sqrt[3]{16}-\sqrt[3]{4}}=\sqrt{3}$. | 6.14. Let's square the expression on the left side, using the fact that $\sqrt{\sqrt[3]{16}-\sqrt[3]{4}}=\sqrt{\sqrt[3]{4}(\sqrt[3]{4}-1)}=\sqrt[3]{2} \sqrt{\sqrt[3]{4}-1}$. We get:
$$
\begin{aligned}
(\sqrt[3]{4}-1)(\sqrt[3]{2}+1)^{2}= & (\sqrt[3]{4}-1)(\sqrt[3]{4}+1+2 \sqrt[3]{2})= \\
& =\sqrt[3]{16}-1+2 \sqrt[3]{8}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,275 |
6.15. Prove the following Ramanujan identities:
a) $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$
b) $\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}=\frac{1}{3}(\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25})$;
c) $\sqrt[6]{7 \sqrt[3]{20}-19}=\sqrt[3]{\frac{5}{3}}-\sqrt[3]{\frac{2}{3}}$
d) $\sqrt... | 6.15. a) It is not difficult to verify that $(1-\sqrt[3]{2}+\sqrt[3]{4})^{3}=9(\sqrt[3]{2}-1)$.
b) It is not difficult to verify that $(\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25})^{2}=9(\sqrt[3]{5}-\sqrt[3]{4})$. Moreover, the inequality $\sqrt[3]{2}+\sqrt[3]{20}>\sqrt[3]{25}$ is easily verified by cubing both sides.
c) I... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,276 |
6.16. Prove that the following numbers are irrational:
a) $\sqrt{p}$, where $p$ is a prime number;
b) $\sqrt{p_{1} \ldots p_{k}}$, where $p_{1}, \ldots, p_{k}$ are distinct prime numbers;
c) $\sqrt{\frac{p_{1} \ldots p_{k}}{p_{k+1} \ldots p_{n}}}$, where $p_{1}, \ldots, p_{n}$ are distinct prime numbers. | 6.16. a) Suppose that $\sqrt{p}=r / s$ is an irreducible fraction. Then $r^{2}=p s^{2}$, so $r$ is divisible by $p$ (here we use problem 4.8). Therefore, $p s^{2}$ is divisible by $p^{2}$. Hence, $s$ is divisible by $p$, which contradicts the irreducibility of the fraction $r / s$.
b) From the equality $r^{2}=p_{1} \l... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,277 |
6.17. Prove that the number $\sqrt{2}+\sqrt[3]{3}$ is irrational. | 6.17. Suppose that $\sqrt{2}+\sqrt[3]{3}=r$, where $r$ is a rational number. Then $(\sqrt[3]{3})^{3}=(r-\sqrt{2})^{3}$, i.e., $3=r^{3}-3 r^{2} \sqrt{2}+6 r-2 \sqrt{2}$. Therefore, $\sqrt{2}=\frac{r^{3}+6 r-3}{3 r^{2}+2}$ is a rational number, which is impossible. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,278 |
6.18. Determine whether the number is rational
$$
\alpha=\sqrt[3]{4+\sqrt{15}}+\sqrt[3]{4-\sqrt{15}}
$$ | 6.18. Answer: irrational. Let $\alpha_{1}=\sqrt[3]{4+\sqrt{15}}$ and $\alpha_{2}=$ $=\sqrt[3]{4-\sqrt{15}}$. Then $\alpha_{1}^{3}+\alpha_{2}^{3}=8$ and $\alpha_{1}^{3} \alpha_{2}^{3}=1$, so $\alpha_{1} \alpha_{2}=1$. Therefore, $\left(\alpha_{1}+\alpha_{2}\right)^{3}=\alpha_{1}^{3}+\alpha_{2}^{3}+3 \alpha_{1} \alpha_{2... | irrational | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,279 |
6.19. a) Prove that if the numbers $a, b, \sqrt{a}+\sqrt{b}$ are rational, then the numbers $\sqrt{a}$ and $\sqrt{b}$ are also rational.
b) Prove that if the numbers $a, b, c, \sqrt{a}+\sqrt{b}+\sqrt{c}$ are rational, then the numbers $\sqrt{a}, \sqrt{b}$, and $\sqrt{c}$ are also rational.
c) Prove that if the number... | 6.19. a) Let $\sqrt{a}+\sqrt{b}=r$ be a rational number. If $r=0$, then $a=b=0$. Therefore, we will assume that $r \neq 0$. Squaring both sides of the equation $\sqrt{a}=r-\sqrt{b}$, we get $a=r^{2}-2 r \sqrt{b}+b$, and thus $\sqrt{b}=\frac{r^{2}+b-a}{2 r}$ is a rational number.
b) Let $\sqrt{a}+\sqrt{b}+\sqrt{c}=r$ b... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,280 |
6.20. Let $p_{1}, \ldots, p_{k}$ be distinct prime numbers. Prove that the number $\sqrt{p_{k}}$ cannot be represented as the sum of a rational number and numbers $\sqrt{p_{i_{1}} \ldots p_{i_{s}}}$, where $1 \leqslant i_{1}<\ldots<i_{s} \leqslant k-1$, with rational coefficients.
## 6.5. Conjugate Numbers | 6.20. We will use induction on $k$. First, consider the case $k=2$. Suppose that $\sqrt{p_{2}}=a+b \sqrt{p_{1}}$, where the numbers $a$ and $b$ are rational. Problem 6.16 v) shows that $a b \neq 0$. Squaring both sides, we get $p_{2}=a^{2}+2 a b \sqrt{p_{1}}+b^{2} p_{1}$, which implies that the number $\sqrt{p_{1}}$ is... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,281 |
6.21. a) Let $p$ be a prime number. Prove that no number can be represented in two different ways as $m+n \sqrt{p}$, where $m$ and $n$ are rational numbers.
b) Prove the analogous statement for representations of the form $m+n \sqrt{p}$, where $p$ is a product of distinct prime numbers.
Thus, for each number $z$ of t... | 6.21. Let $m+n \sqrt{p}=m_{1}+n_{1} \sqrt{p}$, where the numbers $m, n, m_{1}$, and $n_{1}$ are rational. If $m_{1} \neq m$ or $n_{1} \neq n$, then $\sqrt{p}=\frac{m-m_{1}}{n_{1}-n}$. This equality cannot hold, since $\sqrt{p}$ is an irrational number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,282 |
6.22. Prove that if $(a+b \sqrt{p})^{n}=A_{n}+B_{n} \sqrt{p}$, where $p-$ is the product of distinct prime numbers, and the numbers $a, b, A_{n}, B_{n}$ are rational, then $(a-b \sqrt{p})^{n}=A_{n}-B_{n} \sqrt{p}$. | 6.22. We apply induction on $n$. It is easy to verify that $\bar{z}_{1} \bar{z}_{2} = \overline{z_{1} z_{2}}$, i.e., if $(a+b \sqrt{p})(c+d \sqrt{p})=A+B \sqrt{p}$, then $(a-b \sqrt{p}) \times$ $\times(c-d \sqrt{p})=A-B \sqrt{p}$. Therefore, if $(a+b \sqrt{p})\left(A_{n}+B_{n} \sqrt{p}\right)=$ $=A_{n+1}+B_{n+1} \sqrt{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,283 |
6.23. Find the first digit after the decimal point of the number $(2+\sqrt{3})^{1000}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 6.23. Answer: 9. Let $(2+\sqrt{3})^{n}=A_{n}+B_{n} \sqrt{3}$, where $A_{n}$ and $B_{n}$ are natural numbers. According to problem 6.22, $(2-\sqrt{3})^{n}=A_{n}-B_{n} \sqrt{3}$. Therefore, $(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}$ is a natural number. But $2-\sqrt{3} \approx$ $\approx 0.2679 < 0.3$, so $(2-\sqrt{3})^{1000} \a... | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 38,284 |
6.24. Prove that for rational numbers $x, y, z$ and $t$ the equality
$$
(x+y \sqrt{2})^{2}+(z+t \sqrt{2})^{2}=5+4 \sqrt{2}
$$
cannot hold. | 6.24. If for rational numbers $x, y, z$ and $t$ the given equality holds, then according to problem 6.22 for the same numbers the equality
$$
(x-y \sqrt{2})^{2}+(z-t \sqrt{2})^{2}=5-4 \sqrt{2} .
$$
must also hold. But $5-4 \sqrt{2}<0$, while $(x-y \sqrt{2})^{2}+(z-t \sqrt{2})^{2} \geqslant 0$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,285 |
6.25. Prove that for natural numbers $m$ and $n$, the equality $(5+3 \sqrt{2})^{m}=(3+5 \sqrt{2})^{n}$ cannot hold. | 6.25. Indeed, let $(5+3 \sqrt{2})^{m}=(3+5 \sqrt{2})^{n}$. Then according to problem $6.22$, $(5-3 \sqrt{2})^{m}=(3-5 \sqrt{2})^{n}$. A contradiction can now be reached in different ways. First, one can notice that $|5-3 \sqrt{2}|1$. Second, one can multiply the equations $(5+3 \sqrt{2})^{m}=(3+5 \sqrt{2})^{n}$ and $(5... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,286 |
6.26. a) Prove that $(\sqrt{2}-1)^{n}=\sqrt{k}-\sqrt{k-1}$, where $k$ is some natural number.
b) Let $m$ and $n$ be natural numbers. Prove that $(\sqrt{m}-\sqrt{m-1})^{n}=\sqrt{k}-\sqrt{k-1}$, where $k$ is some natural number. | 6.26. a) If $(\sqrt{2}+1)^{n}=x \sqrt{2}+y$, then according to problem 6.22, $(\sqrt{2}-1)^{n}=(-1)^{n}(1-\sqrt{2})^{n}=(-1)^{n}(y+x \sqrt{2})=(-1)^{n}\left(\sqrt{y^{2}}-\sqrt{2 x^{2}}\right)$.
## 84
In this case, $y^{2}-2 x^{2}=(y+x \sqrt{2})(y-x \sqrt{2})=(1+\sqrt{2})^{n}(1-\sqrt{2})^{n}=(-1)^{n}$. Thus, if $n$ is ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,287 |
6.27. Prove that for any natural $n$ the number $\left[(1+\sqrt{3})^{2 n+1}\right]$ is divisible by $2^{n+1}$ and is not divisible by $2^{n+2}$.
## 6.6. Farey Sequence | 6.27. The number $(1+\sqrt{3})^{2 n+1}+(1-\sqrt{3})^{2 n+1}$ is an integer, and $-1 < (1-\sqrt{3})^{2 n+1} < 0$. Therefore, $\left[(1+\sqrt{3})^{2 n+1}\right]=(1+\sqrt{3})^{2 n+1}+(1-\sqrt{3})^{2 n+1}$. The equality $(1 \pm \sqrt{3})^{2}=2(2 \pm \sqrt{3})$ shows that
\[
\begin{aligned}
& (1+\sqrt{3})^{2 n+1}+(1-\sqrt{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,288 |
6.28. Arrange in ascending order all rational numbers that lie between zero and one and whose denominators do not exceed $n$. Let $a / b$ and $c / d$ be two consecutive numbers in this sequence. Prove that $|b c-a d|=1$.
The sequence of numbers from problem 6.28 is called the Farey sequence and is denoted by $F_{n}$. | 6.28. We can assume that $a / b < c / d$. The inequality $\frac{a}{b} < \frac{a}{b-1} \leqslant \frac{a+1}{b}$, which holds when $a+1 \leqslant b$, shows that $b \neq d$, i.e., the denominators of two adjacent fractions cannot be the same.
We will prove the required statement by induction on $n$. For $n=3$, we get the... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,289 |
6.29. Let $a / b < x / y < c / d$ be three consecutive fractions in the Farey sequence. Prove that $\frac{x}{y}=\frac{a+c}{b+d}$. | 6.29. According to problem $6.28 b x-a y=1$ and $c y-d x=1$. Solving this system of linear equations with respect to $x$ and $y$, we get $x=\frac{a+c}{b c-a d}$ and $y=\frac{b+d}{b c-a d}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,290 |
6.31. In the Farey sequence $F_{n}$, compute the fractions adjacent to $1 / 2$.
78 Chapter 6. Rational and Irrational Numbers | 6.31. Answer: $\frac{1}{2} \pm \frac{1}{2 n}$ for odd $n$; $\frac{1}{2} \pm \frac{1}{2 n-2}$ for even $n$.
For odd $n$, the numerator and denominator of the fraction $\frac{n \pm 1}{2 n}$ can be reduced by 2; for even $n$, the same is true for the fraction $\frac{n-1 \pm 1}{2 n-2}$. Therefore, the fractions specified ... | \frac{1}{2}\\frac{1}{2n}foroddn;\frac{1}{2}\\frac{1}{2n-2}forevenn | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,292 |
6.32. a) Prove that the sum of the denominators of the fractions in the Farey sequence is twice as small as the sum of the numerators.
b) Prove that the sum of the fractions in the Farey sequence is twice as small as the number of its terms. | 6.32. a) If the fraction $p / q$ is in the Farey sequence, then the fraction $(q-p) / q$ is also in it, since $\gcd(q-p, q) = \gcd(p, q)$. Therefore, $\sum p = \sum (q-p)$, i.e., $2 \sum p = \sum q$, as required.
b) Similarly to a), we get $\sum \frac{p}{q} = \sum \frac{q-p}{q}$, i.e., $2 \sum \frac{p}{q} = \sum 1$. T... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,293 |
6.33. Prove that the number of members in the Farey sequence $F_{n}$ is equal to $\sum_{q=2}^{n} \varphi(q)$, where $\varphi$ is the Euler's function.
## 6.7. Problems with Integer Parts | 6.33. The fraction $p / q$ is part of the Farey sequence if $1 \leqslant p < q$ and $\operatorname{gcd}(p, q)=1$. For a given $q$, the number of terms of the form $p / q$ is $\varphi(q)$. Therefore, the total number of terms is $\sum_{q=2}^{n} \varphi(q)$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,294 |
6.34. Let $\alpha$ and $\beta$ be positive numbers. Prove that the following conditions are equivalent:
(1) each natural number appears exactly once among the numbers $[\alpha],[\beta],[2 \alpha],[2 \beta],[3 \alpha],[3 \beta], \ldots ;$
(2) $\frac{1}{\alpha}+\frac{1}{\beta}=1$, and $\alpha$ and $\beta$ are irrationa... | 6.34. First, let's prove that (1) implies (2). The number of values of \( x \) for which \([x \alpha] \leqslant n\) is \( n / \alpha + \lambda \), where \( 0 \leqslant \lambda \) are natural numbers. But then \([q \alpha] = p = [(p - q) \beta]\). This leads to a contradiction, so \(\alpha\) is irrational.
Now, let's p... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,295 |
6.35. Let $\alpha_{1}, \ldots, \alpha_{k}$ be positive numbers with the following property: each natural number appears exactly once among the numbers $\left[\alpha_{1}\right], \ldots,\left[\alpha_{k}\right],\left[2 \alpha_{1}\right], \ldots$ $\ldots,\left[2 \alpha_{k}\right],\left[3 \alpha_{1}\right], \ldots,\left[3 \... | 6.35. We can consider that $\alpha_{1}\left[p \alpha_{1}\right]+1$. The last inequality is equivalent to $[p+1+(p+1) \delta]>[p+p \delta]+1$, i.e., $[(p+1) \delta]>[p \delta]$. Thus, $p \delta<q \leqslant(p+1) \delta$ for some natural number $q$.
It is clear that $(p+m-1) \delta=p \delta+(m-1) \delta<q+1$ and $(p+m+1)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,296 |
7.1. In one glass there are 5 spoons of tea, in the other 5 spoons of milk. One spoon of milk was poured from the second glass into the first, and then one spoon of tea with milk was poured back into the second glass. What turned out to be more: tea in the milk or milk in the tea $?$ | 7.1. From the second glass, the amount of milk that decreased is exactly the same as the amount of tea that was added to it. Therefore, the amount of tea in the milk is exactly the same as the amount of milk in the tea. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,297 |
7.2. A pedestrian left point $A$ for point $B$, which are 5 km apart. At the same time, a cyclist left point $B$ towards the pedestrian, with a speed twice that of the pedestrian. Upon meeting the pedestrian, he turned around and rode back to $B$. Upon reaching $B$, the cyclist turned again and rode towards the pedestr... | 7.2. A n s w e r: 10 km. We can assume that the cyclist is always moving in one direction (the length of the path does not change from this).
His speed is twice the speed of the pedestrian, so in the time it takes the pedestrian to walk 5 km, the cyclist will travel 10 km. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,298 |
7.3. A flock of geese was flying over the lakes. On each lake, half of the geese and another half-goose landed, while the rest continued flying. All the geese landed on seven lakes. How many geese were in the flock? | 7.3. Answer: 127. Let's mentally add a bird, for example, a duck, that flies with the flock all the way to the last lake. Then, at each lake, exactly half of the birds land, and on the seventh lake, two birds land - the goose and the duck. This means there were $2^{7}$ birds in the flock, of which $2^{7}-1=127$ were ge... | 127 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,299 |
7.4. Two cyclists set off simultaneously towards each other from points $A$ and $B$ and met 70 km from $A$. Continuing to move at the same speeds, they reached $A$ and $B$ and turned back. They met for the second time 90 km from $B$. Find the distance from $A$ to $B$. | 7.4. Answer: 120 km. Let $x$ be the distance from $A$ to $B$. By the first meeting, both cyclists together had traveled $x$ km, and by the second meeting, $-3x$ km. Therefore, the cyclist who started from $A$ had traveled $3 \cdot 70 = 210$ km by the second meeting. On the other hand, he had traveled $x + 90$ km. Thus,... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,300 |
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