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108. The number $16^{n}-15 n-1$ is divisible by 225.
Hint. Use the method of mathematical induction. | 108. $A_{n}=16^{n}-15 n-1$, for $n=1$ the statement is true. Assume that it is true for some other value of $n$. We will prove that the statement is true for $n+1$. $A_{n+1}=16^{n+1}-15(n+1)-1=16 \cdot 16^{n}-15 n-16=16^{n}-$ $-15 n-1+15\left(16^{n}-1\right)=A_{n}+15\left(16^{n}-1\right)$ is divisible by 225. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,698 |
112. If $n$ is even, then $3^{n}+63$ is divisible by 72. Is the converse statement true? | 112. $3^{2 k}+63=9^{k}+63$ is divisible by 9; moreover, $9^{k}+$ $+63=9^{k}-1+64$ is divisible by 8, since $9^{k}-1$ is divisible
by $9-1$. Thus, $3^{n}+63$ for even $n$ is divisible by 72.
Conversely. If $3^{n}+63$ is divisible by 72, then it is divisible by 9 and 8; for $n>1$ the number $3^{n}+63$ is obviously divis... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,699 |
114. The number $5^{2 n+1} \cdot 2^{n+2}+3^{n+2} \cdot 2^{2 n+1}$ is divisible by 19 for $n \geqslant 0$. | 114. $20 \cdot 50^{n}+18 \cdot 12^{n}=19 \cdot 50^{n}+19 \cdot 12^{n}+50^{n}-12^{n}$ is divisible by 19, since $50^{n}-12^{n}$ is divisible by $50-12=2 \cdot 19$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,701 |
116. The number $\frac{10^{n}-1}{81}-\frac{n}{9}$ is an integer for integer $n>0$. | 116. $A=\frac{10^{n}-1-9 n}{81}=\frac{(9+1)^{n}-1-9 n}{81}=\frac{81 k}{81}=k$ (by the binomial theorem).
Alternatively: $\quad A=\frac{1}{9}\left(\frac{10^{n}-1}{10-1}-n\right)=\frac{1}{9}\left[\left(10^{n-1}-\right.\right.$ $\left.-1)+\left(10^{n-2}-1\right)+\ldots+(10-1)\right]$, all differences $10^{n-1}-$ $-1,10^{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,702 |
117*. The square of a prime number, different from $2,3,5$, when divided by 30, gives a remainder equal to 1 or 19. | 117. \( p^{2} = 30m + r \), where \( r < 30 \). Considering that \( p \) is not divisible by \( 2, 3, 5 \), and that the square of an odd number ends only in the digits \( 1, 5, 9 \), we conclude that \( r \) can only be \( 1, 11, 19, 29 \). However, \( r \neq 11 \), because \( p^{2} - 11 = p^{2} - 1 - 10 \) is not div... | 1;19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,703 |
118. Find the remainders of $a^{4}$ divided by 5 for all possible integer values of $a$. | 118. If $a$ is divisible by 5, the remainder is zero.
If $a$ is not divisible by 5, then $a^{4}-1$ is divisible by 5, therefore, the remainder of $a^{4}$ divided by 5 is 1. | 0or1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,704 |
120. If the numbers $a^{5}$ and $b^{5}$ ( $a>0$ and $b>0$ ) have the same units digit, then $a-b$ is divisible by 10. Find the numbers $a$ and $b$, if $a^{2}-b^{2}=1940$. Under the same conditions, find $a$ and $b$, if $a^{2}-b^{2}=1920$. | 120. Since $a^{5}$ and $b^{5}$ end with the same digit as $a$ and $b$, $a-b$ ends with zero, hence $a-b$ is divisible by 10. From the system of equations $\left\{\begin{array}{l}a^{2}-b^{2}=1940 \\ a-b=10 k\end{array}\right.$ we find $a=\frac{97}{k}+5 k, b=\frac{97}{k}-5 k$. Since $b>0$, then $k<\sqrt{19.4}, k=1 ; 2 ; ... | a_{1}=101,b_{1}=91,a_{2}=58,b_{2}=38,a_{3}=47,b_{3}=17,a_{4}=44,b_{4}=4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,705 |
121*. The number $3^{p}-2^{p}-1$ is divisible by $42 p$, if $p-$ is a prime number greater than 3. | 121. Since $3^{p}-1$ is an even number, then $A=3^{p}-2^{p}$ -1 is divisible by 2. Since $2^{p}+1$ is divisible by $2+1$, then $A$ is divisible by 3.
Moreover, by Fermat's theorem, we conclude that $3^{p}-3$ and $2^{p}-2$ are divisible by $p$, hence $3^{p}-2^{p}-1=\left(3^{p}-3\right)-$ - (2 $\left.2^{p}-2\right)$ is ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,706 |
123. The number $n^{2} \pm 3 n+5$ is never divisible by 121 for any integer $n$. | 123. Let $n^{2} \pm 3 n+5$ be divisible by 121, i.e., $n^{2} \pm 3 n + 5 = 121 t$, from which $n=\frac{\mp 3 \pm \sqrt{11(44 t-1)}}{3}$; therefore, $44 t-1$ must be divisible by 11, which is impossible. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,707 |
124. Solve the equation: $x^{y}+1=z$ in prime numbers. | 124. One of the numbers $x^{y}$ and $z$ is even, and the other is odd. If $z$ is even, then being prime, it equals 2; but then $x$ and $y$ can only equal one or $x$ can be any number and $y$ equals zero; therefore, $x$ and $y$ will not be prime numbers. It remains to assume that $x^{y}$ is even, then $x=2$ and $2^{y}+1... | 2;2;5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,708 |
125. If $2^{n}=10 a+b$, then $a b$ is a multiple of 6 ( $n>3 ; 0<b<10$ ). | 125. From the equality $2^{n}=10 a+b$ it is clear that $b$ is an even number, and since $0<b<10$, then $b=2 ; 4 ; 6 ; 8$; therefore, $a b$ is divisible by 2. If $b=6$, then $a b$ is divisible by 6. If $b=2$, then $a=\frac{2^{n}-2}{10}=\frac{2^{n-1}-1}{5}$ must be an integer and $n-1=$ $=4 t$ (see No. 94); therefore, $2... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,709 |
126. If $a+b$ is divisible by an odd number $n$, then $a^{n}+b^{n}$ is divisible by $n^{2}$. | 126. Let $a+b=n k$, where $k$ is an integer, then $a^{n}+b^{n}=a^{n}+(n k-a)^{n}=n^{n} k^{n}-c_{n}^{1} n^{n-1} k^{n-1} a+\ldots+$ $+n^{2} a^{n-1} k$, since $n$ is odd. From this, it is clear that $a^{n}+$ $+b^{n}$ is divisible by $n^{2}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,710 |
127*. The numbers $2^{p}-1$ and $2^{q}-1$ are coprime if and only if $p$ and $q$ are coprime.
## Chapter III.
## COMMON DIVISOR
A divisor of the number $a$ is a divisor of the number $ap$ (but not vice versa). A common divisor of the numbers $a$ and $b$ is a common divisor of the numbers $a+b$ and $a-b$ (but not vic... | 127. 128) Let \(2^{p}-1\) and \(2^{q}-1\) be coprime numbers. We will prove that \(p\) and \(q\) are coprime. Assume the opposite, i.e., \(p=m r_{1}, q=m r_{2}\), then \(2^{p}-1=2^{m r_{1}}-1=\left(2^{m}\right)^{r_{1}}-1\) is divisible by \(2^{m}-1\), similarly \(2^{q}-1\) is divisible by \(2^{m}-1\); therefore, the nu... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,711 |
128. Prove that two consecutive numbers are coprime.
untranslated text remains the same as requested. However, the task was to translate the text, which I have done. Let me know if you need anything else! | 128. gcd$(a, a+1)=$gcd$(a, a+1-a)=$gcd$(a, 1)=1,$(a, a+$ $+1)=1$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,712 |
129. The numbers $a, a+1, 2a+1$ are pairwise coprime. | 129. $\mathrm{D}(a, 2a+1)=\mathrm{D}(a, a+1)=1$ (see No. 128), $(a, 2a+$ $+1)=1$; $\mathrm{D}(a+1,2a+1)=\mathrm{D}(a+1, a)=1,(a+1,2a+$ $+1)=1$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,713 |
130. If $(a, b):=1$, then $(a \pm b, a b)=1$ and $(a+b, a-b)$ is equal to 1 or 2. | 130. A common divisor of the numbers $a \pm b$ and $a b$ must be a divisor of each of the numbers $a$ and $b$, but this contradicts the condition $(a, b)=1$. Therefore, $(a \pm b, a b)=1$. $(a+b, a-b)=D(2 a, 2 b)$ is equal to 2 or 1, since by the condition $(a, b)=1$. | 1or2 | Number Theory | proof | Yes | Yes | olympiads | false | 40,714 |
132. $\left(\frac{a(a+1)}{2}, 2 a+1\right)=1$.
Translating the text above into English while preserving the original text's line breaks and format, the result is as follows:
132. $\left(\frac{a(a+1)}{2}, 2 a+1\right)=1$. | 132. $D (11 a+2 b, \quad 18 a+5 b)=$ $D [5(11 a+2 b), 2(18 a+$ $+5 b)]=D[5(11 a+2 b), \quad 5(11 a+2 b)-2(18 a+5 b)]=$ $=D[5(11 a+2 b), 19 a]$. The common divisor can be 1, 19, or a divisor of the number $a$; but the latter is excluded, since in that case this divisor would be part of $b$, which contradicts the conditi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,716 |
139. Find two coprime numbers \(a\) and \(b\) that satisfy the equation:
\[
\frac{5a + 7b}{6a + 5b} = \frac{29}{28}
\] | 139. After simplifications, we get $\frac{a}{b}=\frac{3}{2}$, but since by condition $(a, b)=1$, then $a=3, b=2$. | =3,b=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,720 |
140. Prove that the fraction $\frac{a^{3}+2 a}{a^{4}+3 a^{2}+1}$ is irreducible. | 140. First method. D $\left(a^{3}+2 a, a^{4}+3 a^{2}+1\right)=D\left[a\left(a^{3}+2 a\right), a^{4}+3 a^{2}+1\right]=$ D $\left(a^{3}+2 a, a^{2}+1\right)=$ D $\left[a^{3}+2 a, a\left(a^{2}+1\right)\right]=$ $=D\left(a^{2}, a^{2}+1\right)=1$, therefore, the given fraction is irreducible.
Second method. The given fracti... | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,721 |
144. Find two three-digit numbers, the sum of which is divisible by 252, and the ratio of which is $2: 1$. | 144. Answer. $168 m$ and $84 m$, where $1<m<6$. | 168m84mwhere1<<6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,722 |
145. Find two numbers if their sum is 432, and their greatest common divisor is 36. | 145. Given $x=36 x_{1}$ and $y=36 y_{1}$, and $\left(x_{1}, y_{1}\right)=$ $=1$. We find $x_{1}+y_{1}=12$, hence $x_{1}=1, y_{1}=11 ; x_{2}=$ $=5, y_{2}=7$.
Answer. $x_{1}=36, y_{1}=396 ; x_{2}=180, y_{2}=252$. | x_{1}=36,y_{1}=396;x_{2}=180,y_{2}=252 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,723 |
146. Find two numbers if their product is 864, and their greatest common divisor is 6. | 146. Answer. $x_{1}=6, y_{1}=144 ; x_{2}=18, y_{2}=48$. | x_{1}=6,y_{1}=144;x_{2}=18,y_{2}=48 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,724 |
147. If the least common multiple of numbers $a$ and $b$ is denoted by $[a, b]$, then there is a formula:
$$
[a, b]=\frac{a b}{(a, b)}
$$
Proof. A common multiple of numbers $a$ and $b$ is divisible by $a$, hence it has the form $a k$; therefore, $\frac{a k}{b}$ must be an integer. Denoting $(a, b)=d$, we have $a=d a... | 147. From the system $\left\{\begin{array}{l}x^{2}+y^{2}=468 \\ d+\frac{x y}{d}=42\end{array}\right.$, by accepting $x=d x_{1}$ and $y=d y_{1}$, we get $\left\{\begin{array}{l}x_{1}{ }^{2}+y_{1}{ }^{2}=\frac{468}{d^{2}} \\ 1+x_{1} y_{1}=\frac{42}{d}\end{array}\right.$, and $(x_{1}, y_{1})=1$.
We find that $d=1 ; 2 ; 3... | 1218 | Number Theory | proof | Yes | Yes | olympiads | false | 40,725 |
149. Find two numbers if their sum is 60, and the sum of their greatest common divisor and least common multiple is 84.
Chapter IV.
## THE NUMBER OF ALL DIVISORS OF A GIVEN NUMBER AND THEIR SUM
The number of all divisors of a given number $N=P_{1}^{\lambda_{1}} P_{2}^{\lambda_{2}} \ldots \ldots P_{n}^{\lambda}$, inc... | 149. Answer. 24 and 36. | 2436 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,726 |
150. $N=p^{\lambda_{1}} q^{\lambda_{2}}$, where $p$ and $q$ are different prime numbers; $N^{2}$ has 15 divisors. How many divisors does the number $N^{3}$ have? | 150. Since $N^{2}=p^{2 \lambda_{1}} q^{2 \lambda_{2}}$, then $\left(2 \lambda_{1}+1\right)\left(2 \lambda_{2}+1\right)=15$, from which $\left\{\begin{array}{l}2 \lambda_{1}+1=3 \\ 2 \lambda_{2}+1=5\end{array}, \lambda_{1}=1, \lambda_{2}=2\right.$.
Given that $N^{3}=p^{3 \lambda_{1}} q^{3 \lambda_{2}}$, the number of a... | 28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,727 |
152. One number has 21 divisors, and the other has 10 divisors. Find these numbers if their greatest common divisor is 18 and they have no prime divisors other than 2 and 3. | 152. The sought numbers have the form $A=2^{x} \cdot 3^{y}$ and $B=$ $=2^{z} \cdot 3^{u}$.
According to the condition, we get $\left\{\begin{array}{l}(x+1)(y+1)=21 \\ (z+1)(u+1)=10\end{array}\right.$ and $(A, B)=2 \cdot 3^{2}$, from which we find: $x=6, y=2, z=1, u=4$.
Answer. 576 and 162. | 576162 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,729 |
153*. The number $N$ has the form $p^{\alpha} q^{\beta} r^{\gamma}$, where $p, q, r$ are prime numbers, and $p q - r = 3, p r - q = 9$; furthermore, the numbers $\frac{N}{p}, \frac{N}{q}, \frac{N}{r}$ have respectively 20, 12, and 15 fewer divisors than the number $N$. Find the number $N$.
14 | 153. From the system $\left\{\begin{array}{l}p q-r=3 \\ p r-q=9\end{array}\right.$, we find $(p+1)(r-q)=6$, hence $\left\{\begin{array}{l}p+1=3 \\ r-q=2\end{array}, p=2 ;\left\{\begin{array}{l}2 q-r=3 \\ r-q=2\end{array}, q=5, r=7\right.\right.$.
The combination $p+1=6$ and $r-q=1$ does not yield the required results.... | 857500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,730 |
154. a) Find the number if the product of all its divisors is 108.
b) Find the number if the product of its divisors is 5832. | 154. b) $5832=2^{3} \cdot 3^{6}$; therefore, the required number $A=2^{x} 3^{y}$, where $x<3 ; y<6$. Based on № 151, we find:
$$
\left(2^{x} \cdot 3^{y}\right)^{\frac{1}{2}(x+1)(y+1)}=2^{3} \cdot 3^{6}
$$
From the system $\left\{\begin{array}{l}\frac{x(x+1)(y+1)}{2}=3 \\ \frac{y(x+1)(y+1)}{2}=6\end{array}\right.$, by... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,731 |
155. Find the number if the product of its divisors is $3^{30} \cdot 5^{40}$. | 155. Answer. $N=3^{3} \cdot 5^{4}$. | 3^{3}\cdot5^{4} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,732 |
156. Find the sum of the exponents in all divisors of the number $N=p^{\alpha} q^{\beta}$. | 156. The divisors of the number $N$ will be the members of the product $\left(1+p+p^{2}+\ldots+p^{\alpha}\right)\left(1+q+q^{2}+\ldots+q^{\beta}\right)$.
Therefore, the sum of the exponents in all divisors is $[1+2+3+\ldots+(\alpha-1)+\alpha]+[1+2+3+\ldots+$ $+\alpha+(\alpha+1)]+[2+3+\ldots+(\alpha+1)+(\alpha+2)]+\ldo... | \frac{1}{2}(\alpha+1)(\beta+1)(\alpha+\beta) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,733 |
157. Generalize the results of the previous problem to the number $N=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{n}^{\alpha_{n}}$. | 157. We need to show that the sum $s$ of the exponents in all divisors of the number $N=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{n}^{\alpha_{n}}$ is equal to
$$
\frac{1}{2}\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \ldots\left(\alpha_{n}+1\right)\left(\alpha_{1}+\alpha_{2}+\ldots+\alpha_{n}\right)
$$
F... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,734 |
159. Find the relationship between the prime numbers $p_{1}, q_{1}$, $p_{2}, q_{2}$, if it is known that the numbers $p_{1} q_{1}$ and $p_{2} q_{2}$ have the same sum of divisors. | 159. Answer. $\left(p_{1}+1\right)\left(q_{1}+1\right)=\left(p_{2}+1\right)\left(q_{2}+1\right)$. | (p_{1}+1)(q_{1}+1)=(p_{2}+1)(q_{2}+1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,736 |
160. Find a number of the form $2^{l} 3^{m}$, if the sum of all its divisors is 403. | 160. $\left(2^{l+1}-1\right)\left(3^{m+1}-1\right)=2 \cdot 13 \cdot 31$. This is only possible when $\left\{\begin{array}{l}2^{l+1}-1=31 \\ 3^{m+1}-1=26\end{array}\right.$
Answer. 144. | 144 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,737 |
162. Find a perfect number of the form $p q$, where $p$ and $q$ are prime numbers. | 162. $\frac{\left(p^{2}-1\right)\left(q^{2}-1\right)}{(p-1)(q-1)}-p q=p q, \quad q=1+\frac{2}{p-1}, \quad$ from here $p_{1}=2, p_{2}=3, q_{1}=3, q_{2}=2$.
Answer. 6 .
42 | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,739 |
163. Prove that there does not exist a perfect number of the form $p^{n}$. | 163. From the equation $\frac{p^{n+1}-1}{p-1}-p^{n}=p^{n}$ we find $p^{n+1}-$ $-2 p^{n}+1=0(1)$
Since the integer roots of an equation with integer coefficients must be divisors of the constant term, equation (1) has a unique integer root 1. From this, we conclude that there does not exist a perfect number of the form... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,740 |
164. Find a perfect number of the form $p^{2} q$. | 164. From the equation $\left(p^{2}+p+1\right)(q+1)=2 p^{2} q$, we find $q=1+\frac{2(p+1)}{p^{2}-p-1}$, from which we obtain:
1) $p^{2}-p-1=1, p=2, q=7$
2) $p^{2}-p-1=2$, no solutions.
Moreover, the numbers $p+1$ and $p^{2}-p-1$ have no common divisors, since $\gcd \left(p+1, p^{2}-p-1\right)=\gcd[p(p+1), p^{2}-p-1]=\... | 28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,741 |
165. Prove that there does not exist a perfect number of the form $p^{3} q$. | 165. From the equation $\frac{p^{4}-1}{p-1}(q+1)=2 p^{3} q$ we find $q=$ $=1+\frac{2\left(p^{2}+p+1\right)}{p^{3}-p^{2}-p-1}$.
The fraction $\frac{p^{2}+p+1}{p^{3}-p^{2}-p-1}$ is irreducible, since the reciprocal fraction $\frac{p^{3}-p^{2}-p-1}{p^{2}+p+1}$ is irreducible; indeed, $\frac{p^{3}-p^{2}-p-1}{p^{2}+p+1}=p-... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,742 |
166. Find a perfect number of the form $16 p$.
Note: The translation maintains the original text's line breaks and format. | 166. From the equation $31(p+1)=32 p$ we find $p=31$. | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,743 |
167. Find a perfect number of the form $2^{n} p$.
Note. So far, no odd perfect number has been found, and it has not been proven that it does not exist.
## Chapter $V$.
## THE NUMBER OF NUMBERS LESS THAN $n$ AND COPRIME WITH IT
Let the number of numbers less than $n$ and coprime with it be denoted by the symbol $\v... | 167. Answer. $2^{n}\left(2^{n+1}-1\right)$, if $2^{n+1}-1$ is a prime number. | 2^{n}(2^{n+1}-1) | Number Theory | proof | Yes | Yes | olympiads | false | 40,744 |
168. $\varphi(p)=p-1$, where $p-$ is a prime number. | 168. Each number less than a prime number $p$ is coprime with it; therefore, $\varphi(p)=p-1$. | \varphi(p)=p-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,745 |
169. $\varphi\left(p^{\lambda}\right)=p^{\lambda}-p^{\lambda-1}$. | 169. Let's write down the sequence of numbers from 1 to $p^{\lambda}$
$1,2,3, \ldots, p, p+1, p+2,2 p, \ldots, 3 p, \ldots, p^{\lambda-1}, \ldots, p(1)$.
Numbers that are not coprime with $p^{\lambda}$ must be divisible by $p$; from the sequence (1), we can see that there will be a total of $p^{\lambda-1}$ such numbe... | p^{\lambda}-p^{\lambda-1} | Number Theory | proof | Yes | Yes | olympiads | false | 40,746 |
171. Find $\varphi(360)$. | 171. $\varphi(360)=\varphi\left(2^{3} \cdot 3^{2} \cdot 5\right)=360\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)=$ $=96$. | 96 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,748 |
179. Find $N=p^{3}$, if $\varphi(N)=100$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 179. From the equation $p^{3}\left(1-\frac{1}{p}\right)=100$ we get $p^{2}(p-1)=$ $=25 \cdot 4$, from which it is clear that $p=5$.
We can also do it this way: $p^{3}-p^{2}-100=0, p^{3}-125-\left(p^{2}-25\right)=$ $=0,(p-5)\left(p^{2}+4 p+20\right)=0$, from which $p=5$.
Answer. 125. | 125 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 40,754 |
180. Find $N=p q$, if $\varphi(N)=120$, and the numbers $p$ and $q$ are twin prime numbers, i.e., $p-q=2$. | 180. From the system of equations $\left\{\begin{array}{l}p q\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right)=120 \\ p-q=2\end{array}\right.$ we find $p=13, q=11$.
Answer. 143. | 143 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,755 |
182. Show that if $(a, b)=1$, then $\varphi(a) \varphi(b)=$ $=\varphi(a b)$.
## Chapter VI.
## INDEFINITE EQUATIONS | 182. Let $a=p_{1}^{\lambda_{1}} p_{2}^{\lambda_{2}} \ldots p_{n}^{\lambda_{n}}$ and $b=q_{1}^{\mu_{1}} q_{2}^{\mu_{2}} \ldots q_{m}^{\mu^{\mu} m}$,
$$
\begin{gathered}
\text { then } \varphi(a)=p_{1}^{\lambda_{1}} \ldots p_{n}^{\lambda_{n}}\left(1-\frac{1}{p_{1}}\right) \ldots\left(1-\frac{1}{p_{n}}\right), \varphi(b)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,757 |
183. Solve the equation in integers:
$$
7 x+5 y=62
$$ | 183. $7 x+5 y=62, y=\frac{2(31-x)}{5}-x ; \quad$ therefore, $\frac{31-x}{5}$ is an integer; let $\frac{31-x}{5}=t$, then $x=31-$ $-5 t, y=2 t-(31-5 t)=7 t-31$.
Thus, all integer solutions of the equation are represented by the formulas $x=31-5 t$ and $y=7 t-31$, where $t$ is any integer common to both formulas. | 31-5t,7t-31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,758 |
184. Solve the equation in integers:
$$
5 x-17 y=3
$$ | 184. $5 x-17 y=3, x=4 y+\frac{3(1-y)}{5}$;
$$
y=1-5 t, x=4-17 t
$$ | 4-17t,\,1-5t | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,759 |
185. Find the numbers that give a remainder of 4 when divided by 19, and a remainder of 1 when divided by 11. | 185. Denoting the quotient respectively by $x$ and $y$, we get the equation $19 x+4=11 y+1$, from which $x=11 t+1$, $y=19 t+2$; therefore, the desired number is $209 t + 23$, where $t \geqslant 0$.
Answer. $23, 232, 441, \ldots$. | 209+23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,760 |
186. Find the integer negative solutions of the equation $5 x-3 y=2$. | 186. $\left\{\begin{array}{l}x=1+3 t \\ y=1+5 t\end{array}\right.$, where $t<0$. | 1+3t,1+5twhere<0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,761 |
187. For which integer values of $x$ will the number $\frac{5 x+2}{17}$ be an integer? | 187. Answer. $3,20,37, \ldots,-14,-31,-48, \ldots$. | 3+17k | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,762 |
189. $2 x+3 y+5 z=15$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
189. $2 x+3 y+5 z=15$. | 189. $x=7-y-2 z-\frac{y+z-1}{2}, \frac{y+z-1}{2}=t \quad$ (integer); $y=2 t-z+1 ; x=6-z-3 t$; by assigning arbitrary integer values to $z$ and $t$, we can form the table:
| $z$ | 0 | 0 | 1 | -1 |
| :---: | :---: | :---: | :---: | :---: |
| $x$ | 6 | 3 | 2 | 7 |
| $y$ | 1 | 3 | 2 | 2 | | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,764 | |
190. $23 x-53 y+80 z=101$.
The above text has been translated into English, maintaining the original text's line breaks and format. | 190. Answer. $x=53 t-15 z+9, y=23 t-5 z+2$. | 53-15z+9,23-5z+2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,765 |
191. $8 x+7 y+5 z=79$. | 191. Answer. $x=y+5 t-2, z=19-3 y-8 t$. | x=y+5t-2,\quadz=19-3y-8t | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,766 |
192. Solve the system of equations in integers:
$$
\left\{\begin{array}{l}
3 x-2 y+4 z+2 t=19 \\
5 x+6 y-2 z+3 t=23
\end{array}\right.
$$ | 192. Excluding $t$, we find: $x+18 y-16 z=-11$.
Answer. $x=16 z-18 y-11, t=28 y-26 z+26$. | x=16z-18y-11,\quad=28y-26z+26 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,767 |
193. Find the values of $x$ for which the numbers $\frac{x-3}{7}$ and $\frac{x-2}{5}$ are both integers. | 193. $\left\{\begin{array}{l}\frac{x-3}{7}=y \\ \frac{x-2}{5}=z\end{array}\right.$
Answer. $x=35 t+17$. | 35+17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,768 |
195. For what values of $x$ will the numbers $\frac{x-3}{7}, \frac{x-2}{5}$, and $\frac{x-4}{3}$ be integers simultaneously? | 195. $\left\{\begin{array}{l}\frac{x-3}{7}=y \\ \frac{x-2}{5}=z \\ \frac{x-4}{3}=u\end{array}\right.$ from this $\left\{\begin{array}{l}7 y-5 z=-1 \text { (1) } \\ 5 z-3 u=2\end{array}\right.$
From (1) we find $y=5 t+2, \quad z=7 t+3$; substituting into (2), we get $35 t-3 u=-13$, hence
$$
\left\{\begin{array}{l}
t=3... | 105t_{1}+52 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,769 |
196. For what values of $x$ will the numbers $\frac{2 x+1}{3}, \frac{3 x+1}{4}$, and $\frac{4 x+1}{5}$ be integers simultaneously? | 196. Answer. $x=60 t+1$. | 60t+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,770 |
197. Prove that if $a x + b$ for some $x = a$ is divisible by $m$, then it will be divisible by $m$ for all values of $x = \alpha + m t$, where $t$ is any integer, and only for these values, if $a$ and $b$ are coprime with $m$. | 197. If $\frac{a \alpha+b}{m}$ is an integer, then $\frac{a(\alpha+m t)+b}{m}=$ $=\frac{a \alpha+b}{m}+a t$ is also an integer.
Any integer value $x$, for example $x=\beta$, for which $\frac{a x+b}{m}$ is an integer, can be found from the formula $x=a+m t$ for some integer $t$.
Indeed, $\beta=\alpha+m t$, hence $t=\f... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,771 |
198. If the number $a x^{n}+b x^{n-1}+\ldots+l$ is divisible by $k$ for some $x=a$, then it is divisible by $k$ for values of $x=a+k t$, where $t$ is any integer. Compare with the previous problem. Show that the number $a$ can always be considered less than $k$. | 198. We find $a(\alpha+k t)^{n}+b(a+k t)^{n-1}+\ldots+l=a \alpha^{n}+$ $+b a^{n-1}+\ldots+l+k m$ (by the binomial theorem); but since by the condition $a a^{n}+b a^{n-1}+\ldots+l$ is divisible by $k$, the theorem is proved. Note that in the general formula $x=\alpha+k t$ we can always make $\alpha$ less than $k$, for e... | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,772 |
199. If $a x+b$ is divisible by a prime number $p$ at $x=\alpha$ and $x=\beta$, and the difference $\alpha-\beta$ is not divisible by $p$, then $a$ and $b$ are multiples of $p$ and, consequently, $a x+b$ is divisible by $p$ for any integer $x$. | 199. Let $\left\{\begin{array}{l}a \alpha+b=m p \\ a \beta+b=n p\end{array}\right.$, hence $a(\alpha-\beta)=(m-n) p$; therefore, $a(\alpha-\beta)$ is divisible by $p$; but since by condition $\alpha - \beta$ is not divisible by $p$, then $a$ is divisible by $p$; therefore, $b$ is also divisible by $p$, which follows fr... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,773 |
200. If $a x^{2}+b x+c$ is divisible by a prime number $p$ for three values of $x$, namely $\alpha, \beta, \gamma$, and none of the differences $\alpha-\beta, \beta-\gamma, \gamma-\alpha$ is divisible by $p$, then $a, b, c$ are divisible by $p$ and, consequently, $a x^{2}+b x+c$ is divisible by $p$ for any integer $x$.... | 200. Let $\left\{\begin{array}{l}a \alpha^{2}+b \alpha+c=m p \\ a \beta^{2}+b \beta+c=n p \\ a \gamma^{2}+b \gamma+c=q p\end{array}\right.$ (1),
from this $\left\{\begin{array}{l}(\alpha-\beta)[a(\alpha+\beta)+b]=(m-n) p \\ (\alpha-\gamma)[a(\alpha+\gamma)+b]=(m-q) p,\end{array}\right.$
therefore $a(\alpha+\beta)+b$ ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,774 |
201*. The number $1 \cdot 2 \cdot 3 \cdot \ldots \cdot(p-1)+1$ is divisible by $p$ if and only if $p$ is a prime number (Wilson's theorem). | 201. Consider the polynomial $f(x)=$
$=x(x-1)(x-2) \ldots[x-(p-1)]-\left(x^{p}-x\right)(1)$, where $p$ is a prime number greater than 2.
The degree of this polynomial is $p-1$; moreover, $x^{p}-x$ is divisible by $p$ for any $x$; further, we note that $f(0)$, $f(1), f(2), \ldots, f(p-1)$ are divisible by $p$. Therefo... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,775 |
202*. Two odd numbers $n$ and $n+2$ are prime, i.e., twin primes if and only if the number $4[(n-1)!+1]+n$ is divisible by $n(n+2)$.
## Chapter VII.
## INDETERMINATE EQUATIONS OF HIGHER DEGREES | 202. First, let's prove the necessity of this condition. Let \( n \) and \( n+2 \) be prime numbers. Then, by Wilson's theorem (see № 201), the number \( (n-1)! + 1 \), and consequently, \( 4[(n-1)! + 1] + n \) is divisible by \( n \). Similarly, \( (n+1)! + 1 \) is divisible by \( n+2 \); but \( n(n+1)\{4[(n-1)! + 1] ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,776 |
203*. If $x^{2}-a$ is divisible by a prime number $p$ for some value of $x$, then $a^{\frac{p-1}{2}}-1$ is divisible by $p$, if $a$ is not divisible by $p$. | 203. Let $x^{2}-a$ be divisible by a prime number $p$. Since by condition $a$ is not divisible by $p$, then $x$ is also not divisible by $p$.
$x^{p-1}-a^{\frac{p-1}{2}}=\left(x^{2}\right)^{\frac{p-1}{2}}-a^{\frac{p-1}{2}}$ is divisible by $x^{2}-a$, and therefore by $p$; but $x^{p-1}-a^{\frac{p-1}{2}}=\left(x^{p-1}-1\r... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,777 |
209. Solve the equation $x^{2}-11 y=4$ in integers. | 209. $\frac{x^{2}-4}{11}=y$; we directly verify that when $x= \pm 2$, the number $x^{2}-4$ is divisible by 11; therefore, $x=11 t \pm 2$, $y=t(11 t \pm 4)$. Thus, the integer solutions of the given equation are contained in the formulas
$$
\left\{\begin{array}{l}
x=11 t \pm 2 \\
y=t(11 t \pm 4)
\end{array}\right.
$$
... | {\begin{pmatrix}11\2\\(11\4)\end{pmatrix}.} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,780 |
210. Find the integer values of $x$ for which $x^{2}+5 x-12$ is divisible by 6. | 210. The problem reduces to finding integer values of $x$ for which $x^{2}+5 x$ is divisible by 6. First, we find those values of $x$ for which $x^{2}+5 x$ is divisible by 3. For this, we note that for $x=0$ and $x=1$, the number $x^{2}+5 x$ is divisible by 3; therefore, $x=3 t$ and $x=3 t+1$. From these values of $x$,... | 3t3t+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,781 |
211. For what integer values of $x$ is the number $x^{2}+2 x+6$ divisible by 15? | 211. First, let's find the integer values of $x$ for which $x^{2}+2 x+6$ is divisible by 3, which is equivalent to the divisibility of $x^{2}+2 x$ by 3; we find $x=3 t$ and $x=3 t+1$. From these values of $x$, we take those for which $x^{2}+2 x+6$ or $x^{2}+$
$+2 x+1$ is divisible by 5. Substituting $x=3 t$ into the la... | 15t-615t+4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,782 |
213. For what integer values of $x$ is the number $9 x^{2}+29 x+$ +62 divisible by 16? | 213. We immediately notice that for any integer $x$ (whether $x=2t$ or $x=2t+1$), the expression $9x^2 + 29x + 62$ is divisible by 2. Substituting $x=2t$ into it, we find after simplification that $t+1$ is divisible by 2; hence, $t=2t_1-1$, from which $x=4t_1-2$. Substituting $x=2t+1$, we find $t=2t_1$; hence, $x=4t_1+... | 16t_3+616t_3+5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,784 |
216. On the curve $x^{2}+5 x+1=3 y$, find the points with integer coordinates in the second quadrant. | 216. Answer. $x=3 t-1, \quad y=3 t^{2}+3 t-1, \quad$ where $t<-1$. | 3-1,\quad3^{2}+3-1,\quad<-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,787 |
219. For what integer values of $n$ will the number $7 n+2$ be a perfect square? | 219. $x^{2}-2$ is divisible by 7, hence $x=7 t \pm 3$; therefore, $n=7 t^{2} \pm 6 t+1$. | 7^{2}\6+1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,790 |
220. The number $7 t+3$ cannot be a perfect cube for any integer $t$.
The number $7 t+3$ cannot be a perfect cube for any integer $t$. | 220. $x^{3}-3$ must be divisible by 7; but, substituting the values $x=0, \pm 1, \pm 2, \pm 3$, we will see that this is not the case. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,791 |
221. For what integer values of $n$ is the number $6 n+2$ a perfect cube? | 221. Answer. $n=36 t^{3}+36 t^{2}+12 t+1$. | 36^{3}+36^{2}+12+1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,792 |
222. Given a sequence of polynomials:
\[
\begin{gathered}
P_{1}=1, P_{2}=x, P_{3}=x \cdot P_{2}-P_{1}, P_{4}=x \cdot P_{3}- \\
-P_{2}, \ldots, P_{n}=x \cdot P_{n-1}-P_{n-2 .}
\end{gathered}
\]
1) For which integer values of \( x \) is the number \( P_{5} \) divisible by 5?
2) Is \( P_{5} \) divisible by 10?
3) Show t... | 222. 223) $P_{5}=x^{4}-3 x^{2}+1 ; x=5 t \pm 2$.
2) $P_{5}$ cannot be divisible by 10, since for any integer $x$ the number $x^{4}-3 x^{2}+1$ is odd.
3) $P_{6}=x(x-1)(x+1)\left(x^{2}-3\right)$ is divisible by 6, since the product of three consecutive numbers is divisible by $1 \cdot 2 \cdot 3=6$.
4) $x=7 t \pm 2$. | 5t\2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,793 |
223. For what values of $n$ is the number $n^{2}-1$ the product of four consecutive natural numbers? | 223. Let $\quad n^{2}-1=x(x+1)(x+2)(x+3), \quad\left(x^{2}+\right.$ $+3 x)^{2}+2\left(x^{2}+3 x\right)+1=n^{2}, \quad x^{2}+3 x+1=n, \quad x=$ $=\frac{-3+\sqrt{5+4 n}}{2} ;$ hence, $5+4 n=z^{2}$; from this it is clear that $z^{2}-5$ and, therefore, $z^{2}-1$ is divisible by 4; and since $z^{2}-1=(z+1)(z-1)$, it is imme... | (+1)-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,794 |
224. Show that if $x^{2}+y^{2}=z^{2}$, then $xy$ is divisible by 12. | 224. First, note that $x$ and $y$ cannot both be odd (see No. 41). It is easy to show that if only one of them is even, then it is divisible by 4. Moreover, if we assume that neither of them is divisible by 3, then $x^{2}-1$, $y^{2}-1$, and $x^{2}+y^{2}-2$ are divisible by 3; $z^{2}-2$ is divisible by 3, which is impos... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,795 |
226. Show that among the terms of an arithmetic progression with the first term equal to 5 and a common difference of 3, there are no perfect squares and there are infinitely many perfect cubes. Generalize for terms of the form $x^{2m}$ and terms of the form $x^{2m+1}$.
Solve the following equations in integers (Probl... | 226. The general term of the progression has the form $3 n+2$. The equation $3 n+2=x^{2}$ has no integer solutions, while the equation $3 n+2=x^{3}$ has integer solutions. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,797 |
228. $2 x^{2}+3 x y+y^{2}=35$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
228. $2 x^{2}+3 x y+y^{2}=35$. | 228. $(2 x+y)(x+y)=35$, notice that each solution $(x, y)$ corresponds to a solution $(-x,-y)$. We need to consider the systems:
$$
\left\{\begin{array}{l}
2 x+y=1 \\
x+y=35
\end{array},\left\{\begin{array}{l}
2 x+y=35 \\
x+y=1
\end{array},\left\{\begin{array}{l}
2 x+y=5 \\
x+y=7
\end{array},\left\{\begin{array}{l}
2 ... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,799 | |
229. How many integer solutions does the equation $x y + a x + b y = c$ have, where $a, b, c$ are integers? | 229. The equation $x y + a x + b y = c$ has as many integer solutions as the number of divisors (positive and negative) of the number $a b + c$. | + | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,800 |
230. $(x+y)^{2}-(x+y)=150+2 x$. | 230. $x+y=\frac{1 \pm \sqrt{8 x+601}}{2}$; therefore, $8 x+601=$ $=t^{2}, t^{2}-1$ is divisible by 8; from this it is clear that $t$ is any odd number, $t=2 t_{1}+1$. Thus, $x+y=$ $=\frac{1 \pm t}{2}, x+y=t_{1}+1$
Answer. $\left\{\begin{array}{l}x=\frac{t_{1}\left(t_{1}+1\right)}{2}-75 \\ y=76-\frac{t_{1}\left(t_{1}-1... | {\begin{pmatrix}\frac{t_{1}(t_{1}+1)}{2}-75\\76-\frac{t_{1}(t_{1}-1)}{2}\end{pmatrix}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,801 |
231*. $x^{2}+x y+y^{2}+x+y-5=0$, if one solution is known: $x=1 ; y=1$. | 231. Let $x=x_{1}+1$ and $y=y_{1}+1$, we get the equation $x_{1}^{2}+y_{1}^{2}+x_{1} y_{1}+4 x_{1}+4 y_{1}=0, x_{1}=0, y_{1}=0$.
Let $\frac{x_{1}}{y_{1}}=\frac{m}{n}$, where $m$ and $n$ are coprime numbers; we get $\left(m^{2}+m n+n^{2}\right) x_{1}^{2}+4 m(m+n) x_{1}=0$; discarding the obvious solution $x_{1}=0$, we ... | (1,1),(1,-3),(-3,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,802 |
232. $x^{2}+x y+y^{2}=x^{2} y^{2}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
232. $x^{2}+x y+y^{2}=x^{2} y^{2}$. | 232. Let's rewrite the given equation as:
$$
(2 x y+1)^{2}=[2(x+y)]^{2}+1
$$
Thus, $[2(x+y)]^{2}+1$ is a perfect square; but since $(2 x y+1)^{2}$ is also a perfect square, and there cannot be two squares differing by one, except for 1 and 0 (see № 11), we conclude that $x+y=0$ and $2 x y+1= \pm 1$; solving the syste... | (0,0),(-1,1),(1,-1) | Algebra | proof | Yes | Yes | olympiads | false | 40,803 |
233. $x^{2}+x y+y^{2}+2 x-3 y-3=0$, if it is known that the solution is: $x=1 ; y=2$. | 233. Answer. $(1,2),(1,0),(-5,2),(-5,6),(-3,0)$. Hint. Take $x=x_{1}+1$ and $y=y_{1}+2$. | (1,2),(1,0),(-5,2),(-5,6),(-3,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,804 |
234. $x^{2}-x y+y^{2}-x+3 y-7=0$, if it is known that the solution is: $x=3 ; y=1$. | 234. Assuming $x=x_{1}+3$ and $y=y_{1}+1$, we get:
$$
x_{1}=-\frac{2 m(2 m+n)}{m^{2}-m n+n^{2}}, \quad y_{1}=-\frac{2 n(2 m+n)}{m^{2}-m n+n^{2}}
$$
The greatest common divisor (GCD) $D \left(m^{2}-m n+n^{2}, m\right)=D\left(n^{2}, m\right)=$ $= D (n, m)=1 ;$ $D \left(m^{2}-m n+n^{2}, 2 m+n\right)=$ $D \left(m^{2}-m n... | (3,1),(-1,1),(3,-1),(-3,-1),(-1,-5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,805 |
236. $x^{2}+y^{2}=z^{2}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
236. $x^{2}+y^{2}=z^{2}$. | 236. Leaving aside the zero solution, we can write the given equation as:
$$
\left(\frac{x}{z}\right)^{2}+\left(\frac{y}{z}\right)^{2}=1
$$
Let $\frac{x}{z}=\xi$ and $\frac{y}{z}=\eta$, we get the equation:
$$
\xi^{2}+\eta^{2}=1(2)
$$
An obvious solution is $\xi=0, \eta=1$. Let $\xi=x_{1}, \eta=y_{1}+1$. We get the... | x:y:-2n:(^{2}-n^{2}):(^{2}+n^{2}) | Number Theory | proof | Yes | Yes | olympiads | false | 40,807 |
237. $x^{2}+y^{2}=2 z^{2}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
237. $x^{2}+y^{2}=2 z^{2}$. | 237. Let's represent the equation in the form $x^{2}-z^{2}=z^{2}-y^{2}$, or $\frac{x-z}{z-y}=\frac{z+y}{x+z}$.
By setting $\frac{x-z}{z-y}=\frac{m}{n}$, we obtain the final formulas $x=\left(m^{2}-2 m n-n^{2}\right) t, y=\left(m^{2}+2 m n-n^{2}\right) t$,
$$
z=\left(m^{2}+n^{2}\right) t
$$
as well as $x=\frac{m^{2}-... | notfound | Number Theory | proof | Yes | Yes | olympiads | false | 40,808 |
238. $x^{2}+2 y^{2}=z^{2}$. | 238. Answer. $x=\left(m^{2}-2 n^{2}\right) t, y=2 m n t, z=\left(m^{2}+2 n^{2}\right) t$;
$$
x=\frac{m^{2}-2 n^{2}}{\Delta}, y=\frac{2 m n}{\Delta}, z=\frac{m^{2}+2 n^{2}}{\Delta}
$$ | (^{2}-2n^{2}),2n,(^{2}+2n^{2}) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,809 |
241. An angle of the triangle is $60^{\circ}$. Find the sides of the triangle, knowing that they are expressed as integers. | 241. Denoting the sides forming the angle $60^{\circ}$ by $x$ and $y$, and the third side by $z$, we get the equation:
$$
x^{2}-x y+y^{2}=z^{2}
$$
Solving by one of the previous methods, we get:
$$
\begin{gathered}
x=\left(m^{2}-n^{2}\right) t, y=m(m-2 n) t, z=\left(m^{2}-m n+n^{2}\right) t \\
x=\frac{m^{2}-n^{2}}{\... | (^{2}-n^{2}),(-2n),(^{2}-n+n^{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 40,811 |
242. An angle of the triangle is equal to $120^{\circ}$. Find the sides of the triangle, knowing that they are expressed as integers. | 242. The problem reduces to solving the equation $x^{2}+x y+$ $+y^{2}=z^{2}$
Since this equation can be obtained from the equation of the previous problem by substituting $-y$ for $y$, we will finally find:
$$
\begin{gathered}
x=\left(m^{2}-n^{2}\right) t, y=(2 n-m) m t, z=\left(m^{2}-m n+n^{2}\right) t \\
x=\frac{m^... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 40,812 |
243. Two medians of a triangle are perpendicular to each other. Find the sides of the triangle, knowing that they are expressed in whole numbers. | 243. Denoting the sides corresponding to the medians specified in the condition by \(x\) and \(y\), and the third side by \(z\), and using the property of the intersection point of the medians, we obtain the equation \(x^{2} + y^{2} = 5 z^{2} (1)\).
To solve equation (1), rewrite it as \(x^{2} - 4 z^{2} = z^{2} - y^{2... | x=22,y=19,z=13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 40,813 |
244. One median of a triangle is twice as long as another. Find its sides if they are expressed as integers. | 244. Denoting the sides corresponding to the given medians by $x$ and $y(y>x)$, and the third side by $z$, we obtain the equation:
$$
2 z^{2}=2 y^{2}-3 x^{2} \quad(1)
$$
We can represent it as $\frac{2 y+2 z}{x}=\frac{3 x}{y-z}=\frac{m}{n}$, which gives us $x=4 m n, y=m^{2}+6 n^{2}, z=m^{2}-6 n^{2}$.
From the condit... | 16,22,10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 40,814 |
245. One angle of the triangle is twice as large as another. Find the sides of the triangle, knowing that they are expressed as integers. | 245. $A=2B$. Let $BC=a, AC=b, AB=c$. Obviously $a>b$.
Draw the bisector $AD$ of angle $A$, then $\angle CDA = \angle DAB + \angle DBA = 2B; \angle CDA = \angle CAB$. By the property of the angle bisector in a triangle, as well as considering the similarity of triangles, we find $c=\frac{a^2-b^2}{b}$ (1). From (1) and ... | =mn,b=n^2,=^2-n^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 40,815 |
246. The lateral surface area of a rectangular parallelepiped is equal to the sum of the areas of the bases. Find its dimensions, if they are expressed as integers. | 246. Denoting the sides of the base by $x$ and $y$, and the height by $z$, we obtain the equation:
$$
x z + y z = x y
$$
Answer. $x = n(n-m), y = m n, \quad z = m(n-m) \quad$ up to similarity ( $m < n$ ). | x=n(n-),y=n,z=(n-) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 40,816 |
247. Find a parallelogram with integer sides and diagonals. | 247. Denoting the sides by $x$ and $y$, and the diagonals by $z$ and $u$, we get the equation:
$$
2\left(x^{2}+y^{2}\right)=z^{2}+u^{2}
$$
Answer. $\quad x=2 r^{2}-2 s^{2}+2 s t-t^{2}, \quad y=2 r t-4 r s, \quad z=$ $=2 r^{2}+2 s^{2}-4 s t+t^{2}, u=2 r^{2}+2 s^{2}-t^{2}$.
Example. $r=7, s=1, t=3$; then $x=93, y=14$,... | 93,14,97,u=91 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 40,817 |
252. Find the sides of a rectangle given that its perimeter and area are numerically equal to each other, and the sides are expressed in whole numbers. | 252. We have the equation $2 x+2 y=x y$. Let's represent it as $\frac{1}{x}+\frac{1}{y}=\frac{1}{2}$. Find the equal values of $x$ and $y$: $x=4, y=4$. Next, each of the positive fractions $\frac{1}{x}$ and $\frac{1}{y}$ must be less than $\frac{1}{2}$, hence, $x>2$, $y>2$. Moreover, one of these fractions, for example... | (4,4),(3,6),(6,3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 40,822 |
253. Find the dimensions of a rectangular parallelepiped if they are expressed as integers, and the total surface area and volume are numerically equal. | 253. We obtain the equation $x y z=2(x y+x z+y z)$, or
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2}
$$
Hint. First find equal values of $x, y, z$, then consider the case of two equal unknowns, finally the general case. As a result, we get 10 solutions: $(6,6,6),(5,5,10),(4,8,8),(3,12,12),(3,7,42),(3$, $8,24),(3... | (6,6,6),(5,5,10),(4,8,8),(3,12,12),(3,7,42),(3,8,24),(3,9,18),(3,10,15),(4,5,20),(4,6,12) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 40,823 |
254. The circuit consists of two conductors connected in parallel. Find the resistance of the circuit and of each conductor separately, if it is known that these resistances are expressed as integers. | 254. Denoting the resistance of the circuit by $z$, and the resistances of the conductors by $x$ and $y$, we obtain the equation:
$$
\frac{1}{z}=\frac{1}{x}+\frac{1}{y} \text { (see No. 246). }
$$ | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,824 |
255. Find the rational solutions of the equation $x^{2}$ $-3 y^{2}=1$.
| 255. $(x+1)(x-1)=3 y \cdot y, \frac{x+1}{y}=\frac{3 y}{x-1}=k$;
$$
x=\frac{k^{2}+3}{k^{2}-3}, y=\frac{2 k}{k^{2}-3}
$$
where $k$ is an arbitrary rational number. | \frac{k^{2}+3}{k^{2}-3},\frac{2k}{k^{2}-3} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,825 |
256. The equation $x^{2}-D y^{2}=1$, where $D$ is a natural number that is not a perfect square, is called Pell's equation. Show that if $x_{0}, y_{0}$ is an integer solution to this equation, then the numbers determined by the formula $x+y \sqrt{D}=\left(x_{0}+y_{0} \sqrt{D}\right)^{k}$, where $k=1,2,3, \ldots$, are a... | 256. From the equality $x+y \sqrt{D}=\left(x_{0}+y_{0} \sqrt{D}\right)^{k}$ we obtain $x-y \sqrt{D}=\left(x_{0}-y_{0} \sqrt{D}\right)^{k}$.
After multiplying these equalities, we find $x^{2}-D y^{2}=1$.
Further, we have $x_{k}=\frac{\left(x_{0}+y_{0} \sqrt{D}\right)^{k}+\left(x_{0}-y_{0} \sqrt{D}\right)^{k}}{2}$,
$$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,826 |
257. Find several integer solutions of the equation $x^{2}$ $-3 y^{2}=1$, knowing the solution: $x=2, y=1$. | 257. We have $x+y \sqrt{3}=(2+\sqrt{3})^{k} ; \quad$ taking $\quad k=2$, we find $x+y \sqrt{3}=7+4 \sqrt{3}$, from which $x=7, y=4$. Next, $x=4 \cdot 7-2=26, y=4 \cdot 4-1=15 ; \quad x=4 \cdot 26$ $-7=97, y=4 \cdot 15-4=56$ and so on. (see No. 256). | 7,4;26,15;97,56 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,827 |
262. Find a triangular number whose square is also a triangular number. | 262. From the equation $\left[\frac{x(x+1)}{2}\right]^{2}=\frac{y(y+1)}{2}$, we find:
$$
y=\frac{-1+\sqrt{2 x^{2}(x+1)^{2}+1}}{2}
$$
Let $x(x+1)=z$, then $2 z^{2}+1=t^{2}, t^{2}-2 z^{2}=1$.
Solving this Pell's equation, we get an infinite set of values $z=2,12,70,408 \ldots$. From the equation $x(x+1)=2$, we have $x... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,830 |
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