The number of positive integral solutions of | vedclass

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Question

(B) Given the equation $\frac{1}{x} + \frac{1}{y} = \frac{1}{2025}$.This can be rewritten as $\frac{x+y}{xy} = \frac{1}{2025}$, which implies $xy - 20

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) Given the equation $\frac{1}{x} + \frac{1}{y} = \frac{1}{2025}$.This can be rewritten as $\frac{x+y}{xy} = \frac{1}{2025}$, which implies $xy - 2025x - 2025y = 0$.Adding $2025^2$ to both sides, we get $xy - 2025x - 2025y + 2025^2 = 2025^2$.This factors as $(x - 2025)(y - 2025) = 2025^2$.Let $X = x - 2025$ and $Y = y - 2025$. Then $XY = 2025^2$.Since $2025 = 3^4 	imes 5^2$, we have $2025^2 = 3^8 	imes 5^4$.The number of divisors of $2025^2$ is $(8+1)(4+1) = 9 	imes 5 = 45$.Since $x, y > 0$, we must have $x > 2025$ and $y > 2025$, so $X, Y > 0$.Thus, the number of positive integral solutions is equal to the number of divisors of $2025^2$, which is $45$.

The number of positive integral solutions of $\frac{1}{x} + \frac{1}{y} = \frac{1}{2025}$ is

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