Given positive integers x and y such that x not equal to y and 1/x + 1/y = 1/80, what is the smallest possible value for x + y?

Guest Jun 5, 2021

#1**+1 **

If 1/x + 1/y = 1/80 and x ≠ y ≠ 0, we can multiply both sides by 80xy to get 80x + 80y = xy

Moving all terms to one side, we have xy - 80x - 80y = 0

Adding 6400 to both sides, we can factor, giving (x-80)(y-80) = 6400.

With x and y both being integers, we realize that x-80 and y-80 must be a pair of factors of 6400 giving a diophantine equation. To minimize x+y, we want x and y to be as close as possible, and therefore, want x-80 and y-80 to be as close as possible. To achieve this, we want to find two factors of 6400 to be as close as possible, and let x-80 and y-80 equal them. Besides x=y= 160 (x-80=y-80=80), which is not possible due to the fact that x≠y, we have x-80 = 64 y-80 = 100.

In this, x=144, y=180, so x+y = 324

EnchantedLava68 Jun 5, 2021