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?

 Jun 5, 2021

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

 Jun 5, 2021

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