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 Dec 21, 2021

#1**0 **

This problem has a nice solution.

Notice that 1/x + 1/y = (x+y)/(xy). Thus, (xy)/(x+y) = 80. Mutliplying by x+y, we have xy = 80x+80y. This simplifies to xy - 80x - 80y = 0.

Factoring out an x from xy-80x, we get x(y-80) - 80y = 0. A factoring tactic called SFFT will help here, to turn -80y into -80(y-80). This means we must add 80^2 = 6400 to both sides, getting x(y-80) -80y+6400 = 6400. This turns into x(y-80)-80(y-80)=6400, so (x-80)(y-80)=6400.

The sum x+y is minimized when x and y are the closest together, which you can estimate by taking √6400, or 80. The closest such x and y are 64 and 100, giving an answer of **164.**

tinfoilhat Dec 22, 2021