How many ordered pairs of integers (a,b) are there such that 1/a+1/b=1/200?

Answer 1

multiply both sides by 200ab
200b+200a = ab
200b - ab = -200a
b(200-a) = -200a
b = -200a/(200-a) = 200a/(a-200)
or
a = 200b/(b-200)

let a = 201, then b = 40200
let a = 199 , then b = -39800
let a = 202 , then b = 20200
let a = 198 , then b = -19800
let a = 203 , then b = not an integer !
let a = 197 , then b = not an integer!
let a = 204 , then b = 10200
let a = 205 , then b = 8200
let a = 206 , then b = not an integer, ahhh!
...
let a = 240, then b = 1200
let a = 600 , then b = 240000
let a = 10200 , then b = 204

we have to have (a-200) be divisible into 200a
obviously we cannot use a = 200, but we can go up or down from 200 leaving divisors of ±1 , ±2, ±4, ±5, ±8, ±10, .. as long as that divisor divides evenly into 200a
that is, 200 = 2x2x2x5x5

Wolfram says that there are 69 of these

http://www.wolframalpha.com/input/?i=%28x%2By%29%2F%28xy%29+%3D+1%2F200

Answer 2

This problem has been posted on Brilliant as a current problem in this weeks' Geometry problem set.

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Please refrain from telling the OP the answer.

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