+100 Given positive integers x and y such that x≠y and \(\frac1x+\frac1y=\frac1{18}\), what is the smallest possible value for x+y?
+128408 x, y are > 18
Let z = 18
Let x = z+ a
Let b = z + b
So
1/ (z + a) + 1 / (z + b) = 1 / z
(z + b + z + a) / [ (z + a) (z + b) = = 1/z
( 2z + a + b) / [ (z + a) (z + b) ] = 1 /z cross-multiply
2z^2 + az + bz = ( z + a) ( z + b)
2z^2 + az + bz = z^2 + az + bz + ab simplify
z^2 = ab
18*2 = ab
324 = ab
Factors of 324 =
1 | 2 | 3 | 4 | 6 | 9 | 12 | 18 | 27 | 36 | 54 | 81 | 108 | 162 | 324
ab will be minimized when a = 12 and b = 27
So
x = a + z = 12 + 17 = 30
y = b + z = 27 + 18 = 45
Smallest possible value for x + y = 30 + 45 = 75
