+501 In the equation $\frac{1}{j} + \frac{1}{k} = \frac{1}{3}$, both $j$ and $k$ are positive integers. What is the sum of all possible values for $k$?
+129852 1/j + 1/k = 1/3
Let 3 = z
j and k must be > 3
So
Let j = z + m and k = z + n
So we have
1/ ( z + m) + 1/ ( z + n) = 1/z
(z +n + z + m) 1
____________ = ____ cross-multiply
(z + m) (z + n) z
(m + n + 2z) z = (z + m) (z + n)
2z^2 + zm + zn = z^2 + zm + zn + mn
z^2 = mn
3^2 = mn
9 = mn
Either m = 9 and n = 1 or m = 1 and n = 9
If m = 9 then j = z + m ⇒ j = 3 + 9 = 12
And n = 1 then k = z + n ⇒ k = 3 + 1 = 4
So (j, k) = ( 12, 4) or (4, 12)
So....sum of all k's = 12 + 4 = 16
