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**jimmyjimmyjimmy** Hi,

Let n be a positive integer.

Suppose that the sum of all positive divisors of n (besides n itself) is a multiple of n. (e.g. 1+2+3 = 6k where n=6 and k=1 in this case)

Suppose too that n is a product of two distinct primes.

Show that n must equal 6.

I started by letting n = pq for distinct primes p and q.

Then we know 1+p+q = kn = kpq for some k.

This problem just seems so simple, but I can't help but think the proof is really complicated. I would appreciate any hints. Thanks.