Determine all possible value(s) of a positive integer N that satisfies the following conditions:

(i) N is the product of three prime numbers, whose squares sum to 2331.

(ii) The sum of the positive divisors of N ( including 1 and N) is 10560.

N is the product of three primes, p,q,r.  Then the sum of the divisors of N is 1+p+q+r+p*q+p*r+q*r+p*q*r = (1+p)(1+q)(1+r) = 10560

10560 has 11 as a factor, so testing multiples of 11 looking for a number one greater than a prime gives 4*11-1=43, 10*11-1=109, 12*11-1=131, etc.  The first two are too large for the sum of squares to equal 2331, so 43 is one of the factors.

Then, 10560/44 = 240 = (1+q)(1+r) and 2331-43^2 = 482 = q^2+r^2.  Which makes q and r equal 11 and 19.

N = 43*19*11 = 8987.

Posted by Brian Smith on 2009-10-18 14:44:38