List

**all** the integers N, below 10000, such that the sum of their proper divisors (i.e. N excluded) equals their product.

Counterexample:

take** 12: s(1,2,3,4,6)=16; p(1,2,3,4,6)=144; **

so 12 is not on the list.

Assume that N has two divisors greater than 2, call them f and g with f<g.

Then f+g+3<3*g<=f*g implies the product of the divisors will always be greater than the sum. Adding more large divisors will only increase the difference.

Therefore N must have at most one divisor f greater than 2. Then the possible sets of divisors of N are {1}, {1,f}, {1,2}, {1,2,f}.

The first set implies that N is a prime number.

The second set and third sets fail the sum=product requirement.

The fourth set implies 3+f = 2*f, which makes f=3 and N=6.

Then the solution set for this puzzle consists of 6 and all prime numbers.

*Edited on ***October 20, 2017, 11:49 am**