[LongGone]

Originally Posted by Reincarnate

can anyone find the problem in my earlier list? should equal 1356, not 1355 (according to the problem description)

Since you are one off from the actual answer (changing from odd to even), it implies that one set of (p^a)(q^b) should be degrouped, or grouped (e.g. 5*19=95(odd), or 5^a+19^b=even). Swapping pairs of number products will not change the odd/even parity
The main concern are the primes <10 in this case (because obviously you can't multiply two primes >10 that will result in <100 product)
So the usable primes are 2,3,5,7
Since you used up all 4 and ended up with 1355; If 1356 was not a mistake, that means that you have to degroup one of those primes and leave one as a natural power of itself (which is counterintuitive)
...So either the 13 people who solved didnt bother checking Co(100), or I'm missing something rofl
(or a rare case of a possible product of three primes, which idk if its plausible)

For a possible general approach, to try to maximise sum, we can start by getting rid of all the small primes
So e.g. for Co(100), start with 11,13,17,19. With your current solution we got: 99 (3,11), 95 (5,19), and 92 (2,23), 91 (7,13). Could there be a possible solution using 17 in place of 23?