if ur answer doesnt end w "OHHH" u lose

What I'm doing is this: Start with from the bottom with sums, and see which sums make the second statement possible.

Example: For the sum 11 there are four possibilities: 2+9, 3+8, 4+7, 5+6. In order for statement 2 to be possible, there has to be more than one way to get the product of all of these via multiplication. So 2*9=18, but you can also get that from 3*6. 3*8=24 but you can also get that from 2*12 or 6*4. 4*7=28 but you can get that from 2*14. 5*6=30, but you can also get that from 3*10 or 2*15. So 11 would be a candidate sum.

For 13 you get 2*11=22, but there is no other way to get 22 so 13 is not a candidate.

This still gets very cumbersome as the sums get bigger though, and from gabes post I would guess, there's a simpler way.

Once you have the candidate sums, write down the possible products for all of them. Hopefully you should be able to tell which products makes it possible for A to know the numbers, and hopefully knowing that that helps B to know the numbers, you can figure out what the numbers are.

I started working on a very similar method to mute but didn't get far with it because the number of possible sums for the higher numbers is so large. One thing I have just realised though is that you can immediately eliminate any sum that can be made by summing 2 primes, because the product of them will be unique. This means that any prime +2, +3, +5, etc can't be the correct sum, which cuts down on the possibilities quite significantly.

I already suspected, that all even numbers were out, but having really given much thought to proving why, but you can make any even sum by adding 3, 5 or 7 to another prime.

except for 98, which is 79+19

3 and 14