Very hard (few sentence) logic problem

S has 136 (and makes the first statement)
P has 135

From P's perspective, these are the possibilities:

N1 N2 S
---------
1 135 136
3 45 48
9 15 24
5 27 32

As you pointed out, S can't make the first statement if he has a prime+1. Of 136, 48, 24, and 32, only 136 is not a prime+1.

QED

O.....M......G!!!!!


It's TechnoGuyRob!!!!!


It's been about a year and a half!!!


<do I seem excited?>

This wasn't really that hard - took about half an hour. When I was done, I realized it should have been faster.

You can pretty quickly rule out P as having made the first statement - too many sums for a given product.

So, what you're looking for are two numbers (x,y) that add up to 136 such that when their product is expressed as the product of exactly two numbers (p,q), p+q is a prime+1 for all but one case.

This should be clearer:

x+y=136
p = x*y
Say p can be expressed as a product of exactly two numbers in N ways:
p1 * q1 (s1 = p1 + q1)
p2 * q2 (s2 = p2 + q2)
.
.
.
pN * qN (sN = pN + qN)

Then a solution exists when all but one of s1 through sN is a prime+1. Since 136 is always in the list of sN, it must be the only one that isn't a prime+1 (because 135 isn't prime).

It turns out that there is only one case to consider:
(1,135)

Why?

p can't be prime because it is the product of two numbers that sum to 136 (and 135 isn't prime).
(1,p) is obviously one of the N ways of factoring p into two factors.
p+1 can't be a prime+1 because p isn't prime.

Therefore, you will always have two sums (136 and p+1) in the sN list that aren't prime+1, unless 136 is p+1.

Fortunately, as shown in my previous post, it works for p+1=136, otherwise, there wouldn't be a solution.

Aw geez, we made that list too, and messed up checking if 47 is a prime. Bah.

Anyways, thanks, I don't think we would have ever tried that again and ended up just never solving it.

And hey Lash! I remember you both! From way back when...