Michael Kleber wrote

Q: Is it always possible to partition the integers from 2 to 2n-1 into two sets of size n-1, such that the products of two elements, one from each set, all distinct integers >=2n? Seems like the answer is "yes with high probability", but can someone change that to just "yes"?

Looks like "no" is the real answer Suppose so... Let S,T be the two sets, with 2 in S. Let 2<a<n. It must be that a isin S. Given a, then 2a<=2n-1 and hence in one of the sets. If a isin T then we can make 2a two different ways. The first with 1*2a (no matter where 2a lives) and 2*a since 2 and a are not in the same set. So S={1,2,3,...,n-1,r} for some r>=n. Now for n of any size, there are a couple of instances where 2a and 2a+1 are in S and (because we don't know r) at least one case where 2(2a) and 2(2a+1) are in T. This gives two ways to get 4a(2a+1). The cheese ball way to get the job done is to set S={1,2,...,n} and T={1,p1,...,p(n-1)} where the pi are the n -1 smallest primes bigger than n. Mark