AoPS Wiki talk:Problem of the Day/September 13, 2011

Since $7,999,999,999=8,000,000,000-1=2000^3-1^3$ , we can use the difference of cubes factorization: $2000^3-1^3=(2000-1)(2000^2+2000(1)+1^2)=(1999)(4,002,001)$ and since we are given that the original number has at most 2 prime factors, both of these must be prime. The larger is $\boxed{4,002,001}$ .