AoPS Wiki talk:Problem of the Day/September 13, 2011
Revision as of 17:04, 13 September 2011 by Negativebplusorminus (talk | contribs) (Created page with "Since <math>7,999,999,999=8,000,000,000-1=2000^3-1^3</math>, we can use the difference of cubes factorization: <cmath>2000^3-1^3=(2000-1)(2000^2+2000(1)+1^2)=(1999)(4,002,001)</c...")
Since , we can use the difference of cubes factorization: and since we are given that the original number has at most 2 prime factors, both of these must be prime. The larger is .