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...")
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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 $\boxed{4,002,001}$.