Difference between revisions of "AoPS Wiki talk:Problem of the Day/September 13, 2011"
(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...") |
(No difference)
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Latest revision as of 18:04, 13 September 2011
Since 
, we can use the difference of cubes factorization:
![\[2000^3-1^3=(2000-1)(2000^2+2000(1)+1^2)=(1999)(4,002,001)\]](http://latex.artofproblemsolving.com/a/d/1/ad173097a71d675cf05e3a8bf38851a31c681555.png)
and since we are given that the original number has at most 2 prime factors, both of these must be prime. The larger is 
.
