AoPS Wiki talk:Problem of the Day/September 17, 2011
Revision as of 07:55, 17 September 2011 by Negativebplusorminus (talk | contribs) (Created page with "We know that <cmath>(999+1)^3=999^3+3(999)^2+3(999)+1</cmath> by the Binomial Theorem. Since our given expression is one less than that, we compute <cmath>(999+1)^3-1=1,000,000,...")
We know that
![\[(999+1)^3=999^3+3(999)^2+3(999)+1\]](http://latex.artofproblemsolving.com/6/3/5/6355cf6f4ca3a5a1c65b9cefd10d753700695d4f.png)
by the Binomial Theorem. Since our given expression is one less than that, we compute
![\[(999+1)^3-1=1,000,000,000-1=\boxed{999,999,999}\]](http://latex.artofproblemsolving.com/4/1/3/413ac2246750c452312fdac74cfb9fc4b6a914e0.png)
