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Mock AIME 6 2006-2007 Problems/Problem 1

Revision as of 01:32, 6 September 2023 by Abbywong (talk | contribs)

We note that the required sum is equal to $(2^0+2^1+2^2+2^3+2^4)(3^0+3^1+3^2+3^3+3^4)=(31)(121)=3751$, which is $\framebox{751}$ mod 1000. ~AbbyWong

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