Difference between revisions of "Mock AIME 2 2010 Problems/Problem 12"

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==Problem==
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Let<cmath>K = \sum_{a_1=0}^8\sum_{a_2=0}^{a_1}...\sum_{a_{100}=0}^{a_{99}}\binom{8}{a_1}\binom{a_1}{a_2}...\binom{a_{99}}{a_{100}}.</cmath>Find the sum of digits of <math>K</math> in base-100.
 
==Solution==
 
==Solution==

Latest revision as of 13:04, 31 December 2023

Problem

Let\[K = \sum_{a_1=0}^8\sum_{a_2=0}^{a_1}...\sum_{a_{100}=0}^{a_{99}}\binom{8}{a_1}\binom{a_1}{a_2}...\binom{a_{99}}{a_{100}}.\]Find the sum of digits of $K$ in base-100.

Solution