Mock AIME 2 2010 Problems/Problem 12: Difference between revisions
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==Problem== | |||
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![\[K = \sum_{a_1=0}^8\sum_{a_2=0}^{a_1}...\sum_{a_{100}=0}^{a_{99}}\binom{8}{a_1}\binom{a_1}{a_2}...\binom{a_{99}}{a_{100}}.\]](http://latex.artofproblemsolving.com/3/e/0/3e0e64ceec082106b3bb53ef6c103c57d39fd3ee.png)
Find the sum of digits of 
in base-100.
