2000 AIME II Problems/Problem 7

Problem

Given that

$\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10!}=\frac N{1!18!}$

find the greatest integer that is less than $\frac N{100}$ .

Mutiplying each side by $19!$ , We have

$\binom{19}{2}+\binom{19}{3}+\binom{19}{4}+\binom{19}{5}+\binom{19}{6}+\binom{19}{7}+\binom{19}{8}+\binom{19}{9}=19N$

Since we know that

$\binom{19}{0}+\binom{19}{1}+...+\binom{19}{9}=\frac{1}{2}\cdot(2^{19})=2^{18}$

Hence, we know that the above equation equates to

2000 AIME II (Problems • Answer Key • Resources)
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