2000 AIME II Problems/Problem 7

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Problem

Given that $\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10!}=\frac...$ find the greatest integer that is less than $\frac N{100}$ .

Solution

Multiplying both sides by $19!$ yields:

$\frac {19!}{2!17!}+\frac {19!}{3!16!}+\frac {19!}{4!15!}+\frac {19!}{5!14!}+\frac {19!}{6!13!}+\frac {19!}{7!12!}+\frac {19!}...$

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

Recall the identity $2^{19} = \sum_{n=0}^{19} {19 \choose n}$ . Since ${19 \choose n} = {19 \choose 19-n}$ , it follows that $\sum_{n=0}^{9} {19 \choose n} = \frac{2^{19}}{2} = 2^{18}$ .

Thus, $19N = 2^{18}-\binom{19}{1}-\binom{19}{0}=2^{18}-19-1 = (2^9)^2-20 = (512)^2-20 = 262124$ .

So, $N=\frac{262124}{19}=13796$ and $\left\lfloor \frac{N}{100} \right\rfloor =\boxed{137}$ .

2000 AIME II (Problems • Resources)
Preceded by Problem 6	Followed by Problem 8
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2000 AIME II Problems/Problem 7

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