2000 AIME II Problem 7

Question

Given that

\begin{aligned} &\frac{1}{2!17!} + \frac{1}{3!16!} + \frac{1}{4!15!} \\ &\quad {}+ \frac{1}{5!14!} + \frac{1}{6!13!} + \frac{1}{7!12!} \\ &\quad {}+ \frac{1}{8!11!} + \frac{1}{9!10!} = \frac{N}{1!18!}, \end{aligned}

find the greatest integer that is less than

\frac{N}{100}.

Answer

Multiply both sides by

19!

so that each term on the left becomes a binomial coefficient

\binom{19}{k}

By symmetry

\sum_{k=0}^{9}\binom{19}{k} = 2^{18};

subtract the

k = 0

and

k = 1

terms

Multiply both sides by

19!.

Each term on the left becomes

\frac{19!}{k!\,(19 - k)!} = \binom{19}{k}

for

k = 2, \ldots, 9,

while the right side becomes

19N.

Since

\binom{19}{k} = \binom{19}{19 - k},

the first half of the binomial row sums to

\sum_{k=0}^{9} \binom{19}{k} = \frac{2^{19}}{2} = 2^{18},

so

\begin{aligned} \sum_{k=2}^{9}\binom{19}{k} &= 2^{18} - \binom{19}{0} \\ &\quad {}- \binom{19}{1} \\ &= 262144 - 20 \\ &= 262124. \end{aligned}

Hence

N = \frac{262124}{19} = 13796,

so

\frac{N}{100} = 137.96,

and the greatest integer less than this is

137.

Problem 7 in other years