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Difference between revisions of "2000 AIME II Problems/Problem 7"
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So, <math>N=\frac{262124}{19}=13796</math> and <math>\left\lfloor \frac{N}{100} \right\rfloor =\boxed{137}</math>.   
 
So, <math>N=\frac{262124}{19}=13796</math> and <math>\left\lfloor \frac{N}{100} \right\rfloor =\boxed{137}</math>.   
  
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== See also ==
 
 
{{AIME box|year=2000|n=II|num-b=6|num-a=8}}
 
{{AIME box|year=2000|n=II|num-b=6|num-a=8}}

Revision as of 19:34, 18 March 2008

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 $\frac N{100}$.

Solution

Multiplying both sides by $19!$ yields:

$\frac {19!}{2!17!}+\frac {19!}{3!16!}+\frac {19!}{4!15!}+\frac {19!}{5!14!}+\frac {19!}{6!13!}+\frac {19!}{7!12!}+\frac {19!}{8!11!}+\frac {19!}{9!10!}=\frac {19!N}{1!18!}$.

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

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

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

2000 AIME II (Problems • Answer Key • Resources)
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