2000 AIME II Problems/Problem 7: Difference between revisions

Revision as of 07:17, 25 February 2008

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 $\frac N{100}$ .

@@ Line 3: / Line 3: @@
 == Solution ==
-Mutiplying each side by <math>19!</math>, We have <center><math>\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</math></center> Since we know that <center><math>\binom{19}{0}+\binom{19}{1}+...+\binom{19}{9}=\frac{1}{2}\cdot(2^{19})=2^{18}</math>
-</center> Hence, we know that the above equation equates to
 == See also ==
 {{AIME box|year=2000|n=II|num-b=6|num-a=8}}

2000 AIME II (Problems • Answer Key • Resources)
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