Difference between revisions of "2000 AIME II Problems/Problem 7"

Revision as of 07:16, 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 $\frac N{100}$ .

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

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 == Solution ==
-{{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)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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