Difference between revisions of "2000 AIME II Problems/Problem 7"
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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 == | == 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 08:16, 25 February 2008
Problem
Given that
![$\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10!}=\frac N{1!18!}$](http://latex.artofproblemsolving.com/1/6/d/16d26af464ce4b8d84de3933589ddc93c5d5b57e.png)
find the greatest integer that is less than .
Solution
Mutiplying each side by , We have
![$\binom{19}{2}+\binom{19}{3}+\binom{19}{4}+\binom{19}{5}+\binom{19}{6}+\binom{19}{7}+\binom{19}{8}+\binom{19}{9}=19N$](http://latex.artofproblemsolving.com/6/4/8/648a58fa636f38f51313918088ebfe8cb05259a2.png)
Since we know that
![$\binom{19}{0}+\binom{19}{1}+...+\binom{19}{9}=\frac{1}{2}\cdot(2^{19})=2^{18}$](http://latex.artofproblemsolving.com/d/5/d/d5d4a258c3e72a933bc0a38fee8263c4ba59a385.png)
Hence, we know that the above equation equates to
See also
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 | ||
All AIME Problems and Solutions |