Difference between revisions of "2000 AIME II Problems/Problem 7"
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== Solution == | == Solution == | ||
| + | Multiplying both sides by <math>19!</math> yields: | ||
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| + | <math>\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!}</math>. | ||
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| + | <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>. | ||
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| + | Thus, <math>19N = \frac{2^{19}}{2}-\binom{19}{1}-\binom{19}{0}=2^{18}-19-1 = (2^9)^2-20 = (512)^2-20 = 262124</math>. | ||
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| + | So, <math>N=\frac{262124}{19}=13796</math> and <math>\left\lfloor \frac{N}{100} \right\rfloor =\boxed{137}</math>. | ||
== 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 16:58, 25 February 2008
Problem
Given that
find the greatest integer that is less than 
.
Solution
Multiplying both sides by 
yields:
.
.
Thus, 
.
So, 
and 
.
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 | ||
