Difference between revisions of "2000 AIME II Problems/Problem 7"
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Thus, <math>19N = 2^{18}-\binom{19}{1}-\binom{19}{0}=2^{18}-19-1 = (2^9)^2-20 = (512)^2-20 = 262124</math>. | Thus, <math>19N = 2^{18}-\binom{19}{1}-\binom{19}{0}=2^{18}-19-1 = (2^9)^2-20 = (512)^2-20 = 262124</math>. | ||
− | 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>. |
== See also == | == See also == |
Revision as of 14:13, 30 May 2017
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
Multiplying both sides by yields:
Recall the identity . Since
, it follows that
.
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 |
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