Difference between revisions of "2000 AIME II Problems/Problem 7"
Ilovemath04 (talk | contribs) (→Solution) |
Hashtagmath (talk | contribs) m (→Solution) |
||
Line 9: | Line 9: | ||
<cmath>\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.</cmath> | <cmath>\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.</cmath> | ||
− | Recall the [[combinatorial identity| | + | Recall the [[combinatorial identity|Combinatorial Identity]] <math>2^{19} = \sum_{n=0}^{19} {19 \choose n}</math>. Since <math>{19 \choose n} = {19 \choose 19-n}</math>, it follows that <math>\sum_{n=0}^{9} {19 \choose n} = \frac{2^{19}}{2} = 2^{18}</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>. | Thus, <math>19N = 2^{18}-\binom{19}{1}-\binom{19}{0}=2^{18}-19-1 = (2^9)^2-20 = (512)^2-20 = 262124</math>. |
Revision as of 23:51, 7 January 2019
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 Combinatorial 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.