2000 AIME II Problems/Problem 7
Problem
Given that
find the greatest integer that is less than 
.
Solution
Multiplying both sides by 
yields:
![\[\frac {19!}{2!17!}+\frac {19!}{3!16!}+\frac {19!}{4!15!}+\frac {19!}{5!14!}+\frac {19!}{6!13!}+\frac {19!}{7!12!}+\frac {19!}{8!11!}+\frac {19!}{9!10!}=\frac {19!N}{1!18!}.\]](http://latex.artofproblemsolving.com/5/1/2/512c6a00ae7e316e65fee4284136007504350d26.png)
![\[\binom{19}{2}+\binom{19}{3}+\binom{19}{4}+\binom{19}{5}+\binom{19}{6}+\binom{19}{7}+\binom{19}{8}+\binom{19}{9} = 19N.\]](http://latex.artofproblemsolving.com/e/3/e/e3e8008bba65f49823f5d64af9b96e7d8ec7dedc.png)
Recall the Combinatorial Identity 
. Since 
, it follows that 
.
Thus, 
.
So, 
and 
.
Solution 2
Let 
Applying the binomial theorem gives us 
Since 
After some fairly easy bashing, we get 
as the answer.
~peelybonehead
Solution 3 (Brute Force)
Convert each denominator to and get the numerators to be 
(refer to note). Adding these up we have 
therefore is the desired answer.
Note:
Notice that each numerator is increased each time by a factor of 
etc. until 
. If you were taking the test under normal time conditions, it shouldn't be too hard to bash out all of the numbers but it is priority to be careful.
~SirAppel
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. 
