Difference between revisions of "2014 AMC 12B Problems/Problem 23"
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<cmath>\equiv (-1)^k\dbinom{k+2}{2} \mod 2017</cmath> | <cmath>\equiv (-1)^k\dbinom{k+2}{2} \mod 2017</cmath> | ||
Therefore | Therefore | ||
| − | <cmath>\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k= | + | <cmath>\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017</cmath> |
This is simply an alternating series of triangular numbers that goes like this: <math>1-3+6-10+15-21....</math> | This is simply an alternating series of triangular numbers that goes like this: <math>1-3+6-10+15-21....</math> | ||
After finding the first few sums of the series, it becomes apparent that | After finding the first few sums of the series, it becomes apparent that | ||
Revision as of 21:24, 22 November 2015
Problem
The number 2017 is prime. Let 
. What is the remainder when 
is divided by 2017?
Solution
Note that 
. We have for 
![\[\dbinom{2014}{k}\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mod 2017\]](http://latex.artofproblemsolving.com/f/c/4/fc4df4114bc4b5853db7f0594b0bbd7bfe8016d5.png)
![\[\equiv (-1)^k\dbinom{k+2}{k} \mod 2017\]](http://latex.artofproblemsolving.com/7/2/9/729ef0f98878d86fed96a55dbb02a73ed0678d89.png)
![\[\equiv (-1)^k\dbinom{k+2}{2} \mod 2017\]](http://latex.artofproblemsolving.com/b/7/c/b7c09f87ee812d9d79beda60968eac083ee34880.png)
Therefore
![\[\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017\]](http://latex.artofproblemsolving.com/d/4/7/d47dfd0d24e879bd6c9e1a9605813cb6028178b1.png)
This is simply an alternating series of triangular numbers that goes like this: 
After finding the first few sums of the series, it becomes apparent that
![\[\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv -\left(\frac{n+1}{2} \right) \left(\frac{n+1}{2}+1 \right) \mod 2017 \textnormal{ if n is odd}\]](http://latex.artofproblemsolving.com/d/7/5/d75b035918520665cb9d7cdc0d1331d161bb85c8.png)
and
![\[\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv \left(\frac{n}{2}+1 \right)^2 \mod 2017 \textnormal{ if n is even}\]](http://latex.artofproblemsolving.com/0/5/2/052ac2dbf84700f78612633b43c921b35cbd9a98.png)
Obviously, 
falls in the second category, so our desired value is
![\[\left(\frac{62}{2}+1 \right)^2 = 32^2 = \boxed{\textbf{(C)}\ 1024}\]](http://latex.artofproblemsolving.com/a/4/4/a44175c5e1f5de2bd10f00f3187f9ec2351a684b.png)
See also
| 2014 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 22 |
Followed by Problem 24 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
