Difference between revisions of "2014 AMC 12B Problems/Problem 23"

Revision as of 12:26, 22 February 2014

Problem

The number 2017 is prime. Let $S = \sum \limits_{k=0}^{62} \dbinom{2014}{k}$ . What is the remainder when $S$ is divided by 2017?

$\textbf{(A) }32\qquad \textbf{(B) }684\qquad \textbf{(C) }1024\qquad \textbf{(D) }1576\qquad \textbf{(E) }2016\qquad$

Solution

Note that $2014\equiv -3 \mod2017$ . We have for $k\ge1$ $\dbinom{2014}{k}\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mod 2017$ $\equiv (-1)^k\dbinom{k+2}{k} \mod 2017$ $\equiv (-1)^k\dbinom{k+2}{2} \mod 2017$ Therefore $\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=1}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017$ This is simply an alternating series of triangular numbers that goes like this: $1-3+6-10+15-21....$ After finding the first few sums of the series, it becomes apparent that $\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv -\left(\frac{n+1}{2} \right) \left(\frac{n+1}{2}+1 \right) \mod 2017 \textnormal{ if n is odd}$ and $\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv \left(\frac{n}{2}+1 \right)^2 \mod 2017 \textnormal{ if n is even}$ Obviously, $62$ falls in the second category, so our desired value is $\left(\frac{62}{2}+1 \right)^2 = 32^2 = \boxed{\textbf{(C)}\ 1024}$

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 23: / Line 23: @@
 <cmath>\left(\frac{62}{2}+1 \right)^2 = 32^2 = \boxed{\textbf{(C)}\ 1024}</cmath>
+== See also ==
 {{AMC12 box|year=2014|ab=B|num-b=22|num-a=24}}
 {{MAA Notice}}

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Difference between revisions of "2014 AMC 12B Problems/Problem 23"

Revision as of 12:26, 22 February 2014

Problem

Solution

See also