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

## 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 $k\ge1$ $$\dbinom{2014}{k}\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mod 2017$$ $$\equiv (-1)^k\dbinom{k+2}{k} \mod 2017$$ Therefore $$\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017$$ 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 $$\left(\frac{62}{2}+1 \right)^2 = 32^2 = \boxed{\textbf{(C)}\ 1024}$$

### Sidenote

Another way to finish, using the fact that $\dbinom{k+2}{2} = 1 + 2 + \dots + (k+1)$: \begin{align*} \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} &\equiv \sum \limits_{k=1}^{63}(-1)^k (1 + 2 + \dots + k) \\ &\equiv 1 - (1+2) + (1+2+3) - (1+2+3+4) + \dots + (1 + \dots + 63) \\ &\equiv 1 + 3 + 5 + \dots + 63 \\ &\equiv \boxed{1024} \mod 2017 \end{align*}