Question

数 $2017$ 是质数。设 $S = \sum_{k=0}^{62} \binom{2014}{k}$ 。 $S$ 除以 $2017$ 的余数是多少？

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

$32$

$684$

$1024$

$1576$

$2016$

答案：C

知识点：组合模运算裂项相消

难度评级：2560

解答：

在模 $2017$ 下，恒等式 $\binom{2014}{k} \cdot k! \cdot (2014-k)! = 2014!$ 结合 $2016 \cdot 2015 \cdots (2015-k)$ $\equiv (-1)^k (k+2)!$ ，得到因此 $\binom{2014}{k} \equiv (-1)^k \binom{k+2}{2}$ 。 $\begin{gathered} 2\binom{2014}{k} \equiv (-1)^k \\ {}\cdot (k+2)(k+1) \pmod{2017}, \end{gathered}$

于是 $\begin{gathered} S \equiv \sum_{k=0}^{62} (-1)^k \binom{k+2}{2} \\ = 1 \\ {}+ \sum_{k=1}^{31}\left[\binom{2k+2}{2} - \binom{2k+1}{2}\right] \\ = 1 + \sum_{k=1}^{31}(2k+1). \end{gathered}$

剩下的和为 $3 + 5 + \cdots + 63 = 1023$ ，所以 $S \equiv 1 + 1023$ $= 1024 \pmod{2017}$ 。

所以正确答案是 C。

Working modulo $2017,$ the identity $\binom{2014}{k} \cdot k! \cdot (2014-k)! = 2014!$ together with $2016 \cdot 2015 \cdots (2015-k)$ $\equiv (-1)^k (k+2)!$ leads to $\begin{gathered} 2\binom{2014}{k} \equiv (-1)^k \\ {}\cdot (k+2)(k+1) \pmod{2017}, \end{gathered}$ so $\binom{2014}{k} \equiv (-1)^k \binom{k+2}{2}.$

Then $\begin{gathered} S \equiv \sum_{k=0}^{62} (-1)^k \binom{k+2}{2} \\ = 1 \\ {}+ \sum_{k=1}^{31}\left[\binom{2k+2}{2} - \binom{2k+1}{2}\right] \\ = 1 + \sum_{k=1}^{31}(2k+1). \end{gathered}$

The remaining sum is $3 + 5 + \cdots + 63 = 1023,$ so $S \equiv 1 + 1023$ $= 1024 \pmod{2017}.$

Thus, the correct answer is C.

2014 AMC 12B 第 23 题

答案：C

难度评级：2560

解答：

其他年份的第 23 题