Problem:
The number 2017 is prime. Let S=∑k=062​(2014k​). What is the remainder when S is divided by 2017 ?
Answer Choices:
A. 32
B. 684
C. 1024
D. 1576
E. 2016
Solution:
Let n=(2014k​). Note that 2016⋅2015≡(−1)(−2)=2 (mod2017) and 2016⋅2015⋯(2015−k)≡(−1)(−2)⋯(−(k+2))=(−1)k(k+ 2)!(mod2017). Because n⋅k!⋅(2014−k)!=2014!, it follows that
n⋅k!⋅(2014−k)!⋅((2015−k)⋯2015⋅2016)⋅2≡2014!⋅2015⋅2016⋅(−1)k(k+2)!(mod2017)​
Thus
2n⋅k!⋅2016!≡(−1)k(k+2)!⋅2016!(mod2017)
Dividing by 2016 ! â‹…k !, which is relatively prime to 2017 , gives
2n≡(−1)k(k+2)(k+1)(mod2017)
Thus n≡(−1)k(k+22​)(mod2017). It follows that
S≡k=0∑62​(−1)k(k+22​)=1+k=1∑31​((2k+22​)−(2k+12​))=1+k=1∑31​(2k+1)=322=1024​(mod2017).​
The problems on this page are the property of the MAA's American Mathematics Competitions