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 ! ⋅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