Efficient way to find the remainder when $2001+ 2002+ 2003+ … + 2015+ 2016$ is divided by $2017$? I can think of a couple of ways

*

*Notice that unit digit of the first $8$ terms can be added in the last $8$ terms to make them $2017$. Now add the first $8$ terms without their unit digits (i.e. $2000*8$) and find a remainder on that.

*Sum of AP series = $4017*8$, now find the remainder

But both would take a lot of computation.
Is there any more efficient way of doing it manually?

This is equal to:
2016(2017)(1/2)-(2000)(2001)(1/2)
2016(2017)(1/2)≡0(mod 2017)
2000(2001)(1/2)=(1000)(2001)≡(1000)(-16)≡-16000 (mod 2017)
-16000 (mod 2017)≡-136(mod 2017)≡1881(mod 2017)
