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?

Welcome to MSE:
$2001 \equiv -16 \mod (2017)$
$2002 \equiv -15 \mod (2017)$
$2003 \equiv -14 \mod (2017)$
$\cdot$
$\cdot$
$\cdot$
$2016 \equiv -1 \mod (2017)$
Summing these relations we get:
$2001+2002+2002+2003+ . . . 2016 \equiv-\frac{16(1+16)}2=-136 \mod(2017)\equiv 1881 \mod(2017)$
