TITLE: Doubt regarding divisibility of the expression: $1^{101}+2^{101} \cdot \cdot \cdot +2016^{101}$
QUESTION [11 upvotes]: In an interesting contest question I recently encountered, I chanced upon a question I couldn't solve.

$$\sum^{2016}_{i=1}i^{101}$$ is divisible by: (a)2014 (b)2015 (c)2016 (d)2017

How would I go about solving this question in a no-nonsense fashion?

REPLY [5 votes]: On the one hand,
$$
\sum^{2016}_{i=1}i^{101}
=\sum^{1008}_{i=1}i^{101}
+\sum^{1008}_{i=1}(2017-i)^{101}
$$
The two sums cancel mod $2017$ because $101$ is odd.
On the other hand,
$$
\sum^{2016}_{i=1}i^{101}
=\sum^{1007}_{i=1}i^{101}
+\sum^{1007}_{i=1}(2016-i)^{101}
+1008^{101}
$$
The two sums cancel mod $2016$ because $101$ is odd.
Finally, $1008^{101}=\dfrac{2016}{2}\cdot 1008^{100}$ is a multiple of $2016$ because $1008$ is even.
You can apply the same argument for $2014$ and $2015$ and you'll get lots of cancellation but a nonzero remainder.