Remainder? Part-2

Just pair the terms up, and use the fact that $a+b|a^n+b^n$ for odd positive $n$ (NB the notation $x|y$ means " $x$ divides $y$ "):

$2013|1^{2013}+2012^{2013}$

$2013|2^{2013}+2011^{2013}$

$2013|3^{2013}+2010^{2013}$

...etc. So the whole sum is a multiple of $2013$ , and the remainder is $\boxed0$ .

For positive integers $a$ , $b$ , and $n$ , where $n$ is odd, we can write $\displaystyle a^n+b^n = (a+b) \left(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + ab^{n-2} + b^{n-1}\right)$ . This means that $a^n+b^n$ is a multiple of $a+b$ . Similarly, for $S = 1^{2013} + 2^{2013} + 3^{2013} + \cdots + 2012^{2013}$ , we can pair the end terms together $1^{2013} + 2012^{2013}$ , $2^{2013} + 2011^{2013}$ , $3^{2013} + 2010^{2013}$ , $\cdots, 1006^{2013} + 1007^{2013}$ . Since every one of the 1006 pair sums is divisible by $k+2013-k=2013$ , $S$ is also divisible by $2013$ and the remainder is $\boxed 0$ .