Sum Of 2013 Powers

Note that $k^{2013}\equiv -(2014-k)^{2013}\pmod{2014}$ (this is just a fancy general way to state the next two statements and so on). Thus, $1^{2013}+2013^{2013}\equiv 0\pmod{2014}$ $2^{2013}+2012^{2013}\equiv 0\pmod{2014}$ and so on.

After all this cancellation, we are left with finding $1007^{2013}\pmod{2014}$ . Since $1007^{2013}\equiv 1\pmod{2}$ and $1007^{2013}\equiv 0\pmod{1007}$ , we conclude that $1007^{2013}\equiv \boxed{1007}\pmod{2014}$

For n odd:

${ a }^{ n }+{ b }^{ n }=({ a }+{ b })({ a }^{ n-1 }-{ a }^{ n-2 }{ b }+...-{ a }{ b }^{ n-2 }+{ b }^{ n-1 })$

As 1+2013=1014, 2+2012=2014....:

${ 1 }^{ 2013 }+{ 2 }^{ 2013 }+...+2012^{ 2013 }+{ 2013 }^{ 2013 }=$

${ 1 }^{ 2013 }+{ 2013 }^{ 2013 }+{ 2 }^{ 2013 }+{ 2012 }^{ 2013 }+...+{ 1005 }^{ 2013 }+{ 1008 }^{ 2013 }+{ 1007 }^{ 2013 }$ $\equiv { 1007 }^{ 2013 }\quad (mod\quad 2014)$

$1007\times 1007=1014049\equiv { 1007 }\quad (mod\quad 2014)\quad$ $\longrightarrow \quad { 1007 }^{ 2013 }\equiv { 1007 }\quad (mod\quad 2014)$

Note that 2014 is the product of three primes 2, 19, 53, the latter two being regular. This and the fact that 2013 is odd implies that the sum of the first n 2013th powers = n^2(n+1)^2 p(n) where p(n) has rational coefficients with no factors of 19 or 53 in their denominator (regular primes do not divide Bernouilli denominators). Thus the answer is an odd multiple of 19 53=1007.

Outline: Since $2014=2×19×53$ , we will calculate $x^{2013} \pmod {2}, \pmod{19},\pmod{53}$ respectively and combine them to $x^{2013} \pmod{2014}$ in the end.

Mod 2 Odd x give 1, even x give 0. There are 1007 odd numbers in the range 1..2013, which is odd, so $\sum_{x=1}^{2013}x^{2013}=1 \pmod 2$

Mod 19 $2013=105×19+18$

Taking x mod 19 $\sum_{x=1}^{2013}x^{2013} \equiv 105( \sum_{x=1}^{19}x^{2013})+ \sum_{x=1}^{18}x^{2013}$ .

Observe that for odd t: $(p-a)^t \equiv (-a)^t \equiv -a^t \pmod p$ so that for any odd power t and any odd prime p: $1^t+2^t+...+(p-2)^t +(p-1)^t \equiv 0$ .

Also observe $p^t \equiv 0^t \equiv 0 \pmod p$ .

So $\sum_{x=1}^{2013}x^{2013} \equiv 105×0+0 \equiv 0 \pmod {19}$

Mod 53 Here too, 2013 is 1 short of a multiple of 53. (Of course, because 53 is a factor of 2014.)  Along the same lines we find $\sum_{x=1}^{2013}x^{2013} \equiv 0 \pmod {53}$

Conclusion Combining the modulo information, the sum is an odd multiple of 19 and of 53, which must be $19×53=\boxed{1007}$ .

This was a neat problem and not as immediate as the question it was based on. Took quite a lot of observation.