What is the remainder when
$\Large 1^{2013}+2^{2013}+...+2012^{2013}+2013^{2013}$
is divided by $\Large 2014$

Question

What is the remainder when
$\Large 1^{2013}+2^{2013}+...+2012^{2013}+2013^{2013}$
is divided by $\Large 2014$

ganeshie8 · Answer

hint :  $1^{2003} +  2013^{2013} \equiv 1^{2003} +  (-1)^{2013}  \equiv  0 \mod  2014    $

ganeshie8 · Answer

group and $a$ and $2014-a$ bases, and notice that the sum of each group produces $0$   :


$$(1^{2013} +  2013^{2013}) + (2^{2013} + 2012^{2013}) + \cdots    1012^{2013}$$

$$(1^{2013} +  (-1)^{2013}) + (2^{2013} + (-2)^{2013}) + \cdots    1012^{2013} \mod 2014$$

$$(0) + (0) + \cdots    1012^{2013} \mod 2014$$

$$  1012^{2013} \mod 2014$$

anonymous · Answer

$\Large \color{red}{The\ solution:}$
Note that $\large k^{2013}\equiv -(2014-k)^{2013}\ (mod\ \ 2014) $ (this is just a fancy general way to state the next two statements and so on). Thus,
$\large 1^{2013}+2013^{2013} \equiv 0\ (mod\ \ 2014)$
$\large 2^{2013}+2012^{2013} \equiv 0\ (mod\ \ 2014)$
and so on.
After all this cancellation, we are left with finding $\large 1007^{2013}\ (mod\ \ 2014)$
Since $\large 1007^{2013} \equiv 1\ (mod\ \ 2)$and  $\large 1007^{2013} \equiv 0\ (mod\ \ 1007)$,we conclude that
$\Large 1007^{2013} \equiv 1007\ \ \ \ (mod\ 2017)$

ganeshie8 · Answer

nice <3

anonymous · Answer

:D

anonymous · Answer

But, I got a mistake: 
$\Large 1007^{2013} \equiv 1007\ \ \ \ (mod\ 2014)$

anonymous · Answer

what is 0 mod means

ganeshie8 · Answer

haha i see you have fast forwarded 3 years :)

ganeshie8 · Answer

$\large   a \equiv   0 \mod m$

means

$a$  is divisible by $m$

ganeshie8 · Answer

A more accurate definition :

$\large a \equiv   b \mod  m$

means 

$a-b$ is divisible by $m$

anonymous · Answer

$ \color{green}{Fermat's\ little\ theorem}$ tells us that if $\large gcd(a,b)=1$ then $\large a^{p-1} \equiv 1\ (mod\  p)$ [the $\large p$ here is a prime number].