Finding out the remainder if $f(2001) + f(2002) + f(2003) + ......+ f(2014) + f(2015)$ is divided by 3, where f(y) = $y^\text{y}$.

Question

Source: Regional math olympiad of BD.

Define the function $f(y) = y^y$. Let $a=f(2001) + f(2002) + f(2003) + ..... f(2014) + f(2015)$. If $a$ is divided by $3$, then what is the remainder of the above condition?

I couldn't figure out the total sum of $2001^\text{2001} + 2002^\text{2002} +.....+ 2014^\text{2014}+ 2015^\text{2015}$. Let's suppose that the remainder is 'x'. So we can describe the term as $a$ $\equiv$ $x\mod3$. Then, we have to determine the value of x but I failed to calculate te value of 'a'. So, I need a geometrical approach to solve the problem.

Note: I need only the value of 'a' and the process of finding it out.

Excuse my error and thank you in advance.

Answer 1

Just like Christopher Marley's comment, what you got there is the same as: $$0^{2001}+1^{2002}+2^{2003}+0^{2004}+...+2^{2015} \equiv 13 \equiv 1 (\mod 3)$$