The problem:

> 
> Find the remainder of $3^{41}+7^{41}$
>  when divided by $13$.
> 
> 
> 

My approach is by utilizing the cyclicity of remainders for examples $3^1,3^2,3^3,3^4,3^5 \text{ and }3^6$ when divided $13$ gives $3,9,1,3,\text{ and }9 $ respectively,so we can see that the cycle repeats after $3$ steps.Hence the remainder of $3^{41}$ when divided by $13$ will be same as $3^2$ i.e $9$

For $7$ the repetition will occur in $7^{13}$ which means the cycle is of step $12$, hence it will be same as $7^5$ which gives $11$.

So the final answer is remainder of $(11+9)$ divided by $7$ which is $7$.

But as you might noticed, the computation of step for $7$ is quite a bit tedious (we have to check till $7^{13}$).

But, considering the fact that this problem is been taken from a exam which requires solution within $2$ mints, I am quite sure that there might be an easier method for finding the answer but I can't think of any other easier method, so could anybody suggest me a tricky method?