Find $r$ for $3^{574} + 26^{1347}\equiv r\mod 5$ Even though I'm looking to solve the problem, could anyone also point me out any sort of "strategy" to solve similar exercises? 
I know the things I must make use of most are the congruence properties and Fermat's little theorem, but I have a hard time applying them and finding out how to solve a problem.
Any help is greatly appreciated.
 A: Start with $3^2 \equiv -1 \mod 5$ and $26 \equiv 1 \mod 5$.
A: HINT:
As I've commented in your previous Question, dealing $1$ or $−1$ is the easiest to in case modulus operation, our target should be to reach at $\pm1$
In general, we can use Fermat's Little Theorem when modulus is prime like $5$
and Carmichael Function or Euler's Totient Theorem for composite modulus.
But other trick as the one below often come handy 
$\displaystyle26\equiv1\pmod5\implies 26^n\equiv1^n$ for integer $n\ge0$
and $\displaystyle3^2=9\equiv-1\pmod5$ and $574=2\cdot287$
A: You can also do it this way ,
$3^4 \ \equiv \ 1 \ (mod \ 5)$ 
$3^{572} \ \equiv \ 1 \ (mod \ 5)$ 
$3^{574} \ \equiv \ 4 \ (mod \ 5) \ \ \ \ \ \ \ $ (1)
$ 26 \ \equiv \ 1 \ (mod \ 5)$ 
$ 26^{1347} \ \equiv \ 1 \ (mod \ 5)$ 
Now you can add the congruences 
$ 3^{574} \ + \ 26^{1347} \ \equiv \ 4 \ + \ 1 \ (mod \ 5)$ 
$ 3^{574} \ + \ 26^{1347} \ \equiv \ 0 \ (mod \ 5)$ 
Alternatively , for positive integer n ,
$3^{4n - 2}$ ends in 9 , 574 = 4*144 - 2
$26^n $ ends in 6 
So their sum ends in 5
