Find $r$ for $3^{574} + 26^{1347}\equiv r\mod 5$

Question

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.

Answer 1

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 $574=2\cdot287$

Answer 2

Start with $3^2 \equiv -1 \mod 5$ and $26 \equiv 1 \mod 5$.

Answer 3

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