elementary number theory - What will be the remainder when $2^{31}$ is divided by $5$?

The question is given in the title:

Find the remainder when $2^{31}$ is divided by $5$.

My friend explained me this way:

$2^2$ gives $-1$ remainder.

So, any power of $2^2$ will give $-1$ remainder.

So, $2^{30}$ gives $-1$ remainder.

So, $2^{30}\times 2$ or $2^{31}$ gives $3$ remainder.

Now, I cannot understand how he said the last line. So, please explain this line.

Also, how can I do this using modular congruency?

Answer

Your friend is wrong in one statement. Indeed you have $2^2 \equiv -1 \pmod 5$. But this implies that $(2^2)^n \equiv (-1)^n \pmod 5$ not that any power of $ 2^2 $ will give $ -1 $ remainder.

However you get $$(2^2)^{15} = 2^{30} \equiv (-1)^{15} = -1 \pmod 5.$$ And therefore $2^{31} \equiv 2 \cdot 2^{30} = -2 = 3 \pmod 5$.