Find the rest of the division when $23^{84292}$ is divided by $7$, is the procedure and the result correct?

Question

I want to know if the procedure I have followed in order to get the result for the next problem is correct. The problem is this:

Find the rest of the division when $23^{84292}$ is divided by $7$.

As $7$ is a prime number and $23$ is not divided by $7$ Small Fermat's Theorem can be used. So, I know,

$23^{6}\equiv1\pmod7$ and $23^{6\cdot 14048} = 23^{84288} \equiv1\pmod7$

Then,

$23^{84288}\cdot 23^{4}=23^{84292}$

And,

$23 \equiv 2\pmod7$

So, finally,

$23^{84288}\cdot 23\cdot 23\cdot 23\cdot 23=23^{84292} \equiv 16\pmod7$

So,

$23^{84292} \equiv 2\pmod7$

Is the procedure and the result all correct?

Answer 1

Yes, that is correct. $\tiny{\text{(Converting to an answer.)}}$

By the way, you could have saved a little bit of work by noting $ 23 \equiv 2 \pmod 7 $ from the start. This way, you could write $2$ for every $23$.

Answer 2

We can reduce further as follows: $$\text{As Ahaan has identified, }23\equiv2\pmod7,$$

$$\implies 23^{84292}\equiv2^{84292}\pmod7$$

$$\text{Again as }2^3=8\equiv1\pmod 7\text{ and } 84292\equiv1\pmod3,$$

$$2^{84292}\equiv2^1\pmod7$$

Answer 3

Notice $23^3 \equiv 2^3 \equiv 1 \pmod{7}$, and $84292 \equiv 1 \pmod{3}$. This means $$23^{84292} \equiv 23 \cdot 23^{84291} \equiv 23 \equiv \boxed{2} \pmod{7}$$