What is $1111^{2222}\pmod9$?

Question

Find the remainder upon dividing $1111^{2222}$ by $9$.

I'd like to first confirm that my solution is correct. Since $\varphi(9)=6$ and $2222=6\times370+2$, we have by Euler's theorem

$$1111^{2222}\equiv1111^2\mod9$$

From $1111=9\times123+4$, it follows that

$$1111\equiv4\mod9\implies1111^{2222}\equiv1111^2\equiv16\equiv7\mod9$$

Is this solution correct? (The reasoning, that is, not the answer which I've verified with a calculator.) This is the first of several exercises related to Fermat's little/Euler's theorem and would just like to know if anything is funky about what I applied or at which point I applied it.

Second, more out of curiosity: Are there any other (perhaps quicker?) solutions?

Answer 1

Your solution is correct. A quicker solution maybe which is somewhat equivalent would be:

$$ 1111^{2222}\equiv 4^{2222} \pmod{9} $$

since $1111\equiv 10^3+10^2+10+1\equiv 1+1+1+1\equiv 4 \pmod{9}$ as told in the comments.

We know that $4^3 \equiv 64\equiv 1 \pmod 9$ and $$4^{2222}\equiv 4^{3\cdot 740 + 2}\equiv (4^3)^{740}\cdot 4^2\equiv 1^{740}\cdot 16\equiv 16 \equiv 7 \pmod 9$$

Answer 2

Your answer and your work are correct. You can use Wolfram Alpha to check your answer.

As for quicker solutions, none come to mind, but there are tricks among the mathematics of congruences that can help, such as the rule mentioned in the comments by @Daniel Schepler that a number is congruent to the sum of its digits mod $9$.