The textbook gives the answer as:
By Fermat’s little theorem, we know that $7^{10} ≡ 1 \pmod{11}$, and so $(7^{10})^k ≡ 1 \pmod{11}$, for every positive integer $k$.
Therefore, $7^{222} = 7^{22\cdot10 + 2} = (7^{10})^{22}7^2 ≡ (1)^{22}\cdot 49 ≡ 5 \pmod{11}$.
Hence, $7^{222}\mod{11} = 5$.
The part that I do not understand is $(7^{10})^{22}7^2 ≡ (1)^{22}\cdot 49 ≡ 5 \pmod {11}$.
Can someone please explain to me why can we say:
$(7^{10})^{22}7^2 ≡ (1)^{22}\cdot49$
Best Answer
If we have $$a \equiv b \pmod{c}$$ then we can also state that $$a^n \equiv b^n \pmod{c}$$ in your case, take $a= 7^{10}$, $b=1$ and $c=11$. Noting that you can write $$7^{222} = (7^{10})^{22 + 2}$$ the result follows. Since $$(7^{10})^{22} = 1^{22} \pmod{11}$$
The reason that we can say $a \equiv b \pmod{c} \implies a^n \equiv b^n \pmod{c}$ is from using the well known result that if $a \equiv b \pmod{c} \implies ad \equiv be \pmod{c}$ then taking $d = a$ and $e=b$ yields $a^2 \equiv b^2\pmod{c}$ and we continue by induction on $n$ to get the result.