This is the problem I am currently working on
Find this value: $(7^3\bmod{23})^2\bmod{11}$
Here's my work:
$$\begin{align*} &(7^3\bmod{23})^2\bmod{11}\\ &64\bmod{11}=9 \end{align*}$$
This made sense to me because $64 = 11(5) + 9$.
Here is my instructor's answer
$$\begin{align*}(32^3\bmod{13})^2&\bmod{11}\\(32768\bmod{13})^2&\bmod{11}\\(8)^2&\bmod{11}\\12&\bmod{11}\\&1\end{align*}$$
We both reduced down to $64\bmod{11}$ but I don't understand why she reduced that down to $12\bmod{11}$. It's a lot more intuitive to me to just evaluate $64\bmod{11}$ and get $9$ as the answer. She got 1 as her final answer. Did I miss something?
Working your problem, $$\begin{align}(7^3 \bmod 23)^2 \bmod 11 &\equiv (7\cdot 49 \bmod 23)^2 \bmod 11 \\ &\equiv (7\cdot 3 \bmod 23)^2 \bmod 11 \\ &\equiv (21 \bmod 23)^2 \bmod 11 \\ &\equiv 21^2 \bmod 11 \\ &\equiv 441 \bmod 11 \\ &\equiv 1 \bmod 11 \\ \end{align}$$
Working a version pf your instructor's problem, since I'm almost certain you have changed the second modular base... although probably not the first, since there aren't any other primes $> 8$ that would result in $32^3 \equiv 8$... $$\begin{align}(32^3 \bmod 13)^2 \bmod 13 &\equiv (6^3 \bmod 13)^2 \bmod 13 \\ &\equiv (36\cdot 6 \bmod 13)^2 \bmod 13 \\ &\equiv (10\cdot 6 \bmod 13)^2 \bmod 13 \\ &\equiv (60 \bmod 13)^2 \bmod 13 \\ &\equiv 8^2 \bmod 13 \\ &\equiv 64 \bmod 13 \\ &\equiv 12 \bmod 13 \\ \end{align}$$