<s>[INST] What is wrong with my solution of finding remainder of $50^{(51^{52})}$ when divided by 11? I used the following method using remainder theorem. (I used method from here: Find the remainder of $128^{1000}/153$.)
$$\begin{align}
  (50^{{51}^{52}})/11 & = (50^{2652})/11  \implies \\
(6^{2652})/11 & = (36^{1326})/11 \implies \\
(3^{1326})/11 &= (27^{442})/11 \implies \\
(5^{442})/11 & = ({25}^221)/11 \implies \\
(3^{221})/11
\end{align}$$
Now, 
3^1/11 implies remainder 3

3^2/11 implies remainder 9

3^3/11 implies remainder 5

3^4/11 implies remainder 4

3^5/11 implies remainder 1

3^6/11 implies remainder 3

So modulo 11, powers of 3 repeat with period 5
so 221 is 1 modulo 5
so $50^{{51}^{52}}/11$ should become finally $3^1/11$ implies remainder 3 (my answer). Actual answer is 6.
What is wrong with this method?
 [/INST] Since $50\equiv 6\pmod{11}$, you should start by looking at powers of $6$ modulo $11$. If you do, you'll find that $50^{10}\equiv6^{10}\equiv 1\pmod{11}$. That means that when computing $50^x\pmod{11}$ we can discard all multiples of $10$ in $x$ and just concentrate on the remainder mod $10$. For example, if we needed to know $50^{672}\pmod{11}$ we could write
$$
672=67\cdot10+2
$$
and so
$$
50^{672}= 50^{67\cdot 10+2}= (50^{67\cdot 10})(50^2)= (50^{10})^{67}(50^2)\equiv(1)^{67}(50^2)\equiv 50^2\pmod{11}
$$
To recap, we now know
$$
50^{\,x}\equiv 50^{\,x\ \bmod{10}}\pmod{11}
$$
In this problem, we have $x=51^{52}$ which is good news since modulo $10$ we have
$$ 
51^{52}\equiv 1^{52} \equiv 1 \pmod{10}
$$
Putting the two display lines above together, we have
$$
50^{51^{52}} \equiv 50^{\,51^{52}\ \bmod {10}}\equiv50^1\equiv 6^1\equiv 6\pmod{11}
$$
As Thomas suggested in his comment to your post, there's a cleaner way to do this if you know Fermat's Little Theorem.
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