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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?

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user2409011
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    Presumably, the problem meant $50^{\left(51^{52}\right)}$ and not $\left(50^{51}\right)^{52}$? – Thomas Andrews Aug 07 '13 at 16:18
  • Yes, the first one 50^(51^52). I am new, I do not know how to raise the power properly as you did. – user2409011 Aug 07 '13 at 16:22
  • Yeah, figured that out, MJD. I think if you calculated $50^{\left(51^{52}\right)}$ you'd get the answer is $6$. You got the correct answer for $\left(50^{51}\right)^{52}$ – Thomas Andrews Aug 07 '13 at 16:25
  • (From an advanced theory, $51^{52}\equiv 1\pmod{10}$ and thus $50^{51^{52}} \equiv 50^{1}\equiv 6\pmod {11}$. But if you don't know Fermat's Little Theorem, that isn't going to help you. – Thomas Andrews Aug 07 '13 at 16:28
  • @ThomasAndrews Please refer to the edited question details by MJD. I think I am not sure if it is 50^(51^52) or (50^51)^52. I have taken it as 50^(51^52). For this answer should be 6. But I am getting 3. – user2409011 Aug 07 '13 at 16:33
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    The reason you are getting $3$ is that you are starting with $3$, for some reason I have not figured out. Note that $50\equiv 6\pmod{11}$. So you should have been taking powers of $6$. But there is no work to do, since $51^{52}\equiv 1\pmod{10}$. – André Nicolas Aug 07 '13 at 16:38
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    your mistake is in the first line, where you get $2652$, so you did $51*52$ instead of $51^{52}$ – Denis Aug 07 '13 at 16:47
  • @user2409011 No, you haven't taken it to be $50^{51^{52}}$, because $50^{51^{52}}\neq 50^{51\cdot 52}$. You've computed $\left(50^{51}\right)^{52}$. – Thomas Andrews Aug 07 '13 at 17:13

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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.

Rick Decker
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