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First of all i have created a sequence of remainder of an expression just like below

  1. $50^{1}$ 50% 11 =6(remainder)
  2. $50^{2}$ 6*50% 11 =6*6%11=36%11=3(remainder)
  3. $50^{3} $ 3*50%11 =3*6%11=20%11=7(remainder)
  4. $50^{4}$ 7*50% 11 =7*6%11=42%11=9(remainder)
  5. $50^{5} $ 9*50% 11 =9*6%11=54%11=10(remainder)

  6. $50^{6} $ 10*50% 11=10*6%11=60%11=5(remainder)

  7. $50^{7} $ 5*50% 11=5*6%11=30%11=8(remainder)

  8. $50^{8} $ 8*50% 11=8*6%11=48%11=4(remainder)
  9. $50^{9} $ 4*50% 11=4*6%11=24%11=2(remainder)
  10. $50^{10} $ 2*50% 11=2*6%11=12%11=1(remainder)

Then I turned an expression

into ${{50^{51}}^{52}}$ which is ${50}^{2652}$

${50}^{2652}$ $\div$ 11 = ${{50}^{10}}^{265}$ $*$ 50 $*$ 50 $\div$11 =1$*$ 6$*$ 6$\div$11=36$\div$11=3(remainder).

indeed the remainder would be 3 but answer for the same question in the book is 6.

so clarify this if i m correct or wrong.

avk avk
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    You've already asked a similar question once, and already made the mistake of assuming $a^{(b^{c})}=(a^b)^c$ once. Now, you have not responded to any of the comments and answers there, telling you about this, so the real question is - do you even bother to read the answers to your questions??? – barak manos Nov 16 '16 at 18:46
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    And strangely enough, you've opened a different account with the exact same username, specifically for this question. I must say that you're kind of (ab)using this website in very bad/poor manner (for your own good, as well as for the good of those who spend their efforts trying to assist you). – barak manos Nov 16 '16 at 18:51
  • Exactly,@barakmanos I respect your comment .one final thing i wanna know how to treat this kind of exponent ${{50^{51}}^{52}}$. – avk avk Nov 16 '16 at 18:56
  • No way sir, i am sorry for the inconveniece but this question really sucks me. @barakmanos – avk avk Nov 16 '16 at 18:59
  • Also asked at http://mathoverflow.net/q/254830/12357 where it was closed. – JRN Nov 17 '16 at 04:12

2 Answers2

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By Euler's theorem, since $\gcd(11,50)=1$, we get $$50^{51^{52}}\equiv 50^{51^{52}\pmod {10}}\pmod{11}$$

$$\equiv 50^{1^{52}}\equiv 50^{1}\equiv 50\equiv 6\pmod{11}$$

user236182
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$51^{52}\equiv_{10} 1^{52}\equiv_{10} 1$ or $51^{52}=10k+1$ for some $k$.

Therefore, $$50^{51^{52}}=50^{10k+1}=50*(50^{10})^k.$$ By Fermat's theorem: $a^{p-1}\equiv_{p} 1$ for $a$ and prime $p$.

Set $a=50$ and $p=11$ in order to get: $50^{10}\equiv_{11} 1$.

Therefore, $(50^{10})^k\equiv_{11} 1$ and finally

$$50^{51^{52}}=50^{10k+1}\equiv_{11} 50\equiv_{11} 6.$$

alans
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