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$$(((3!)^{5!})^{7!})^{9!...}$$ when divided by 11 what will be the reminder?

Hint is appreciated


Sorry I do not know how to start this problem, so I have not shown my efforts!

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  • $\begingroup$ Do you know of any tools for reducing a power modulo an integer? $\endgroup$
    – anon
    May 14, 2015 at 5:30
  • $\begingroup$ @anon no I have no idea! :( $\endgroup$
    – Freddy
    May 14, 2015 at 5:32
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    $\begingroup$ You've never seen Fernat's little theorem or Euler's theorem before? If not, where did you get the problem from? (Because certainly any class, notes or textbook would cover tools for how to solve this problem before giving it...) $\endgroup$
    – anon
    May 14, 2015 at 5:35
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    $\begingroup$ What exactly do the "..." mean? $\endgroup$
    – 2'5 9'2
    May 14, 2015 at 5:42
  • $\begingroup$ It's implying that the sequence of residues $3!$, $3!^{5!}$, $(3!^{5!})^{7!}$, $((3!^{5!})^{7!})^{9!}$, $\cdots$ stabilizes presumably. This should have been stated up-front in the question of course. $\endgroup$
    – anon
    May 14, 2015 at 5:46

3 Answers 3

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Now we see that $(3!, 11)=(6,11)=1$. Hence by Fermat's theorem we have $(3!)^{10}\equiv 1[11]$ and hence $(3!)^{10m}\equiv 1[11]$. Moreover $10$ divides $5!$ so that $(((3!)^{5!})^{7!})^{...}\equiv 1[11]$.

Required reminder is 1.

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$$(3!,11)=1$$ By Fermat's Little Theorem, $$3^{11-1}\equiv1\pmod{11}$$

and $10|5!$

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see, the above number can be written like this

$(3!)^{5!*7!*9!*\dots}$

first look at the power of 3!, it will be a number ending with a lot of zeroes let that number be n hence $n(\mod 10)=0$ $\implies n=10*k$, $k$ is a positive integer Now, we have to find $6^n (\mod 11)=$?

by applying Fermat's little theorem $6^{10}(\mod 11)=1$ $=R[(6^{10})^k/11]$ $=R[1^k/11]$ $=1$.

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