what is remainder when $(((3!)^{5!})^{7!})^{9!...}$ is divided by 11

Question

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

Hint is appreciated

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

Answer 1

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.

Answer 2

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

and $10|5!$

Answer 3

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 $6^{10}(\mod 11)=1$ $=R[(6^{10})^k/11]$ $=R[1^k/11]$ $=1$.