Finding the remainder of a factorial raised to more factorials when divided by 11

Question

Find the remainder when $$3!^{{{5!}^{...}}^{2013!}}$$ is divided by 11.

My answer is 5, but the answer given by the answer sheet is 1. How did this happen?

I tried getting the remainders when several powers of 6 is divided by 11 since $3!=6.$ I actually got a pattern: 6, 3, 7, 9, 10, 5 and restarts to 6. Since the immediate exponent raising 6 is 120 which is divisible by 6, I reasoned that the remainder should be 5.

I know my solution does not make sense. I really do not know what to do. Can anyone help?

Answer 1

Your approach is fine, you just made a calculation mistake: $6\cdot 5=30\equiv 8\bmod 11$, not $6$.

You can also consider Fermat's Little Theorem: $a^{p-1}\equiv 1\bmod p$ for any prime number $p$ and any integer $a$ relatively prime to $p$. Since is $6$ is relatively prime to $11$, we have $6^{10}\equiv 1\bmod 11$, and therefore $6$ raised to any power divisible by $10$ is also equivalent to $1$ modulo $11$.

Answer 2

Hint: The key insight is that $(5!)^{\cdots}$ is a multiple of $10$.