What is the last non-zero digit of $((\dots(((1!)!+2!)!+3!)!+\dots)!+1992!)!$?

Question

What is the last non-zero digit of $((\dots(((1!)!+2!)!+3!)!+\dots)!+1992!)!$?

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Clarification of the given expression:

Let $A_1=(1!)!$

To get $A_2$, we add $2!$ to $A_1$ then we take the factorial of that expression. So, $A_2=((1!)!+2!)!$

To get $A_3$, we add $3!$ to $A_2$ then we take the factorial of that expression. So, $A_3=(((1!)!+2!)!+3!)!$

To get $A_4$, we add $4!$ to $A_3$ then we take the factorial of that expression. So, $A_4=((((1!)!+2!)!+3!)!+4!)!$

and so on.

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Now we need to find the last non-zero digit of $A_{1992}$. I am not asking for a solution, but asking for some hints how shall I begin.

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Your help would be appreciated. THANKS.

Answer 1

Let $a_n$ denote the last nonzero digit of $n!$; we need only consider $n\ge5$.

• Prove $a_n\in S:=\{2,\,4,\,6,\,8\}$.
• Prove $a_{10k+5}=a_{10k+14}$.
• Determine the order of the permutation $\sigma$ of $S$ for which $a_{10k+15}=\sigma(a_{10k+14})$.
• Find a fairly small upper bound $p$ on the period of $a_n$.
• Convince yourself $a_p,\,\cdots,\,a_{2p-1}$ is a period.
• Explain why the original problem's answer is $a_p$, then compute it.