Fooling around with factorials - with Thanks to Chew-Seong Cheong

Number Theory Level 4

Let $S(n)$ denote the number of digits of $n$ in decimal representation. What is the largest integer $n$ satisfying $S(n!) +1= n$ ?

The answer is 21.

1 solution

Chew-Seong Cheong
Aug 22, 2015

Thanks, Alisa. I guess I have to provide a solution. But I can only offer a numerical one.

We note that $S(n) = \left \lfloor \log_{10} n \right \rfloor + 1$ , where $\lfloor x \rfloor$ is the greatest integer function. Therefore, $S(n!) + 1 = \left \lfloor \log_{10} n! \right \rfloor + 2$ . The following spreadsheet shows the that there are four $n$ 's that satisfy the equation. And the largest one is $\boxed{21}$ . For $n>21$ , $S(n!) + 1 > n$ .

Moderator note:

It is unlikely to have a non-numerical approximation solution to this problem.

One could use stirling's formula as bound.

Xuming Liang - 5 years, 9 months ago

I actually thought of that but it work probably involved similar computation work.

Chew-Seong Cheong - 5 years, 9 months ago

Fooling around with factorials - with Thanks to Chew-Seong Cheong

The answer is 21.

1 solution

Moderator note:

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