Problem on my calculator

$\displaystyle \begin{aligned} 9.99999999 \times 10^{99} & \geq n! & \small \color{#3D99F6} \text{Assuming the calculator has } \leq 9 \text{ significant digits} \\ 10^{100} & > n! & \small \color{#3D99F6} \text{Stirling's formula} \\ 10^{100} & > \sqrt{2 \pi n} \left( \frac ne \right) ^n & \small \color{#3D99F6} \text{Take the natural log of both sides} \\ 100 \ln 10 & > \ln \sqrt{2\pi} + \ln \sqrt{n} + n \ln n - n \ln e \\ 100 \ln 10 - \ln \sqrt{2\pi} & > \left(\frac 12 + n\right)\ln n - n \\ 229.34 & > \left(\frac 12 + n\right)\ln n - n \end{aligned}$

The LHS now contains more manageable numbers, so we can do some checking with Kalpok's calculator (:

Note that $e^4 < n < e^5$ :

$\displaystyle n = e^4 \Longrightarrow \left(\frac 12 + e^4\right) \cdot 4 - e^4 = 165.79 < 229.34 \\ n = e^5 \Longrightarrow \left(\frac 12 + e^5\right) \cdot 5 - e^5 = 596.15 > \ 229.34$

Clearly, $n$ must be closer to $e^4$ than to $e^5$ , so we make further approximations for $n$ . By testing some rational exponents of $e$ between $4$ and $5$ , we might find that $e^{4.25} = 70.1$ sets a very precise upper bound for $n$ :

$\displaystyle n = e^{4.25} \Longrightarrow \left(\frac 12 + e^{4.25}\right) \cdot 4.25 - e^{4.25} = 229.97 > 229.34$

Since $n$ is an integer, we test the integers $n \leq 70$ :

$\displaystyle \begin{aligned} n & = 70 \Longrightarrow \left(\frac 12 + 70\right) \ln 70 - 70 = 229.51 \ {\color{#D61F06}{\cancel{<}}} \ 229.34 \\ n & = 69 \Longrightarrow \left(\frac 12 + 69\right) \ln 69 - 69 = 225.27 \ {\color{#3D99F6}{<}} \ 229.34 \end{aligned}$

Hence, $n = \boxed{69}$ .

$-$

An alternate solution would be to find the number of digits in $n!$ :

$d(n!) = \big \lfloor \log_{10} n! \big \rfloor + 1$

We see that $n$ must be quite large for $\lfloor{n!\rfloor} = 10^{100}$ , so $(n + 1)!$ must have more digits than $n!$ . Since $d\left(10^{100}\right) = 101$ and $10^{100}$ is the smallest number with $101$ digits, we have

$\displaystyle \begin{aligned} d(n!) < 101 \leq d\bigg((n + 1)!\bigg) \tag{1} \end{aligned}$

Stirling's formula says that for any positive integer $n$ , we have the bounds $\sqrt{2\pi}\ n^{n+1/2}e^{-n} \le n! \le e\ n^{n+1/2}e^{-n}$

so for $\left \lfloor \log_{10} n! \right \rfloor + 1$ ,

$\Bigg \lfloor \log_{10} \sqrt{2\pi} + \left(n + \frac 12\right) \log_{10} n - n \log_{10} e \Bigg \rfloor + 1 \ \leq \ \big \lfloor \log_{10} n! \big \rfloor + 1 \ \leq \ \Bigg \lfloor \log_{10} e + \left(n + \frac 12\right) \log_{10} n - n \log_{10} e \Bigg \rfloor + 1$

Combining this result with our previous inequality, $(1)$ , and doing some checking, we find that $69!$ has $99 < 101$ digits and $70!$ has $101 \leq 101$ digits.

Hence, $n = \boxed{69}$ .