Estimate the digits

Given any positive integer $n,$ the number of digits in $n$ is $\lfloor \log_{10}{n} \rfloor +1.$ We can compute the exact number of digits in $100!$ by taking the log base 10. We can then apply the log sum/product rule to re-write this value as a sum:

$\log_{10}{100!}=\sum\limits_{k=1}^{100}\log_{10}{k}$

The exact value can be obtained with a computer, but our goal is just to estimate. We obtain a minimum and maximum value this number could be. Make the following observations:

• For $k<10,$ $\log_{10}{k}<1.$
• For $10 \le k <100,$ $1 \le \log_{10}{k}<2.$
• For $k=100,$ $\log_{10}{k}=2.$

We can round down each $\log_{10}{k}$ to the nearest integer to obtain a minimum estimated sum:

$(0\times 9) + (1 \times 90) + (2 \times 1)=92.$

We can likewise obtain a maximum estimated sum by rounding up each $\log_{10}{k}$ to the nearest integer:

$(0 \times 1) + (1 \times 9) +(2 \times 90)=189.$

The actual value of the sum must be between 92 and 189. It doesn't make much sense to estimate 100 because that is so close to our minimum. And so, our estimation is 150.

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The actual number of digits in $100!$ (computed from the sum of logarithms above) is 158.

Relevant wiki: Stirling's Formula

The number of digits of $100!$ is given by:

$\begin{aligned} N & = \lfloor \log_{10} {\color{#3D99F6}100!} \rfloor + 1 & \small \color{#3D99F6} \text{Using Stirling's formula } n! \approx \sqrt{2\pi n}\left(\frac ne \right)^n \\ & = \left \lfloor \log_{10} {\color{#3D99F6}\sqrt{2\pi \cdot 100}\left(\frac {100}e\right)^{100}} \right \rfloor + 1 \\ & = \left \lfloor \frac 12 \log_{10}2 + \frac 12 \log_{10} \pi + 1 + 100(2) - 100 \log_{10} e \right \rfloor + 1 \\ & = \lfloor 157.970 \rfloor + 1 \\ & = 158 \end{aligned}$

Note that 158 is the exact answer and the nearest option is $\boxed{150}$ .