How many digits are there in 100!? How many digits are there in 100 factorial?
How does one calculate the number of digits?
 A: Suppose that $x$ is a positive, $n$-digit integer.  Note that,
$$n-1=\log_{10}(10^{n-1}) \leq \log_{10}(x) < \log_{10}(10^n) = n.$$
Thus, you might compute, the floor of $\log_{10}(100!)$ and add $1$.  But
\begin{align}
\log_{10}(100!) 
 &= \sum_{k=1}^{100} \log_{10}(k) \approx \int_{1}^{100}\log_{10}(t)dt \\
 &= \left.\frac{t\ln(t)-t}{\ln(10)}\right|_1^{100} \approx 157.005
\end{align}
Thus, the answer ought to be 158.

The reason this can be expected to work is that the integral can be approximated via Riemann sums.  In fact,
$$\sum_{i=1}^{n} \log_{10}(i) < \int_1^n \log_{10}(t)dt < \sum_{i=2}^n\log(10,i) = \sum_{i=1}^n\log(10,i),$$
as illustrated for $n=20$ in the following picture:

Thus, 
$$\sum_{i=1}^{n} \log_{10}(i) - \sum_{i=1}^n\log(10,i) < 
 \int_1^n \log_{10}(t)dt - \sum_{i=1}^n\log(10,i) < 0.$$
But, the term on the left is just $\log_{10}(100)=2$. So, the integral is a lower bound for the sum that cannot be more than two off of the actual value.  Accounting for the fact that the integral is nearly mid-way between the sums, the error should less than one, which is why we hit he answer exactly.
A: Stirling's formula gives a good approximation: $$n!\approx\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$
That is messier, but it helps because I can take the logarithm:
$$\log(n!)\approx \log(\sqrt{2\pi n})+n\log\left(\frac{n}{e}\right)$$
That was the common base-ten logarithm.
The number of digits in $n!$ equals the next integer above $\log(n!)$.
A: If I were writing a computer program to do it: 
$$\left\lfloor\log_{10}(100!)\right\rfloor+1=\left\lfloor\sum_{i=1}^{100}\log_{10}(i)\right\rfloor+1 = 158$$
$\lfloor\log_b(x)\rfloor+1$ calculates the smallest integer power of $b$ which is greater than x. i.e. $$b^{\lfloor\log_b(x)\rfloor} \le x < b^{\lfloor\log_b(x)\rfloor+1}.$$ This is the same as the number of digits in a base-$b$ representation of $x$.
A: Using a four-operations calculator, you can work as follows:


*

*start from $2$,

*multiply by increasing integers,

*every time the product exceeds $10$, shift the comma (divide by $10$) and count the shift.


The number of digits will be the number of shifts plus one.
$$\begin{align}
&2&0\\
&6&0\\
&2.4&1\\
&1.20&2\\
&7.20&2\\
&5.040&3\\
&4.0320&4\\
&\dots&\dots\\
&9.3326215\dots&157
\end{align}$$
Actually, you are computing $100!$ in the scientific notation.
