# Finding The Number of Digits

Some problems ask you to find the **number of digits** of an integer or variable. For example, the number 23 has two digits, and the number 1337 has four. But how many digits does an integer $x$ have, if $x^2$ has more than 2 but less than 4 digits?

There's not yet enough information to determine exactly what $x$ is, but a range is ascertainable. If all that's known is that $x$ is an integer and that $x^2$ produces an integer that's $3$ digits long, then $x$ is on the range $[10,31]$ as $9^2=81, 10^2 = 100, 31^2=961,$ and $32^2=1,024$. $_\square$The wiki will discuss further and more rigorous ways to conduct this analysis.

Note that the numbers with precisely one digit are those integers in the range $[1,9]$, the numbers with precisely two digits are those integers in the range $[10, 99]$, and the numbers with precisely three digits are those integers in the range $[100, 999]$, and so on. The two digit numbers are shown below:

However, determining the number of digits of an extremely large number can be somewhat tricky and is explored below.

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## Large Numbers

A number will have precisely $j$ digits if and only if it is in the range $I_j = [10^{j-1}, 10^{j} - 1]$. For instance, the number $5,000,000$ has $7$ digits and is in the range $[10^{7-1},10^7-1] = [\text{1,000,000}, \text{ 9,999,999}].$

Given an integer $n$, one can determine $j$, the number of digits in $n$, by working with the inequality

$n\in I_j \implies 10^{j-1} \le n \le 10^j - 1.$

Taking the base 10 logarithm gives

$j-1 \le \log_{10} n < j,$

so $j-1 = \left\lfloor \log_{10} n \right\rfloor,$ where $\lfloor x \rfloor$ is the floor function, denoting the greatest integer less than or equal to $x$. Thus, $j = \left\lfloor \log_{10} n \right\rfloor + 1$.

For any positive integer $n$, the number of digits in $n$ is $\left\lfloor \log_{10} n \right\rfloor + 1$.

For example, take $n = 5,000,000$, then $\left\lfloor \log_{10} 5000000 \right\rfloor = 6,$ meaning that we expect this number to have $\left\lfloor \log_{10} 5000000 \right\rfloor +1 = 7$ digits, which in fact it does.

## Examples

How many digits does $100!$ have?

Using Stirling's formula $n! \approx \sqrt{2\pi n}\left(\frac ne \right)^n,$ we can get the number of digits of $100!$ as follows:

$\begin{aligned} N & = \lfloor \log_{10} {\color{#3D99F6}100!} \rfloor + 1 \\ & = \left \lfloor \log_{10} {\color{#3D99F6}\sqrt{2\pi \cdot 100}\left(\frac {100}e\right)^{100}} \right \rfloor + 1 \\ & = \left \lfloor \frac 12 \log_{10}2 + \frac 12 \log_{10} \pi + 1 + 100(2) - 100 \log_{10} e \right \rfloor + 1 \\ & = \lfloor 157.970 \rfloor + 1 \\ & = 158.\ _\square \end{aligned}$

**Cite as:**Finding The Number of Digits.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/finding-digits-of-a-number/