# 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 \(32^2=1,024\).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.

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

#### Contents

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

For any natural number \(n\), the number of digits in \(n\) is \(\lfloor \log_{10} (n) \rfloor + 1\).

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

## Examples

How many digits does \(100!\) have?

We must compute \(\lfloor \log_{10} (100!) \rfloor + 1\). 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 \(\lfloor \log_{10} (100!) \rfloor + 1\): \[\log_{10} (\sqrt{2\pi})) + \left(n + \frac{1}{2} \right) \log_{10} (n) -\frac{n}{\ln(10)} \le \log_{10} (n!) \le 1 + \left(n + \frac{1}{2} \right) \log_{10} (n) -\frac{n}{\ln(10)} \] Using a calculator with \(n=100\), this implies the numerical inequality \(101.4 \le \log_{10}(100!) \le 102\). It follows \(\lfloor \log_{10} (100!) \rfloor = 101\), hence \(100!\) has 102 digits.

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

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