# Number of Binary Digits in Power of 10

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## Contents

## Theorem

Let $n$ be a positive integer.

Expressed in binary notation, the number of digits in the $n$th power of $10$:

- $10^n$

is equal to:

- $\ceiling {n \log_2 10}$

where $\ceiling x$ denotes the ceiling of $x$.

## Examples

### $1000$

When expressed in binary notation, the number of digits in $1000$ is $10$.

## Proof

Let $10^n$ have $m$ digits when expressed in binary notation.

By the Basis Representation Theorem and its implications, a positive integer $x$ has $m$ digits if and only if:

- $2^{m - 1} \le x < 2^m$

Thus:

\(\displaystyle 2^{m - 1}\) | \(\le\) | \(\, \displaystyle 10^n \, \) | \(\, \displaystyle <\, \) | \(\displaystyle 2^m\) | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle m - 1\) | \(\le\) | \(\, \displaystyle \map {\log_2} {10^n} \, \) | \(\, \displaystyle <\, \) | \(\displaystyle m\) | ||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle m - 1\) | \(\le\) | \(\, \displaystyle n \log_2 10 \, \) | \(\, \displaystyle <\, \) | \(\displaystyle m\) |

Because a power of $10$ cannot equal a power of $2$, it will always be the case that:

- $m - 1 < n \log_2 10 < m$

and so:

- $m - 1 < n \log_2 10 \le m$

Hence from Integer equals Ceiling iff Number between Integer and One Less:

- $m = \ceiling {n \log_2 10}$

$\blacksquare$

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $3 \cdotp 321 \, 928 \ldots$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $3 \cdotp 32192 \, 8 \ldots$