Number of Binary Digits in Power of 10
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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:
\(\ds 2^{m - 1}\) | \(\le\) | \(\, \ds 10^n \, \) | \(\, \ds < \, \) | \(\ds 2^m\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds m - 1\) | \(\le\) | \(\, \ds \map {\log_2} {10^n} \, \) | \(\, \ds < \, \) | \(\ds m\) | |||||||||
\(\ds \leadsto \ \ \) | \(\ds m - 1\) | \(\le\) | \(\, \ds n \log_2 10 \, \) | \(\, \ds < \, \) | \(\ds m\) |
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): $3 \cdotp 32192 \, 8 \ldots$