Number of Bits for Decimal Integer

Theorem

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $n$ have $m$ digits when expressed in decimal notation.

Then $n$ may require as many as $\ceiling {\dfrac m {\log_{10} 2} }$ bits to represent it.

Proof

Let $d$ be the number of bits that may be needed to represent $n$.

Let $n$ have $m$ digits.

Then:

$n \le 10^m - 1$

and so:

 $\displaystyle d$ $=$ $\displaystyle \ceiling {\map {\log_2} {\paren {10^m - 1} + 1} }$ Number of Digits to Represent Integer in Given Number Base $\displaystyle$ $=$ $\displaystyle \ceiling {\map {\log_2} {10^m} }$ $\displaystyle$ $=$ $\displaystyle \ceiling {m \log_2 10}$ $\displaystyle$ $=$ $\displaystyle \ceiling {\dfrac m {\log_{10} 2} }$ Reciprocal of Logarithm

$\blacksquare$

Examples

$14$ Digits

A positive integer $n \in \Z_{>0}$ which has $14$ digits when expressed in decimal notation may require $47$ bits to represent in binary.