Number of Digits in Number

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Theorem

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

Let $b \in \Z_{>1}$ be an integer greater than $1$.

Let $n$ be expressed in base $b$.

Then the number of digits $d$ in this expression for $n$ is:

$d = 1 + \floor {\log_b n}$

where:

$\floor {\, \cdot \,}$ denotes the floor function

$\log_b$ denotes the logarithm to base $b$.

Proof

Let $n$ have $d$ digits when expressed in base $b$.

Then $n$ is expressed as:

$n = \sqbrk {n_{d - 1} n_{d - 2} \dotsm d_1 d_0}$

where:

$n = \ds \sum_{k \mathop = 0}^{d - 1} n_k b^k$

Thus:

$b^{d - 1} \le n < b^d$

Thus we have:

By Integer equals Floor iff Number between Integer and One More:

$d - 1 \le \log_b n < d \iff \floor {\log_b n} = d - 1$

and the result follows.

$\blacksquare$