# Proof: How many digits does a number have? $\lfloor \log_{10} n \rfloor +1$

I read recently that you can find the number of digits in a number through the formula $\lfloor \log_{10} n \rfloor +1$ What's the logic behind this rather what's the proof?

Suppose that $n$ has $d$ digits; then $10^{d-1}\le n<10^d$, because $10^d$ is the smallest integer with $d+1$ digits. Now take logs base $10$: $\log_{10}10^k=k$, so the inequality becomes $d-1\le\log_{10}n<d$. If you now take the floor (= integer part) of $\log_{10}n$, throwing away everything to the right of the decimal point, you get $\lfloor\log_{10}n\rfloor=d-1$. Thus, $d=\lfloor\log_{10}n\rfloor+1$. This isn’t quite what you posted, but it’s the integer part of it, and clearly the number of digits must be an integer.
• @Emmad: It’s not about the number of digits in $\log_{10}8$; the result is that $8$ has $\lfloor\log_{10}8\rfloor+1$ digits, which is true, since $\log_{10}8\approx0.90309$. Commented Nov 6, 2012 at 23:38
• @NetMage: That’s correct, but $\Bbb Z^+$ is generally the domain of interest, and $0$ is inherently a special case. If one really wants to include $0$, one can use $d=\lceil\log_{10}(n+1)\rceil$; the argument that justifies it for $n\in\Bbb Z^+$ doesn’t extend to $n=0$, but the formula works for $n=0$ anyway. Commented Mar 24, 2022 at 4:41
• @jruota: By definition $\lfloor\log_{10}n\rfloor$ is the unique integer $k$ such that $k\le\log_{10}n<k+1$. We already know that $d-1\le\log_{10}n<d$; setting $k=d-1$, we have $k\le\log_{10}n<k+1$ and hence $\lfloor\log_{10}n\rfloor=k=d-1$. Commented Aug 22, 2023 at 5:38
let's consider the binary base-2 system. Any decimal number (say 8) can be represented as $2^3$. So if you take $log_2(8)$ you get 3, which represents the number of bits to represent 8. Thus, any number between 8 and 16 can be represented with $\lceil log_2(N)\rceil$ bits and so on. You can easily extend this to digits.
• Your formula doesn’t work when $N$ is a power of $2$. It should be $\lfloor\log_2N\rfloor+1$. Commented Nov 6, 2012 at 23:40