Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 21 down vote accepted

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.

share|cite|improve this answer
So, given N=8 for example, how many digits would there be in log(8)? Thank you. – NoChance Nov 6 '12 at 23:35
@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 $\log_{10}8\approx0.90309$. – Brian M. Scott Nov 6 '12 at 23:38
I see now, thanks. – NoChance Nov 6 '12 at 23:42
@Emmad: You’re welcome. – Brian M. Scott Nov 6 '12 at 23:43

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.

share|cite|improve this answer
Your formula doesn’t work when $N$ is a power of $2$. It should be $\lfloor\log_2N\rfloor+1$. – Brian M. Scott Nov 6 '12 at 23:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.