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?
Proof: How many digits does a number have? $\lfloor \log_{10} n \rfloor +1$
2 Answers
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
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.
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1Your formula doesn’t work when $N$ is a power of $2$. It should be $\lfloor\log_2N\rfloor+1$. – 2012-11-06