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Math / Count number of digits with Logarithm

Suppose that a number $n$ 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 of this relation:

$$lo{g}_{10}({10}^{d-1})\le lo{g}_{10}n<lo{g}_{10}({10}^d)$$

This becomes:

$$d-1\le lo{g}_{10}n<d$$

If you now take the integer part of $lo{g}_{10}n$, throwing away everything to the right of the decimal point, you will get $\lfloor lo{g}_{10}n\rfloor =d-1$. Thus, $d=\lfloor lo{g}_{10}n\rfloor +1$.

For example, $234$ is a $3$ digits number, take the logs base 10:

$$lo{g}_{10}(234)\approx 2.369215$$

If we take the integer part and throw away everything to the right of the decimal point, it’s $2$.

The calculation can be done programmatically like this:

#include <math>
 
int digits = log10(n) + 1;

Source: Proof: How many digits does a number have? - StackExchange Mathematics^◹