We know the number of digits of a number. Which is it? Assume we know the number of digits of a number $n \in \mathbb N$ has in some different bases:
$$b_1,b_2,\cdots,b_n$$
What is the best approximation or the best "narrowing down" we can make of which number $n$ is?
 A: First of all, the number of digits of a number $n$ in base $k$ representation is, of course,
$$\lfloor \log_k n\rfloor+1$$
Which means that if you were given the number of digits $d$ in a number's base $k$ representation, you could determine that the number $n$ satisfies
$$k^{d-1}\le n \lt k^d$$
And so if you were given this information about $n$ in for multiple bases $$k_1,k_2,...,k_a$$
then you would need to find the solution, or the "overlap", of the system of inequalities
$$k_1^{d-1}\le n \lt k_1^d$$
$$k_2^{d-1}\le n \lt k_2^d$$
$$...$$
$$k_a^{d-1}\le n \lt k_a^d$$
in order to find all possible values of $n$.
For example:
Suppose we are given that $n$ has:


*

*$9$ digits in binary 

*$5$ digits in ternary

*$3$ digits in decimal


Then we can conclude that
$$512\le n\lt 1024$$
$$243\le n\lt 729$$
$$100\le n\lt 1000$$
and so $n$ can be any number in the interval
$$512\le n \lt729$$
If we are given lengths $d_1,...,d_a$ in bases $k_1,...,k_a$, we can only narrow down the exact values of $n$ if and only if, for some $d_i,k_i$ and $d_j, k_j$,
$$k_i^{d_i-1}=k_j^{d_j}-2$$
and if no other $d_v, k_v$ satisfy
$$k_v^{d_v-1} \gt k_i^{d_i-1}$$
$$k_v^{d_v} \lt k_j^{d_j}$$
because if this is the case, then we could narrow our conditions down to
$$k_i^{d_i-1}\le n \lt k_j^{d_j}-1$$
$$k_i^{d_i-1}\le n \lt k_i^{d_i-1}+1$$
and conclude that
$$n=k_i^{d_i-1}$$
But, of course, this can only happen if $n$ is a perfect power.
