What is the largest possible value of $\omega(n)$ (the number of distinct prime factors of $n$) , if $n$ is a $30$-digit number containing only zeros and ones in the decimal expansion.
I checked those numbers divisible by $2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17$ and the maximum value was $\omega(n)=16$ , for example for $$n=100011001000111001010100000110$$
But I did not check all those numbers , so a larger value is still possible. Is there another method than simple brute force to find such a maximum for a larger number of digits , say , $40$ ?
Bonus question : A variant is that the number must be squarefree , as the above example is. Does this make a difference or is the maximum the same , no matter how many digits we give ?
Using brute force I found the following numbers with $\omega(n)=16$:
These are provably the only 30-digit numbers that satisfy $\omega(n)\geq 16$.
Using my brute-force algorithm (going through all binary numbers of length 30) it would be possible to extend the program to 40 digits. But that would take about a day on a good computer that I don't have...
Update: I ran the program for 35 digit numbers and found the following numbers with $\omega(n)\geq 17$ (note that there are 3 with $\omega(n)=18$, none of those are squarefree).
As a side note, these values also give all the $\omega(n)\geq 17$ values for smaller lengths of the number. Just remove $0$'s from the end.
These are (once again) all of the solutions.