Given a number $n$ and an Interval $I = [ \; \lfloor n^{1/4} \rfloor, \lfloor n^{(1/3) \rfloor \;} ]$, can we say anything about the distribution of $\{ n \mod b \;\;| \; b \in I \}$?


1. In particular, if I wanted the residue to be close to any region in $[0, b-1]$, say close to the "top" of the residue classes around $b-1$ could I choose a $b$ for which that is the case without doing much work? By not doing much work I mean not having to try every single $b$ in the interval $I$ until I find one.
2. Could I somehow choose better $b$'s in my search than just randomly trying them; or even better, just choose the right $b$ in one go.


The factorization of $n$ is known if that helps any.


Heuristically, it seems most residues tend to be close to the "top", but I just need to $b$'s where I will know where the residues will land without having to try all of them.



Residues distribution modulo an interval
Note that $n \equiv b-1 \pmod b$ implies $b | n+1$, $n \equiv b-2 \pmod b$ implies $b | n+2$, and so on. Therefore you can factor $n+1$, $n+2$, and so on, until one of them has a factor in $I$. You can do something similar for other target regions.


While I don't have a proof, empirically it seems like you only need to factor $O(1)$ numbers on average, because around half of the numbers $n$ I've checked have a divisor in the range $[n^{\frac14},n^{\frac13}]$.


