Given some large integer, is there way to construct a modulus such that the remainder that is below some threshold? The proper context is integers too large for factorization to be computationally feasible, but, for the sake of illustration, the question is given an integer, such as 228483159, is there a procedure for constructing a modulus, such as 9834, that will yield a small remainder:
 228483159 ≡ 3 (mod 9834)

Thanks!
 A: Yes. If the integer is $n$, choose a modulus like $n-5$ so the remainder is at most $5$. 

If you also want the modulus to be small (in addition to the remainder), there's at least one quick way to get that. If you want the modulus to be less than one $k^\text{th}$ of the integer $n$, you can start by dividing $n$ by $k$ and taking the floor. 
In your example, if you want the remainder to be less than $10$ and the modulus to be less than $n/10$, this would give:
$$\left\lfloor\frac{228483159}{10}\right\rfloor = 22848315$$
If you take the original integer modulo $2284315$, the remainder is less than $10$ (specifically, it's $9$). 
Say you want a remainder less than $100$ instead. We get
$$228483159 \equiv 59 \pmod{2284831}$$

Note that this is equivalent to noticing that $100$ is a factor of $228483159 - 59$. In general, you can use anything instead of the easy powers of $10$ I've used as examples above. 
If $r$ is the maximum residue you want, $n$ is the integer you want to divide by, then the modulus $k$ can be $n-r$, or any factor of $n-r$. 
