Assume we have unknown number $d$ and an oracle which can tell us in one step if for any residual $x$ the equation $(1)$ $d \equiv x \textrm{ mod } p$ , with $p$ prime, holds. The important thing here is that we don't know $d$, that means in the worst case we have to ask the oracle $p$ times until we have the residual $x$ of $d$ over $p$. Now assume we have another equation $(2)$ $d \equiv n \textrm{ mod } q$, such that $q>p$, $p$ and $q$ are coprime and we know the number $n$. Is it possible to use the second equation to reduce the number of queries to the oracle to get a solution to the first equation.
An Example:
Let $d$ be unkown and the first equation be $d \equiv x \textrm{ mod } 631$ and the second equation be $d \equiv 26108 \textrm{ mod } 28608$. If i ask the oracle for every $x \in {1,...,630}$ it tells me that for $x = 485$ the first equation holds. Is it possible to get an answer faster?
Since $p$ and $q$ are coprime, the system of equations below always has a solution: $$d \overset{p}{\equiv} x$$ $$d \overset{q}{\equiv} n$$ for any pair of $(x,n)$. So the answer is No, there is no shortcut.
In general case, if $(p,q)=e$, then $(n,e)|x$.