Floor function within a congruence

In essence, the floor function is causing problems. Is there any way to get the inner linear expression, outside of the floor function?

$\lfloor(a_1x_1+...+a_nx_n)/d\rfloor \equiv b\pmod m$, for $a_i,b,d,m \in \Bbb Z$ and unknown $x_j\in\Bbb Z$.

Ideally into form $a'_1x_1+...+a'_nx_n \equiv b'\pmod {m'}$

I've been tinkering with fractional parts but keep getting stuck on what to do with the $\mod 1$ term since this is all within a congruence.

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What is $expr$? Is it $exp(r) = e^r$? Being as concrete as possible will likely lead to the best answer in this case. –  JavaMan Jan 22 '13 at 5:58
No it's actually a linear expression, i.e. $a_1x_1+...+a_kx_k$ where $a_i$ are constants, not necessarily integers, and $x_j$ are variable integers. –  Joe Jan 22 '13 at 6:27
Are the constants $a_i$ at least rational numbers, or can they be irrational? If they're irrational then this is probably hopeless. –  Greg Martin Jan 22 '13 at 7:14
Yes, the constants $a_i$ are rational. Actually, they're integers with a single division. I just restated the problem to make this more clear. –  Joe Jan 22 '13 at 15:33