Some Russian guy (if he wasn't Russian, then I'm Stalin!) and I came up with this corollary during class in first year (Math 135). I ended up proving this in the exam and using it to make some of the questions trivial to solve.
I don't have the proof here, but I'll get around to writing it up.
Anyway, the Chinese Remainder Theorem is:
If GCD(m1, m2) = 1 then, for any choice of integers a1 and a2, the simultaneous congruences
x == a1 (mod m1) and x == a2 (mod m2)
have a solution. Moreover, if x = x0 is one solution, the complete solution is
x == x0 (mod m1 m2).
If GCD(m1, m2) = 1 then there exists two integers a1 and a2, where
(m2 ± m1) x == (a1 m2 ± a2 m1) (mod m1 m2)
More general case
If GCD(mi, mj) = 1 then, for any choice of integers a1, a2, ... , an the simultaneous congruences
x == a1 (mod m1)
x == a2 (mod m2)
...
x == an (mod mn)
have a solution. Moreover, if x = x0 is one solution, the complete solution is
x == x0 (mod m1 m2 ... mn).
Therefore the corollary is
(m1 ± m2 ± ... ± mn) x == (a1 Pm / m1 ± a2 Pm / m2 ± ... ± an Pm / mn) (mod Pm)
Where
Pm = m1 m2 ... mn
