Modulo expansion with divisor multiples Is there any general expansion for 'a mod mn' ?
mod is modulo operation: http://en.wikipedia.org/wiki/Modulo_operation
Eg: we have 
(a + b ) mod n = ((a mod n) + (b mod n)) mod n
(a x b ) mod n = ((a mod n) x (b mod n)) mod n
Similarly, Is there any simplification/expansion for :
a mod (m x n) = ?

Thanks
Edit: Added example for clarification

There is no really simple rule here. There is a not so simple one, which may not be what you want, but here we go: This rule assumes that $m$ and $n$ have no common factor, i.e., that $\gcd(m,n)=1$. In this case you can use the extended Euclidean algorithm (Bézout's identity) to find integers $u$ and $v$ so that $$mu+nv=1.$$
Now assume $$\begin{aligned}a&\equiv r\pmod{m},\\a&\equiv s\pmod{n}.\end{aligned}$$
So there are integers $p$, $q$ so that $a=pm+r$ and $a=qn+s$. Multiply the former equation by $n$ and the latter by $m$ to get
$$\begin{aligned}am&\equiv rm\pmod{mn},\\an&\equiv sn\pmod{mn}.\end{aligned}$$
From this you get
$$a=a\cdot 1=amu+anv\equiv rmu+snv\pmod{mn}$$
which is a connection between $a\bmod mn$ on one hand and $r=a\bmod m$ and $s=a\bmod n$ on the other.
This reduction is quite useful in applications such as RSA encryption, where you need to compute large powers modulo a product of two large primes. It is enough to compute the power modulo each factor and then combine the results.
