How to reduce congruence systems with moduli not coprime? I'm given a system of congruences and want to apply the Chinese remainder theorem (CRT) on it, but the greatest common divisor (GCD) of the moduli is not $1$.
Well, I need to reduce the system, such that the GCD is $1$, but I don't know how the "reducing" works. I mean I need an explanation.
I found an example, but don't get how they magically transform it from
$$x \equiv 1 \mod 108$$
$$x \equiv 13 \mod 40$$
$$x \equiv 28 \mod 225$$
to
$$x \equiv 1 \mod 27$$
$$x \equiv 5 \mod 8$$
$$x \equiv 3 \mod 25$$
I see that 
$$108 = 2^2 \times 3^3$$
$$ 40 =5 \times 2^3$$
$$225=3^2 \times 5^2$$
but not sure when and how they used it... thanks for the help! 

Reducing is trivial:
If $a \equiv b \mod n$ then $a \equiv b \mod k$ for all $k|n$.
Just think about it......
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$x \equiv 1 \mod 108 \implies x= 108k + 1 = 27(4k) + 1 \implies x \equiv 1 \mod 27$.
$x \equiv 13 \mod 40 \implies x = 40k + 13 = 8(5k) + 13 \implies x \equiv 13 \equiv 5 \mod 8$. etc.
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It's going the other way that requires (minor) stipulation.
If $a \equiv b \mod k$ then $a \equiv b + ck \mod n$ for any $n$ a multiple of $k$ and $c$ is some integer; (exactly which integer is not necessarily known).
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So $x \equiv 1 \mod 3^3*2^2$
$x \equiv 13 \mod 2^3*5$
$x \equiv 28 \mod 5^2*3^2$
Pick the largest of the mutually prime factors: $3^3;2^3;5^2$ 
So 
$x \equiv 1 \mod 3^3$
$x \equiv 13\mod 2^3$
$x \equiv 28 \mod 2^5$.
