Originally Posted by
hatsoff the watered-down version of the crt (which is all you need for stuff like this) tells us that there is a natural map from $\displaystyle \mathbb{z}/p_1^{\alpha_1}\mathbb{z}\times\cdots\times\mathbb{ z}/p_n^{\alpha_n}\mathbb{z}$ to $\displaystyle \mathbb{z}/p_1^{\alpha_1}\cdots p_n^{\alpha_n}\mathbb{z}$, for distinct primes $\displaystyle p_1,\cdots,p_n$ and positive integers $\displaystyle \alpha_1,\cdots,\alpha_n$, and it is an isomorphism.
To apply the theorem here, we have an element $\displaystyle (1,2,3)\in\mathbb{z}/2^3\mathbb{z}\times\mathbb{z}/5^2\mathbb{z}\times\mathbb{z}/3^4\mathbb{z}$, and we want to find the associated element in $\displaystyle \mathbb{z}/2^3 5^2 3^4\mathbb{z}$. By the euclidean algorithm we can find linear combinations
$\displaystyle 1(25\cdot 81)-253(8)=1$
$\displaystyle 12(8\cdot 81)-311(25)=1$
$\displaystyle 32(8\cdot 25)-79(81)=1$
each of these equalities corresponds to a value 1, 2 or 3 (mod 8, 25 and 81, respectively). Take the first term in the left of each expression, multiply it by the corresponding value 1/2/3, and then add them all together. We get:
$\displaystyle 1(25\cdot 81)(1)+12(8\cdot 81)(2)+32(8\cdot 25)(3)=36,777$
and this is the solution mod $\displaystyle 8\cdot25\cdot81=16,200$. Reducing we get
$\displaystyle x\equiv 4,377\pmod{16,200}$.
This equivalence class captures all the solutions to the system by the crt.