Question: > 
> Writing $\left(\mathbf Z\big/2\oplus\mathbf Z\big/4\oplus\mathbf Z\big/8\right)\oplus\left(\mathbf Z\big/9\oplus\mathbf Z\big/27\right)$ in a different way
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If $M$ is equal to above how can I find the ideals,(of course of the form $m\mathbf Z$) $I\_1,\dots I\_n$ such that $I\_1\supset I\_2\supset \dots I\_n$ and $M\simeq\mathbf Z\big/I\_1\oplus\dots\oplus\mathbf Z\big /I\_n$

(there should be only $1$ way of selecting such ideals, but I don't know how)

Answer: 
We have:

$$M \cong Z/2 \oplus( Z/4 \oplus Z/9) \oplus (Z/8 \oplus Z/27) \cong Z/2 \oplus Z/36 \oplus Z/216$$
By Chinese Remainder Theorem if $I\_k$ and $I\_j$ are relatively prime ideals in a commutative ring $R$ then $R/I\_kI\_j \simeq R/I\_k \oplus R/I\_k$ (and one may generalize by induction). In your case there are many ways to split up $M$. You could write $$M \simeq \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/72\mathbb{Z} \oplus \mathbb{Z}/108\mathbb{Z}$$