$15 \cdot 4$ Set
$M=\{u \mid u=12 m+8 n+4 l$, where $m, n, l \in Z\}$,
$N=\{u \mid u=20 p+16 q+12 r$, where $p, q, r \in Z\}$

Then their relationship is
(A) $M=N$.
(B) $M \nsubseteq N, \quad N \nsubseteq M$.
(C) $M \subset N$.
(D) $M \supset N$.
(China High School Mathematics League, 1989)
[Solution] For any element $u$ in $N$, we have
$$
u=20 p+16 q+12 r=12 r+8(2 q)+4(5 p) \in M,
$$

thus
$$
N \subseteq M \text {. }
$$

On the other hand, for any element $u$ in $M$, we have
$$
u=12 m+8 n+4 l=20 n+16 l+12(m-n-l) \in N \text {, }
$$

thus
$$
M \subseteq N \text {. }
$$

Therefore, $M=N$.
So the answer is $(A)$.