Twin divisors?

If $a$ is evenly divisible by $b,$ then there is an integer $k$ such that $k • b = a.$

$12$ and $21$ are factors of $36$ and $63,$ respectively $-$ meaning that $36$ and $63$ are evenly divisible by $12$ and $21.$ Thus, if a number $n$ is divisible by the latter two $-$ that is, there is some integer that when multiplied by $36$ and $63$ gives you $n -$ then $n$ must also be divisible by the former two.

Lemma:

$ab \mid n \Rightarrow b \mid n, \quad a, b, n \in \mathbb{Z}$

Proof:

$ab \mid n \Rightarrow \exists k \in \mathbb{Z} : n = kab \Rightarrow \exists m \in \mathbb{Z} : n = mb \Rightarrow b \mid n$ . $\square$

Now, as corollary of the lemma we get, $36 \mid x \wedge 63 \mid x \Rightarrow 12 \mid x \wedge 21 \mid x, \quad x \in \mathbb{Z}$ .

Hence, the answer is $\boxed{\text{Yes}}$