Set logic when there is no elements given

Question

Suppose $A$, $B$ and $C$ are finite sets with the following properties:

1. $B$ has four more elements than $A$.
2. $C$ has twice as many as $B$.
3. $A \cap B$ has three times as many elements as $B \cap C$.
4. $A$ and $C$ have no elements in common.

Show that $|A\cup B\cup C|$ is an integer multiple of $4$.

Answer 1

Remember the exclusion-inclusion formula:

$$ |A \cup B \cup C| = |A| + |B| + |C| - |A\cap B| - |A \cap C| - |B \cap C| + |A\cap B \cap C | $$

Now, given is that $|B| = |A| + 4$, |C| = 2 |B| and $|A \cap B| = 3|B \cap C| $. Clearly we have $|A \cap C | = |A \cap B \cap C | = 0$. Thus

$$ | A \cup B \cup C| = |A| + 3(|A| + 4) - 3|B\cap C| - 4|B \cap C| = 4|A|+12-|B \cap C|-4$$

Hence,

$$ |A \cup B \cup C| = 4\underbrace{( |A| + 2 - |B \cap C| )}_{\in \mathbb{Z}}$$