Q:

intersection and cartesian product

Let $  \mathbb{N}\supseteq A, B $.
$|B \times A| = 35,$ and $|A\times B \setminus B\times B| = 14$.
Find $|A \cap B|$.
I figured out that $|B|$ must be 5 and $|A|$ must be 7, but that I got stuck.

A:

Following the hint, we have $$|A\times B| = 35 \qquad |(A\setminus B) \times B|=14$$Using that $|A\times B|=|A|\cdot |B|$, and the previous equations, we conclude that $|B|$ has to divide both $14$ and $35$.
Now, note that $|A\setminus B|= |A|-|A\cap B|$.

