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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.

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    hint:$$|A\times B \setminus B\times B| = 14\\ |A\times B - B\times B| = 14\\ |(A-B)\times B| = 14$$ – 2017-02-03
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    The hint justifies my assumption about the groups' size, but how can I figure out the intersection? – 2017-02-03

2 Answers 2

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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 $|A\setminus B|= |A|-|A\cap B|$.

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    The last line is what I needed. Thank you! – 2017-02-03
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$$|A\times B \setminus B\times B| = 14\\ |A\times B - B\times B| = 14\\ |(A-B)\times B| = 14$$ $$|A\times B - B\times B|=|A\times B -(A\times B)\cap (B\times B)|= \\ |A\times B -(A\cap B)\times B)|=\\ |A\times B |-|(A\cap B)\times B)|=14\\\to 35-|(A\cap B)\times B)|=14\\|(A\cap B)\times B)|=21=3\times7$$note that $A \cap B \subset B \to |A \cap B|\leq |B| \to |A\cap B|=3,|B|=7$