Anyone notice that the inclusions are all each prime factors of the three factors (2, 3, and 5)

Anyone notice that the inclusions are all each prime factors of the three factors (2, 3, and 5)

@TheBladeDancer
Hmm... I think I can answer this question.
It’s because of the numbers Prof Loh selected.
Remember \(A \cap B \cap C\) is \( \frac{30}{2 \times 3 \times 5} = 1\)?
\(A \cap B\), is \( \frac{30}{2 \times 3}\), which leaves out a \(5\) from the denominator and thus making the answer \(5\), as it is basically \( \frac{2 \times 3 \times 5}{2 \times 3}\) and we cancel out the common \(2 \times 3\).
It is the same with the other two: \(B \cap C\) leaves out the \(2\), and \(C \cap A\) leaves out the \(3\).
Hope that helped!PS sorry for the very, very, very late answer. In fact I think you had already finished Module 3

Thank you to @RZ923 for answering.