Anyone notice that the inclusions are all each prime factors of the three factors (2, 3, and 5)
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Anyone notice that the inclusions are all each prime factors of the three factors (2, 3, and 5)
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@The-Blade-Dancer
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
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Thank you to @RZ923 for answering.
