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I've reviewed this question and a little stuck on the formula presented: Integers divisible by 4 but not by 3 and 16 and was hoping someone could provide some additional details as to the logic being applied. I've tried to write it out as follows:

Let $A = 3|n$ , $B = 4|n $ and $C = 3 \bullet 4 | n$ where $|C| = |A \cap B| = \lfloor\frac{|A|}{4}\rfloor = \lfloor\frac{|B|}{3}\rfloor$

I understood the formula as follows: $|A\cup \overline B| = |A| - |A \cap B| $ but can I transform this into the format: $|A\cup \overline B| = |A| + |\overline B| - |A \cap \overline B| $ or vice versa?

I tried the following to show show that $|A \cup B| = |A| + |B| - |A \cap B| = |A \cap \overline B| - |\overline B|$ $= |\overline{\overline{A \cap \overline B}}| - |\overline B|$ $= |\overline{\overline A \cup B| } - |\overline B|$

$|U| - |\overline A \cup B| - |\overline B| = |U| - |\overline A| - |B| - |\overline A \cap B| - |\overline B|$ ... which I can't seem to finish.

Elliott de Launay
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1 Answers1

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You've got some issues with your approach. Let me address them point by point.

Let $A = 3|n$ , $B = 4|n $ and $C = 3 \bullet 4 | n$ where $|C| = |A \cup B| = \lfloor\frac{|A|}{4}\rfloor = \lfloor\frac{|B|}{3}\rfloor$

One issue here is a bit of a nitpick: you seem to want $A,B,C$ to be statements and sets at the same time. Rather, I think you mean $A=\{n\in U:3\mid n\},B=\{n\in U:4\mid n\},C=\{n\in U:12\mid n\},$ where $U=\{n\in\Bbb Z^+:100\le n\le 999\}.$ Another issue may simply be a typo: we should have $C=A\cap B,$ not $C=A\cup B.$

I understood the formula as follows: $\left|A\cup \overline B\right| = |A| - |A \cup B| $ but can I transform this into the format: $|A\cup \overline B| = |A| + |\overline B| - |A \cap \overline B| $ or vice versa?

Here again, I suspect a typo: we should have $\left|A\cap \overline B\right| = |A| - |A \cap B|,$ assuming that $\overline B$ refers to the relative complement of $B$ in $U.$ Your second formula here is correct, but we aren't interested in the numbers that are divisible by $3$ or not divisible by $4,$ so it isn't relevant.

I tried the following to show show that $$\begin{eqnarray}|A \cup B| &=& |A| + |B| - |A \cap B|\\ &=& \left|A \cap \overline B\right| - \left|\overline B\right|\\ &=& \left|\overline{\overline{A \cap \overline B}}\right| - \left|\overline B\right|\\ &=& \left|\overline{\overline A \cup B}\right| - \left|\overline B\right|\\ &=& |U| - \left|\overline A \cup B\right| - \left|\overline B\right|\\ &=& |U| - \left|\overline A\right| - |B| - \left|\overline A \cap B\right| - \left|\overline B\right|...\end{eqnarray}$$ which I can't seem to finish.

This again doesn't seem to be relevant, but I'll talk about it, anyway. It is true that $|A \cup B| = |A| + |B| - |A \cap B|,$ since we're dealing with finite sets. By the formula $|A\cap \overline B| = |A| - |A \cap B|,$ we should then conclude that $$|A\cup B| = |A\cap \overline B|+|B|,$$ instead.

You've correctly applied DeMorgan's laws to $A\cap\overline B$, so that $$\left|\overline{\overline A \cup B}\right| - \left|\overline B\right| = |U| - \left|\overline A \cup B\right| - \left|\overline B\right|,$$ but then went off the rails again. Since $\left|\overline A \cup B\right|=\left|\overline A\right|+|B|-\left|\overline A\cap B\right|,$ we should instead find that $$|U| - \left|\overline A \cup B\right| - \left|\overline B\right|=|U|-\left|\overline A\right|-|B|+\left|\overline A\cap B\right|-\left|\overline B\right|.$$

Now, instead, we'll use the formula $$|A\cap \overline B| = |A| - |A \cap B|,$$ or more simply, $$|A\cap \overline B| = |A| - |C|.\tag{1}$$ All that's left is to calculate $|A|$ and $|C|.$ I'll take care of $A,$ and I'll leave $C$ to you.

Just to spoil it, I'll let you know that $$|A|=\frac{|U|}3=\frac{900}3=300.$$ But why is that? Note that in each of the sets $$\{100,101,102\},\{103,104,105\},\{106,107,108\},$$ there is exactly one element that is divisible by $3,$ and these sets have no elements in common More generally, for any integer $n,$ the set $\{99+3n-2,99+3n-1,99+3n\}$ has exactly one element divisible by $3,$ and if $m$ is an integer with $m\ne n,$ then $$\{99+3n-2,99+3n-1,99+3n\}\cap\{99+3m-2,99+3m-1,99+3m\}=\emptyset.$$ Now, note that $$U=\bigcup_{n=1}^{300}\{99+3n-2,99+3n-1,99+3n\},$$ so the number of elements in $A$ is the same as the number of $3$-element sets we've partitioned $U$ into, which is $\frac{900}3=300.$

Cameron Buie
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