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I am having trouble understanding this exercise from my study guide:

Let $M = \{5k + 1 \mid k \in \mathbb{Z}, 0 \le\ k \le\ 240\}$. How many integers in $M$ are neither odd nor multiples of $6$?

Given Answer:

Let $A = \{x \in M \mid x \text{ is odd}\}$ and $B = \{y \in M \mid y \text{ is a multiple of 6}\}$.

We want to find $|M - (A \cup B)|$.

We see that a multiple of $6$ is even; hence $A \cap B = \emptyset$.

Why is this equal to the empty set?

Verify that,

I don't understand how to get the sets for $A$ and $B$

$$A = \{10k + 1 \mid 0 \le k \le 23\} \qquad\text{and}\qquad B = \{30m + 6 \mid 0 \le m \le 7\}.$$

Hence, $$|A \cup B| = 24 + 8 = 32,$$ and

Why $24$ and $8$?

$$|M - (A \cup B)| = |M| - |A \cup B| = 241 - 32 = 209 $$

Why $241$ and $32$?

Thanks for any help.

  • My latex does not seen to be formatting properly. Does anyone know why? –  Oct 07 '13 at 01:16
  • You didn't enclose it in dollar signs: $...$ for inline math, $$...$$ for displayed math. – Zev Chonoles Oct 07 '13 at 01:17
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    I've improved your question's formatting; apologies if I changed your meaning. You can see here how I edited your question.

    Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown.

     – Zev Chonoles Oct 07 '13 at 01:22
  • Thank you Zev. As you can tell, I am new to MathJax. –  Oct 07 '13 at 01:25
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    I'm pretty sure the expression defining $M$ is wrong. It should be $$M={5k+1,|,k\in\mathbb{Z},,0\leq 5k+1\leq 240}.$$ – triple_sec Oct 07 '13 at 01:32
  • I checked the exercise and I have it stated as it is in the study guide. But this study guide is full of errors, so you may be right in your assumption. –  Oct 07 '13 at 01:51
  • Something is definitely wrong. If $M$ is as originally stated in the study guide, then $$M={1,,6,,11,,16,,21,,26,,\ldots,,1\mathord{,}191,,1\mathord{,}196,,{1\mathord,201}}.$$ Then, indeed, $M$ contains 241 elements as stated. But in this case, 121 of them are odd (slightly more than half of them), not 24. – triple_sec Oct 07 '13 at 02:33

1 Answers1

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To summarise the comments, especially those from triple_sec,

  • There are $241$ elements of $M_{\text{big}}=\{5k + 1 \mid k \in \mathbb{Z}, 0 \le\ k \le\ 240\}$
  • There are $48$ elements of $M_{\text{small}}=\{5k+1\,|\,k\in\mathbb{Z},\,0\leq 5k+1\leq 240\}$
  • About half of the elements are odd and about a sixth are multiples of $6$; none are both since any multiple of $6$ must be even, so the intersection of odd numbers and multiples of $6$ is empty
  • $121$ elements of $M_{\text{big}}$ are odd and another $40$ elements of $M_{\text{big}}$ are multiples of $6$; $121+40=161$
  • $24$ elements of $M_{\text{small}}$ are odd and another $8$ elements of $M_{\text{small}}$ are multiples of $6$; $24+8=32$
  • $80=241-161$ elements of $M_{\text{big}}$ are neither odd nor multiples of $6$
  • $16=48-32$ elements of $M_{\text{small}}$ are neither odd nor multiples of $6$
Henry
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