12
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Jack and Jill are students of Mr. Hill. Mr. Hill's birthday is on $M$ month, $D$ day. Neither student knows when Mr. Hill's birthday is, but both know it is one of the dates below. Mr. Hill told Jack the $M$ value and Jill the $D$ value. Then Mr. Hill asked both of them when is his birthday.

Mr. Hill's birthday is one of the days below (format: month/day):

3/4 3/5 3/8 6/4 6/7 9/1 9/5 12/1 12/2 12/8

  • Jack: I don't know, and Jill surely doesn't know either.
  • Jill: At first I didn't know either, but now I do.
  • Jack: Oh, then I know it too.

Based on the above conversation between Jack and Jill, determine Mr. Hill's birthday. Give your step by step reasoning.

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The answer is

9/1

  • Day 2 only shows up once (12/2), and day 7 only shows up once (6/7). Days 1, 4, 5, 8 each show up at least twice. Jack says that Jill surely doesn't know, and therefore neither day 2 nor day 7 can be combined with Jack's month M. This means that $M \ne 12$ and $M \ne 6$.

  • After Jack's first statement, the list of dates has been reduced to 3/4 3/5 3/8 9/1 9/5.

  • Jill (knowing D) can deduce M from the new list. This means $D \ne5$, but the other values 1,4,8 are still possible. The list reduces to 3/4 3/8 9/1.

  • Jack (knowing M) can deduce D from the new list. This means $M \ne 3$ and onlly leaves one possible answer stated above.

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  • $\begingroup$ How did you come to the conclusion that M != 12? After you take out 12/2 there are still 12/1 and 12/8 left. $\endgroup$ – Ben Black Apr 1 '15 at 20:43
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    $\begingroup$ @BenBlack If M were to equal 12, it would leave Jack uncertain of whether Jill could know Mr. Hill's birthday. In other words, if Jack was given 12, then he'd see 12/2 as a possibility, meaning that if Jill were given 2, she'd know Mr. Hill's birthday without any input from Jack. $\endgroup$ – itriedacrab Apr 1 '15 at 21:36
  • $\begingroup$ Yes. That rules out 12/2 but not 12/1 or 12/8 $\endgroup$ – Ben Black Apr 1 '15 at 22:37
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    $\begingroup$ Think of it as a set. If Jack has 12, then D is either 1, 2, or 8. Since Jack doesn't know, and Jack knows Jill doesn't know, it means that he doesn't have 12. (Because) If he did have 12, he'd have to consider the possibility of D being 2 (a unique day), making it impossible for him to know whether or not Jill knows Mr. Hill's birthday. You can eliminate the entire month of December based on the fact that it has a unique day D. $\endgroup$ – itriedacrab Apr 1 '15 at 23:07
  • $\begingroup$ I was typing up a response to this and then it hit me, I realize now why you would think of it as a set. Well played. $\endgroup$ – Ben Black Apr 2 '15 at 13:22
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I disagree with Martin's answer. My answer would be:

3/4

  • Day 2 only shows up once (12/2), and day 7 only shows up once (6/7). Days 1, 4, 5, 8 each show up at least twice. Jack says that Jill surely doesn't know, therefore $D \ne 2$ and $D \ne 7$:

    3/4 | 3/5 | 3/8 | 6/4 | ̶6̶/̶7̶ | 9/1 | 9/5 | 12/1 | ̶1̶2̶/̶2̶ | 12/8

  • He still doesn't know the answer from this (we're still on his first statement saying neither of them knows). Only one month shows up once, 6. Since he still doesn't know, $M \ne 6$:

    3/4 | 3/5 | 3/8 | ̶6̶/̶4̶ | ̶6̶/̶7̶ | 9/1 | 9/5 | 12/1 | ̶1̶2̶/̶2̶ | 12/8

  • Jill says she didn't know before but now does. At this point day 4 is the only day showing up once (3/4). So she can deduce that $M = 3$.

  • Now that Jack knows Jill is positive of her day (4 is the only possible answer that can lead to that) he can deduce that $D = 4$

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    $\begingroup$ I came up with this answer yesterday and couldn't convince myself that it was wrong. $\endgroup$ – blakeoft Apr 2 '15 at 12:18
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    $\begingroup$ Martin's answer is actually the correct one. I disagreed at first but then the following conversation in the comments on his answer made it make sense. $\endgroup$ – Ben Black Apr 2 '15 at 13:22
  • $\begingroup$ I think I see, although I didn't think about it as a set. Jack wouldn't be able to decide between 3/4 and 3/5 if the birthday was actually 3/4 even though the first two quotes from Jack and Jill would still hold. $\endgroup$ – blakeoft Apr 2 '15 at 15:15
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Jack: I don't know implies that Mr. Hill could have told him any month, so $${3/.} \bigvee {6/.} \bigvee {9/.} \bigvee {12/.} \bigvee {3/.} = \text{true} $$ This is true because it's all possibilities.

Jack: And Jill surely doesn't know implies that either Jill wasn't told $12/2$ or $6/7$ or Jack wasn't told $12/.$ or $6/.$. Either of these two can make Jack certain that Jill doesn't know. So

$$\overline{12/. \bigvee 6/.} \bigwedge \overline{12/2 \bigvee 6/7} = \overline{12/.} \bigwedge \overline{6/.}$$

Combining the two result together, Jack has successfully eliminated month $12$ and month $6$ and we are left with.

$$3/4 \bigvee 3/5 \bigvee 3/8\bigvee 9/1\bigvee 9/5 = \overline{12/.} \bigwedge \overline{6/.}$$

Now they both know the date without being aware of the other's value. Jill knows the date without knowing Jack's month and Jack knows the date without knowing Jill's day.

Jill knows implies that it's not day $5$ i.e $\overline{3/5 \bigvee 9/5} $ which brings us to $$\left(3/4 \bigvee 3/5 \bigvee 3/8\bigvee 9/1\bigvee 9/5\right) \bigwedge \overline{3/5 \bigvee 9/5} = \left(3/4 \bigvee 3/8\bigvee 9/1\right) $$

Jack knows implies that it's not month $3$ i.e $\overline{3/4 \bigvee 9/8} $ which brings us to $$\left(3/4 \bigvee 3/8\bigvee 9/1 \right) \bigwedge \overline{3/4 \bigvee 3/8} =9/1 $$

NB: ${3/.} = 3/4 \bigvee 3/5 \bigvee 3/8 $

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protected by leoll2 Apr 21 '15 at 16:02

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