My friend once told me a logic puzzle along the lines of:

A student asks a teacher what the ages of the teacher's three kids are. The teacher said, "the product of their ages is 72, and the sum of their ages is your [the student's] house number." The student thought about this for a while, then came back to the teacher and said, "I don't have enough information!" The teacher replied, "Oh, and the oldest plays piano."

What are the ages of the kids?

How do I even go about solving this problem?

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    $\begingroup$ If two people have the same mother, they can't have exactly the same age (biological reasons). But if you assume that the "not the same age" equals "differ at least one year", than I like the puzzle. $\endgroup$ – martijnn2008 May 17 '14 at 20:26
  • $\begingroup$ @martijn Age is typically recorded as an integer value, not a decimal one. But, point taken. $\endgroup$ – user20 May 17 '14 at 20:27
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    $\begingroup$ As @martijnn2008 says, even among twins, there's an older and a younger. Should be some other way to disambiguate. Can be difficult to eliminate special cases like twins with different birthdays (eg, just before and after midnight). $\endgroup$ – igelkott May 19 '14 at 12:06
  • $\begingroup$ (But, in actuality, there really isn't much I can do. Edit the riddle? I guess... I just don't think it's a problem) $\endgroup$ – user20 May 19 '14 at 14:07
  • $\begingroup$ One of the ways I saw a puzzle book do it was to state that the oldest one is at least a year older than the others. $\endgroup$ – Joe Z. May 20 '14 at 14:39

This has since become one of my favorite logic puzzles! I highly recommend that you make an effort to solve this one first. It's a lot of fun.

We know the kids' ages must multiply to 72. Therefore, we need to find all combinations of three numbers that multiply to 72. 72 has the prime factors 2*2*2*3*3. For convenience, I've added a list of their sums for the student's house number.

1 + 1 + 72 = 74
1 + 2 + 36 = 39
1 + 3 + 24 = 28
1 + 4 + 18 = 23
1 + 6 + 12 = 19
1 + 8 + 9  = 18
2 + 2 + 18 = 22
2 + 3 + 12 = 17
2 + 4 + 9  = 15
2 + 6 + 6  = 14
3 + 3 + 8  = 14
3 + 4 + 6  = 13

Here we see a list of possible ages and house numbers. Now, of course, we know that the student comes back and tells the teacher "I need more information." If the student didn't need more information, then the student's house number would appear once in this list.

However, because the student needs more information, we know that the student's house number must appear more than once in this list. The only combinations for which this is the case are 2+6+6=14 and 3+3+8=14.

The teacher tells the student that the oldest plays piano, implying that there must be just one oldest child. Therefore, the answer can't be 2, 6, and 6, because there is no oldest. The answer must, therefore, be 3, 3, and 8.

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    $\begingroup$ Why is every1 answering there own questions. Just give me a day or two I am not behind my computer 24/7 and I need time to think. $\endgroup$ – martijnn2008 May 17 '14 at 20:27
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    $\begingroup$ @martijnn2008 Self-answering is encouraged on SE, and especially in private beta, as long as the question and answer are high-quality (which these are). Nothing prevents you from adding another answer; multiple answers are encouraged as well, if you have something new to add! $\endgroup$ – WendiKidd May 17 '14 at 21:37
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    $\begingroup$ @WendiKidd While I agree that self-answering is (and should be!) encouraged on SE in general, it does feel a bit weird on Puzzling.SE: "Hey, here's an interesting puzzle" "Cool, let me .." "Oh, btw, here's the answer." $\endgroup$ – SQB May 19 '14 at 8:39
  • $\begingroup$ People who answer the own answers on other SE sites for the most part didn't know the answer and learned how to solve it. This is not SE. Why would I not just get a puzzle book and copy every question and answer? $\endgroup$ – blankip Jun 28 '14 at 6:54
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    $\begingroup$ @blankip The community has since decided that we agree with you - it is not good form to self-answer if you already know the answer. However, we were exploring the limits of these types of questions at the time, which is why there are a few out there with self-answers. $\endgroup$ – user20 Jun 28 '14 at 6:56

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