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#129890 07/01/04 04:39 PM
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So after a false start (and the need to think a bit more cleverly), I think I got it. Am I right in assuming that there is, however, only one circumstance in which the product and sum can be the same using two different sets of three numbers, hence the blue-eyed tiebreaker?


#129891 07/01/04 06:19 PM
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still lost.



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#129892 07/01/04 09:26 PM
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As far as I can tell, shanks, that's it exactly.

I'll post an explanation of the problem (as I see it) a little later on, under Answers to Puzzles. Assuming I can find it again!

#129893 07/02/04 12:19 AM
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I got a tip via PM(thanks!!), buthow do we know the apt number? what am I missing?



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#129894 07/02/04 12:30 AM
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It doesn't really matter, any more than the blueness of the eyes matters. It's just that there is still not enough information to solve the puzzle even after knowing the sum...

Full answer is posted in I and A, "Answers to puzzles", located here: http://wordsmith.org/board/showflat.pl?Cat=&Board=announcements&Number=131223
at the bottom (today, anyway).

But it's not whited-out or anything, so only look if you really want to know.

#129895 07/02/04 12:41 AM
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huh



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#129896 07/02/04 12:03 PM
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From a strictly logical point of view, even if there is a set of twins who are older than the third daughter, one of the twins will be older than the other. Not to mention that saying that one of them has blue eyes does not rule out the possibility that two, or even all three, have blue eyes. This means that even if Jet doesn't know which of the twins is older, if they both have blue eyes, then the older of them will.

However, if we assume that we have enough information* then there is a unique answer.

*You have a sphere. You drill a hole right through the center and measure the cylinder that the hole forms. How much volume is left in the sphere?


BTW: If you'd gone threaded, you could point directly to the post and not the whole thread, thus avoiding searching for the post as it becomes far up from the bottom.

http://wordsmith.org/board/showthreaded.pl?Cat=&Board=announcements&Number=131223

#129897 07/02/04 01:06 PM
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one of the twins will be older

nine minutes in the case of our twins...

left in the sphere

is this one of them x-y=z thangs?



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#129898 07/02/04 01:25 PM
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>one of the twins will be older

does anyone know of twins whose birth-dates are celebrated separately??
-ron (reject-a-nit) obvious


#129899 07/02/04 02:21 PM
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From a strictly logical point of view, even if there is a set of twins who are older than the third daughter, one of the twins will be older than the other.

True of course, which is the reason I stipulated that the ages were reckoned in years. Indeed you don't even need to have twins to have two kids the same age; they could have been born eleven months apart...

The "hole-in-the-sphere" problem is nifty (whatever that means, but generally positive). If the diameter of the hole isn't given, and the problem is presented as having a unique solution, then the diameter must be irrelevant and you can choose any convenient figure you like. Including zero. But you do need the diameter of the original sphere!

The proof that the size of the hole is indeed irrelevant is left as an exercise for the reader...all you mathematicians out there listening?


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