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Title: Daughters' ages Post by spooked_001 on Feb 19th, 2004, 1:23pm does anyone know the answer to this riddle? Local Berkeley professors Dr. Demmel and Dr. Shewchuk bump into each other on Telegraph Ave. They haven't seen each other since Vietnam. Shewchuk hey! how have you been? Demmel great! i got married and i have three daughters now Shewchuk really? how old are they? Demmel well, the product of their ages is 72, and the sum of their ages is the same as the number on that building over there.. Shewchuk right, ok ... oh wait ... hmm, i still don't know Demmel oh sorry, the oldest one just started to play the piano Shewchuk wonderful! my oldest is the same age! How old are the daughters? thanks if you answer. :) |
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Title: Re: Daughters' ages Post by towr on Feb 19th, 2004, 2:07pm There are allready two threads about this riddle (but different wording), but coinsidering the search function isn't do it's job (again) here's a breakdown of how to deal with the puzzle. First find the divisors of 72 72 ------ 72|1 36|2 24|3 18|4 12|6 9 |8 8 |9 6 |12 4 |18 3 |24 2 |36 1 |72 Then write the possible ages for the daughters (in order, starting with the oldest), and the corresponding sum of the ages ages sum 72, 1, 1 -> 74 36, 2, 1 -> 39 24, 3, 1 -> 27 18, 4, 1 -> 23 18, 2, 2 -> 22 12, 6, 1 -> 19 12, 3, 2 -> 17 9, 8, 1 -> 18 9, 4, 2 -> 15 8, 3, 3 -> 14 6, 6, 2 -> 14 6, 4, 3 -> 13 Since the sum of the ages doesn't help Shewchuk, it must mean that there are still multiple sets of ages possible, so it can be only either (8, 3, 3) or (6, 6, 2) and the last clue then solves it, only for (8, 3, 3) is there an 'oldest' one.. (If on the other hand one wanted to be original, the final clue could be 'the youngest just had her birthday', and you'd have the other answer) |
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Title: Re: Daughters' ages Post by towr on Feb 19th, 2004, 2:23pm Here's a selection of other threads, in case you didn't find my solution helpfull =) Daughters' ages (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1027806330;start=0) Daughters' Ages (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1067893063;start=0) How old are they ? (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_easy;action=display;num=1061834054;start=0) 3 daughters problem (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_easy;action=display;num=1046188926) How to solve 3 Daughters problem? (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1027805208;start=0) |
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Title: Re: Daughters' ages Post by Sameer on Feb 20th, 2004, 4:23pm Doesn't this problem always elude everyone? Even though there are so many threads here is another version of it. A rich Mathematician (?) had two sons: Blaise and Charles (hehe just on another note which famous people are these?). The rich mathematician dies and leaves a will with the lawyer. The lawyer goes to both sons and gives them each one positive integer (both of which could be equal or not). He says to them that their numbers multiply to either 8 or 16. Without asking each other the value of number, the one that figures out other's number wins. The two sons start talking: Blaise: I have no clue which number you have Charles: Same here Blaise: You will have to give me a hint. Charles: Same here Blaise figures out Charles' number and wins the rich fortunes. What number does Charles have? |
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Title: Re: Daughters' ages Post by John_Gaughan on Feb 20th, 2004, 8:39pm on 02/20/04 at 16:23:59, Sameer wrote:
:: [hide]Blaise Pascal and Charles Babbage.[/hide] :: |
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Title: Re: Daughters' ages Post by Sameer on Feb 23rd, 2004, 6:59am nice .. now how about the actual problem? :P |
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Title: Re: Daughters' ages Post by John_Gaughan on Feb 23rd, 2004, 7:35am :: [hide]Since these are integers, the possibilities are 1, 2, 4, 8, 16. Blaise: I have no clue which number you have Blaise does not have 1 or 16. Charles: Same here Charles does not have 1 or 16. This leaves 2, 4, and 8 as possibilities, but 8 only if the numbers multiply to 16. Blaise: You will have to give me a hint. Charles: Same here The numbers are 2 and 4, 4 and 4, or 2 and 8. Since neither one can definitively pin it down, and 8 only appears once, that disqualifies that solution, leave 2 and 4, or 4 and 4. Charles cannot determine the answer, so he must have a 4. If he had a 2, he would know Blaise has a 4. We do not know Blaise's number, but based on Charles' uncertainty, we can figure his out. Blaise tells the lawyer that Charles has a 4, and he walks away with the money. [/hide] :: |
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Title: Re: Daughters' ages Post by towr on Feb 23rd, 2004, 7:45am on 02/23/04 at 07:35:12, John_Gaughan wrote:
Quote:
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Title: Re: Daughters' ages Post by John_Gaughan on Feb 23rd, 2004, 9:06am on 02/23/04 at 07:45:58, towr wrote:
Because he has no clues. If he has 1, he knows the other must have 8 or 16. With a hint, he could find it easily. The integers could be negative, but then there is no way to solve this puzzle because with the implied information given, we cannot narrow it down any more. There would be two solutions, and no way to choose between them. By imposing the arbitrary (and common) restriction that the numbers must be positive, there is a solution. |
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Title: Re: Daughters' ages Post by towr on Feb 23rd, 2004, 9:45am on 02/23/04 at 09:06:01, John_Gaughan wrote:
with 4 he knows the other must have 2 or 4 with 8 he knows the other has 1 or 2. So he allways has a clue.. |
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Title: Re: Daughters' ages Post by John_Gaughan on Feb 23rd, 2004, 11:29am towr, I stand by my original answer until you prove me wrong ;-) |
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Title: Re: Daughters' ages Post by Eigenray on Feb 23rd, 2004, 12:11pm If Blaise had a [hide]1, then when Charles says the first time he doesn't know, Blaise would then know that Charles had an 8, but at that point he still doesn't know. Similarly, if Charles had a 1, then the first time Blaise says he doesn't know, Charles would know Blaise had an 8. Now, when Charles says the first time that he doesn't know, both people know nobody has a 1. So if either had an 8, they'd know the other was a 2, but they both admit they still don't know after that point. Then when Charles says he doesn't know the second time, both people know the only possibilities are {2,4} or {4,4}. If Charles had a 2, he'd have known Blaise had a 4. Therefore at this point, everybody knows Charles has a 4[/hide]. |
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Title: Re: Daughters' ages Post by Sameer on Feb 23rd, 2004, 12:55pm Yep John nailed it and EigenGray gave a nice explanation as to why not '1'. And as far as negatives are concerned that's my bad. They were each given 'positive' integers. I will edit my post to reflect that. |
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Title: Re: Daughters' ages Post by Nigel_Parsons on Apr 10th, 2004, 2:27pm Sameer: No need to re-word the question to avoid negative integers. Blaise & Charles have been told that the numbers multiply to either 8 or 16. In order for their brother to have a negative integer, their own would also need to be negative. They are thus aware that this is not the case. |
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Title: Re: Daughters' ages Post by Sameer on Apr 12th, 2004, 9:24am Nigel of course the "brother" would know but since "you" are supposed to answer the riddle you have two posibilties - negative and positive, and hence the need to mention they are positive numbers. |
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Title: Re: Daughters' ages Post by bagoftricks on Nov 22nd, 2004, 1:48am charles got screwed :'( |
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Title: Re: Daughters' ages Post by Icarus on Nov 22nd, 2004, 5:20pm Not really - If he had started the conversation, he would have been the one able to figure it out first. So it boils down once again to "you snooze, you lose". Indeed, if Charles had reason to believe the contest was intended to be fair, he should have realized right at the start that Blaise had "4" as well. |
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Title: Re: Daughters' ages Post by Brad711 on Dec 11th, 2004, 9:41am Did anyone ever actually answer the Daughters' Ages problem. You just explained, but never gave a streightforward answer. |
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Title: Re: Daughters' ages Post by Icarus on Dec 11th, 2004, 6:48pm on 02/19/04 at 14:07:33, towr wrote:
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Title: Re: Daughters' ages Post by eemayank on Dec 29th, 2004, 10:09pm why it cant be 2, 6, 6 |
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Title: Re: Daughters' ages Post by BNC on Dec 29th, 2004, 11:10pm on 12/29/04 at 22:09:27, eemayank wrote:
Because then there is no "oldest one" |
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Title: Re: Daughters' ages Post by Padzok on Jan 4th, 2005, 11:51pm on 12/29/04 at 23:10:59, BNC wrote:
What if the birthdays were 2 Jan 99 and 31 Jan 98? They would both be 6 as we speak, but one is older. |
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Title: Re: Daughters' ages Post by BNC on Jan 5th, 2005, 12:07am on 01/04/05 at 23:51:29, Padzok wrote:
You are right in this case... but the riddle usually states either precise ages or rounded to nearest integer ages. |
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