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spooked_001
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Daughters' ages
« on: Feb 19th, 2004, 1:23pm » |
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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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towr
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Re: Daughters' ages
« Reply #1 on: Feb 19th, 2004, 2:07pm » |
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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
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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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| « Last Edit: Feb 19th, 2004, 2:08pm by towr » |
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Sameer
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Re: Daughters' ages
« Reply #3 on: Feb 20th, 2004, 4:23pm » |
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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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| « Last Edit: Feb 23rd, 2004, 12:56pm by Sameer » |
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John_Gaughan
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Re: Daughters' ages
« Reply #4 on: Feb 20th, 2004, 8:39pm » |
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on Feb 20th, 2004, 4:23pm, Sameer wrote:| A rich Mathematician (?) had two sons: Blaise and Charles (hehe just on another note which famous people are these?). |
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:: Blaise Pascal and Charles Babbage. ::
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Sameer
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Re: Daughters' ages
« Reply #5 on: Feb 23rd, 2004, 6:59am » |
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nice .. now how about the actual problem? 
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"Obvious" is the most dangerous word in mathematics.
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Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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John_Gaughan
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Re: Daughters' ages
« Reply #6 on: Feb 23rd, 2004, 7:35am » |
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::
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.
::
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| « Last Edit: Feb 23rd, 2004, 7:36am by John_Gaughan » |
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towr
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Re: Daughters' ages
« Reply #7 on: Feb 23rd, 2004, 7:45am » |
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on Feb 23rd, 2004, 7:35am, John_Gaughan wrote:Since these are integers, the possibilities are 1, 2, 4, 8, 16.
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and -1 -2 -4 -8 -16
Quote:| Blaise does not have 1 or 16. |
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Why not 1?
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| « Last Edit: Feb 23rd, 2004, 7:48am by towr » |
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John_Gaughan
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Re: Daughters' ages
« Reply #8 on: Feb 23rd, 2004, 9:06am » |
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on Feb 23rd, 2004, 7:45am, 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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towr
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Re: Daughters' ages
« Reply #9 on: Feb 23rd, 2004, 9:45am » |
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on Feb 23rd, 2004, 9:06am, John_Gaughan 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. |
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if he has 2 he knows the other must have 4 or 8
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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John_Gaughan
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Re: Daughters' ages
« Reply #10 on: Feb 23rd, 2004, 11:29am » |
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towr, I stand by my original answer until you prove me wrong 
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Eigenray
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Re: Daughters' ages
« Reply #11 on: Feb 23rd, 2004, 12:11pm » |
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If Blaise had a 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.
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| « Last Edit: Feb 23rd, 2004, 12:28pm by Eigenray » |
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Sameer
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Re: Daughters' ages
« Reply #12 on: Feb 23rd, 2004, 12:55pm » |
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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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--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
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Nigel_Parsons
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Re: Daughters' ages
« Reply #13 on: Apr 10th, 2004, 2:27pm » |
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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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Sameer
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Re: Daughters' ages
« Reply #14 on: Apr 12th, 2004, 9:24am » |
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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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"Obvious" is the most dangerous word in mathematics.
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Proof is an idol before which the mathematician tortures himself.
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bagoftricks
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Re: Daughters' ages
« Reply #15 on: Nov 22nd, 2004, 1:48am » |
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charles got screwed 
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Icarus
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Re: Daughters' ages
« Reply #16 on: Nov 22nd, 2004, 5:20pm » |
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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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Brad711
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Re: Daughters' ages
« Reply #17 on: Dec 11th, 2004, 9:41am » |
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Did anyone ever actually answer the Daughters' Ages problem. You just explained, but never gave a streightforward answer.
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Icarus
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Re: Daughters' ages
« Reply #18 on: Dec 11th, 2004, 6:48pm » |
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on Feb 19th, 2004, 2:07pm, towr wrote:
and the last clue then solves it, only for (8, 3, 3) is there an 'oldest' one. |
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I wonder: Which is larger
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eemayank
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why it cant be 2, 6, 6
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BNC
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Re: Daughters' ages
« Reply #20 on: Dec 29th, 2004, 11:10pm » |
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on Dec 29th, 2004, 10:09pm, eemayank wrote:
Because then there is no "oldest one"
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Padzok
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Re: Daughters' ages
« Reply #21 on: Jan 4th, 2005, 11:51pm » |
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on Dec 29th, 2004, 11:10pm, BNC wrote:
Because then there is no "oldest one" |
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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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BNC
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Re: Daughters' ages
« Reply #22 on: Jan 5th, 2005, 12:07am » |
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on Jan 4th, 2005, 11:51pm, Padzok 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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You are right in this case... but the riddle usually states either precise ages or rounded to nearest integer ages.
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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