wasaunders@watdragon.UUCP (Alec Saunders) (01/27/86)
This is a puzzle which someone told me the other day. I have no solution as
yet - perhaps some of you creative types can come up with one.
Two friends are walking down the street. One says to the other "Do
you have any children?". The other replies "Yes - three sons".
The first asks "How old are they?", to which the second replies
"The sum of their ages is thirteen, the product of their ages
is as old as you are. The oldest weighs 61 pounds."
How old are the three sons? For that matter how old is the friend?
And what does the eldest weight have to do with anything???
Alec Saunders - University of Waterloo CS.
P.S. No discussions of weight and mass please - assume common usage.ags@pucc-h (Dave Seaman) (01/28/86)
In article <292@watdragon.UUCP> wasaunders@watdragon.UUCP (Alec Saunders) writes: > Two friends are walking down the street. One says to the other "Do > you have any children?". The other replies "Yes - three sons". > > The first asks "How old are they?", to which the second replies > > "The sum of their ages is thirteen, the product of their ages > is as old as you are. The oldest weighs 61 pounds." You did not state the problem correctly. The problem as stated as twelve different answers. However, the following problem has exactly one answer: The first asks "How old are they?", to which the second replies "The sum of their ages is thirteen, and the product of their ages is equal to your age." The first says, "I can't tell their ages from that." The second adds, "The oldest weighs 61 pounds." The first says, "Now I know their ages." -- Dave Seaman pur-ee!pucc-h!ags
tps@homxb.UUCP (T.SCHROEDER) (01/29/86)
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ One possible answer would certainly be sons of ages 1, 5, and 7. 1+5+7 = 13 1x5x7 = 35, a reasonable age for a friend of a father of a 7-year old The weight of the eldest must be used to establish a reasonable upper bound for the children's ages. I doubt this solution is unique (certainly it isn't if you expand the horizons of "reasonable"). I can't see where any more constraints might come from, though. Tim. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
msb@dciem.UUCP (Mark Brader) (01/29/86)
> "The sum of their ages is thirteen, the product of their ages > is as old as you are. The oldest weighs 61 pounds." Of course the intention here is that when you see "the oldest", you're supposed to deduce that the two oldest ones have ages that are a different number of years. This is silly. Even among twins one is older than the other; and the person could have non-twin sons born *almost* a year apart but whose age in years is the same at the moment. This is a silly verbal trick and nobody should pose problems that depend on it. Mark Brader
wasaunders@watdragon.UUCP (Alec Saunders) (01/29/86)
In article <1783@dciem.UUCP> msb@dciem.UUCP (Mark Brader) writes: >> "The sum of their ages is thirteen, the product of their ages >> is as old as you are. The oldest weighs 61 pounds." > >Of course the intention here is that when you see "the oldest", >you're supposed to deduce that the two oldest ones have ages that >are a different number of years. This is silly. Even among twins >one is older than the other; and the person could have non-twin >sons born *almost* a year apart but whose age in years is the same >at the moment. > >This is a silly verbal trick and nobody should pose problems that >depend on it. > >Mark Brader I admit it is a complication I hadn't thought of, and when I sat down to work out the problem originally I did not take into account that two of the children might have the same age. That remaining - do you have any suggestions on how to approach it? Alec Saunders
hopp@nbs-amrf.UUCP (Ted Hopp) (01/29/86)
> This is a puzzle which someone told me the other day. I have no solution as > yet - perhaps some of you creative types can come up with one. > > Two friends are walking down the street. One says to the other "Do > you have any children?". The other replies "Yes - three sons". > > The first asks "How old are they?", to which the second replies > > "The sum of their ages is thirteen, the product of their ages > is as old as you are. The oldest weighs 61 pounds." > > How old are the three sons? For that matter how old is the friend? > And what does the eldest weight have to do with anything??? > > Alec Saunders - University of Waterloo CS. > P.S. No discussions of weight and mass please - assume common usage. Let's examine the clues. The trick, of course, is to recognize a clue for what it is. "Three sons" "The sum of their ages is thirteen" From this, we know the ages are a partition of 13 into three positive integers. There are fourteen partitions (ignoring permuations). I won't bother listing them here. "the product of their ages is as old as you are" This doesn't tell us much yet, except that the friend now knows this. "The oldest weighs 61 pounds" Aha! This gives it all away. There are two vital clues here, neither of which has to do with weight. The first clue is: knowing the sum AND the product of the ages isn't enough information (at least in the world of logic puzzles), otherwise this new information wouldn't have been offered. Well, we can form the product of each of the partitions of 13, and we find that the only partitions that do not have unique products are (1,6,6) and (2,2,9), both of which multiply to 36. (Therefore, the friend is 36 years old.) The second clue is: there is an oldest son. The solution, then is (2,2,9). As a check, 61 pounds is a reasonable weight for a 9-year old boy. -- Ted Hopp {seismo,umcp-cs}!nbs-amrf!hopp
bulko@ut-sally.UUCP (Bill Bulko) (01/29/86)
wasaunders@watdragon.UUCP (Alec Saunders) writes: > > Two friends are walking down the street. One says to the other "Do > you have any children?". The other replies "Yes - three sons". > > The first asks "How old are they?", to which the second replies > > "The sum of their ages is thirteen, the product of their ages > is as old as you are. The oldest weighs 61 pounds." > > How old are the three sons? For that matter how old is the friend? > And what does the eldest weight have to do with anything??? > The weight of the oldest child is probably meant to be used as a constraint for his age. However, I think the information given is inadequate for a solution unless we make a few (reasonable) assumptions: (1) that the children are at least one year apart from each other in age (since there was no mention of twins); (2) that the man's age is appropriate for the age of his children; and (3) the friend's age is roughly equal to that of the man (a *major* assumption). Solutions for the problem producing an appropriate age for the father are: 9/3/1 (with the friend's age 27), 8/4/1 (32), and 7/5/1 (35). You'd have to consult a table of average heights and weights to figure out which age (9, 8, or 7) best fits a weight of 61 pounds. _______________________________________________________________________________ "In the knowledge lies the power." -- Edward A. Feigenbaum "Knowledge is good." -- Emil Faber Bill Bulko Department of Computer Sciences The University of Texas {ihnp4,harvard,gatech,ctvax,seismo}!sally!bulko _______________________________________________________________________________
ins_aaaw@jhunix.UUCP (Adlai A. Waksman) (01/30/86)
Here's a puzzle I heard in high school. It's quite similar to the one Alec Saunders posted. THE CENSUS-TAKER'S PROBLEM (or, the Chocolate Pudding Puzzle) A census taker was interviewing a man at the door of his house. He soon asked, "Do you have any children?" "Yes, three daughters." "How old are they?" Now the man, being a bit of an eccentric, decided to pose the answer in the form of a puzzle. "The product of their ages is 72, and the sum of their ages is the house number." The census taker produced a sheet of paper, did some arithmetic, but finally gave up. "Oh, yes," the man recalled, "my oldest daughter likes chocolate pudding." The census taker immediately knew the daughters' ages. WHAT ARE THEY? While you're at it, you can probably figure out the house number, as well as what chocolate pudding has to do with all this. NOTE: Babies can like chocolate pudding, teenagers can like chocolate pudding, elderly women can like chocolate pudding. Also, the man and his wife might have unbelievable longevity and fertility.... I'll post a spoiler in a week or so (unless someone beats me to it). -- Adlai Waksman Mathematical Sciences, The Johns Hopkins University, Baltimore, MD 21218 UUCP: seismo!umcp-cs \ AT&Tnet: (301) 889-8498 ihnp4!whuxcc > !jhunix!ins_aaaw Bitnet: INS_AAAW@JHUNIX allegra!hopkins / INS_AAAW@JHUVMS Arpa: ins_aaaw%jhunix.BITNET@wiscvm.WISC.EDU "Synergy is the essence of it all . . . and nobody played guitar." -- L. Fast
bonomi@ssc-vax.UUCP (Scott Bonomi) (01/30/86)
> This is a puzzle which someone told me the other day. I have no solution as > yet - perhaps some of you creative types can come up with one. > > Two friends are walking down the street. One says to the other "Do > you have any children?". The other replies "Yes - three sons". > > The first asks "How old are they?", to which the second replies > > "The sum of their ages is thirteen, the product of their ages > is as old as you are. The oldest weighs 61 pounds." > > How old are the three sons? For that matter how old is the friend? > And what does the eldest weight have to do with anything??? > > Alec Saunders - University of Waterloo CS. > P.S. No discussions of weight and mass please - assume common usage. ages are 1,5 and 7, the children must be of different ages (assume same mother) the weight of the oldest limits it (I assume the child is rathe thin) other permutationns of the younger ages are possible, but most age the friend rapidly, this gives his age as 35, not unreasonable for the friend oa a person with a 7 year old son. -- scott
jacob@renoir.berkeley.edu.BERKELEY.EDU (& Butcher) (01/30/86)
I just saw this puzzle in a book recently. It was stated slightly differently: "I will tell you some things about my sons; stop me when you know their ages." When stated this way, the third statement has very definite significance. ~jacob
pz@emacs.UUCP (Paul Czarnecki) (01/30/86)
<eat it> In article <2578@pucc-h> ags@pucc-h.UUCP (Dave Seaman) writes: [Dave corrects the previous poster...] >The first asks "How old are they?", to which the second replies > > "The sum of their ages is thirteen, and the product of their > ages is equal to your age." > >The first says, "I can't tell their ages from that." >The second adds, "The oldest weighs 61 pounds." > >The first says, "Now I know their ages." > >Dave Seaman pur-ee!pucc-h!ags *** SPOILER WARNING *** *** answer follows *** When I first saw this problem I thought is was impossible. I also thought that I had to apply some *reasonable* constraints. Well, after looking at some of the other solutions posted so far I realized that everybody so far has been making one BIG mistake. And you only have to apply one real life constraint. You see, the first friend *knows* his own age! (we however don't know it, yet) a exhaustive search yields: kids' friend's ages age 1 1 11 11 ; *maybe* we can throw this one out! 1 2 10 20 ; okay, so he was precocious 1 3 9 27 1 4 8 32 1 5 7 35 1 6 6 36 2 2 9 36 2 3 8 48 2 4 7 56 2 5 6 60 3 3 7 63 3 4 6 72 ; somewhere here we *maybe* could start 3 5 5 75 ; throwing things away. But I really don't 4 4 5 80 ; want the AARP and Dr. Ruth down my back. since friend1 knows his own age and he *still* can't decide we know the the answer must be one of: 1 6 6 36 2 2 9 36 Nows here's the part I don't like in a logic puzzle, a 6 year old is more likely to weigh 61 pounds than a 9 year old is. (is this true? I don't know!) therefore the answer is 1 6 6. pZ -- -- You pretend I'm him and I'll pretend you're her. Paul Czarnecki Uniworks, Inc. decvax!{cca,wanginst!infinet}!emacs!pz 20 William Street emacs!pz@cca-unix.ARPA Wellesley, MA 02181 (617) 235-2600
wet@whuxlm.UUCP (William Taylor) (01/30/86)
> The first asks "How old are they?", to which the second replies > > "The sum of their ages is thirteen, and the product of their > ages is equal to your age." > > The first says, "I can't tell their ages from that." > The second adds, "The oldest weighs 61 pounds." > > The first says, "Now I know their ages." > > -- > Dave Seaman pur-ee!pucc-h!ags The friend knows how old he is. Since he doesn't have enough information to solve the puzzle at first, there must be two or more sets of three numbers which add up to 13 and when multiplied together equal the friend's age. In fact, there are only two sets of numbers that add up to thirteen and when multiplied together give you the same number. These are: (9 2 2) and (6 6 1). But the friend does not know the ages of the kids yet. So the dad tells him that the oldest weighs 61 pounds. Now granted in real life, one of the six year olds is older than the other, but in years, they are the same age. So, with (6 2 2) there is no oldest. Thus, the oldest is 9 and the other two are both two years old. Buddy Taylor AT&T Bell Labs whlmos!wet
brianu@ada-uts.UUCP (01/31/86)
>> you have any children?". The other replies "Yes - three sons". >> The first asks "How old are they?", to which the second replies > "The sum of their ages is thirteen, and the product of their > ages is equal to your age." >The first says, "I can't tell their ages from that." >The second adds, "The oldest weighs 61 pounds." >The first says, "Now I know their ages." >Dave Seaman pur-ee!pucc-h!ags The revised problem has only one solution: 2 year old twins and a 9 year old. Reasoning: There are 14 triplets whose sum is 13. Their products range from 11 (1,1,11) to 80 (4,4,5). Presumably the second person knows his (her) own age and should be able to deduce the correct answer. However, the product of 36 comes up twice (1,6,6 and 2,2,9) and is the only one that comes up more than once. Since the person is unable to deduce the answer his age must be 36. When the other says that the oldest weighs 61 pounds we eliminate 1,6,6 because there would be no "oldest". Threefore the answer is 2,2,9. Brian Utterback Intermetrics Inc. 733 Concord Ave. Cambridge MA. 02138. (617) 661-1840 UUCP: {cca!ima,ihnp4}!inmet!faust!brianu LIFE: UCLA!PCS!Telos!Cray!I**2
jens@moscom.UUCP (Jens Fiederer) (02/04/86)
The argument that we could have twins whose age is only minutes apart, and still have an "oldest" child does not strike me as reasonable. Either we are reckoning age as an integer value, in which case two twins have the same age, or we are looking at the real numbers, in which case the father would have had to say, "AT THE SOUND OF THE TONE, the children's ages will sum to 13, and their product will be your age!....BEEP" Azhrarn