If you've seen it before, keep quiet. Otherwise let's see who's quick to solve it. Send me the answer in a PM and I'll tell you if you're right so as not to spoil it for others until say, a day or two has passed. So here is the problem:

A maths teacher is invited round for dinner to a friend's house. The friend has 3 young children that the teacher hasn't met yet. The maths teacher asks his friend how old the kids are. His friend answers: "You're a maths teacher, I'll give you a clue, work it out! The clue is that the product of their ages is 36."
The teacher thinks for a bit and then asks for another clue. "Ok 2nd clue is that the sum of their ages is the same as our house number." The maths teacher goes outside looks at the house number, comes back in and says: "Ok but I still need another clue." "3rd clue: the oldest one plays the violin." The maths teacher then says: "Ah ok, now I know their ages."

The question is, what are the ages of the 3 children?
My cousin gave me this question. It's a bit tricky but sit down with a bit of paper and have a go and you should be able to do it too [img]smile.gif[/img]

[ 09-07-2003, 06:37 PM: Message edited by: Vaskez ]

Eh... don't we need to have the house number like the teacher..?. not fair he gets to see the number and we don't... [img]smile.gif[/img] .

Hehe, no you don't need it but you do need the information that he still didn't get the answer after knowing the house number. :D


BTW.....looking forward to becoming the half-orc Jorath? :D

[ 09-06-2003, 03:22 PM: Message edited by: Vaskez ]

God, I suck at math. I have absolutely NO idea what that violin is doing in this riddle.

say you had 2 * 3 * 6 = 36. i doubt a 6 year old is going to play a violin :D

<font color="orange">Would I be correct in assuming no two have the same ages?</font>

say you had 2 * 3 * 6 = 36. i doubt a 6 year old is going to play a violin :D </font>[/QUOTE]Actually, it's a hint to tell you that there is in fact only one "oldest" child among the three.
Figured it out within two minutes, btw (anyone using a simple piece of paper should be able to, I suppose). Can't be bothered to PM the answer, though. ;)

Why would you be able to assume something like this? There is nothing about it in the problem, and its possible for two children to have the same age.

I don't think there's any reason to make such a radical assumption - the third hint should enlighten you a bit on this matter, though (also see previous post).

Ah Grojlach, always the smartass :D But you are right about the 3rd clue. I took longer than 2 minutes to figure it out :( but I did eventually [img]smile.gif[/img]
However I am not sure if I am to believe that Groj is correct unless he does PM the answer ;)

[ 09-06-2003, 03:59 PM: Message edited by: Vaskez ]

I personally would say that the problem is not solvable in the strictest sense.

No where in the problem does it state that all the children are from the same woman, and no where does it state that the children can have an age difference of less then one year. (For example, If I were born on Sept 30 1981, and my brother were born August 24 1982 we would both be the same age (in years) today, but I don't think anyone would dispute the fact that I was the older brother).

<font color="orange">Oo, I got it, I think. Not too bad. My problem was I was doing the 3rd step before the second. Doh! [img]tongue.gif[/img] </font>

*sigh*
Very well then...

Well if you want to be REALLY nitpicky then you can't assume they are from the same woman, but who cares? It doesn't change the fact that they are 3 kids and the facts about their ages. Also, we are talking about ages rounded to the nearest year. I don't see the reason to make the post that you did since it doesn't effect the problem. This problem can be solved with logic and nothing else, no assumptions need to be made. There is only one answer which fits all the logic given in the question, no more or less information is needed.

In that extraordinarily nitpicky case, you can use the fact that the father names the third hint as the one to actually discern which answer it concerns as a fourth hint.
Else that third hint would have been rather obsolete, no? ;) Like Vaskez said, simple logic.

Attention: Only Read Below if You want this whole problem spoiled kinda as I managed to work it out in my head while reading it over 3 times and listening to Greek music, I doubt the music was a major factor but I figured I'd let you know the conditions I was under when I came to the conclusion of this problem [img]tongue.gif[/img]

So there are 3 kids, and the oldest is a violin player... I knew 6 and 7 year old violin players ... so that doesnt help much

but given these facts the only logical answer is that the house number was the product of the numbers which when multiplied are 36 so there is only one answer [img]tongue.gif[/img] 1 x 1 = 1 x 36 = 36 s

o you add that up and you get 1 + 1 + 36 which is 38 which is their house number and since the oldest one plays the violin the childrens ages are 10, 10, and 18 [img]tongue.gif[/img] but ah! not so! Because the sum does not equal 36

so we try again [img]tongue.gif[/img]

Alright 1 X 3 X 12 = 36


Bingo! [img]tongue.gif[/img] the violin player is 12, and their house number is 16

but then again. . .

If it was an incredibly YOUNG violin player, like the kind i've met... the children could be

1 x 4 x 9 so the violin player might be 9

OR the two younger kids could be 2 and the violin player 9

OR the violin player could be 4 and the kids could each by 3 whichi s highly doubtable... [img]tongue.gif[/img]

Anyway I've said everything I can come up with at this point...

Attention: Only Read Below if You want this whole problem spoiled kinda as I managed to work it out in my head while reading it over 3 times and listening to Greek music, I doubt the music was a major factor but I figured I'd let you know the conditions I was under when I came to the conclusion of this problem [img]tongue.gif[/img]

So there are 3 kids, and the oldest is a violin player... I knew 6 and 7 year old violin players ... so that doesnt help much

but given these facts the only logical answer is that the house number was the product of the numbers which when multiplied are 36 so there is only one answer [img]tongue.gif[/img] 1 x 1 = 1 x 36 = 36 s

o you add that up and you get 1 + 1 + 36 which is 38 which is their house number and since the oldest one plays the violin the childrens ages are 10, 10, and 18 [img]tongue.gif[/img] but ah! not so! Because the sum does not equal 36

so we try again [img]tongue.gif[/img]

Alright 1 X 3 X 12 = 36


Bingo! [img]tongue.gif[/img] the violin player is 12, and their house number is 16

but then again. . .

If it was an incredibly YOUNG violin player, like the kind i've met... the children could be

1 x 4 x 9 so the violin player might be 9

OR the two younger kids could be 2 and the violin player 9

OR the violin player could be 4 and the kids could each by 3 whichi s highly doubtable... [img]tongue.gif[/img]

OR one kid could be 1, another kid 2, and the oldest one 18, which would make their house number 21 . . .

But 18 is a little late to start learning the violin, it's best to start practicing at an early age for an instrunment. But I guess there are those who decide to start whent hey turn 18... I'm a Sax player and I started when I was 10...

Anyway I've said everything I can come up with at this point...

You're going about it the wrong way DK. Only two people have PM'd me a correct answer so far: Seraph and Grojlach. Seraph was first but Groj had already posted that he knew the answer before Seraph PM'd me so I'm not sure who the real quickest is :D

I've been told the answer now in pm but I still must say that this question isn't necessarily solvable as it assumes a reasonably high level of intelligence from a math teacher ;)

Hmm good point H, I never really thought about that. Maybe it's a maths teacher from the old days when they still could actually think logically. :D

Assumptions, 1 yr old is considered a baby not a young kid, and teenager is too old to be considered a kid.

1x4x9 sum 14 --impossible
1x3x12 sum 16 --impossible
1x2x18 sum 21 --impossible
1x6x6 sum 13 --impossible
3x3x4 sum 10 --possible w/child prodegy
3x2x6 sum 11 --possible

So we have a 50/50 chance of getting it right
as was mentioned before, any kid can play a violin, and in all of the cases the children will always have a 'oldest' since even in the case of twins one would come seconds before the other.

the house number is different for each of the ways, so the teacher should have figured it out at the 2nd clue.

Euh Vasky, why does he need the third clue? Hasn't he seen the children or is he simply unable to spot the difference between a 2 year old and a 6 year old? Math teachers these days...

LOL, try solving the problem and you'll see why he needs it. I'm afraid he's not the stupid one at the moment. [img]tongue.gif[/img] ;)

Gabrielles I see how you decided on which were impossible but you are not using maths or logic much. You didn't even LIST the combination which is the correct answer. Try finding the rest of the possible age combinations that give a product of 36. Another clue is that you don't actually need to assume that an 18 year old is not a kid - the solution still works if you don't assume anything about how kids are classed based on age.

[ 09-07-2003, 02:38 PM: Message edited by: Vaskez ]

I have it solved already, that is why I was wondering. :rolleyes: [img]tongue.gif[/img]
If the teacher had seen the children and been any good at estimating ages (and there are very few people who can't tell the difference between 2 year olds and 6 year olds) then he wouldn't have need of the third hint. We would since we don't have that additional bit of information.

At this moment I am wondering how or why there is only one answer possible. Why aren't there many answers?

Legolas if you solve the problem CORRECTLY you'll probably see why the 3rd clue is needed ;)
Zuvio - trust me, there is only one possible answer that fits the clues in the question. Obviously the ages of the kids are in years (nothing more accurate).

Hmm maybe I should just post the solution and put people out of their misery?

I have had quite a few PMs but still only Grojlach and Seraph managed to solve it!

No, I have to agree with Legolas. WE definitely need the clue, but if he did, obviously he was very poor at judging ages. Might have been better if the teacher was guessing their ages before he saw them ;)

Here's a hint. Figure out what the last clue must mean and come up with the only answer it CAN'T be from that. Then use that answer to get the real answer from the second clue ;)

Yeah ok I see your point, well maybe he was guessing their ages before he saw them? Who said he saw them? I just said that the friend had 3 kids and the teacher asked how old they were. Doesn't say anything about him seeing them [img]tongue.gif[/img]

Hehe I've edited the original question now...stop being so bloody nitpicky you lot! It's supposed to be a problem that can be worked out through logic not by judging ages! [img]tongue.gif[/img]

[ 09-07-2003, 06:38 PM: Message edited by: Vaskez ]

Aww Vasky... it's so cute to see you go through all that trouble just to boost your popularity by having all those people PM you! ;)
(or maybe I'm just feeling too stupid for your question and decided to distract attention from that fact.... :D )

Hmmm maybe I just didn't wanna spoil the answer for people by getting everyone to post their solution here? [img]tongue.gif[/img] I know I didn't want any help when I was solving this question. I think the reason for your post was the sentence in brackets :D

Do you now? ;)
Well I'm highly insulted you'd judge my intelligence as lowly as that!
Just for that, I'm not even going to PM you my steenkin correct answer as I was about to do..... [img]graemlins/angelwings.gif[/img]

Bah that trick's so old it didn't even make it into the book to be the oldest in the book [img]tongue.gif[/img] :D
Besides, if I really wanted to be popular I wouldn't put on my grouchy act. :D Besides that, who wants to be popular with a bunch of people he's never gonna meet anyway ;)

[ 09-07-2003, 06:56 PM: Message edited by: Vaskez ]

You know that's a lie, your grouchy act is your way of making chicks adore you! In fact, is that Cloudy I hear rushing in to cuddle aaaalll your grouchiness away?! [img]tongue.gif[/img] ;)

Alright alright! So I'm too tired and disinterested to solve the problem. Or maybe I'm just stupid. I'll go off to sulk then. :D

Hmmm I'm intrigued. Could you make a list of all the chicks that "adore me"? :D I think it's gonna be a short/non-existant list....might have to change tactics :D

Ah it's ok Mel, I'll come and comfort you when I'm in Amsterdam and then I'll explain the solution :D

hrm i evidently missed a permutation in my first try
2x2x9 sum 13

Now, i can only assume since my logic was refuted in PM that playing the violin has some other meaning
and the only meaning i can think of is perhaps that 9 and 6 are the same number just one upside down in relation to the other so could be said to be playing the violin i suppose, not that that makes any sense at all.

However,
violin has 6 letters in it, and upside down that 6 is a 9
so i suppose....
it could mean that the answer is either 2x2x9 or 1x4x9 since both have upside down 6's

LOL Gabrielles, guessing doesn't count. In logic questions there is always one answer that you can come to and be SURE it's the right answer because logic rules out all the others. Well I'm at work right now but I'll post my worked solution when I get home (in about 5 hours).

OK here's how I solved it:

First you should write out the possible ages that give a product of 36.
To do this, find all the numbers that divide 36:
36, 18, 12, 9, 6, 4, 3, 2, 1

now write out the possible combinations of ages of the kids:

1,1,36 sum = 38
1,2,18 sum = 21
1,3,12 sum = 16
1,4,9 sum = 14
1,6,6 sum = 13
2,2,9 sum = 13
2,3,6 sum = 11
3,3,4 sum = 10


The second clue says that even though the maths teacher knew the house number he could not work out their age knowing that the sum of their ages is the house number. This means that the number must be ambiguous i.e. at least 2 possible age combinations add up to the house number.
The only possible combinations that do not give a unique total are 1,6,6 and 2,2,9 so the answer must be one of those. The 3rd clue says that the oldest one plays the violin - not plural so there must only be one oldest one i.e. it cannot be 1,6,6 so their ages MUST be 2,2,9. No other answer can be correct.

there is an oldest 'one' in the case of 1 6 6
in cases of twins people seem to note which one came out a few seconds before, and so say hes/shes the older one.

But this time is insignificant when taking the unit of measure (whole years) and level of persision (0 decimal places) in considderation. Jeesh, like Vask said, way to be too nit-picky ... :rolleyes:

Thanks NS. Gabrielle, Shaddap! :D