Who can solve this problem?

[Seraph]

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).

[True_Moose]

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]

[Grojlach]

Originally posted by Vaskez: Ah Grojlach, always the smartass 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

*sigh*
Very well then...

[Vaskez]

Originally posted by Seraph: 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).

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.

[Grojlach]

Originally posted by Seraph: 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).

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.

[/)eathKiller]

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...

[/)eathKiller]

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...

[Vaskez]

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

[The Hierophant]

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

[Vaskez]

Originally posted by The Hierophant: 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.