# Weird Maths Problem

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Our teacher set us this problem and said he wont give any answers until somone gets it. No one has any idea how to even begin it.

Help?

"How old are your three children?" the mathematics master asks his former student. He is told their ages add to 13, and multiply to give the number on his study door (which they can both see). "I will need to know more", the master says. After a few moments reflection. "The eldest one is learning to play the violin", replies his former pupil. "Ah! in that case I can now give you their ages", the master tells him, and does so correctly.

How does he know? What are their ages? What is the number on the door?

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I'm not sure if you are looking for a solution or just a few pointers. The bit about the violin is pretty clearly a red herring, though the third statement as a whole is not; think about what other information the master can gain from this statement. Also important is the fact that the two bits of information given first by the student are _not enough_ to determine the answer on their own, but with the information from the third statement, it is possible.

Edited by Josy
Removal of solution upon reading the rules. Many apologies.
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If this is indeed a mathematical problem and not some weird "haha" problem, here's how I'd start it, at least:

We have 3 children we don't know their ages, but we do know that they add up to 13:

x + y + z =13

We have another number on the door (let's call it R) that *we* don't know, but the student does know, so he has another equation:

xyz=R

What you're missing is another equation (3 variables, 3 equations), and the value of R. It seems to me that the statement about the violin is your clue, and it might be a cultural answer (do people start studying the violin at a certain age in your country, or perhaps it *means* another thing, like the fact he studies the violin means he is a twin, or something?). If you figure that out, you can solve it.

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If this is indeed a mathematical problem and not some weird "haha" problem, here's how I'd start it, at least:

We have 3 children we don't know their ages, but we do know that they add up to 13:

x + y + z =13

We have another number on the door (let's call it R) that *we* don't know, but the student does know, so he has another equation:

xyz=R

What you're missing is another equation (3 variables, 3 equations), and the value of R. It seems to me that the statement about the violin is your clue, and it might be a cultural answer (do people start studying the violin at a certain age in your country, or perhaps it *means* another thing, like the fact he studies the violin means he is a twin, or something?). If you figure that out, you can solve it.

[hide]

Or the fact that there is an eldest child, (potentially) implying there aren't 2 eldest children if you only consider integer ages.

[/hide]

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Unless I'm missing something, there could still be a whole mess of different ages.

[hide]6-5-2 would be just as likely as 7-5-1 or 8-4-1 etc. There is an oldest child, but that doesn't narrow the range down by much.[/hide]

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Ha, cracked it. And there are no 'red herrings', although the sentence about the violin is disguised. All three pieces of information are critical. As this is homework, I hesitate to give too big a clue, but start by explicitly writing out all allowable combinations of ages and their products. ie 1*2*10=20.

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Unless I'm missing something, there could still be a whole mess of different ages.
[hide]Yes, I missed it for a while as well but the ambiguity of if there's an oldest child or not is the key[/hide]
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Do I need to know something special about violins to solve this question?

[hide]'Cause I wrote the combinations and solved the multiplications and I don't see it.. I either forgot one of the multiplications or I am missing something here with the violin. Meh, hint?[/hide]

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No, they could equally well have said something like the eldest child is at the doctor's today or celebrated their birthday last week.

You might need to know something about mathematicians. They only think it's a proper problem (ie worth asking) if there's a unique solution.

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So is there a unique solution or not?

Yes there is.

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Thanks guys!

I logged on last night not really expecteing a lot of responses.

I finished it without reading any spoilers.

(and it seems so obvious now!)

thanks alot!!

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Meh, now help the riddly-challenged!!

pvt msg me, please, I'm dying with this one, I'm missing something for sure but I can't see it....

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now as you are over with the problem so please post the solution for me. i think im going wrong somewhere but i really dont know where so now please help me!

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Yea - I don't get it either.

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...I now have a long list of numbers that add to 13 and multiply to give a bunch of stuff. What am I supposed to see that I'm not seeing?

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Well, since the op has apparently solved the problem, I will give a clue that almost, but not quite, gives it away. With the table of room numbers and ages in front of you, ask yourself the following question.

How was it possible for the professor, who _did_ know the room number, to _not_ know the answer until the clue about the eldest child?

The answer is right in front of you. (assuming you didn't miss a age/room combination.)

Thanks

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Thanks

Yeah, ditto

Thanks btw, all you guys who sent me some more clues on pvt. I completely didn't think about that above issue... nice riddle!

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i'm stumped by this. i've made a chart of the possible combinations and nothing jumps out at me as being the obvious answer.

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I dont know...

Edited by Zebbygoss
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Ok, I'll put you out of your misery. Any room number that appears once on the list would allow the professor to know immediately the answer, because he knows the room number. So...

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atlast i did some thing! now will any one pm me what is the correct answer

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• 6 months later...

i worked on this for about an hour and i think i got the correct answer. could someone PM me?

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I'm assuming since this is so old that insane aline got it etc?

kevinalm has really given the fundamental help away without literally spelling it out. Take what you know.. write down all permutations.. Why does the person after seeing the housenumber still need more info?

There is one ambiguity however which is why maybe some people wonder... the riddle assumes the oldest* child is not* a twin (or a triplet).

with that ambiguity pointed out, it now is pretty easy to figure out.

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