The best maths puzzle ever!!

PopeOfEruke

One day walking in a street, Ralf meets his old math teacher. Happy to meet him, he says hello to the teacher but he doesn't remember that his teacher answers everything with a puzzle.

"How are you doing, professor? It's been a long time since we don't meet! Are your daughters ok? How old are they now?"

Professor replies "Multiplying the three ages you get 36"

"But that's not enough to know the ages!"

Professor says : "So, add up the ages and you'll get the number of that house."

Ralf starts calculating, but still can't figure out.

The professor seeing that he wouldn't be able to find the answer says "The oldest girl plays the piano..."

That's what Ralf needed to know to figure out the problem.

How old is each girl?

Simon

Yes, very clever

(2, 2 & 9 ?)

Leolaia

I think this one is hard because we don't know the house number, and so we have to consider all the possibilities to know which two had to be disambiguated by the last clue. And I think the purpose of the last clue is to show that there is only one oldest girl, not two of them (e.g. twins).

Looks like Simon got the right answer. One could have 3 x 2 x 6, but that does not involve twins, but there are also 1 x 6 x 6 and 2 x 2 x 9, and it is the latter one that is disambiguated from 1 x 6 x 6 by the final clue. Apparently, the house number is 13, which is why the guy did not guess 3 x 2 x 6 (which yields 11 as the house number). There may be other possibilities, I'm not all that strong in math.

PopeOfEruke

Yes, 2, 2, and 9 is the normally accepted answer.

But if you really want to be pedantic, the other possibilty of 1, 6, and 6 *could* also work, if the two 6 year olds are not twins!

One kid could have been born in early January and the other kid in say, late November. So if the puzzle took place in December of a certain year, both kids would be 6 but the one with the January birthday would be the 'eldest'.

But as I said, it's normally assumed the two 6-year olds are twins hence the information about the piano solves the puzzle.

Twitch

My first thought was 3, 3 and 4 which still solves the last clue, if the child was a prodigy :-)

The second clue as a clue is moot.

As for the last clue, I don't see where it's inferred that the youngest two have to be twins by stating that there is a single eldest. This only disqualifies 6,6,1 Other combos that then work are 2x3x6, 12x3x1 18x2x1, even 36x1x1

But then, this doesn't solve the puzzle,....

Jourles

I would love to see a riddle/puzzle get posted one day that doesn't have the answer already on the web.

Leolaia

Twitch....36 x 1 x 1 .... that is one heck of an age gap for siblings.

About the others, I think you miss the point about the address number. That clue implies that the correct answer shares its sum with an incorrect answer. The results 12 x 3 x 1, 18 x 2 x 1, and 36 x 1 x 1 each have individually unique sums. It is 1 x 6 x 6 and 2 x 2 x 9 that have the same sum (13), thus the correct answer must be one of these two.

PopeOfEruke

Twitch

this is the important clue in the puzzle:

Ralf starts calculating, but still can't figure out.

As Leolaia points out, if it was any of the combinations you mentioned, Ralf would have worked it our because they all produce a unique house number.

It's only the duplicate set of 3 numbers adding up to the house number (which Ralf knows of course) that causes the inability to solve the puzzle without the 3rd clue from the professor.

Pope

SirNose586

2, 2, and 9. Nice. I had at first eliminated that choice, because I thought, "Who builds houses with the number 13?"

I narrowed it down to 2, 3, and 6 or 3, 3, and 4. I didn't like the second one, because I don't know any 4 year old prodigies. I went with the second one because I thought, "Well, 6 seems okay for a kid to be playing piano." Too subjective, I know.

tijkmo

traditionally streets dont have a number 13

this can surely be the only reason that the house is significant at all...ruling out 2-2-9

The best maths puzzle ever!!

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