Two Friends (another puzzle)

[toasty]

Ok, I've got one.

Two old (and clever) friends from school meet in the street, they haven't seen each other in over 30 years.


The 1st says to the 2nd: "how are you?"
2nd guy: "I'm married and i have three sons now"
1st guy "How old are they?"
2nd guy: "The product of their ages is 72, the sum of their ages is the same as the number on that building.."
1st: "right, ok.... hmm, i still don't know"
2nd: "oh sorry, the oldest one just started to play the piano"
1st: "Lovely! my eldest is the same age!"


How old are the sons?
-Dan

 

[markie]

There sons not daughters,

 

[Thermo]

he has sons not daughters

 

[toasty]

Well spotted, but it isn't a trick, that was a typo, now fixed.

 

[bethrob99]

Possible combinations

9 4 2
8 3 3
18 2 2 - bit late to start playing the piano but still possible

Where does the building number come in to it?

 

[bethrob99]

?

 

[toasty]

on the right lines rob..

 

[johnny_t]

Possible combinations

9 4 2
8 3 3
18 2 2 - bit late to start playing the piano but still possible

Where does the building number come in to it?

Some more cominations....

12 3 2 (building = 17)
6 4 3 (building = 13)
24 3 1 (building = 28)
18 4 1 (building = 23)
18 2 2 (building = 22)


ermm.... I don't know where I'm going with this.....

 

[empip]

The ages of the three children must be 3, 3, and 8, and the address is 14.
Take all combinations of three numbers having product of 72
Their sums all differ except two, hence the confusion with house number.- like which triple ???

The sum value 14, is shared by two combos 3, 3, 8. / 2, 6, 6, Only one of these has an elder kiddy ... hence 3 3 8 and no.14
Bingo.. Give me the money Mabel !

 

[bethrob99]

We're looking for two product combinations with the same sum. One will have twins with higher age, the solution will have single highest age

You beat me to it Empip

 

[johnny_t]

I suppose that if he can work it out exactly, and we know that there is an oldest, then it must be something with one unique solution i.e the oldest can't be 18, otherwise the others could be 2 and 2 or 4 and 1.

So the only one I can think of is 36 1 1, which can't be right, 'cos they saw each other 30 years ago

So.........

EDIT: Oh, right, erm....what empip said

 

[ban-all-sheds]

So the only one I can think of is 36 1 1, which can't be right, 'cos they saw each other 30 years ago

Not to mention the fact that the product of 36 1 1 is not 72....

 

[johnny_t]

So the only one I can think of is 36 1 1, which can't be right, 'cos they saw each other 30 years ago

Not to mention the fact that the product of 36 1 1 is not 72....

That too.........

Actually, using some of my earlier logic, the combinations are...

72 1 1
36 2 1
18 4 1
18 2 2
12 6 1
12 3 2
9 8 1
9 4 2
8 3 3
6 4 3
6 6 2

, so the only oldest ages with a unique 'lower ages' solution are 72, 36 and 8, and given that they saw each other 30 years ago, we can work out that its the 8 3 3 without the need for adding the house number into the equation.

 

[toasty]

empip, you are a clever bloke and yes well done!

You would have got an extra point in my interview if you'd pointed out that two children with the same age (6 + 6) would still always have one older than the other, but I think for the purposes of this question the answer you gave was more obvious and I think correct.

-Dan

 

[ban-all-sheds]

so the only oldest ages with a unique 'lower ages' solution are 72, 36 and 8, and given that they saw each other 30 years ago, we can work out that its the 8 3 3 without the need for adding the house number into the equation.

[being ridiculous]
What about the following possibilities. It could be that 30 years ago friend #2 had a 6 year old son, and may or may not have been married, but
1) The 1st friend, despite being clever, has a poor memory, and could not remember the age of #2's eldest
2) 30 years ago #2 did not know he had a 6-year old son, but has subsequently re-met and married the mother
3) 30 years ago #2 omitted to tell #1 about his child
[/being ridiculous]