Two MIT math graduates bump into each other at Fairway on
the upper west side. They hadn't seen each other in over 20
years.
The first grad says to the second: "how have you been?"
Second: "Great! I got married and I have three daughters
now"
First: "Really? how old are they?"
Second: "Well, the product of their ages is 72, and the sum
of their ages is the same as the number on that building
over there.."
First: "Right, ok.. oh wait.. hmmmm.., I still don't know"
second: "Oh sorry, the oldest one just started to play the
piano"
First: "Wonderful! my oldest is the same age!"
Problem: How old are the daughters?
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Answer / jaspreet
We know the guy trying to work out the ages can see the
number. We therefore know that *he* knows exactly what
number it is. We don't know what the number is, but that
doesn't matter. The important information revealed here is
that the guy knows what the number is.
Why is that important information? Well, we know that even
though he knows this number, that's insufficient information
for him to work out the ages.
So what information do we have at this stage? We know the
product of the ages is 72. And we know that even if you know
the sum of the ages as well as the product, that's not
enough to work out the individual ages.
I believe these are all the possible ages that give a
product of 72 where the numbers are all under 20, and I've
shown the sum of the ages next to them:
1, 8, 9 : 18
1, 6, 12 : 19
1, 4, 18 : 23
8, 3, 3 : 14
12, 2, 3 : 17
18, 2, 2 : 22
3, 4, 6 : 13
2, 4, 9 : 15
2, 6, 6 : 14
We can actually eliminate most of these now. They can't be
1, 8, and 9, because those add up to 18, and, significantly,
that's the *only* combination that adds up to 18. Remember
the guy knows the sum total (even if we don't). If it had
been 18, he'd have worked out from that information alone
that the ages were 1, 8, and 9.
You can eliminate all the combinations with a unique sum. If
those had been the ages, the guy would not have needed that
last clue.
In fact there are only two possible outcomes: 2, 6, and 6,
or 8, 3, and 3. Of all the sets of 3 numbers less than 20
that have a product of 72, these are the only ones with a
non-unique sum.
now since in the combination of 6,6,2 the "oldest" daughter
could not be as we have 2 daughters of equal age the answer
must be 8,3,3.
| Is This Answer Correct ? | 143 Yes | 15 No |
Answer / vikram raj
72 = 2*2*2*3*3
as A is elder to B, B is elder to C.
The ages must be
A=9
B=4
C=2
| Is This Answer Correct ? | 4 Yes | 15 No |
Answer / helen
It's actually pretty simple...
The guy does not know the answer, if he knows only the sum &
product. he does know the answer if he knows the sum &
product & that the largest number is unique.
Therefore, there must be only one combination where the
largest number is unique, and one or more combinations where
the largest number is not unique. In other words, x^2 + y =
72, where x > y.
There is only one set of numbers that works: {6,6,2}.
Therefore, they are talking about Building 14.
So you need x+y+z=14, and x*y*z=2*2*2*3*3
We've made the leap of logic and it is just math from here.
| Is This Answer Correct ? | 6 Yes | 30 No |
Answer / dj
The Second grad specifies that - "Great! I got married and
I have three daughters NOW"
This means that the third daughter was recently born and
hence she will be less than a year - which we consider as 1.
So that leaves us with the two daughters whose ages
multiply to 72.
The factors of 72 are : 72&1, 36&2, 24&3, 18&4, 12&6, 8&9
Now if the MIT Grads left school around their twenties, it
means they are in their forties now. Which means that even
if they got married as soon as they graduated and had a
child, there would be a twenty yrs difference between the
father and the dughter. Hence &2&1, 36&2 are omitted.
Leaving us with 18&4, 12&6, 8&9.
Now they say the right age to start piano lessons is 12 so
that leaves us with 12&6. Hence the daughters are ages 12,
6 &1.
| Is This Answer Correct ? | 11 Yes | 58 No |
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