From: chris@questrel.com (Chris Cole)
Newsgroups: rec.puzzles
Subject: Re: Story Puzzle
Date: 8 Oct 92 17:20:41 GMT
Organization: Questrel, Inc.

In article <int470u.718541096@aurora.cc.monash.edu.au> int470u@aurora.cc.monash.edu.au (Wes Jones) writes:
[ sum and product ages problem deleted]
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From the FAQL, the answer is:
********
logic/children.p
********
A man walks into a bar, orders a drink, and starts chatting with the
bartender.  After a while, he learns that the bartender has three
children.  "How old are your children?" he asks.  "Well," replies the
bartender, "the product of their ages is 72."  The man thinks for a
moment and then says, "that's not enough information."  "All right,"
continues the bartender, "if you go outside and look at the building
number posted over the door to the bar, you'll see the sum of the
ages."  The man steps outside, and after a few moments he reenters and
declares, "Still not enough!"  The bartender smiles and says, "My
youngest just loves strawberry ice cream."

How old are the children?

A variant of the problem is for the sum of the ages to be 13 and the
product of the ages to be the number posted over the door.  In this
case, it is the oldest that loves ice cream.

Then how old are they?



********
logic/children.s
********
First, determine all the ways that three ages can multiply together to get 72:

72  1  1        (quite a feat for the bartender)
36  2  1
24  3  1
18  4  1
18  2  2
12  6  1
12  3  2
9  4  2
9  8  1
8  3  3
6  6  2
6  4  3

As the man says, that's not enough information; there are many possibilities.
So the bartender tells him where to find the sum of the ages--the man now knows
the sum even though we don't.  Yet he still insists that there isn't enough
info.  This must mean that there are two permutations with the same sum;
otherwise the man could have easily deduced the ages.

The only pair of permutations with the same sum are 8 3 3 and 6 6 2, which both
add up to 14 (the bar's address).  Now the bartender mentions his
"youngest"--telling us that there is one child who is younger than the other
two.  This is impossible with 8 3 3--there are two 3 year olds.  Therefore the
ages of the children are 6, 6, and 2.

Pedants have objected that the problem is insoluble because there could be
a youngest between two three year olds (even twins are not born exactly at
the same time).  However, the word "age" is frequently used to denote the
number of years since birth.  For example, I am the same age as my wife,
even though technically she is a few months older than I am.  And using the
word "youngest" to mean "of lesser age" is also in keeping with common parlance.
So I think the solution is fine as stated.

In the sum-13 variant, the possibilities are:

11  1  1
10  2  1
9  3  1
9  2  2
8  4  1
8  3  2
7  5  1
7  4  2
7  3  3
6  6  1
6  5  2
6  4  3

The two that remain are 9 2 2 and 6 6 1 (both products equal 36).  The
final bit of info (oldest child) indicates that there is only one
child with the highest age.  This cancels out the 6 6 1 combination, leaving
the childern with ages of 9, 2, and 2.