Date: Sat, 7 Jan 2006 12:40:23 +0000
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First, determine all the ways that three ages can multiply together to
get 72 (quite a feat for the bartender):

<pre>
 72  1  1
 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
</pre>

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 product-36 variant, the possibilities are:

<pre>
 36  1  1
 18  2  1
 12  3  1
 9  4  1
 9  2  2  (sum = 13)
 6  6  1  (sum = 13)
 6  3  2
 4  3  3
</pre>

In the sum-13 variant, the possibilities are:

<pre>
 11  1  1
 10  2  1
 9  3  1
 9  2  2  (product = 36)
 8  4  1
 8  3  2
 7  5  1
 7  4  2
 7  3  3
 6  6  1  (product = 36)
 6  5  2
 6  4  3
</pre>

In both cases, the two that remain are 9 2 2 and 6 6 1.  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
children with ages of 9, 2, and 2.
