Let's start by letting N be the number on the front of the bus (known to both magicians, but not to us).

*Magician 1: "The sum of the ages of my children is the number on the front
of this bus,
and the product of their ages is twice my age."*

Suppose C_{1},...,C_{k} are the children's ages,
arranged for convenience
in decreasing order (repetitions allowed), and let A be the age of the
first magician. (The number of children and the various ages are
unknown to the second magician, as well as to us.) The list
(C_{1},...,C_{k}) is called a *partition of N* because
N = C_{1}+...+C_{k}. The number k is the *length*
of the partition.
Because we know 2A = C_{1}...C_{k}, at least one of the
children has even age. Let us call C_{1}...C_{k}
the *product*
of the partition. A biological assumption that should be inserted
here is that the age of a parent is greater than the ages of any of the
children. With this in mind, let us call a partition
(C_{1},...,C_{k})
of N *admissible* if: (i) at least one C_{i} is even, and
(ii) half the product is greater than C_{1} (by convention,
the maximum
C_{i}).

*Magician 2: "Well, that's not very informative. Aside from the number
on the front of this bus, I know practically nothing.
If you were to tell me your
age and the number of children you have, could I then work out their
ages?"*

The second magician not only knows what N is, but knows a lot about numbers and is a lightening-quick calculator.

*Magician 1: "No, you couldn't."*

This is a key answer for Magician 2. It means that there are two
distinct admissible partitions of N, of the same length, that yield
2A for their respective products. So, given the first magician's
age, plus the number of children, there is still not enough information
to determine the
children's ages. Call a number M *ambiguous relative to N*
if M is the product of two
distinct admissible partitions of N of the same length.

*Magician 2: "Maybe not, but at least I now know
*your* age."*

This is a key answer for *us*. It means Magician 2
knows N to have just one number M that is ambiguous relative to N.
Let us call such a number N *magical.*

*The problem is to determine the number on the front of the bus.*

The only way this problem can be well posed is that there is exactly one magical number N. This will be the number on the bus. And for us to have solved the problem completely, we need two things: (i) a magical number N, and (ii) a proof that no other number is magical.

Let us first try to show (ii). Suppose N and M are given so that M is
ambiguous relative to N. Then there are two distinct admissible partitions
(C_{1},...,C_{k}) and (D_{1},...,D_{k}),
of the same length k, so that
M = C_{1}...C_{k} = D_{1}...D_{k}.
Let us now consider the two partitions
(C_{1},...,C_{k},1) and (D_{1},...,D_{k},1)
of N+1, both of length k+1.
These are obviously admissible partitions, with product M, so we
know that M is ambiguous relative to N+1. Next, consider the two
partitions (C_{1},...,C_{k},1,1) and
(D_{1},...,D_{k},1,1) of N+2, both of
length k+2, witnessing the fact that M is ambiguous relative to N+2.
The pair (C_{1},...,C_{k},2) and
(D_{1},...,D_{k},2) (reshuffle if either C_{k} or
D_{k} is 1) are also distinct admissible partitions of N+2
that witness
the fact that 2M is ambiguous relative to N+2. This means that N+2
(as well as all greater numbers) fails to be magical because of there being
two numbers ambiguous relative to it. So if N is the least magical number,
then no number besides N+1 can be magical. Thus, while we haven't proved
uniqueness; at least we know that if there is a magical number N, then
there can be only one other magical number, either N+1 or N-1.

It is now time, unfortunately, to try to guess a magical number. And
here is where we must use psychological considerations; i.e., that
our number N cannot be too big. Otherwise we have a computational
nightmare on our hands. What I did was try 14 first (just a guess);
and, before long, I found that 40 = 10.2.2 = 8.5.1 and
36 = 9.2.2.1 = 6.6.1.1. This tells us that all numbers greater than or
equal to 14 fail to be magical, and really narrows down our search.
Trying 13, we get 36 = 9.2.2 = 6.6.1 and 48 = 6.2.2.2.1 = 4.4.3.1.1,
so 13 isn't magical either. But trying 12, we hit pay dirt.
48 = 6.2.2.2 = 4.4.3.1, and no other pair of admissible partitions of the
same length give the same product. So 12 indeed is a magical number.
And, by the argument above, only 11 is a possible candidate for a second
magical number. But the number of possible checks is fairly small, and
it turns out that no number is ambiguous relative to 11. Thus 12 must
be the number on the front of the bus, and 48/2 = 24 must be the age
of Magician 1, the parent of four children. (But we *cannot*
know their ages.)

[If you got this far, or if you solved the puzzle yourself without
my help, you have a right to take a certain amount of pride. What I
want to add is the observation that it takes a vastly greater amount
of cleverness
to *set* puzzles such as this than it takes to solve them.
I am told that this puzzle owes
its authorship to John H. Conway, the inventor of the game of LIFE,
surreal numbers, and a host of other amazing intellectual adventures.]