The number on the front of the bus is 12; and, by the way, the first magician has 4 children and is 24 years old. Here's why:

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 C1,...,Ck 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 (C1,...,Ck) is called a partition of N because N = C1+...+Ck. The number k is the length of the partition. Because we know 2A = C1...Ck, at least one of the children has even age. Let us call C1...Ck 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 (C1,...,Ck) of N admissible if: (i) at least one Ci is even, and (ii) half the product is greater than C1 (by convention, the maximum Ci).

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 (C1,...,Ck) and (D1,...,Dk), of the same length k, so that M = C1...Ck = D1...Dk. Let us now consider the two partitions (C1,...,Ck,1) and (D1,...,Dk,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 (C1,...,Ck,1,1) and (D1,...,Dk,1,1) of N+2, both of length k+2, witnessing the fact that M is ambiguous relative to N+2. The pair (C1,...,Ck,2) and (D1,...,Dk,2) (reshuffle if either Ck or Dk 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 = = 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 = =, so 13 isn't magical either. But trying 12, we hit pay dirt. 48 = =, 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.]