I think this also makes a good tie-in for teaching algebra.

Once you figure out that the most significant digits should be the largest ones, and the 3rd and 4th largest digits should be in the 2nd most significant place, you end up with a situation like this:

      A C E
    *   B D

To make it more clear what to do next, you can just append a zero onto the number "BD" and still solve the same problem, because instead of multiplying "ACE" * "BD", you are multiplying "ACE" * "BD0" = ("ACE" * "BD") * 10:

      A C E
    * B D 0

Now, to figure out which of the two greatest digits are A and B, and which of the next two greatest are C and D, you can apply the identity (x + y) * (x - y) = x^2 - y^2 to this. Since "ABC" * "BD0" = (x + y) * (x - y), then equating "ABC" = x + y and "BD0" = x - y, you can solve to get x = ("ACE" + "BD0") / 2, which is the same number no matter the order between A and B, or between C and D. Then maximizing (x + y) * (x - y) means minimizing y^2; or, making the two numbers "ACE" and "BD0" as close together as possible, which leads to the given solution.