# Problem 06

## Problem 06

If two classical dices are thrown, the sum has generating function

F(x) = x^{2} + 2x^{3} + 3x^{4} + 4x^{5} + 5x^{6} + 6x^{7} + 5x^{8} + 4x^{9} + 3x^{10} + 2x^{11} + x^{12} = x^{2} (1 + x)^{2} (1 + x + x^{2})^{2} (1 - x + x^{2})^{2}.

One can write other positive integers on the 12 faces to obtain the same sum distribution. Find the (unique) other way.

### Solution

We are looking for a_{1}, ..., a_{6}, b_{1}, ..., b_{6}, such that F(x) = (x^{a1} + x^{a2} + ... + x^{a6})(x^{b1} + x^{b2} + ... + x^{b6}) = G(x)H(x).

Since we need positive integers, x must divide both factors. Since G(1) = H(1) = 6, the factors 1 + x and 1 + x + x^{2} must also appear exactly once in both G(x) and H(x). Hence, the only choice is whether to put the two factors 1 - x + x^{2} together or not.

The first option yields G(x) = H(x) = x (1 + x) (1 + x + x^{2}) (1 - x + x^{2}) = x^{1} + x^{2} + x^{3} + x^{4} + x^{5} + x^{6}, which yields the classical dice. The second option yields the unique other solution; one dice with {1, 3, 4, 5, 6, 8} and one dice with {1, 2, 2, 3, 3, 4}:

G(x) = x (1 + x) (1 + x + x^{2}) (1 - x + x^{2})^{2} = x^{1} + x^{3} + x^{4} + x^{5} + x^{6} + x^{8},

H(x) = x (1 + x) (1 + x + x^{2}) = x^{1} + x^{2} + x^{2} + x^{3} + x^{3} + x^{4},

### Remarks

Students often wonder why they should study generating functions. Artificial counting problems don't convince them. And I can't blame them. Alas, most routine exercises on this subject are just counting problems. This is a different kind of problem, with for most an unexpected result. I googled this and it turns out to be known as the

Sicherman dice.