Problem 129 Solution

The following table shows the frequency of each total.

Two dice totals
Die 1 Die 2
1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12

Note that the manner of systematically taking the sums of each set of faces is the same as adding exponents when taking the product of (x6+x5+x4+x3+x2+x1)2 = x12+ 2x11+ 3x10+ 4x9+ 5x8+ 6x7+ 5x6+ 4x5+ 3x4+ 2x3+ x2. The coefficient for xn corresponds to the number of ways a total of n can be thrown.

The challenge is to find two other polynomials whose product is also x12+ 2x11+ 3x10+ 4x9+ 5x8+ 6x7+ 5x6+ 4x5+ 3x4+ 2x3+ x2. Let's start by factoring each standard die.

x6+x5+x4+x3+x2+x1 =
x*(x5+x4+x3+x2+x1+1) =
x*(x6-1)/(x-1) =
x*(x3-1)*(x3+1)/(x-1) =
x*(x-1)*(x2+x+1)*(x+1)*(x2-x+1)/(x-1) =

So the product of two dice is
(x6+x5+x4+x3+x2+x1)2 =
(x*(x2+x+1)*(x+1)*(x2-x+1))2 =

So we want to divide these terms into two sets, one for each die. There are lots of ways to do this so lets try to narrow down our options. We know each face has to have at least one dot. In other other words the polynomial for each die must not have a 1=x0 term, which would correspond to a face with zero dots. So we have to put on x on each side, otherwise one side would be left with a 1.

We also know the number of faces on each die is six. So each die must have six factors. If we let x=1 then the polynomial for each die should equal 6. Let's look at the value of each factor when x=1.

x: 1
x+1: 2
x2-x+1: 1
x2+x+1: 3

So we want to divide these up into two groups such that the product of the terms equals 6. We already know we have to put one x on each side. There is only one way to get to a product of 6 with these constaints: 2*3 = (x+1)*(x2+x+1).

So we know each side must contain x*(x+1)*(x2+x+1). With the (x2-x+1) we are free to do as we wish. Putting one set on each die results in two standard dice. So putting two sets on one side and zero on the other results in:

x*(x+1)*(x2+x+1) and x*(x+1)*(x2+x+1)*(x2-x+1) 2 =
x + 2x2 + 2x3 + x4 and x + x3 + x4 + x5 + x6 + x8.

So the two dice are {1,2,2,3,3,4} and {1,3,4,5,6,8}.

I'd like to thank Nick's Math Puzzles for helping with the solution to this problem. Nick's solution can be found at There are also two alternative answers for 8-sided dice and one for 35-sided dice. For more on that visit Mathnerds of do a search on "Sicherman Dice".

Michael Shackleford, A.S.A., March 4, 2003