Christmas Streak 10/88: Two Strange To Be Strange

We need to use generating functions .

A normal dice can be expressed as $(x+x^2+x^3+x^4+x^5+x^6),$ where throwing two normal dices would be

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

The exponents express the sum; the coefficients express the number of cases.

Then let $f(x)=(x+x^2+x^3+x^4+x^5+x^6)(x+x^2+x^3+x^4+x^5+x^6).$

Let $g(x)$ and $h(x)$ be the generating functions of the two dices, respectively.

$f(x)=g(x)h(x),$ but we all know that the sum of the coefficients of $g(x)$ and $h(x)$ should be $6$ respectively.

Then $g(1)=h(1)=6.$

We all know that both $g(x)$ and $h(x)$ should have $x$ as their factor.

Since $f(x)=(x+x^2+x^3+x^4+x^5+x^6)^2=\{x(x^2+x+1)(x^3+1)\}^2=x^2\cdot (x+1)^2\cdot (x^2-x+1)^2\cdot (x^2+x+1)^2,$

$f(1)=g(1)h(1)=1^2\cdot 2^2\cdot 1^2\cdot 3^2.$

The only way of dividing this (without making $g(x)$ and $h(x)$ equal, of course) is:

$g(x)=x(x+1)(x^2+x+1)$ and $h(x)=x(x+1)(x^2-x+1)^2(x^2+x+1)^2.$

Expanding both expressions, we get

$g(x)=(x^2+x)\{(x^2+x)+1\}=x^4+2x^3+x^2+x^2+x=x^4+2x^3+2x^2+x,$

$h(x)=\{(x^2+x)(x^2-x+1)\}\{(x^2+1+x)(x^2+1-x)\}=(x^4+x)(x^4+x^2+1)=x^8+x^6+x^5+x^4+x^3+x.$

Therefore, the two dices would be $(1,~2,~2,~3,~3,~4)$ and $(1,~3,~4,~5,~6,~8).$

$\therefore~122334\times134568=\boxed{16462241712}.$

The solution with generating functions seems to be the most general. However, one can often find some of the possibilities as follows.

First, I subtract one from each of the two six-sided dice. This does not significantly alter the problem. Instead of rolling a six-sided die, one could also roll a three-sided "die" and a two-sided "die": $\{0,1,2,3,4,5\} = \{0,1,2\} + \{0,3\}\ \ \ \ \ \ \mathbb D_6 = \mathbb D_3 + 3\mathbb D_2.$ Or also, $\{0,1,2,3,4,5\} = \{0,1\} + \{0,2,4\}\ \ \ \ \ \ \mathbb D_6 = \mathbb D_2 + 2\mathbb D_3.$ The sum of two six-sided dice may therefore be written as $\mathbb D_6 + \mathbb D_6 = (\mathbb D_3 + 3\mathbb D_2) + (\mathbb D_2 + 2\mathbb D_3).$ The terms can be shuffled to yield a different pair of $\mathbb D_3/\mathbb D_2$ combinations: $\mathbb D_6 + \mathbb D_6 = (\mathbb D_3 + \mathbb D_2) + (3\mathbb D_2 + 2\mathbb D_3).$ This describes the pair of dice we are looking for. They are known as Sicherman dice . We add the "1" back in: $\mathbb D_A = 1 + \mathbb D_3 + \mathbb D_2 = 1 + \{0,1,2\} + \{0,1\} = \{1,2,2,3,3,4\};$ $\mathbb D_B = 1 + 3\mathbb D_2 + 2\mathbb D_3 = 1 + \{0,3\} + \{0,2,4\} = \{1,3,4,5,6,8\}.$

Note . There are other ways to recombine the terms of $\mathbb D_6 + \mathbb D_6$ , resulting in different set of dice with this property. For instance, with $\mathbb D_6 + \mathbb D_6 = (\mathbb D_3 + 3\mathbb D_2) + (\mathbb D_3 + 3\mathbb D_2) \\ = (\mathbb D_3 + \mathbb D_3) + (3\mathbb D_2 + 3\mathbb D_2)$ we get a nine-sided die and a four-sided die: $1 + \mathbb D_3 + \mathbb D_3 = 1 + \{0,1,2\} + \{0,1,2\} = \{1,2,2,3,3,3,4,4,5\};$ $1 + 3\mathbb D_2 + 3\mathbb D_2 = 1 + \{0,3\} + \{0,3\} = \{1,4,4,7\}.$ Can you find a different solution with a nine-sided die and a four-sided die?

If it's a little late in the evening and the brain is no longer well-equipped for math, brute-forcing is your friend.

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#!/bin/env python  
from itertools import product, combinations_with_replacement  
from collections import Counter  
cgen=lambda a,b: Counter(map(lambda x:x[0]+x[1],product(a,b)))  
c=cgen(range(1,7),range(1,7))  
for a, b in combinations_with_replacement(combinations_with_replacement(range(1,10),6),2):
   if c == cgen(a,b):
      print(str(a) + ' x ' + str(b) + ' = ' +str(int(''.join(str(d) for d in a))*int(''.join(str(d) for d in b))))  

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(1, 2, 2, 3, 3, 4) x (1, 3, 4, 5, 6, 8) = 16462241712  #<--- this is the one.  
(1, 2, 3, 4, 5, 6) x (1, 2, 3, 4, 5, 6) = 15241383936  #<--- normal dice  

I don't know how to properly format source code here. Had to do it in a way to avoid indents. Looks terrible, but will work.