Two six sided dice are marked with 12 different integers, so that any number from 1 to 36 can be derived when both dice are thrown and the showing faceup values are added.
What is the lowest possible value of the highest number on one of the dice faces?
Lovely problem.
Take the known solution: f(x)=(x^1+x^2+x^3+x^4+x^5+x^6)(1+x^6+x^12+x^18+x^24+x^30)
The trick is to repartition this into two factors of six elements, something like (x.x.x.x.x.x)(x.x.x.x.x.x) with each partition having 6 factors. Trial and error suggests some guidelines:
1. Terms including a minus sign have to be multiplied by something similar to eliminate the negative terms.
2. Where possible match up the smallest available factor(s) to the largest.
These guidelines though not rigorous are sufficient for present purposes, particularly as otherwise there are hordes of possible combinations.
Start by combining (1x+x^2) (1+x+x^2) to obtain (1+x^2+x^4), This again matches with (1x^2+x^4) to produce 1+x^4+x^8: similarly, (1x^3+x^6) (1+x^3+x^6)= 1+x^6+x^12, multipled by (1x^6+x^12)= x^24+x^12+1.
We have left x (1+x) (1+x^2). The second two factors cannot be combined as they would produce too many terms, so we have either x(1+x) and (1+x^2) or x(1+x^2) and (1+x). In due course we find: (x^25+x^24+x^13+x^12+x+1)(x^11+x^9+x^7+x^5+x^3+x).
Even though the guidelines are not rigorous, they are good enough in this case to ensure that no smaller solution is possible.
The exponents give the numbers on the dice.
So the two dice are numbered: (1,3,5,7,9,11)( 0,1,12,13,24,25)
Note: 26 and 27 are also possible solutions.
Edited on October 30, 2011, 11:38 am

Posted by broll
on 20111030 11:34:43 