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 face-up values are added.
What is the lowest possible value of the highest number on one of the dice faces?
(In reply to
re(2): Possible solution CONCERNS EVERYBODY by Ady TZIDON)
Helpful. I use my original method, borrow Charlie's numbers and plug them in with Dej's suggestion about adding and subtracting, so that the largest number is 21, as per your post:
(x^16+x^17+x^18+x^19+x^20+x^21)*(x^(-15)+x^(-9)+x^(-3)+3+x^9+x^15)=
x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^10 + x^11 + x^12 + x^13 + x^14 + x^15 + 4 x^16 + 4 x^17 + 4 x^18 + 3 x^19 + 3 x^20 + 3 x^21 + x^25 + x^26 + x^27 + x^28 + x^29 + x^30 + x^31 + x^32 + x^33 + x^34 + x^35 + x^36.
Aren't committees great?
I look forward to seeing the proof that it is minimal in due course.
It's amazing that even though all 36 numbers can be scored, there are still, according to the expansion, apparently 3 or 4 different ways of making each of the numbers from 17 to 21. Seems almost too good to be true...
Edited on October 31, 2011, 10:08 am
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Posted by broll
on 2011-10-31 09:34:40 |