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Five cubic dice were rolled, and the product of the numbers obtained was 432. What is the largest possible sum of five such numbers?

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If we assume these are regular dice being rolled one at a time, the answer would be

6+6+6+2+1=21

...unless, of course,

the method of rolling the dice was left intentionally vague: one could throw three dice at once followed by two single rolls and get 18+6+4=28 instead.

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...unless, of course

Using 'existing' dice: Backgammon uses a doubling dice with value 16 on one of its sides. Mix that with 4 normal dice to get 16+3+3+3+1 = 26

However:

The maximum is unbounded. It seems obvious that it is not intended as a trick question, but you only need (for example) one face with value 1/n (rolled 4 times) and another with value n^4*432.

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  • $\begingroup$ Our answers can be combined for a score of gjb uhaqerq rvtugrra... $\endgroup$ – AxiomaticSystem Oct 19 at 14:26
  • $\begingroup$ Not sure how you got that total, but: Gur onpxtnzzba qbhoyvat qvpr vf n phor nf erdhrfgrq gung bayl tbrf gb guvegl-gjb. Rvtugl-bar vf gur pbzovarq znk $\endgroup$ – Retudin Oct 19 at 15:14
  • $\begingroup$ oeps that should be 64 and 148 $\endgroup$ – Retudin Oct 19 at 15:19
  • $\begingroup$ Ah, a sixth cube must have slipped in while I was calculating $\endgroup$ – AxiomaticSystem Oct 21 at 1:11
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The regular cubic backgammon dice have a number of dots from 1 to 6. The divisors of 432 are 1,2,3,4,6 the best chance to obtain the maximum sum is if the the combination has the most divisors. If we assume we throw one or two dice at a time we obtain the following combinations and their sum.

6,6,6,2,1 the sum is 21

4,4,3,3,3 the sum is 17

6,6,4,3,1 the sum is 20

6,4,3,3,2 the sum is 18

To obtain the maximum sum 6+6+6+6+6=30 we have the least probability. In addition to that $6^5$ is far greater than 432. So I will bet as maximum sum 20.

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