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