An ordinary die is labeled with numbers from 1 to 6. These numbers are arranged on the die such that the sum of the numbers on opposite faces—called the die constant—always equals 7. Label the faces of a special die with six different prime numbers so that the die constant is as small as possible.
Except for 2, all prime numbers are odd. Therefore, 2 cannot be on the prime number die, because then the die constant, K, would be odd for one pair of sides and even for the other two pairs. Thus, all prime numbers on the die are odd, and K is even.
Call the sum of the six numbers on the die S. Because K = S⁄3 and is also an even number, S must be a multiple of 6. The six smallest odd prime numbers are 3, 5, 7, 11, 13 and 17. Their sum is 56, which is not a multiple of 6. Therefore, these prime numbers cannot be the six numbers on the die.
The next smallest possible value is S = 60, which leads to K = 20. But 20 can only be represented as the sum of two different prime number pairs: 3 + 17 and 7 + 13. Similarly, 22 can only be represented in two ways: 3 + 19 and 5 + 17. Only K = 24 leads to the solution: 5 + 19, 7 + 17 and 11 + 13.
Because the larger prime numbers are opposite the smaller ones on the cube, there is a corner where the three sides with the smaller numbers meet. Around this corner, the three numbers 5, 7 and 11 can be arranged either clockwise or counterclockwise. There are no other possibilities. Consequently, the six prime numbers 5, 7, 11, 13, 17 and 19 can only be arranged on the cube in two different ways. Either way is an acceptable answer.
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This puzzle originally appeared in Spektrum der Wissenschaft and was reproduced with permission. It was translated from the original German version with the assistance of artificial intelligence and reviewed by our editors.