Problem 8.3. Masha has three identical dice, on the faces of each of which six different prime numbers are written, with a total sum of 87.

Masha rolled all three dice twice. The first time, the sum of the numbers that came up was 10, and the second time, the sum was 62.

Exactly one of the six numbers did not come up at all. Which one?
# Answer: 17.

Solution. Note that the number 10 can be uniquely represented as the sum of three prime numbers: $10=2+3+5$. This means that the numbers $2,3,5$ are on the dice and they came up the first time.

Note that if the even number 62 can be represented as the sum of three prime numbers, then one of them is even and therefore equals 2. Then the sum of the remaining two is 60. Note that among these two numbers, there cannot be the number 2, 3, or 5, and they must be different (since the numbers 58, 57, 55, and 30 are composite). This means that the dice have two different prime numbers with a sum of 60, greater than 5, and they came up the second time (along with 2).

Thus, each die has the numbers $2,3,5$, as well as two other prime numbers with a sum of 60. Since the sum of all six numbers is 87, the sixth number, which never came up, is $87-2-3-5-60=17$.