The class had 30 students, but a few were absent that day. The teacher gave each student present a cardboard cube and either a red or a blue marking pen, and she told each student to mark the faces of his or her cube like the faces of a die: with numbers 1 through 6 on the faces in such a manner that opposite faces add to 7. This was the only required similarity to a standard die.
The teacher gathered all the cubes and placed all the red-marked ones in one row, and all the blue-marked ones in another row. Both rows stretched from left to right and each cube had its 5 face on top and its 3 face toward the teacher.
In the red row, the sum of the digits on the left sides of the cubes was a perfect square, as was the sum of the digits on the right sides.
In the blue row, the corresponding sums were not squares, but rather prime numbers. The ratio of the larger to the smaller prime was less than 2.
How many students were absent? How many red markers and how many blue markers were handed out? What were the sums involved (the perfect squares and the primes)?
Forgot to note, that the sums for the red (perfect square) were 49; and the sums for the blue (two different primes) were 37 and 47. I infer that the primes must be different from the phrases "but rather prime numberS", and "the primeS" -- otherwise there might have been only two blue, and the side sums might each have been the same prime, viz. 7. (There is no stated requirement here that the ratio of the two blue primes must be between 1.0 and 2.0 -- but without that or a simulacrum the solution would not be unique.) Perhaps saying "a FEW were absent" precludes the smaller solution which would have almost half the class (14) absent.