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)?
(In reply to
Solution by Dej Mar)
Nice analysis, Dej. There are indeed multiple solutions (for blue) if there is not a further constraint on the primes sums. The recent origin of this problem used "the sums were both primes with a ratio between 1 and 2" where "between" eliminated your middle three, and also the first (both 7), leaving only the final row (47/37). Perhaps one could argue that even 8 was more than "a few" absent, leaving your final row as the solution.
(Apparently you were entering your posting while I was typing my second comment, so I had not seen your solution until after I posted my second. Perhaps Charlie will agree that the problem should have had some other way to constrain the blue options -- but he was right to add that these dice in some respects were not standard dice.)
re: Solution
