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)?

Few may be defined as "not very many people or things, but more than two". If we assign the variable a to the number of students that were absent, we should then be able to say that
2 < a.
Many may be defined as "a majority of". The number of students of the class is given as 30. A majority of 30 is greater than [30/2], i.e., 15. Thus
a < 15.
Therefore, the maximum number of cubes that could have been displayed is 27 (= 30 - (2 + 1)), and the minimum number of cubes that could have been displayed is 16 (= 30 - (15 - 1)).

With the quantity of students with red markers as r and the quantity of students with blue markers as b, the equation of the number of students present (also the number of cubes handed out), c, can be expressed as:
c = (30 - a) = (r + b)

The top-side of each cube is given as a 5, the unseen bottom of each cube would have been marked with a 2 (= 7 - 5). The side facing the teacher is given as a 3, its opposite side would then be a 4 (= 7 - 3). Thus, the right and left side of each cube would have been marked with the remaining two numbers, 1 and 6.

Following is a table of left and right face combinations (from 1 to 27 cubes) where the left faces, composed solely of 1's and 6's, total to a perfect square:

Left              Right            Total
Face              Face             Red
Total   1's 6's   Total   1's 6's  Pens
  1     1   0       6     0   1     1
  4     4   0      24     0   4     4
  9     9   0      54     0   9     9
  9     3   1      19     1   3     4
 16    16   0      96     0  16    16
 16    10   1      61     1  10    11
 16     4   2      26     2   4     6
 25    25   0     150     0  25    25
 25    19   1     115     1  19    20
 25    13   2      80     2  13    15
 25     7   3      45     3   7    10
 25     1   4      10     4   1     5
 36    24   2     146     2  24    26
 36    18   3     111     3  18    21
 36    12   4      76     4  12    16
 36     6   5      41     5   6    11
 36     0   6       6     6   0     6
 49    19   5     119     5  19    24
 49    13   6      84     6  13    19
 49     7   7      49     7   7    14    <--
 49     1   8      14     8   1     9
 64    16   8     104     8  16    24
 64    10   9      69     9  10    19
 64     4  10      34    10   4    14
 81    15  11     101    11  15    26
 81     9  12      66    12   9    21
 81     3  13      31    13   3    16
100    10  15      75    15  10    25
100     4  16      40    16   4    20
121     7  19      61    19   7    26
121     1  20      26    20   1    21
144     0  24      24    24   0    24

As can be seen, there is only 1 combination where both the left and right faces total perfect squares. This occurs where the face totals for both are 49, thus r = 14. And with 14 students with red pens, the maximum number of students that could have been handed a blue pen is 13 (= max - r = 27 - 14), and the minimum number of students that could have been handed a blue pen is 2 (= min - r = 16 - 14).

Following is a table of left and right face combinations from 2 (the minimum) to 13 (the maximum) cube where the left faces, composed of solely 1's and 6's, total to a prime:

Left              Right            Total
Face              Face             Blue
Total   1's 6's   Total   1's 6's  Pens
   2     2   0      12     0   2     2
   3     3   0      18     0   3     3
   5     5   0      30     0   5     5
   7     7   0      42     0   7     7
   7     1   1       7     1   1     2    <--
  11    11   0      66     0  11    11
  11     5   1      31     1   5     6    <--
  13    13   0      78     0  13    13
  13     7   1      43     1   7     8    <--
  13     1   3       9     3   1     4
  17     5   2      32     2   5     7
  17    11   1      67     1  11     7    <--
  19     7   2      44     2   7     9
  19     1   3       9     3   1     4
  23     5   3      18     3   5     8
  29     5   4      34     4   5     9
  31     7   4      46     4   7    11
  31     1   5      11     5   1     6    <--
  37     7   5      47     5   7    12    <--
  37     1   6      12     6   1     7
  41     5   6      36     6   5    11
  43     7   6      48     6   7    13
  43     1   7      13     7   1     8    <--
  47     5   7      37     7   5    12    <--
  53     5   8      38     8   5    13
  61     1  10      16    10   1    11
  67     1  11      17    11   1    12    <--


As can be seen, there are multiple solutions where the right face total is also a prime. A number of these, in effect, are mirrored solutions between the left and right face totals. Thus we have the following for (absent, red pens, blue pens, [squares], [primes]:

• 14, 14,  2, [49 & 49], [ 7 &  7]
• 10, 14,  6, [49 & 49], [11 & 31]
•  8, 14,  8, [49 & 49], [13 & 43]
•  4, 14, 12, [49 & 49], [17 & 67]
•  4, 14, 12, [49 & 49], [37 & 47]

With the additional constraint that 'the ratio of the larger to the smaller prime was less than 2', we have but two solutions:

• 14, 14,  2, [49 & 49], [ 7 &  7] ratio: 7/7    = 1 
•  4, 14, 12, [49 & 49], [37 & 47] ratio: 47/37 ~= 1 1/4

Edited on September 25, 2008, 6:29 am

Posted by Dej Mar on 2008-09-24 12:02:21