7
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The 1 x 1 x 2 die with new configuration was created from 2 dice (with opposite faces totalling 7) that were glued together on one of their faces. If this long die is rolled twice, the sum of non-repeated results can be equal to 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16 or 20. What are the glued faces of the 2 dice?

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  • $\begingroup$ Perhaps i'm misunderstanding, but how can 5 be the sum of two rolls? Given that the dice is gonna land with two faces up, the rolls would have to be 2 + 3. In order to get 2, it'd have to be both 1 faces on top, then how would you be able to roll a 3 next time, which would require a 1 and 2 face on top $\endgroup$ – indubitablee Jun 22 '17 at 16:09
  • $\begingroup$ @indubitablee-the die can also land on 1x1 faces though less likely $\endgroup$ – TSLF Jun 22 '17 at 16:20
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    $\begingroup$ @TSLF haha understood, thanks! #gravityRebels #momentumShmomentum $\endgroup$ – indubitablee Jun 22 '17 at 16:22
  • $\begingroup$ What does "sum of non-repeated results" mean exactly? $\endgroup$ – Gareth McCaughan Jun 22 '17 at 16:23
  • $\begingroup$ @GarethMcCaughan Presumably that rolling the same side twice in a row wouldn't be counted. Looking at the results, this must be the case, as if duplicates were counted, the smallest roll is necessarily even. $\endgroup$ – Ian MacDonald Jun 22 '17 at 16:29
5
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The two hidden faces are

3 and 5

The net of the new die is:

+-----+ | 1 4 | +---+-----+---+ | 4 | 2 1 | 2 | +---+-----+---+ | 6 3 | +-----+ | 5 6 | +-----+

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  • $\begingroup$ note the chirality $\endgroup$ – TSLF Jun 22 '17 at 17:00
  • $\begingroup$ How did you work that out? I'm struggling to think of appropriate logic tht isn't just brute force.... $\endgroup$ – Chris Jun 23 '17 at 16:32
  • $\begingroup$ @Chris The 5 couldn't be made from two double faces, as there aren't enough 1s. I guessed that it was made of a double face and a single face. That only left two possibilities, and one happened to be right. The rest was pretty much just following the implications. $\endgroup$ – user27014 Jun 23 '17 at 16:46
  • $\begingroup$ @Chris-5 can be made from a double face and a single face: [22]1],[21]2] & [11]3] $\endgroup$ – TSLF Jun 24 '17 at 18:24
1
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The original faces glued together are indeed as mentioned in user27014’s answer ...

                                                  ___________
                                                 |           |
 ... 3 and 5, producing this long die            |  O     O  |
              with faces 2,3,4,5,9,11, ...       |           |
                                                 |  O     O  |
              ___________ ___________ ___________|___________|
             |           |           |           |           |
             |  O     O  |  O  O  O  |        O  |           |
             |     O     |           |           |     O     |
             |  O     O  |  O  O  O  |  O        |           |
             | . . . . . | . . . . . | . . . . . | . . . . . |
             |           |           |           |           |
             |  O     O  |        O  |           |  O     O  |
             |  O     O  |     O     |     O     |           |
             |  O     O  |  O        |           |  O     O  |
             |___________|___________|___________|___________|
             |           |
             |        O  |                  (the original dice
             |           |                  happen to be mirror
             |  O        |                 images of each other)
             |___________|
 

... derived with a method introduced by demonstration on a typical die.

The two-roll sum of a standard 6-sided die is equivalent to a product of two numbers. Notice how the digits 1,2,3,4,5,6,5,4,3,2,1 from the product 111111×111111 = 12345654321 represent the familiar distribution of roll sums 2,3,4,5,6,7,8,9,10,11,12:

1× 2   2 × 3   3 × 4   4 × 5   5 × 6   6 ×7   5 × 8   4 × 9   3 ×10   2 ×11   1×12  

           111111
         x 111111
     -------------
           111111
          111111
         111111
        111111
       111111
      111111
     =============
      12345654321

The two-roll sum of a newfangled long die can also be represented as the product of two numbers. This time the counts of the sums are not known but their values are, as 5,6,7,8,9,11,12,13,14,15,16,20, so the distribution is (where # = a non-0 count):

# × 5   # × 6   # ×7   # × 8   # × 9   0 ×10   # ×11   # ×12   # ×13   # ×14   # ×15   # ×16   0 ×17   0 ×18   0 ×19   # × 20

Turn that distribution into a 16-digit number, #####0######000#, and adjust the product for two dice in a way that makes that number. Before any adjustment, the point of departure is:

            111111                                            111111
          x 111111                                          x 111111
      -------------                                     -------------
            111111            Remove  ------>                 11111.
           111111              the     ----->                1111.1
          111111                6      ----->               111.11
         111111              possible  ----->              11.111
        111111               repeated ------>             1.1111
       111111                 rolls  ------->            .11111
      =============                                     =============
       12345654321                                        224464422
                                                                   |
                                                                   |
                                               ignore any trailing 0s

Stretch the multiplicands with 0s until the product is correct. The first stretch inserts 000 before the product’s last visible digit.

                                   stretched                         stretched
                                     \   /                              \ /
       111111     -->             111100011     -->              1111000101
     x 111111     -->           x 111100011     -->            x 1111000101
 -------------           -------------------           ---------------------
       11111.                     11110001.                      111100010.
      1111.1                     1111000.1                     1111000.01
     111.11                  111.00011                     111.000101
    11.111                  11.100011                     11.1000101
   1.1111                  1.1100011                     1.11000101
  .11111                  .11100011                     .111000101
 =============           ===================           =====================
   224464422      -->      224220244420002      -->      2242202244220002
          /\                    /     \                  #####0######000#
       want 000                too few #s                   just right

Interpret the multiplicand, 1111000101. Its digits represent the counts of faces, shifted by the value of the lowest face, variable a.

1× a1× (a +1)   1× (a + 2)   1× (a + 3)   0 × (a + 4)   0 × (a + 5)   0 × (a + 6)   1× (a +7)   0 × (a + 8)   1× (a + 9)

As the sum of the two lowest faces is 5 = a + (a +1), a = 2 so the long-die faces are ...

... 2,3,4,5,9,11, ...

... leading to two observations.

  1. As the long-die face 11 can only be 5 + 6 and as opposite faces of the original dice add up to 7, the long-die face opposite of 11 must be (7− 5) + (7− 6) = 2 +1 = 3. Furthermore, as one of the original 1s is now taken, the long-die face 2 cannot be 1+1 and must be an original 2 by itself.

The sum of the glued/hidden faces is the difference between the sum of all original faces minus the sum of faces on the long die.

  1. (1+ 2 + 3 + 4 + 5 + 6) + (1+ 2 + 3 + 4 + 5 + 6) − (2 + 3 + 4 + 5 + 9 +11) = 8 hidden dots combined on the glued faces. So the only possibilities are 2 + 6, 3 + 5 and 4 + 4.

Observations 1 and 2 combine to a single conclusion.

Can the hidden-faces sum be 2 + 6? No, both 2s are already known to be visible.

Can the hidden sum be 4 + 4? No, as that would cause long-die face 9 to be 3 + 6, making its opposite face be (7− 3) + (7− 6) = 4 +1 = 5, but both 4s would already be accounted for in hiding.

Only one possibility remains for the hidden sum: 3 +5.

Ensuing long-die face combinations, as laid out at the top of this solution, work out automatically although uniqueness is not proven.

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    $\begingroup$ I like the "multiplication" interpretation, and the visuals that go along with it! Seems like this could be a very good introduction to generating functions. $\endgroup$ – Deusovi May 2 at 3:43
  • $\begingroup$ This solution only works for gluing Western die and Eastern die? $\endgroup$ – TSLF May 2 at 9:41
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    $\begingroup$ Ah, @TSLF , your creative puzzle is especially sweet for unveiling this kind of solution. Integer multiplication as shorthand for polynomials works neatly for any number of dice, any mix of side counts, and a broad range of conditions on face values. As a bonus not in play this time, desired solutions can be untangled by online factorizers! For the 2-standard-dice example, 111111×111111 = 12345654321 = 3×3×7×7×11×11×13×13×37×37 = (11×11)×(3×7×13×37)×(3×7×13×37) = (121)×(10101)×(10101) means that the same distribution of sums is obtained by one {0,1,1,2} 4-sided die + two {1,3,5} 3-sided dice. $\endgroup$ – humn May 2 at 14:55

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