Dice Rolling Logic Puzzle Walkthrough


3 3
4 4 4
5 5 5 5
6 6 6 6 6
7 7 7 7 7 7
8 8 8 8 8
9 9 9 9
10 10 10
11 11

One 2
Two 3’s
Three 4’s
Four 5’s
Five 6’s
Six 7’s
Five 8’s
Four 9’s
Three 10’s
Two 11’s
One 12

(Back to puzzle)


1.) The clue in “How to play” tells you that each corner is a prime number of which there are only four. That description only fits 5, so the corner cells are all 5.

2.) Cell 6 says that of Cells 1, 8, and 27, one of them is the same number as the cell number, and one of them is the cube root of its cell number. Since 1 and 27 are not possible answers, you know that 8 goes in Cell 8. Since 5 is not the cube root of 1, and since 1 is not a possible answer anyway, you know that Cell 27 must have the cube root of 27, which is 3.

3.) Take a look at the fifth column. The first 5 cell numbers are all prime numbers. That means that, according to Cell 31, the fifth column is in ascending order. It also ends (Cell 35) with a prime number according to Cell 8. The column could go 2,3,4,5,6,7 except that all the 5’s are already known. So the only prime number that can go in Cell 35 that will allow that column to be in ascending order is 11.

4.) Since the fifth column is in ascending order (according to Cell 31), Cell 5 can be no greater than 6. Since all the 5’s are known, Cell 5 is definitely a factor of 12 (either 2,3,4, or 6). So you know that Cell 35 is talking about Cell 5. This means that Cell 2 cannot be a 2, 3, 4, or 6. All the 5’s are already known, and Cell 2 is prime, meaning it must be smaller than the 8 below it (Cell 31’s rule). So Cell 2 can only be 7.

5.) According to Cell 1, there’s no way any 7 could be in the sixth row. All the 5’s are known and one 3 and one 11 are known. This means that the only prime numbers left available to fill the sixth row are the 2, the other 11, and the other 3. This means that Cell 1 is referring to 12. There’s no way more than one number can be adjacent to the same six numbers, and there are six 7’s. So Cell 1 could only possibly refer to a number of which there is only one: 2 or 12. Knowing that 2 is in row 6, Cell 1 must refer to 12. And the only place left to be adjacent to six 7’s (since one of them is already known in Cell 2) is Cell 9. (Note: you can also get the 12 after step 7 by simple process of elimination)

6.) Since the 2 and the other 3 are in Row 6, Cell 5 can either be 4 or 6. Since there won’t be a 7 in the fifth column (because they’re all adjacent to the 12 in Cell 9), Cell 5 can’t be 6 and still keep Cell 31’s rule intact. So Cell 5 must be 4.

7.) All the other cells in the fifth column fall into place since the 5’s are all known and the 7’s are all adjacent to the 12. The column must be 4, 6, 8, 9, 10, 11.

8.) The first column has three consecutive prime cell numbers: 7, 13, and 19. This means that Cells 7, 13, 19, and 25 must be in ascending order. The 12 is known. One 11 is known and the other must also be in the sixth row. So the largest number Cell 25 can be is 10. There can be no 7 in this column, so the largest number Cell 7 can be is 6. The 5’s are all known, and the 2 and unknown 3 are both in the sixth row. Since the 12 in Cell 9 is adjacent to six 7’s, you know that the 4 in Cell 5 is adjacent to at least one 7. Therefore, according to Cell 23, Cell 7 cannot be a 4. Cell 7, therefore, can only be a 6.

9.) Again, the whole first column falls into place: it must be 5, 6, 8, 9, 10,

10.) Since the 2 must be in the sixth row (because there are only three prime numbers left and the sixth row is only prime numbers), it must be in Cell 33, because Cell 9 says it is not adjacent to a 10.

11.) The numbers in the third row currently add up to 16. You also know it must have at least two 7’s. So the row adds up to 30 with two more cells. Therefore the final two cells in the third row must equal 12, according to Cell 33. Since all the 5’s are known, there cannot be three 7’s in the row. Therefore Cells 3,4, and 10 are all 7, according to Cell 1.

12.) The sixth row is going to be 11, 11, 5, 5, 3, 2. So the greatest four numbers are 11, 11, 5, and 5. Since you can only place two more 4’s, the only way for two 11’s and two 5’s each to be adjacent to a 10 and a 4 is for the other 11 to be in Cell 32.

13.) Following step 12 and Cell 4, Cells 26 and 30 are both 4.

13.) The only prime number left for Cell 34 is the other 3.

14.) Back to the third row and Cell 33. The third row is going to be 8, 8, 7, 7, x, y, with x+y=12. Here are the possibilities for 12: 10+2, 9+3, 8+4, 7+5, 6+6. There is no 2 left, no 3 left, no 4 left, no 5 left, and no 7 left. So the only option is 6+6. You know that among Cells 14-16 are two 7’s, so one of those is a 6 and Cell 18 also must be a 6.

15.) Two 9’s are known, and they are in different columns, both of which are complete. Cell 18, therefore, tells you that the remaining two 9’s are in the same column. Cells 14-16 must be two 7’s and a 6; therefore the final two 9’s must go in Cells 22 and 28.

16.) Cell 22 forces Cell 21 to be an 8, for the sake of the 9 in Cell 28.

17.) The first row is complete and has three different numbers. The second row already has four. The third row has two, but is certainly going to add two 7’s and a 6 for a total of three different numbers; therefore Cell 21 is describing the first and third rows. This means that the fourth row cannot have exactly three different numbers. It currently has only two, but there’s only one more 8 and no more 9’s. So the fourth row cannot remain at only two different numbers. So it must add two more different numbers. Both 7’s are in the third row, so the numbers to add to the fourth row are 6 and the final 10. According to Cell 36, the 10 is in Cell 24 and the 6 is in Cell 20. Also, since you know that Cells 14-16 are two 7’s and a 6, you know that the final 8 goes in Cell 12.

18.) Cell 20 tells you that the final 6 is in Cell 16.

19.) Cells 14 and 15 must both be 7’s.

(Back to puzzle)



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