Ways to complete the grid puzzle


I want to complete the cross figure shown here. I would like all eleven answers to be prime numbers and none to have a leading zero. Also I don’t want any digit to appear more than once in any row or column or on either of the two long diagonals.

complete-the-grid-puzzle

Given the starting values shown, how many different ways could I complete the grid?

Reveal Solution
We use 1d for “1 down”, etc., and introduce notation as shown here:

complete-the-grid-solution-key

Evidently \{{\sf C},{\sf J},{\sf M}\}=\{1,3,7\} and we rule out 5 of the possible six permutations as follows:

  • If ({\sf C},{\sf J},{\sf M})=(1,3,7) or ({\sf C},{\sf J},{\sf M})=(1,7,3) or ({\sf C},{\sf J},{\sf M})=(7,3,1) then it quickly follows that 6a=691 and {\sf D}=6. This implies that {\sf E}\in\{1,7\} and this clashes with either 3a or 9d and these cases fail.
  • If ({\sf C},{\sf J},{\sf M})=(3,7,1) then {\sf I=1} and there is no way to choose {\sf F}.
  • If ({\sf C},{\sf J},{\sf M})=(7,1,3) then {\sf F}=4 and {\sf G}\in\{1,9\} which are numbers that neighbour {\sf G}.

We are left with ({\sf C},{\sf J},{\sf M})=(3,1,7) and the grid has been narrowed down to the summary on the right here:

complete-the-grid-solution

To finish the job we look at cases according to 5d. There are 4 possibilities:

  1. If 5d=271 then every square is uniquely defined except {\sf A}\in\{5,8\}.
    So there are two possible grids here.
  2. If 5d=431 then {\sf E}=1 and we can now consider two subcases according to 9d
    1. If 9d=37 then 2d=61 and 7d=67. The grid is unique in this case.
    2. If 9d=97 then 2d\in\{31,61\} and 7d\in\{31,61\}. Four different grids here.
  3. If 5d=571 then the grid is unique except {\sf A}\in\{2,8\}.
  4. If 5d=631 then the grid is uniquely defined except for {\sf H}\in\{3,6\}.

There are eleven ways to complete the grid.

Problem Page Coordinator: Stephen Lee CMath MIMA – Mathematics in Education and Industry
Acknowledgement: The IMA are indebted to Peter Chamberlin FIMA (University of Reading) for creating and supplying Mathematics Today with this puzzle.
First published in Mathematics Today (December 2015)
Published