5
$\begingroup$

My initial question had lots of answers, as we can practically choose the middle square to be anything we want, and work outwards to the exterior edges, with the corners providing the flexibility needed.

So, as a bonus, I have added some more constraints. Same rules as before, with the outside sums only counting 2 sets of minuses.

       8      7      3
  ---    ---    ---    ---
5     13     14     11     5
  ---    ---    ---    ---
7     11     10     13     7
  ---    ---    ---    ---
8     11      9     14     8
  ---    ---    ---    ---
       5      6      9

Can you now solve the grid?

|improve this question
$\endgroup$
  • 1
    $\begingroup$ Actually, there are too many constraints (i.e. there are redundant one), because it's only 16 numbers but 21 constraints for them. So, 5 of them (but it's important which ones) can be safely removed. $\endgroup$ – trolley813 Dec 11 '18 at 10:44
3
$\begingroup$

I think I have the answer

       8      7      3
   3      5      2      1 
5     13     14     11     5
   2      3      4      4 
7     11     10     13     7
   5      1      2      3 
8     11      9     14     8
   3      2      4      5 
       5      6      9

Reasoning

The 8 at the top left hand corner of the grid must be made by the sum of 3 and 5 or 4 and 4. If 5 is on the left, this forces the digit below to be 0 (which is not allowed). If 4 is on the left it forces the digit two below to be 6 (also not allowed). Hence, it must be 3 on the left and 5 on the right. From there, we can easily follow the grid around the edge and then deduce the inside part from remainders.

|improve this answer
$\endgroup$
  • $\begingroup$ bingo! that's what I've got! $\endgroup$ – JMP Dec 11 '18 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.