A	B	C	D	E	F
1		44	44		66
2	33		52			49
3	96				57
4	11	6				42
5	67					12
6	4	48	71	62	98	38

Place the numbers 1 to 36 in the 6x6 grid, in such a way that consecutive numbers cannot be adjacent in any direction (including diagonally), nor can they appear in the same row or column.
Additionally, cells which are diametrically opposite have a difference of +/- 18.
So, for example, if number 2 was in cell 1A then number 20 would be in cell 6F; if it was in 2C then 20 would be in 5D.

The small digits show the sum of the figures in the adjoining horizontal and vertical cells.

The large digits (6 and 12) are starters.

This is similar in principle to X Marks the Spot

1.  From the sum in F6 and the 12 in F5, we know that E6 is 26.

 

2.  By the diametric rule, we know that E3, A2, and B1 are 24, 30, and 8, respectively.

 

3.  From the sum in A4 and the 6 in B4, we know that A3 and A5 are either 4+1 or 3+2. But they can't be 3+2 because then we would be violating one of the consecutive rules. So we know A3 and A5 must be either 4 or 1.

 

4.  From the sum in A6, we know that A5 and B6 must be either 1 or 3. From what we found out in (3) we now know A5 is 1, which means B6 is 3 and A3 is 4.

 

5.  From the sum in A2 and the 4 in A3, we know that A1+B2=29.

 

6.  From the sum in B1 and what we found out in (5), we now know C1 is 15.

 

7.  By the diametric rule, we know that D6, F4, F2 and E1 are 33, 22, 19, and 21, respectively.

 

8.  From the sum in C6 and the 3 and 33 in B6 and D6, we know that C5 is 35.

 

9.  From the sum in A3 and the 30 in A2, we know that B3+A4=66. Of the remaining numbers, the only possibility is 32+34.

 

10. From the sums in A5 and B6, we know that 67=A4+B5+A6 and 48=B5+A6+C6. So 67-48=A4-C6=19. We learned in (9) that A4 can only be 32 or 34. That means C6 can only be 13 or 15. But 15 is already taken. Therefore C6, A4 and B3 are 13, 32 and 34, respectively.

 

11. By the diametric rule, we know that E4, F3, D2 and D1 are 16, 14, 17 and 31, respectively.

 

12. From the sum in C1 and the 8 and 13 in B1 and D1, we know that C2 is 5.

 

13. In (5) we learned that A1+B2=29. Of the remaining numbers, there are only three possible combinations. Two of these combos would be forced to violate one of the consecutive rules: 27+2 would force the 2 into the same column as either 1 or 3 no matter what and 20+9 would force the 9 next to the 8. Therefore the only option is that they are 18+11. B2 cannot be 18 because then it would be in the same row as the 19, which would violate one of the consecutive rules. Therefore, B2 is 11 and A1 is 18.

 

14. By the diametric rule, we know that D5, E5 and F6 are 23, 29 and 36, respectively.

 

15. From the sum in A5 and the 32 in A4, we know that B5+A6=35. Of the remaining numbers, there are only two possible combinations: 7+28 and 10+25. B5 cannot be 7, 28 or 10 because it would violate the consecutive rule due to the 6, 29 and 11, respectively. Therefore B5 is 25 and A6 is 10.

 

16. From the sum in C2 and the three numbers around it so far, we know that C3 is 9.

 

17. By the diametric rule, we know that D4, E2 and F1 are 27, 7, and 28, respectively.

 

18. From the sum in E3 and the three numbers around it so far, we know that D3 is 20.

 

19. By the diametric rule, and the fact that it is the last number available, we know that C4 is 2.

 

Awesome puzzle Josie, as always!

 

Edited on January 4, 2008, 4:48 pm

Posted by nikki on 2008-01-04 16:42:23