Self-referential Puzzle 1 Solution

Suggested Solution

1. N cannot be 2 or greater, as then H would be greater than 9. Thus N = 1. Hence K = 1 and H = 1.

2. P = M+n+O-1 so M+N+O+P = 2(M+N+O)-1, thus the sum of the bottom row is odd and A = 1.

3. E = 2, as otherwise if E = 3 then M =D = 3 and so 3 is repeated in a column. Hence O = 2 (as 4 is not possible - the grid is not big enough)

4. M = 4 would be impossible, as then P = 6 and D = 2, contradicting E. Hence M = 3.

5. Thus P = 5, L = 4, D = 3

6. Notice there are now at most 12 odd numbers (noting clue N). Hence I is at most 5. As it cannot be 1, 2, 3 or 4, we must have I = 5.

7. The clue on C implies that the parity of J matches that of C, hence they must both be odd (as otherwise there would be too many even numbers)

8. F = 5, as it cannot be 1 and else would contradict O.

9. C is either 1, 3, 5, 7 or 9. However 7 contradicts clue O. It cannot be 1 or 3. Thus it is 5 or 9. But B must be at least 8 (as there is an 8). But if it is 8, there is no 9. If it is 9, then C is in the same row. IN either case, C is not 9. Thus C = 5.

10. B cannot be 7, but by clue O we must have J = 7.

11. G cannot be 8 by clue O, so G = 3 and B = 8

So: