From 5, we know that the top row must have two odd numbers and an even number, and the other two rows have all even and all odd in some order.
From 2, we know the left column's sum is even, so the NW corner is odd.
From 3, we know that the right column's sum is odd because double the bottom row (always even) plus the right column is 45, and odd. Therefore, the NE corner is even.
From 5 again, we deduce that N is odd.
Because the middle column is even, the left column must be divisible by 4, by rule 2. The sum of the left is 16, 20, or 24, limited by the possibilities of both the middle and left column.(The middle column can't sum less than 6, and if it did, it would use 1, 2, and 3, making a sum of 12 impossible.)
The right column's sum, based on the sum of the other columns, is limited to 21, 15, or 9. 15 is precluded by rule 4.
First assume 9. The columns' sums left to right are 24, 12, and 9. The rows' sums top to bottom are 18, 9, and 18. The left column must have 7,8, and 9, but none will fit in the middle row. Therefore, our assumption was incorrect, and the right column's sum is 21.
The columns' sums left to right are now 16, 8, and 21. The rows' sums top to bottom are 22, 11, and 12. Notice that all of the middle row is odd now, and all of the bottom row is even from rule 5.
The lowest possible number in the top row is 5, coinciding with the highest possible number in the middle column. 5 is in the N position. 9 is in NW and 8 NE because of the parity limitations I mentioned earlier. 1 is center, 2 is S, 3 is W, 4 is SW, 6 is SE, and 7 is E.
958
317
426
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Posted by Tristan
on 2004-05-16 01:17:06 |
Rambling deductive proof
