Arrange the digits 1-9 in a 3x3 square using each once only according to the following rules:

1. Exactly one prime is directly above a prime two less.



2. Each pair of opposite corners sums to the same square total and exactly two columns share another square total.



3. Exactly one prime is directly to the left of a non-prime two greater.



4. Directly above exactly one square digit is a digit four greater.



5. Directly to the right of exactly one cubic digit is a digit one greater.

Some things should be easily deducible.

1) There is a 5 above a 3 OR a 7 above a 5 (herein denoted by [5/3] OR [7/5])
2) Opposite corners total 9
The column totals are 13, 16, 16 (not necessarily in that order)
3) There is a 2 left of a 4 OR 7 left of a 9 (herein denoted by (2,4) OR (7,9))
4) There is [5/1] OR [8/4]
5) There is (1,2) OR (8,9)

(Restraining my amateur notation for the first part)
Consider the number 5 in a column totaling 16,

If the 5 is above the 3 then the remaining number in that column must be an 8. Since the 8 can’t then be above a 4 then 5 must be above 1 (by rule 4) which is a contradiction.

If the 7 is above the 5 then the remaining number in that column must be a 4. Since there is no 8 in the column 8 can’t then be above a 4 and again, by rule 4, 5 must be above 1 which is another contradiction.

So the 5 is in a column totaling 13.

Clearly the 5 can’t be above the 3 since this would require another 5 in the column, so 7 is above the 5 and the other number in the column is a 1.

Options
[1/7/5]
[7,5,1]

If [1/7/5] is the extreme left or right column the other side must be [4/x/8] and this column must total 16. x = 4 is a contradiction to our remit - that you can’t duplicate a number.
If [1/7/5] is in the middle column there is no room for a (8,9) combination and (1,2) must be true. The 2 in the top-right position would require a 7 bottom-left that’s already spoken for.

So, [7/5/1] appears in the solution.
Again if this were the middle column there is no room for the (8,9) and (1,2) must be true – requiring the extra 7 in the 2’s opposite corner.
So [7/5/1] appears as the extreme right or left column with the opposing left or right column being [8/6/2] (since opposite corners sum to 9 and this column must sum to 16).
[7/5/1] as the left column and the right column as [8/6/2] means (1,2) can’t be true so there must be an (8,9) combination – a contradiction with the 8 to the extreme right of the grid.
So [7/5/1] must be the extreme right column.
The rest fills itself.
Answer;
897
635
241

Posted by Lee on 2003-12-29 13:00:26