Arrange the digits 1-9 in a 3x3 square using each once only according to the following rules:
- Exactly one prime is directly above a prime two less.
- Each pair of opposite corners sums to the same square total and exactly two columns share another square total.
- Exactly one prime is directly to the left of a non-prime two greater.
- Directly above exactly one square digit is a digit four greater.
- Directly to the right of exactly one cubic digit is a digit one greater.
From 1: Columns of
3 5 7
| | |
1 3 5
From 4: Columns of
5 8
| |
1 4
Only possiblity is column:
7
5
1
From 2 we have corner sums of 9 since all other squares are impossible for two locations. From this we find that 9 cannot be in a corner.
From Rule 3 we have the choice of row:
2 - 4
or
7 - 9
If the column (7,5,1) is in center position then from Rule 3 :2-4 is not possible (requires split) and 7-9 puts 9 in a corner where it cannot go. Therefore column (7,5,1) is not in center position.
From Rule 5 we have rows:
1 - 2
or
8 - 9
If column (7,5,1) is the left column, then to meet Rule 2, the right column is (8, x, 2). Neither possibility (1-2, 8-9) for Rule 5 will fit.
Therefore column (7,5,1) is the right column.
Using Rule 2 and drawing the 3x3 square:
8 x 7
x x 5
2 x 1
Rule 5 gives (cube next to number 1 higher)
8 9 7
x x 5
2 x 1
Rule 3 (prime next to non-prime 2 higher) gives
8 9 7
x x 5
2 4 1
and to meet the remaining requirement of Rule 2, the first two columns must total 16.
8 9 7
6 3 5
2 4 1
Solution (no program required)
