[Guest]

Substitute each of the letters by a different digit from 1 to 9 in this 3x3 square such that the six sums formed by the leftmost column, rightmost column, top row, bottom row and the two main diagonals are equal, and the common sum is a positive integer x > 15. Each of the two sums corresponding to the middle column and the middle row is different from x.

A    B    C

D    E    F

G    H    I

[bushindo]

I got 8 answers

Just kidding. I only got one, but mirroring and rotating can produce 8. I include all 8 just in case your solution is one of the 8.

     [,1] [,2] [,3]

[1,]    6    5    7

[2,]    4    3    2

[3,]    8    1    9


     [,1] [,2] [,3]

[1,]    6    4    8

[2,]    5    3    1

[3,]    7    2    9


     [,1] [,2] [,3]

[1,]    7    5    6

[2,]    2    3    4

[3,]    9    1    8


     [,1] [,2] [,3]

[1,]    7    2    9

[2,]    5    3    1

[3,]    6    4    8


     [,1] [,2] [,3]

[1,]    8    4    6

[2,]    1    3    5

[3,]    9    2    7


     [,1] [,2] [,3]

[1,]    8    1    9

[2,]    4    3    2

[3,]    6    5    7


     [,1] [,2] [,3]

[1,]    9    2    7

[2,]    1    3    5

[3,]    8    4    6


     [,1] [,2] [,3]

[1,]    9    1    8

[2,]    2    3    4

[3,]    7    5    6

Edited August 5, 2009 by bushindo

[Guest]

I got 8 answers

Just kidding. I only got one, but mirroring and rotating can produce 8. I include all 8 just in case your solution is one of the 8.

     [,1] [,2] [,3]

[1,]    6    5    7

[2,]    4    3    2

[3,]    8    1    9


     [,1] [,2] [,3]

[1,]    6    4    8

[2,]    5    3    1

[3,]    7    2    9


     [,1] [,2] [,3]

[1,]    7    5    6

[2,]    2    3    4

[3,]    9    1    8


     [,1] [,2] [,3]

[1,]    7    2    9

[2,]    5    3    1

[3,]    6    4    8


     [,1] [,2] [,3]

[1,]    8    4    6

[2,]    1    3    5

[3,]    9    2    7


     [,1] [,2] [,3]

[1,]    8    1    9

[2,]    4    3    2

[3,]    6    5    7


     [,1] [,2] [,3]

[1,]    9    2    7

[2,]    1    3    5

[3,]    8    4    6


     [,1] [,2] [,3]

[1,]    9    1    8

[2,]    2    3    4

[3,]    7    5    6

There is one more solution as below:

[,1] [,2] [,3]

[1,] 9 1 6

[2,] 2 4 3

[3,] 8 5 7

[Guest]

There is one more solution as below:

[,1] [,2] [,3]

[1,] 9 1 6

[2,] 2 4 3

[3,] 8 5 7

This one looks like not correct

Sum of 1st row = 16, 3rd row = 20. These are not equal and also sum of the diagonals are not equal. Probably you have written it as wrongly over here. Can you check again?

[Guest]

Substitute each of the letters by a different digit from 1 to 9 in this 3x3 square such that the six sums formed by the leftmost column, rightmost column, top row, bottom row and the two main diagonals are equal, and the common sum is a positive integer x > 15. Each of the two sums corresponding to the middle column and the middle row is different from x.

A    B    C

D    E    F

G    H    I

1 8 3

9 7 5

2 6 4

so my anwere is as above

[Guest]

you each are missing some part of the OP

that last one was pretty good but the common sum must be greater then 15 and yours is 12

All of the possible ones were found by bushindo

[Guest]

I got 8 answers as well.

648

531

729

657

432

819

729

531

648

756

234

918

819

432

657

846

135

927

918

234

756

927

135

846

Edit: Looking back at the other posts made me realize that I didn't even notice these were mirrors/rotations. I solved this with a program that tried each possibility exhaustively, so these are the only answers.

Edited August 7, 2009 by Kriil