Fill a 3x3 square with 1 to 9 so that the sum of the numbers belonging to any given 2x2 subsquare is the same

Question

the given 3x3 grid is like that
_________________
|__1__|______|___ __|
|_____|______|___7__|
|_____|______|______|

the given 3x3 grid is like that
_________________
|__1__|______|___ __|
|_____|______|___7__|
|_____|______|______|

there are 5 solutions to this.

thanx you kevin!!! :D

thanx you kevin!!! :D

thanxs tom!! :D

i got one myself

153
987
426

the sum for each 2x2 square is 23

wow wow wow!!! thanx tom! : D

hmm.... sorry fredy maybe my qn wasn't clear enough. there is a given grid as shown above with "1" and "7" 's positions fixed.

i guess the last one should add up to 22. i will go work on it...

yeah yeah i was working on the eqn too :D wah thanx you thanx you so grateful!!!

Answer 1

EDITED: So here's how I finally got all 5:

If you label the grid positions like this:

A B C
D E F
G H I

A = 1
F = 7

We know that:

A+B+D+E = B+C+E+F
A+D = C+F
1+D = C+7

D = C+6

and

We know that C and D are greater than 1, less than 10, and not equal to 7

The only 2 possible solutions for C and D are:
C=2 and D=8
C=3 and D=9

D+E+G+H = E+F+H+I
D+G = F+I

8+G = 7+I (if D=8) or
9+G = 7+I (if D=9)

G = I-1 or
G = I-2

I and G possibles are
C=2 and D=8:
G = I-1: (3,4,5,6,9) so G and I = 3,4 or 4,5 or 5,6
B+C+E+F = E+F+H+I
B+C=H+I
3,4: B+2 = H+4 --> B=H+2 (5,6,9) impossible
4,5: B+2 = H+5 --> B=H+3 (3,6,9) possible (6,3 and 9,6)
5,6: B+2 = H+6 --> B=H+4 (3,4,9) impossible

C=3 and D=9:
G = I-2: (2,4,5,6,8) so G and I = 2,4 or 4,6 or 6,8
2,4: B+3 = H+4 --> B=H+1 (5,6,8) possible (6,5)
4,6: B+3 = H+6 --> B=H+3 (2,5,8) possible (5,2 and 8,5)
6,8: B+3 = H+8 --> B=H+5 (2,4,5) impossible

1 6 2
8 9 7 (24)
4 3 5

1 9 2
8 3 7 (21) Kevin's
4 6 5

1 6 3
9 8 7 (24)
2 5 4

1 5 3
9 8 7 (23)
4 2 6

1 8 3
9 2 7 (20)
4 5 6

Answer 2

9 1 8
4 6 5
3 7 2

Answer 3

this was fun

1 9 5
6 2 7
8 4 3


1 8 6
5 3 7
9 4 2

Answer 4

No Tom! Here's other three solutions:
1,8,3
9,2,7
4,5,6 (sum=20)

2,5,3
9,4,8
6,1,7 (sum=20)

3,7,4
6,1,5
8,2,9 (sum=17)

I didn't continue; I guess at least 3 other solutions must be achieved. But may be they are mirror inflection of existing solutions.

Answer 5

1 9 2
8 3 7
4 6 5

The sums are 21

I solved it myself^_^

^_^

Answer 6

Have you given us all of the instructions? I haven't been able to solve it with what you've given us.

Answer 7

Suduko???