719
1 7 9
Now, other answerers have pointed out that the middle number is 5, so the 9 should go on the bottom rather than the bottom-right, in which case the completed magic square looks like this:
672159834
6 1 8 7 5 3 2 9 4
But that's only for a magic square with the numbers 1 to 9. If you thought a little outside the box, you might wonder whether you could do it with other numbers.
We can use some algebra to show that this is still impossible. Let's suppose the number 1 isn't filled in, and the middle number is
x
, like so:
79
7 x 9
One of the properties that all 3-by-3 magic squares have is that the rows, columns, and diagonals add up to 3
3
x
(3 times the middle number). So we can fill in some of the other numbers with respect to
x
, by subtracting 3
3
x
from them:
2−972−79
2x−9 7 x 2x−7 9
And then solve for the other two corner squares:
2−97+2−22−79
2x−9 x−2 7 x 2x−7 x+2 9
And finally solve for the last two edge squares.
2−97+2112−11−22−79
2x−9 11 x−2 7 x 2x−7 x+2 2x−11 9
Huh? It turns out that no matter what the middle number
x
is, if the 7 and 9 are placed where they are, the top square has to be 11. The fact that it was filled out as 1 means that the magic square contradicts itself.
But then again... if you think a little further outside the box, who said that the filled-in squares had to remain unmodified? If you draw another 1 next to the existing 1 in the top square, you get 11 as required.
So... do that. Then, fill in any single number on any square and the rest of the numbers will work themselves out like magic. You can ask your teacher for that can of soda now ;-)
...except not really. Your teacher probably still expects a solution with distinct positive integers (whole numbers that are all different). There exists one, but I'll leave it to you to figure it out on your own since you seem to want to do as much of this as you can by yourself.
Hope this helps!!!
