The numbers 1 through 9 are arranged in a 3 x 3 grid so that each number is in the grid exactly once. They are arranged such that the top row plus the middle row gives the bottom row. If the grid forms another such addition when it is rotated 90 degrees to the left, what is its composition?

(Note: The numbers don't flip, for example 6 doesn't turn into 9.)

(In reply to is this possible? by mike)

A + D doesn't necessarily equal G.  A + D + 1 could also equal G, due to carrying.  The same goes for C + B equalling A.  I denote this as A + D [+1] = G and C + B [+1] = A.

Here's what I have so far:
Rotating

A B C
D E F
G H I

to the left produces

C F I
B E H
A D G

This implies that
( C + F ) mod 10 = I
( B + E [+1] ) mod 10 = H
A + D [+1] = G
( I + H ) mod 10 = G
( F + E [+1] ) mod 10 = D
C + B [+1] = A

A >= 3, since C and B could be 1 and 2.  G is also >= 3, by the same logic.  G must also be >= 4 since A + D [+1] = G, and >= 6 since C + B [+1] + D = G.

The rest has me stymied at the moment. :cS

Posted by JT on 2004-05-11 21:45:42