Imagine a rectangle divided into 3x4 squares, and put a digit in each square.

     +---+---+---+---+
     | a | b | c | d |  A
     +---+---+---+---+
     | e | f | g | h |  B
     +---+---+---+---+
     | i | j | k | l |  C
     +---+---+---+---+
       D   E   F   G

The number abcd is denoted by A, that is, A = 1000a + 100b + 10c + d, and the same for the other 2 horizontal numbers B and C.

The number aei is denoted by D, that is, D = 100a + 10e + i, and the same for the other 3 vertical numbers E, F and G.

Prove that if any 6 of these numbers (A, B, C, D, E, F, G) are divisible by 7, then the last number must also be divisible by 7.

(In reply to re: Proof and Extensions by Charlie)

Right you are, Charlie.  Thanks for the excellent counterexample. 

It fails because 10 mod 2 = 0.  In order to work for modulus m and base 10, we need 10 mod m and 100 mod m and 1000 mod m to all be non-zero.

Let me try a less bold prediction:

"I expect that this will work for any number of rows and columns, and any modulus and any base, if the modulus has at least one prime factor that is not a factor of the base."

In base 10, for instance, it will work for any number of rows and columns, for moduli which have a prime factor that is not 2 or 5. 
It will work for all size grids for modulus 3,6,7,9,11,12,13,14,15,17, etc. 
It will not work for all size grids for modulus 2,4,5,8,10,16,20,25,32,40,50,64,80, etc.

For instance, it fails for 40 because 1000 mod 40 = 0.

Posted by Steve Herman on 2008-11-06 14:20:11