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.

The rows, mod 7, are equal to

     6a + 2b + 3c + d
     6e + 2f + 3g + h
     6i + 2j + 3k + L

The columns, mod 7, are equal to

     2a +3e +i
     2b +3f +j
     2c +3g +k
     2d +3h +L

Mod 7,

    2*row_1 + 3*row_2 + row_3 =
    
    12a + 4b + 6c + 2d  +
    18e + 6f + 9g + 3h  +
     6i + 2j + 3k +  L      =
    
    6*column_1 + 2*column_2 + 3*column_3 + column_4
   
So, if any 6 of the numbers mod 7 = 0, then the 7th must also.

It seems that this is not specific to modulus 7, or base 10, or to the number of rows of columns.  I expect that this will work for any number of rows and columns, and any modulus, and any base.

Edited on November 6, 2008, 2:28 am

Posted by Steve Herman on 2008-11-06 02:26:51