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.

Since only 1 of A-G is not divisible by 7, either all of A,B,C or all of D,E,F,G is divisible by 7. In either case, sum(a-l) is divisible by 7.

So sum(a-l) = the sum of either 2 or 3 distinct subsets of (a-l), each divisible by 7, plus the sum of the remaining numbers. Putting the sum of the remaining numbers as not divisible by 7 leads to a contradiction.

Posted by xdog on 2008-11-04 14:40:41