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 proof by xdog)

"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. "

This seems to imply divisibility by 7 implies the sum of the digits is also divisible by 7. But that's not the case.

For example, in

2  3  6  6
2  1  0  0
4  5  2  2

all are multiples of 7, to fit the puzzle, but none of the sums of digits, including of the whole array, is a multiple of 7.

 

Posted by Charlie on 2008-11-04 16:34:49