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 All divisible by 7 (Posted on 2008-11-04)
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.

 See The Solution Submitted by pcbouhid Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: proof not understood | Comment 3 of 8 |
(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  62  1  0  04  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

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