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 GThe 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.
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Submitted by pcbouhid
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Rating: 3.5000 (2 votes)
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Solution:
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From the numbers A, B and C, We have:
102 * A + 10 * B + C = 105a + 104 * (b + e) + 103 * (c + f + i) + 102 * (d + g + j) + 10 * (h + k) + l
with a corresponding result for 103 * D + 102 * E + 10 * F + G.
Hence we have:
102 * A + 10 * B + C = 103 * D + 102 * E + 10 * F + G
Or
A + B + C = D + E + F + G (mod 10).
Thus, if the seven numbers A, B, C, D, E, F, G contains six numbers, for instance, A, C, D, E, F, G, divisible by 7, say, then it follows B is divisible by 7, too.
So, if all but one of the numbers obtained by reading from left to right or downwards are known to be divisible by 7, then the last number must also be divisible by 7.
This property can be extended to larger rectangles, and to divisibility by 13, or by 17, or in general, by any number that is relatively prime to 10, which means that it must end in 1, 3, 7, or 9 (3, 7, 9, 11, 13, 17, 19, ...).
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