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Unformatted text preview: c even. Having worked out these cases he sees that a solution requires finding a number of the form 199. ..997 which is divisible by 7 . The smallest such is 199997 . This gives the equation 280000a + 28000b + 2800c + 280d +28 e = 199997f. Since the lefthand side is divisible by 4 he gets that f must be 4 or 8 . With f = 4 the solution x = 285714 comes out, while f = 8 gives x = 571428 . Both are solutions, but the first is the smallest. Chirs Greenslade observes that the next solutions are the twelve digit numbers 285714285714 and 571428571428. He conjectures that all possible solutions are made by repeating one of the six digit solutions any number of times. Several solvers noted that the smallest solutions are the repeating parts for the decimal representations for 2/7 = .285714. .. and 4/7 = .571428. .. It might be interesting for you to experiment with other permutations of the smallest solution. this page....
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This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.
 Spring '10
 AlbertoDelgado
 Combinatorics

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