Correct solutions were also received from Bradley student Ray Kremer, as well as Wayne Bosma (Bradley Chemisty Professor), Kyle McCormick, Jeremy Rostand, Brent Young, Chris Greenslade, Thomas Teo.
The most elegant solution was submitted by Thomas Teo. It goes as follows.
He calls his original number x and writes x = abc...uvw, where a through w are the digits of x. Note that x is divisible by 2, so its one's digit must be even. Suppose, for the moment that x is a two digit number. The equation
3(10a + b) = 2(10b + a)
would then have a single digit integer solution; simplifying gives 28a = 17b. Since the left-hand side is divisible by 7, the right is, too; but b is even, which is impossible.
He next looks at a three digit solution. The relevant equations are
3(100a + 10b + c) = 2(100c + 10a + b),
280a + 28b = 197c.
The left hand side is divisible by 7 again, which is impossible with 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 left-hand 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.
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