Correct solutions were submitted by Paul Botham, Philippe Fondanaiche, Jens Voss, Ahron Teitelman.

Write a solution in the form ab/bc = a/c.  I'll ignore the (trivial) solutions having a = b or b = 0 and assume that a, b, c are one-digit numbers in the range 1 £ a, b, c £ 9.  There are then just four (non-trivial) solutions:

Of course, you can find these with a simple computer program which checks all 9³ possibilities, but where's the fun in that?  A little simple algebra reduces the number of cases to check to very small number.

Write the basic equation as

and simplify to read

Suppose that a > b.  Then a/c = ab/bc  > 1 and a > c.  Since the  right-hand side of (***) is then positive, it must be the case that c > b, too.  But then c×(a - b) = 10a×(c - b) > 10×(c - b) > 0, and  a - b > 10(c - b) > 0.  This forces 8 ³ a - b ³ 10(c - b) ³ 10, which is impossible.

So a < b, and arguing as in the previous paragraph, c < b as well.  Since 2 and 5 divide the left-hand side of (***), it follows that either

Suppose the first case holds.  Since c is a one-digit number, it follows that c = 5 and then 2a×(5 - b) = a - b.  This, in turn, gives that a divides a - b, and so a divides b.  These conditions on a, b, c imply that a is one of 1, 2, 3, 4,  which allow us to solve for b in (***).  The second and third examples appear here.

Suppose the second case holds.  Then 5 = a - b which leads again to a being one of  1, 2, 3, 4 from which we can solve for b, c.  The first and fourth examples appear here.

Philippe Fondanaiche and Jens Voss observe that each of the examples forms the first in an infinite chain of such monstrous coincidences.

Additional larger examples abound: 484/847 = 4/7, 654/545 = 6/5, 4324/3243 = 4/3, 4582/1264 = 58/16, 3346/1673 = 34/17, 4455/3564 = 45/36.

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