Find all possible values of a real number x that satisfy this equation:
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10 x – floor(x)
Let x = a + b, where a is an integer and b in the range [0,1).
Then 9(a+b)/10 = a/b, so b cannot be zero.
Solving for a gives a = 9b^2/(10-b), so a must be positive.
And the the numerator is maximized and the denominator minimized as b increases, so the maximum a is slightly below 9 (because b is at most slightly below 1).
So, by inspection, a can only be between 1 and 8.
Substitute all 8 values in the above, and solve a quadratic equation to get 8 solutions, matching Charlie's values.
Edited on February 2, 2015, 1:48 pm