Lets say one hand is at position x=0..1 from 12 o'clock and another is at angle y (entire circle is assumed to have unit circumference).
There are two possibilities:
1) x is hour hand. Then time is 12x hour and y must be at position
y = fractional part of 12x
2) y is hour hand. Then
x = fractional part of 12y
One of these equation is always true, but sometimes both are true, and it such moments we cannot tell which hand is which. At such moments we have two simultaneous equations:
y = fractional part of 12x
x = fractional part of 12y
x = fractional part of 12(fractional part of 12x)
x = fractional part of 12² x
This equation has 143 solutions in form
x = n/143, n=0...142,
out of which 11 are symmetric, that is x=y and confusing the hands does not alter time.
Answer:
264
