時針和分針互換!?

Since the minute hand is at 'a' minutes. The hour hand will be at (45 + a/12) minute position.
When the 2 hands reverse, the hour hand will be at the 'a' minute position and the minute hand will be at the (45+a/12) minute position.
If the minute hand moves backward by (45 +a/12) minutes, the hour hand will move backward by (45 + a/12)/12.
If the minute hand moves backward by (45+a/12) the time must be 1pm, 2pm, 3pm, 4pm etc. That is
a - (45+ a/12)/12 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55 = n
Simplifying the equation, we get
12a - (45 +a/12) = 12n
12a - (540 + a)/12 = 12n
144a - (540 +a) = 144n
144a - 540 - a = 144n
143a = 540 + 144n
a= (540 + 144n)/143.......(1)
For example, n = 20
a = (540 + 144 x 20)/143 = 23.9 = 24.
That is the clock is originally at 9: 24.
When the hands reverse, the time is 4: ( 45 + 24/12) = 4: 47.
So a is given by (1) depending on the value of n, which is a multiple of 5.
For n = 20, x = 7 hour 23 min.