Mgccl's Ivory Towerso tired

While on Google talk 3 in the morning, I received the following problem from Alex. He made it up, but he thought it's possible someone else have thought of that question too.

a) How many valid positions of the minute and hour hands are there on a standard 12-hour clock face, so that if you rearrange the minute and hour hands' position, you would still have a valid time?
(Poorly worded, sorry :P)
b) What is the first such position?
(First after 12:00)

This is in a generalization of "How many valid positions of the minute and hour hand meets?".(which is 11)
Solving the above problem is quite intuitive. Notice the minute hand only meet the hour hand once per 720/11 minutes. It's also valid to reason the hour hand meet the minute hand in every hour, except the 23rd hour, which is counted in the next hour. Both give the correct answer of 11.

I can't think of an approach of this problem similar to the above one. I used something else.
We can map the angle between a hand point at 12 and any other hand by how much of a 1 full revolution is required to move the hand pointing at 12 to the other hand. It can be used to define the position of a hand.
Thus we can create function and , . where and are position of hour and minute hand respectively.
Let and We are trying to find all the intersections between the images of and .
The plot is shown below.

Of course, there is no need to plot it with a computer, it is quite easy to imagine it is going to be a grid because is linear, and each mod does a translation. Thus it is 12 parallel lines for , and is reflection though the line . The out most lines intersects inside the image(if we consider t = 720), there are 144 intersections. Remove the one for t = 720 we arrive the answer 143.
I don't know anymore, since for some t, it generate 2 intersections. For example, maybe at time a, the hands can be flipped and get time b. At time t it generated 2 intersections. but at time b, it might also generated 2 intersections. If that's true, then the answer is 77, else it's 143. I didn't have the time to prove that's not possible. and it's not likely for me to prove it today because I need to go to sleep...
Silly me, it happens as one draw a plot of first, then draw , which only generate 1 intersection for each t.

If we trace the process of generating the plot, one can see 1 line is drawn from origin going up, then the 2nd is drawn, each is a translation of the original. Omitting the point where t=0, the first intersection would be either the leftmost point.
Formula for those 2 lines are
and
and . The solution is minutes after 12.

In fact, the nth one after 12 is minutes after 12. Notice how 143 = 13 * 11. The 13th position is the solution to the trivial hour meet minute hand problem.