Hands of a Clock

Date: 10/10/97 at 22:17:40
From: Porter
Subject: Can we stump ya?

How many times do the hour and minute hands cross in a 12-hour period 
of time?

Date: 10/11/97 at 02:40:49
From: Doctor Luis
Subject: Re: Can we stump ya?

This is an interesting problem. The answer is 11 times.

An even more interesting question would be: at what times does
this happen?

To answer that, consider two separate clocks. Now, these clocks
are special: they don't have two hands, but only one, and, 
furthermore, instead of twelve o'clock they have a zero o'clock. 
The hand of the first clock completes one turn in one hour. 
Also, the hand of the first clock is twelve times as fast as the hand 
of the second clock (which completes one turn in twelve hours).

       
     * 0 *       Measure the angle between the zero o'clock
   *       *     and the hand of the first clock. Call it m.
  *    +    *    
   *    \  *
     *   *

       
     * 0 *
   *     / *
  *    +    *    Similarly measure the angle between the zero o'clock
   *       *     and the hand of the second clock. Call it h.
     *   *

Clearly, m and h are functions of time, so I'll denote them by
m(t) and h(t), respectively.

Now, let both hands start at 0 o'clock at time t=0, so that

   m(0) = h(0) = 0  

The first hand turns at a rate of 1 turn/hr so that after t hrs,

   m(t) = (1 turn/hr * 360 deg/turn) * (t hours)

or 

   m(t) = 360t   

In the same way, the second hand turns at a rate of 1 turn/12 hrs,
which means that the angle h(t) after t hours is,

   h(t) = ( 1/12 turn/hr * 360 deg/turn) * ( t hours)

   h(t) = 30t

This gives us formulas for the angles (in degrees) of the first and
second hands of our clocks at any time t (in hours). Since we want
to know when the hands coincide, that is, when the angles are the
same, we have to set them equal to each other. However, if we do that,
we get,

     m(t) = h(t)
    360t  = 30 t
    330t  = 0
       t  = 0

which is certainly a solution, but there must be other solutions as 
well!What happened? Well, if we look closely at our equations,

  m(t) = 360t
 
  h(t) =  30t

we can notice something interesting: as t goes from 0 to 12 hrs,
h(t) goes from 0 to 360 degrees, but m(t) goes from 0 to 4320 degrees!
(this means that when the second clock completes one turn, the first 
clock will have completed 12 turns).

Clearly, the way we defined m(t), it cannot exceed 360 degrees.
After the first turn, it started from 0 once again, not from 360, and 
so on after the second...

This means that m(t) differs from h(t) by a multiple of 360.

Therefore,

         m(t) = h(t) + 360*N   , where N is an integer
         360t = 30t  + 360*N
         330t =        360*N
                 360
            t = ----- * N
                 330
       or
                12
            t = -- * N
                11

Since 0 <= t < 12 , it follows that 0 <= N < 11 
and since N is an integer N = 0 , 1 , 2 , ... , 10
Thus, there are 11 solutions, and the times (in hrs) are

    0   hrs or  0:00:00 (the same as 12:00:00)
  12/11 hrs or  1:05:27
  24/11 hrs or  2:10:54
  36/11 hrs or  3:16:21
    .
    .
 120/11 hrs or 10:54:33 

We can check that the solutions actually make sense, because after 
the two hands meet at 0, the "hours" hand is lagging behind the 
"minutes" hand, until the "minutes" hand catches up again to the 
"hours" hand in the second turn. We can see that the answer is going 
to be a few minutes after one o'clock, because the "minutes" hand 
points at 0 o'clock when the "hours" hand points at 1 o'clock,
and it will eventually catch up, since it's faster. And so on for the
third turn, fourth turn, ..., tenth turn.

Now, what happens in the eleventh turn? The "minutes" and "hours"
hands both point back at 0 o'clock, so that the solution is the same 
as for the first turn when they met at the beginning. In fact, after 
M turns, the solution will be equivalent to that of N = M mod 11 
number of turns.

 ( M mod 11 just means " the remainder when you divide M by 11 ")

If you have any questions, feel free to ask us again. :-)

-Doctor Luis,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/