At what times of day is the hour hand of a clock pointing precisely at a minute tick-mark and the minute...

In other words, when is the minute hand 6 degrees ahead of the hour
hand, with the hour hand at an angle divisible by 6?

The minute hand moves 1 tick per minute, the hour hand 1/12 tick per
minute, so the minute hand moves 11/12 ticks/minute faster than the
hour hand, so it first gets 1 tick ahead at 12/11 minutes past noon (or
midnight).  It also laps the hour hand (gains another 60 ticks) exactly
once every 720/11 minutes, i.e., 11 times in 720 minutes (12 hours).
You want (for starters) the integer values of

  f(n) = 12/11 + n*720/11 = (1 + n*60)*12/11,

where f(n) tells you how many minutes past noon it is when the minute
hand is exactly 1 tick ahead of the hour hand for the nth time.

f(n) is an integer if and only if 1 + n*60 is divisible by 11.  Let's
do some arithmetic modulo 11:

  60n + 1 = 0 (mod 11)
  60n = -1 (mod 11)
  (-60)n = 1 (mod 11)
  -60 = 6 (mod 11)
  6n = 1 (mod 11)

It helps to know at this point that since 11 is prime, an equation such
as the last will have a unique answer (modulo 11).  We find that the
answer is 2 (trial and error is good enough at this point).  (Of course
what this really means is that anything equal to 2 mod 11 is an answer,
for instance 13, 24, or -9, but those answers are all some multiple of
12 hours away from the answer we're going to get, so they represent the
same position of the hands.)

f(2) = (1 + 2*60)*12/11
     = 121*12/11
     = 132

So the answer is 132 minutes past noon (or midnight), or 2:12 (am or
pm).

To find when the minute hand is exactly 1 tick *behind* the hour hand,
just run the clock backwards.  By symmetry, the event we're looking for
happens just at 132 minutes before noon/midnight, or 9:48.

There are various drawbacks to my method.  One is that it uses concepts
such as modular arithmetic that I don't think are usually well
understood by 6th graders.  On the other hand, the method is much too
slow even with the shortcut of modulo 11.  If I needed the answer to a
question such as this quickly (for a math competition, for example),
I'd simply recall from experience that the relative positions of the
minute and hour hands repeat 11 times every 12 hours, so a given
relation can occur at most once when both hands are on tick marks,
(because 60 is relatively prime to 11), and then I'd just check one
after another all the candidates for the one time that the minute hand
is one tick ahead of the hour hand and on a tick mark.

>how many possible combinations of times are there such that the
>minute-hand and the hour-hand are pointing in exact opposite
>directions (180 degrees between)?

Remembering that the minute hand gains 60 ticks on
the hour hand exactly 11 times every 12 hours, and then repeats the
cycle, any given configuration (such as this one) occurs exactly 11
times.  Only one of these, however (6:00) has the hands on tick marks.

>(i guess another interesting problem would be to figure out how many
>times there are when the hour & minute hands are perpendicular to each
>other...)

Counting both the times the minute hand is ahead and
when it is behind---11 of each kind of conjunction---there are 22.
Again, however, only two of these times (3:00 and 9:00) have the hands
exactly on tick marks.

-- David A. Karr <karr@cs.cornell.edu>

[ Editor: the counts of 11 and 22 are per 12 hours; per day they are 22
and 44, with 6:00 twice, and 3:00 and 9:00 twice each, on tick marks. ]