## April 17, 2003

### One Hundred Interesting Mathematical Calculations, Puzzles, and Amusements: Number 17

One Hundred Interesting Mathematical Calculations, Puzzles, and Amusements: Number 17: The Clock Hands Puzzle

At 12:00:00 the hour and the minute hands of a clock coincide. How much time passes before the next instant that the hour and minute hands coincide?

After an hour the minute hand has gone all the way around the clock... but the hour hand has gone 1/12 of the way around the clock, so it is still ahead of the minute hand. So it is more than an hour.

After 1 + 1/12 hours the minute hand has caught up to the previous position of the hour hand, but by that moment the hour hand has moved an additional 1/144 of the way around the clock. So it is more than 1 1/2 hour.

Continuing the same chain of reasoning, we see that the minute hand catches up to the hour hand after the following amount of hours have elapsed:

1 + 1/12 + (1/12)^2 + (1/12)^3 + (1/12)^4 + (1/12)^5 + ... + (1/12)^n + ...

You have to add up an infinite number of terms to get the answer!

How can you add an infinite number of numbers? Isn't the answer infinity? No, the answer is not infinity. This is a series with a limit. It has an answer. But how can you calculate the answer without taking an infinity of time to add up the infinity of numbers?

Let's approach it from a completely different direction. Suppose you divide 1 by 11/12. The first stage of your division gets you an answer of 1 with a remainder of 1/12:

1 / (11/12) = 1 R(1/12)

Now let's continue the division another step further, and get:

1 / (11/12) = 1 + (1/12) R(1/144)

And another step further:

1 / (11/12) = 1 + (1/12) + (1/44) R(1/1728)

You can see where we are going: keep calculating more and more of these terms in the division of 1 by 11/12, and you find that you get exactly the same series:

1 + 1/12 + (1/12)^2 + (1/12)^3 + (1/12)^4 + (1/12)^5 + ... + (1/12)^n + ...

So when you add up all the terms of the series, they must be equal to 1/(11/12). What is 1/(11/12)? It is (12/11), of course. The answer is 1 1/11 hours. Or, in more familiar terms: 1:05:27 27/99: one hour, five minutes, 27 and 27/99 seconds.

Posted by DeLong at April 17, 2003 11:19 AM | TrackBack

Comments

The conceptually easier way (IMO) to solve the problem is with parametric equations describing the position of the minute and hour hand.

If we turn the clock on its side and allow the hands to rotate counterclockwise (we could fix this with phase shifts and minus signs if it bothers you, but won't really matter), we have:

r_1(t) = (cos 2*pi*t, sin 2*pi*t) [minute hand]

r_2(t) = (cos 2*pi*t/12, sin 2*pi*t/12) [hour hand]

here, t is measured in hours. thus at t=0,12,24, etc. the two position vectors coincide.

All we need, then, is the first solution in the interval 0 < t < 12 to

cos 2*pi*t = cos 2*pi*t/12
sin 2*pi*t = sin 2*pi*t/12

actually, a bit of intuition will tell us that the first solution will be for 1 < t < 2. At that point the hour hand will be stuck between 1 & 2 and the minute hand will be sweeping 'round to intersect. Now you can solve numerically, on your TI or in Matlab or whatever.

Posted by: godlesscapitalist on April 17, 2003 07:26 PM

But this is for a nine-year-old!

:-)

Posted by: Brad DeLong on April 17, 2003 10:09 PM

I guess one could use algebraic topology as well, but here's a solution (for the five-year-olds).

The minute and hour hand will obviously collide again *after* one and before *two* complete revolutions of the minute hand. Let
m
be the number of minutes the hour hand travels. Then
60+m
is the number of minutes the minute hand travels.

The minute hand travels at a rate of 1 (in terms of minutes passed). The hour hand travels at a rate of 5/60 = 1/12 (since it is on the 5 minute mark after 60 minutes). So, using d = rt, that is distance = rate * time,
m = (1/12)t
60 + m = (1)t = t
Thus
60 + (1/12)t = t
60 = (11/12)t
60*(12/11) = t

This is in minutes. In hours you should divide by 60, so in (12/11) hours.

Posted by: Andrew Boucher on April 17, 2003 11:00 PM

There's a cool solution for this that may be suitable for a nine-year-old.

In some equations with infinite series, you can substitute part of the equation for itself. (It's *infinite*, for Pete's sake, so taking away part of it leaves you with an infinite series.) For instance, here we have

t = 1 + 12^-1 + 12^-2 + 12^-3 + ....

Let's rewrite this as

t = 1 + 12^-1(1 + 12^-1 + 12^-2 + ...)

and then we substitute t (the whole thing) for the parenthesized sum:

t = 1 + t/12

Now we can use junior-high-school algebra to solve directly for t = 12/11.

Posted by: John O'Neil on April 18, 2003 04:17 AM

You guys have high expectations of your nine-year olds. I think I was capable of basic algebra at that age, no more. (I was far ahead of my classmates in understanding negative numbers at that age.)I'd say Andrew's is the one most in reach of a "typical" smart 9 year old.

Posted by: rvman on April 18, 2003 08:15 AM

The solution can be demonstrated convincingly without using any algebra at all. Just count the number of times that the two hands coincide in a 12-hour period. You can do this by mentally advancing a clock or even by playing with a real clock, and the answer is 11. "Obviously" the coincidences of the two hands occur at evenly spaced times, so the length of the interval between two of these occurences is 12/11 hours.

Posted by: Daniel Lam on April 21, 2003 10:54 AM

I am not a mathematician, but isn't this the same type of problem as the paradox of Achilles and the Tortoise, by Zeno of Elea? Have you tried Zeno's paradoxes on the kids yet, Brad? It might be a more offbeat way of introducing these types of problems.

Posted by: andres on April 21, 2003 08:55 PM

As I prepared to post this, I noticed that Daniel Lam beat me to the punch, but here goes anyway:

You start at noon. Not counting the noon match-up, how many times will the hands have lined up by the time you reach midnight? Eleven times: Between 1 and 2, 2 and 3, . . . 10 and 11, and midnight itself. The twelve hour period is thus split into 11 intervals, and symmetry guarantees that these intervals are equal. So each one measures 12/11 hours.

Posted by: Derrick Niederman on June 3, 2003 08:33 AM

As I prepared to post this, I noticed that Daniel Lam beat me to the punch, but here goes anyway:

You start at noon. Not counting the noon match-up, how many times will the hands have lined up by the time you reach midnight? Eleven times: Between 1 and 2, 2 and 3, . . . 10 and 11, and midnight itself. The twelve hour period is thus split into 11 intervals, and symmetry guarantees that these intervals are equal. So each one measures 12/11 hours.

Posted by: Derrick Niederman on June 3, 2003 08:34 AM

As I prepared to post this, I noticed that Daniel Lam beat me to the punch, but here goes anyway:

You start at noon. Not counting the noon match-up, how many times will the hands have lined up by the time you reach midnight? Eleven times: Between 1 and 2, 2 and 3, . . . 10 and 11, and midnight itself. The twelve hour period is thus split into 11 intervals, and symmetry guarantees that these intervals are equal. So each one measures 12/11 hours.

Posted by: Derrick Niederman on June 3, 2003 08:35 AM

you could just write a macro in excel to solve this for you... I think a 9 year old can do that nowadays...

Posted by: Dean on December 14, 2003 11:56 PM
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