Surprising results from midday arithmetic

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1st-Dec-2006 05:27 pm

Times of day where a twelve-hour perfectly analog clock's hour and minute hands line up exactly

12:00:00.00
01:05:27.27
02:10:54.54
03:16:21.81
04:21:49.09
05:27:16.36
06:32:43.63
07:38:10.90
08:43:39.18
09:49:05.45
10:54:32.72

What surprised me most about this was the fact that there is no intersection during hour 11. Or, more accurately, there is an intersection at 60 minutes past the start of hour 11, which is both a) not included in my calculations, and b) congruent to 12:00:00.00 above.

What's sad is that this problem has been bouncing on and off in my head since HS calculus, 7 years ago, and calculus is not required to solve it (I probably could have done it in elementary school, arithmetically).

Mood:satisfied

Comments

1st-Dec-2006 11:01 pm (UTC)

mithrigil

Now, what's coolest (and makes perfect sense) is that all the hundredth-seconds are multiples of nine. I knew they'd be of three, but of nine?

4th-Dec-2006 07:56 pm (UTC) - Overbars

xoder

Actually, I used "overbars" on the decimal portions to indicate infinitely repeating decimals. They're elevenths, actually.

1st-Dec-2006 11:06 pm (UTC)

zandperl

I've never bothered to calculate the exact times, but I forget when I figured out there's 11 of them. I guess once I knew that much, I figured calculating them was trival. :-P

It's also like how there's a different number of hours in a sidereal day (23.934) from a (mean) solar day (24), or the many types of months to astronomers.

1st-Dec-2006 11:22 pm (UTC)

omgimsuchadork

I remember that problem! I first encountered it in seventh grade, at pre-school math club. I got the question wrong, and never bothered to work it out on my own, but I remembered that there were eleven times when the hands lined up (the question asked how much time would pass between one line-up and another; 65 5/11 minutes, for a total of eleven times in a twelve-hour period).

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