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Do the 3 hands on a clock ever divide the face of the clock into 3 equal segments, i.e. 120 degrees between each hand?

[Ref: ZJGA]

Answer: No. The full solution is a little complicated...

First let us assume that our clock has 60 divisions. We will show that any time the hour hand and the minute hand are 20 divisions (120 degrees) apart, the second hand cannot be an integral number of divisions from the other hands, unless it is straight up (on the minute).

Let us use h for hours, m for minutes, and s for seconds. We will use =n to mean congruent mod n, therefore 12 =5 7.

We know that m =60 12h, that is, the minute hand moves 12 times as fast as the hour hand, and wraps around at 60. We also have s =60 60m. This simplifies to s/60 =1 m, which goes to s/60 = frac(m) (assuming s is in the range 0 <= s < 60), which goes to s =60 frac(m). So, if m is 5.5, s is 30.

Now let us assume the minute hand is 20 divisions ahead of the hour hand. So m =60 h + 20, so 12h =60 h + 20, 11h =60 20, and, finally, h =60/11 20/11 (read 'h is congruent mod 60/11 to 20/11'). So all values of m are k + n/11 for some integral k and integral n, 0 <= n < 11. s is therefore 60n/11. If s is to be an integral number of units from m and h, we must have 60n =11 n. But 60 and 11 are relatively prime, so this holds only for n = 0. But if n = 0, m is integral, so s is 0.

Now assume, instead, that the minute hand is 20 divisions behind the hour hand. So m =60 h - 20, 12h =60 h - 20, 11h =60 -20, h =60/11 -20/11. So m is still k + n/11. Therefore s must be 0.

But if s is 0, h must be 20 or 40. But this translates to 4 o'clock or 8 o'clock, at both of which the minute hand is at 0, along with the second hand.

Therefore the 3 hands can never be 120 degrees apart, QED

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