Interchangeable positions of minute hand and hour hand occur when the original interval between the two hands is 60/13 minute spaces or a multiple of this.

Could anyone explain this with examples?

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- Jul 7th 2013, 12:18 PMhisajeshClock problem
Interchangeable positions of minute hand and hour hand occur when the original interval between the two hands is 60/13 minute spaces or a multiple of this.

Could anyone explain this with examples? - Jul 7th 2013, 05:57 PMchiroRe: Clock problem
Hey hisajesh.

Can you show us what you have tried? (Hint: Try and formulate the problem in terms of congruence equations and number theory). - Jul 14th 2013, 07:37 PMhisajeshRe: Clock problem
I learnt congruence equation from Congruence Equation -- from Wolfram MathWorld

But I have no clue what can I do with congruence equation. - Jul 15th 2013, 07:40 PMhisajeshRe: Clock problem
- Jul 16th 2013, 10:47 AMebainesRe: Clock problem
I think the question is stated incorrectly. The hour hand and minute hand line up every 12/11 hours, or in other words after the minute hand has made 12/11 revolutions. The new position of the minute hand is therefore 1/11 further along than on the previous alignmemt, which works out to be 60/11 = 5.4545 minutes.

The derivation of the 12/11 figure come from this: the position of the minute hand is $\displaystyle M= 2 \pi t$ where t is in hours, and the position of the hour hand is $\displaystyle H = 2 \pi \frac t {12}$. The hands line up when M is equal to H plus a multiple of $\displaystyle 2 \pi$:

$\displaystyle 2 \pi t = 2 \pi ( \frac t {12})+ k (2 \pi)$

where k = 1, 2, 3... This occurs when

$\displaystyle t = \frac {k (2 \pi)} {2 \pi - \frac {2\pi} {12}} = \frac k {1 - \frac 1 {12}} = \frac {12 }{11}k$

Thus every multiple of 12/11 hours the hands are realigned, at which time the hands are 60/11 "minutes space" further on.