The following is taken from the 2nd round of the 2014 South African Maths Olympiad:

How many times in a 24-hour day do the hands on a 12-hour clock point in exactly the same direction?

I am unsure about the answer which they give as 22.

What answer do you get?

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## Comments

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TopNewestDo they seriously give such easy question in the S.A M.O? I could solve this even 3 years back!

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2nd round 2nd question but there's also a 3rd round and a camp in the junior section

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U must take a look at India's M.O papers at AoPS.. They are far more tougher than this!

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here

u can get themLog in to reply

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24

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I also thought that,but remembered the second hand.

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An intuitive answer that I haven't found justification for:

In \(12\) hours, the minute hand travels \(12\) times the circumference and the hour hand travels \(1\) time, so the minute hand will cross the hour hand \(11\) times, and those crossings are exactly when the two hands point in the same direction. Just multiply this by \(2\) since there are \(24\) hours.

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Thanks for the clear explanation @josh silverman provided more than enough justification

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The angular velocity of the minute hand and hour hand are \(\omega_m = 2\pi\text{ hr}^{-1}\) and \(\omega_h = \frac{2\pi}{12}\text{hr}^{-1}\), respectively.

The first intersection will occur when \(\omega_m t - 2\pi = \omega_h t\), i.e. when \(t = \frac{2\pi}{\omega_m- \omega_h} \approx 1.09\text{ hr}\).

As soon as an intersection happens, we have the same problem again, and the next intersection will take the same amount of time as the first. \(24/1.09\approx 22.02\) which means the crossing will occur 22 times..

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Why isn't it \(\omega_m t - 2\pi = \omega_h t\)? I mean, the LHS shows the distance traveled by the minute hand minus one circumference and the RHS shows the path traveled by the hour hand, which is speed times time.

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Typo, you are right

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That's Brilliant

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