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This question already has an answer here:

I'm quite familiar with "clock puzzles", but only when the puzzle is such that all 3 hands start at 12:00.

I recently tried this one: Over a 12 hour period, how often are AT LEAST 2 of the 3 hands of a clock in the same position?

I'm pretty sure answer is 1436. Can someone confirm? Thank you.

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marked as duplicate by ghosts_in_the_code, Gamow, Dr Xorile, AJL, Ric Nov 7 '15 at 17:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The time from $10:30$ to $22:30$ is a $12$ hour period.

During $12$ hours the hour hand makes $1$ full circle, the minute hand makes $12$, and the second hand makes $720$.

During the $1$ full circle of the hour hand it is passed $12-1=11$ times by the minute hand and $720-1=719$ times by the second hand. Similarly the minute hand is passed $720-12=708$ times by the second hand during the $12$ hour period. Summing it all together gives $11+719+708=1438$.

But we have to consider the time when all $3$ hands meet at $12:00$. This is $1$ overlap which we have counted $3$ times in the previous paragraph. To correct this we need to subtract $2$ which gives the final result $1438-2=1436$.

Edit: Proof that there is no other overlap of all $3$ hands in a $12$ hour period

From the text above we know that the hour and minute hands overlap every $\frac{12}{11}h$ and the minute and second hand overlap every $\frac{12}{708}h$. For all 3 hands to overlap there must be a solution with integer $x$ and $y$ for the following equation:

$\frac{12}{11}x=\frac{12}{708}y$
$\frac{1}{11}x=\frac{1}{708}y$
$708x=11y$

As $11$ is prime and $708$ is not divisible by $11$ the smallest solution is $x=11$ and $y=708$ which shows there can be only one overlap of all $3$ hands in a $12$ hour period.

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  • $\begingroup$ +1 I was on my way to the same answer and saw yours, so deleted mine. Also, I didn't think about the 12:00 overlap. $\endgroup$ – Trenin Nov 6 '15 at 19:46
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If you read the accepted answer on this question, the answer becomes evident.

If revolutions per day are given by:

  • Second hand - 1 revolution per minute or $1440$ per day
  • Minute hand - 1 revolution per hour or $24$ per day
  • Hour hand - 1 revolution per 12 hours or $2$ per day

Then meetings in a day are given by:

  • (a) the minute and hour hands ($24-2=22$ per day)

  • (b) the minute and second hands ($1440-24=1416$ per day)

  • (c) the hour and second hands ($1440-2=1438$ per day)

  • (d) all three hands ($2$ per day)

Now, to answer your question, we want the following value

$$(a)+(b)+(c)-2(d)=22+1416+1438-2(2)=2872$$

We subtracted $2(d)$ since the meeting of all three hands has been counted thrice, once in $(a)$, once in $(b)$ and once in $(c)$, when we want it to be counted only once.

This is answer for one day. If we want $12$ hours, we just divide by $2$. Hence we conclude that:

Your answer is correct; as $2872/2=1436$

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