My wording will not be exactly clear, but this is what I remember.
Suppose you have a clock with minute and hour hands and you switch their places to form another correct time. How many such times can be formed in a clock?
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3@AustinMohr That's not so. If you swap the hour and minute hands at 6:30 you do not get a valid time. – response May 23 '13 at 17:47
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@AustinMohr The way I read that statement is to actually take out the hour hand and swap its place with the minute hand. Thus, for the example I gave, after the swap the hour hand will be pointing at 6 and the minute hand will be at the mid-point between 6 and 7 which does not give a valid time. – response May 23 '13 at 18:33
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2@Austin: It does mean the physical position of the hands; you just seem to be misremembering that position :-) At 6:30, the hour hand is between 6 and 7, and that's not where the minute hand is supposed to be when the hour hand is at 6 -- the minute hand should then be at 0. – joriki May 23 '13 at 18:34
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Relevant: clock related challenge "the hour hand and the minute hand have exactly interchanged their positions. What time did he leave his house?" – MJD May 23 '13 at 18:44
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Also relevant: 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 – MJD May 23 '13 at 18:46
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At $h$ hours and $m$ minutes, the hour hand has position $(h+m/60)/12$ and the minute hand has position $m/60$ (where positions range from $0$ to $1$). If we swap positions, the new hours $h'$ and minutes $m'$ must satisfy $(h'+m'/60)/12=m/60$ and $m'/60=(h+m/60)/12$. Thus we're looking for solutions of
\begin{align} 60h'+m'&=12m\;,\\ 60h+m&=12m' \end{align}
with $0\le h,h'\lt12$, $0\le m,m'\lt60$ and $h,h'$ integer. We can solve for $m$ and $m'$ in terms of $h$ and $h'$ by elimination:
\begin{align} m&=\frac{60h+720h'}{143}\;,\\\\ m'&=\frac{60h'+720h}{143}\;. \end{align}
Then demanding $m,m'\lt60$ yields
\begin{align} h+12h'&\lt143\;,\\ h'+12h&\lt143\;. \end{align}
This is fulfilled by all combinations of $h$ and $h'$ except $h=h'=11$. Thus there are $12\cdot12-1=143$ such times, and you get them by substituting the corresonding integer values of $h,h'$ into the equations for $m,m'$ above.
joriki
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If you are given to find such a time where the hour hand ranges between 5 and 6 and the min hand between 6 and 7 how would you do it? – rahul May 24 '13 at 05:32
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@Rahul: See http://math.stackexchange.com/questions/145537/clock-related-challenge/ for an example. – MJD May 24 '13 at 15:17
