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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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    $\begingroup$ @AustinMohr That's not so. If you swap the hour and minute hands at 6:30 you do not get a valid time. $\endgroup$ – response May 23 '13 at 17:47
  • $\begingroup$ @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. $\endgroup$ – response May 23 '13 at 18:33
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    $\begingroup$ @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. $\endgroup$ – joriki May 23 '13 at 18:34
  • $\begingroup$ Relevant: clock related challenge "the hour hand and the minute hand have exactly interchanged their positions. What time did he leave his house?" $\endgroup$ – MJD May 23 '13 at 18:44
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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.

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  • $\begingroup$ 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? $\endgroup$ – rahul May 24 '13 at 5:32
  • $\begingroup$ Thaks for the answer by the way. $\endgroup$ – rahul May 24 '13 at 5:32
  • $\begingroup$ @Rahul: See math.stackexchange.com/questions/145537/clock-related-challenge for an example. $\endgroup$ – MJD May 24 '13 at 15:17
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A pictorial proof. Count the intersection points...enter image description here

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