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- Jan 26, 2012

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Hint:

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- Thread starter
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- #1

- Jan 26, 2012

- 4,058

Hint:

--------------------

Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Jan 26, 2012

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1) MarkFL

2) Sudharaka

Solution (from Sudharaka):

\[x=6,18,30,42\mbox{ and }y=5,17,29,41\]

Note that the hour hand should be one minute past the hour when the minute mark is twelve minutes past the hour. The hour hand should be two minutes past the hour when the minute mark is twenty four minutes past the hour. The hour hand should be three minutes past the hour when the minute mark is thirty six minutes past the hour. The hour hand should be four minutes past the hour when the minute mark is forty eight minutes past the hour.

Therefore the only possibility among the listed values for \(x\) and \(y\) is, \(x=6\) and \(y=17\). That is initially the time is, \(1.12\) and in the second observation it's \(3.24\). Time that is elapsed in between is therefore, 2 hours and 12 minutes (in fact it could be \((12n+2)\) hours and \(12\) minutes where \(n\in\mathbb{N}=\mathbb{Z}\cup\{0\}\)).

A second solution from MarkFL:

Let $\displaystyle m-h=d$

$\displaystyle 12k-(5H+k)=d$

$\displaystyle 11k-5H=d$

The integer solution is:

$\displaystyle (k,H)=(d+5n,2d+11n)$ where $\displaystyle n\in\mathbb{Z}$

For the first observation, we are told:

$\displaystyle d=6$

The only solution satisfying the given restrictions is $\displaystyle (k,H)=(1,1)$, i.e., $\displaystyle n=-1$.

For the second observation, we are told:

$\displaystyle d=7$

The only solution satisfying the given restrictions is $\displaystyle (k,H)=(2,3)$, i.e., $\displaystyle n=-1$.

And so we find the elapsed time between observations is (3 - 1) = 2 hours and (2 - 1)·12 = 12 minutes.

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