# Riddle for everybody :-)

1. May 5, 2016

### geoffrey159

A teenager is looking at his father's beautiful mechanical swiss made watch, and asks him: 'Dad, is there a time of the day, beside noon and midnight, where the hour and minutes needles are aligned and pointing in the same direction ? '. What would you answer at the nearest second ?

2. May 5, 2016

### andrewkirk

The minute hand overtakes the hour hand at intervals of slightly over an hour. Between noon and midnight the minute hand does twelve full rotations and the hour hand does only one rotation, so the former must pass the latter ten times in-between.

If you divide the number of seconds in twelve hours by eleven, you get the period between passings. You can then use that to specify the ten times, down to the second, when such passings occur.

ie the time of the ten passings is

noon + $k\cdot\frac{12\cdot60\cdot 60}{11}$ seconds, for $k\in\{1,2,...,10\}$

There are of course another ten such times between midnight and noon. They are the same times with pm replaced by am.

3. May 5, 2016

spoiler

4. May 5, 2016

Yes

5. May 5, 2016

### Algr

At the nearest second I would answer "Well..." because one second is to fast to come up with a better answer.

6. May 5, 2016

### rootone

Since high precision mechanical watches are designed to tell the time to the nearest second then 'nearest second' would be accurate.
On the other hand Dad's mechanical clock compared to a Caesium 133 based atomic clock is so wildly innaccurate that you might as well call it random.

7. May 6, 2016

### geoffrey159

I thought the riddle would last a little longer, but @andrewkirk came up with a written answer in 15 minutes !

An alternative answer is the following: there are 60 equal angular sections on the watch. Every hour, the hour needle crosses exactly five angular sections, and between two consecutive hours, it crosses an additional 1/12 of what crosses the minutes needle. A relationship between time $h : m$ and the positions $x_h$ and $x_m$ (in number of angular sections) of the needles is

$x_m = m \quad$ and $\quad x_h = 5(h \text{ mod } 12)+ x_m/12$

The problem consists in finding the pairs $(h,m)$ such that $x_h = x_m$.
This happens whenever $m = \frac{5\times 12 \times (h \text{ mod } 12) } {11}$.
The fractional part of $m$ must be converted in seconds by multiplying it by 60.

So the exact times of needles alignments are :
$h$ hours, $\lfloor \frac{5\times 12 \times (h \text{ mod } 12) } {11} \rfloor$ minutes, and $60\times ( \frac{5\times 12 \times (h \text{ mod } 12) } {11} - \lfloor \frac{5\times 12 \times (h \text{ mod } 12) } {11} \rfloor)$ seconds.

So if you want to see a needle eclipse, take a look at your watch at 1:05:27 !!!