Why do analog clocks have 12 divisions?
What are watch tasks?
......  Clock tasks are brain teasers that deal with the face of a clock and the hands. Ideal clocks are assumed here, i.e. clocks whose hands do not move step by step as usual, but instead move continuously. 
Nine tasks are set Top
1) Pointers are on top of each other
At 12 o'clock the large and small hands point upwards.
When will the two hands be on top of each other again?
Source: (2) Problem 43
2) Pointers form a segment
At 6 o'clock the big hand points up, the small hand down.
When do the two hands form a straight line again?
3) Angle between hands
The time is 12:44 p.m. What angle do the two hands form?
4) Pointers are perpendicular to each other
At 3 o'clock the large and small hands form a right angle.
When do the two pointers form a right angle again?
5) Three pointers are on top of each other
At 12 o'clock the large, small and second hands are on top of each other. When does this event happen again?
6) Swap the pointer
At which hand positions are times indicated that also mean times when the minute and hour hands are exchanged?
Source: (1) Issue 61
7) Quarter division of the dial
......  The dial opposite is divided into four areas. The sum of the numbers there is entered in each. How do you have to divide the clock face so that the same total appears in every area? 
8) Path of the pointer tip
The large hand of a wristwatch is 1cm long. What distance will the pointer tip cover in a year?
9) Direction with the clock
How can you tell the south direction with a watch when the sun is shining?
Source: Volksgut
Further exercises can be found e.g. as exercise 57 to 66 in (1) in chapter Clock puzzles.
1) Pointers are on top of each other  solution Top
At 12 o'clock the large and small hands point up. When will the two hands be on top of each other again?
First solution
Once the large pointer circles the dial, it has covered an angle of 360 °. During this time, the little pointer has moved one digit further, i.e. an angle of 30 °. The big pointer is therefore 12 times faster than the small pointer.
The starting time is 12 o'clock. The large and small hands coincide. With one full revolution of the large hand in one hour, the small hand has moved 1/12 further. To reach it, the large pointer must turn 1/12 further. Then the small pointer has moved on by 1/12 of 1/12, i.e. by 1 / 12². The big pointer has to follow suit. Again the small hand moves 1/12 of 1 / 12² further, the big one follows and so on.
So the big pointer moves 1 + 1/12 + 1 / 12² + 1 / 12³ + ... further.
That is a sum with any number of summands. Mostly it goes beyond all borders. In this case it approaches a limit value.
Since this is a geometric series with a_{0}= 1 and q = 1/12, it is known: s = a_{0}/ (1q) = 12/11.
The large hand therefore moves 12/11 hours further, i.e. 1/11 more than a full turn.
1/11 of 60 minutes are 60/11 min. It is also 60/11 min = 5 min + 27 s + 3/11 s.
Result: At 13:05:27 and 3 / 11s the two hands are again on top of each other.
Second solution
The large pointer runs 12 times faster than the small pointer.
The large pointer sweeps over the angle 360 * t degrees in time t, the small one 30 * t degrees.
The pointers match if 30t = 360t360n or t = (12/11) n applies.
n is the number of revolutions.
The large hand therefore moves 12/11 hours further for n = 1, i.e. 1/11 more than a full turn.
1/11 of 60 minutes are 60/11 min. It is also 60/11 min = 5 min + 27 s + 3/11 s.
Result: At 13:05:27 and 3 / 11s the two hands are again on top of each other.
generalization
For the nth meeting, (1/12) t = t  n or t = (12/11) n applies.
At the following times, the pointers are superimposed.
n = 02 n = 03 n = 04 n = 05 n = 06 n = 07 n = 08 n = 09 n = 10 n = 11  24/11 36/11 48/11 60/11 72/11 84/11 96/11 108/11 120/11 132/11 

2) Pointers form a segment  solution Top
At 6 o'clock the big hand points up, the small hand down.
When do the two hands form a straight line again?
solution
The solution to exercise 1 applies here  slightly modified. The time is 7:05:27 and 3 / 11s. 
3) Angle between pointers  solution Top
The time is 7:17 a.m. Which angle alpha do the two pointers form?
solution
The angle between the vertical pointing upwards and the large pointer is 17 * 6 °. The angle between the vertical and the small pointer is 7 * 30 ° + (17/60) * 30 °. 
alpha = [7 * 30 ° + (17/60) * 30 °] 17 * 6 ° = 116.5 °
generalization
Let the time be hh: mm. The following applies: hh * 30 ° + (mm / 60) * 30 ° mm * 6 ° = hh * 30 ° mm * 5.5 °.
The angle between the pointers is equal to the absolute value of the term: alpha =  hh * 30 ° mm * 5.5 ° .
4) Pointers are perpendicular to each other  solution Top
At 3 o'clock the large and small hands form a right angle.
When do the two pointers form a right angle again?
solution
......  The formula from Exercise 3 is assumed: alpha =  hh * 30 ° mm * 5.5 ° . The following should apply with hh = 03: 90 ° = 5.5 ° * mm 3 * 30 °. This results in 5.5 ° * mm = 180 ° or mm = 180 / 5.5 = 360/11 = 32 + 8/11. Result: The next time is 3:32:43 and 7 / 11s. 
5) Three pointers are on top of each other  solution Top
At 12 o'clock the large, small and second hands are on top of each other. When does this event happen again?
Exercise 1 (second solution) shows that the minute and hour hands are on top of each other at time t = (12/11) n. This follows from the approach 30t = 360t360n.
Now the second hand is added.
Further considerations will follow.
At Dr. Math (URL below) can be read further: "One could ask for the closest bunching not at 12 o'clock. I find that this occurs at about 5: 27: 27.3, when all the hands are within a 1.0014 degree sector. "
6) Swap pointers  solution Top
At which hand positions are times indicated that also mean times when the minute and hour hands are exchanged?
solution
In (1) the hand positions between 3 p.m. and midnight are asked for and 66 time pairs are given for this.
Manfred Börgens offers a comprehensible solution on the Internet (URL below).
He finds 143 time pairs. Below are the 11 positions in which both hands are on top of each other.
If x is the position of the hour hand, then x = (12/143) j (j = 0,1,2, ..., 142).
7) Quarter division of the dial  solution Top
......  The dial opposite is divided into four areas. The sum of the numbers lying there is entered in each. How do you have to divide the clock face so that the same total appears in every area? 
solution
......  The face of a clock, if it is provided with Roman numerals, usually looks like this. The numbers are read from within. The four is written as IIII and not IV. Why this is so is unclear. If you add the values together, you get 1 + 2 + ... + 12 = 78. If you rearrange the digits, you get XXXXVVVVIIIIIIIIIIIIIIIIIIII. That gives the value 40 + 20 + 20 = 80. For a solution, you can also get a total of 80 if you split up certain composite representations of a number. That happens in the next two pictures. 
......  ...  Both solutions can be found in Martin Gardner's book "Mathematischer Karneval" on pages 231ff. under the name Cooks (freely translated: errors). They were given by the two pioneers of entertainment mathematics, Dudeney (left) and Loyd. A flaw in Dudeney's solution is that you have to read the IX from the outside in order for it to become XI. Gardner notes that Loyd missed at least 12 other solutions. 
Annotation:
Here is a place to look at the two "puzzle inventors" Sam Loyd (18411911) from the USA and Henry Ernest Dudeney (18571930) from England.
Both were contemporaries and celebrities in their time. Of the two puzzle experts, Dudeney was the one with the greater math skills; Sam Loyd, originally just a chess expert, was a smarter businessman.
Both were in contact with each other. Dudeney sent a large number of his puzzles to Loyd, but soon stopped when he noticed that Loyd was publishing them under his own name.
The two different solutions to the "quartering the clock face" problem above may be an example of the fact that they often published similar puzzles and that they tried to outdo each other.
You can read more about them in Martin Gardner's book (4):
> Sam Loyd: America's Greatest Puzzle Inventor (Pages 40 to 48)
> Henry Ernest Dudeney: England's greatest puzzle inventor (pages 70 to 77)
On my homepage I go into some famous Dudeney puzzles:
Tangram, cutting up a triangle, flyspider problem, crescent puzzle, send more money.
Sam Loyd was represented on my homepage by the game of fifteen. Recent research has shown that he was not the inventor of this puzzle.
8) Path of the pointer tip  solution Top
The large hand of a wristwatch is 1cm long. What distance will the pointer tip cover in a year?
solution
s = (2 * pi * 1cm) * 60 * 24 * 365 = 33 km (rounded to two places)
9) Cardinal direction with the clock  solution Top
How can you determine the direction south with a watch when the sun is shining?
solution
If you hold the watch so that the hour hand points to the sun and then halve the angle between the hour hand and the imaginary line of the 12 o'clock display, the bisecting line is pointing south.
Watch tasks on the Internet Top
German
Manfred Börgens
Math problems # 44
Wikipedia
Samuel Loyd, Ernest Dudeney
English
Doctor Rob, The Math Forum
Three hands on a clock
Gary Darby
Clock angles
Henry Ernest Dudeney (Project Gutenberg Literary Archive Foundation)
Amusements in Mathematics
Wikipedia
Clock angle problem, Henry Dudeney, Sam Loyd
credentials Top
(1) Henry Ernest Dudeney: Amusements in Mathematics, Project Gutenberg (can be downloaded from the Internet)
(2) Sam Loyd: Mathematical puzzles and games, Cologne 2003 [ISBN 3832110496]
(3) Martin Gardner: Mathematical Carnival, Frankfurt / M, Berlin 1977 [ISBN 3 550 07675 4]
(4) Martin Gardner: Mathematical Riddles and Problems, Braunschweig 1968
URL of my homepage:
http://www.mathematischebasteleien.de/
© 2008 Jürgen Köller
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