Brain Teasers
Swap Time
This puzzle is split into an initial logic based piece and then a rigorous math piece, so everyone can play.
Picture a standard clock with continuously moving hour and minute hands.
A "Swap Time" is a time when you can swap the hour and minute hand, and still get a valid time.
For example, 3:00 is NOT a Swap Time. If we swap the minute hand and the hour hand, then the minute hand points directly at the 3 and the hour hand points directly at the 12. This is not a valid time. When the minute hand points at the three, the hour hand will not be pointing directly at any number.
A) Find one Swap Time.
B) What are the Swap Times between 1:00 and 2:00?
Times will be specified as hours, minutes, and seconds rounded to two decimal places.
Ignore a.m. and p.m. distinctions.
Bonus: How many Swap Times are there in total?
Picture a standard clock with continuously moving hour and minute hands.
A "Swap Time" is a time when you can swap the hour and minute hand, and still get a valid time.
For example, 3:00 is NOT a Swap Time. If we swap the minute hand and the hour hand, then the minute hand points directly at the 3 and the hour hand points directly at the 12. This is not a valid time. When the minute hand points at the three, the hour hand will not be pointing directly at any number.
A) Find one Swap Time.
B) What are the Swap Times between 1:00 and 2:00?
Times will be specified as hours, minutes, and seconds rounded to two decimal places.
Ignore a.m. and p.m. distinctions.
Bonus: How many Swap Times are there in total?
Answer
A) Noon is a Swap Time. Since both hands are pointing at 12, swapping them generates the exact same valid time. In fact, all eleven times when the hands point in the exact samedirection are Swap Times.
B) Let's do some math.
Rule 1) For a valid time, the fraction of the minute hand's progress around the full circle must be equal to the fraction of the hour hand's progress between two numbers. For example, when the minute hand points at the 6, halfway around the clock, the hour hand must be pointed halfway between two numbers.
Rule 2) For a Swap Time, it must also be true that the fraction of the progress of the hour hand around the full circle is equal to the fraction of the progress of the minute hand between two numbers.
Let H = hour hand angle in degrees
Let M = minute hand angle in degrees
Note that there are 360/12 = 30 degrees between numbers on the clock.
Rule 1
H = <angle to number> + <angle past number>
H = 30 * i + <minute hand progress> / 30 : i = 0, 1, 2,..., 11
H = 30 * i + (M / 360) / 30
H = 30 * i + M / 12
Rule 2
M = 30 * j + H / 12 : j = 0, 1, 2,..., 11
Substituting for H:
M = 30 * j + (30 * i + M / 12) / 12
M = 30 * j + 5 * i / 2 + M / 144
143 * M / 144 = 30 * j + 5 * i / 2
M = 144 / 143 * (30 * j + 5 * i / 2) : i = 0,1,2,...,11 j = 0,1,2,...,11
Let s be the number of seconds that have elapsed in the current hour.
Since the minute hand moves at 0.1 degrees per second (360 deg / 3600 s), s = 10 * M
s = 10 * [144 / 143 * (30 * j + 5 * i / 2)] : i = 0,1,2,...,11 j = 0,1,2,...,11
To find all Swap Times between 1:00 and 2:00, set i = 1 and check all values of j.
j = 0, s = 25.17482517, time = 1:00:25.17
j = 1, s = 327.2727273, time = 1:05:27.27
Remaining swap times are
1:10:29.37
1:15:31.47
1:20:33.57
1:25:35.66
1:30:37.76
1:35:39.86
1:40:41.96
1:45:44.06
1:50:46.15
1:55:48.25
Swap Time occur every 12 * 3600 / 143 = 302.0979 seconds.
There are 143 total swap times.
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Comments
You put a lot of work into that!
Great maths equations!! Well done!
I sense that there should be an easier way, but I'm not sure I have it right ... almost, maybe? ...
In each 5 minute period, if the minute hand is thought of as the hour hand, then there will be one moment when the hour hand is at the point where it could be a minute hand. i.e. a swap time occurs!
Now, there will be 12 each hour x 12 hours = 144 times. I think we end up counting 12 noon twice, so subtract 1.
They therefore occur 143 times every 12 hours, or once every 60x60x12 /143 seconds. Start at noon, then keep adding...
I sense that there should be an easier way, but I'm not sure I have it right ... almost, maybe? ...
In each 5 minute period, if the minute hand is thought of as the hour hand, then there will be one moment when the hour hand is at the point where it could be a minute hand. i.e. a swap time occurs!
Now, there will be 12 each hour x 12 hours = 144 times. I think we end up counting 12 noon twice, so subtract 1.
They therefore occur 143 times every 12 hours, or once every 60x60x12 /143 seconds. Start at noon, then keep adding...
Dalfamnest, outstanding! You are absolutely correct. When I started thinking about this, it was not clear to me that non-overlapping Swap Times existed, or that they would be evenly spaced. Your intuition is far better than mine. I had to take the more rigorous path to see the big picture.
I do not think it occurs every 143 second but rather 143 seconds after you have found your first answer.
Thank you
Thank you
exhibit 1: 143 seconds=2 min 23 seconds
We are looking for swap time and if 2min 23sec is added to 10, we will have 12:23sec
Conclusively , one can say 143 sec will be constant, after you have found your first answer
We are looking for swap time and if 2min 23sec is added to 10, we will have 12:23sec
Conclusively , one can say 143 sec will be constant, after you have found your first answer
Realconnect, you are correct. The times are evenly spaced starting at midnight. Since this question asked for the Swap Times between 1 and 2 o'clock, the answers do not start right at 1 o-clock. Good clarification.
A scip model for solving this teaser can be found at
https://gist.github.com/saska-gist/
7cf1ab19a944196a3897d2bf52018bb4
https://gist.github.com/saska-gist/
7cf1ab19a944196a3897d2bf52018bb4
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