SOLUTION: A circle has 29 points arranged in a clockwise manner numbered from 0 to 28, as shown
in the figure below. A bug moves clockwise around the circle according to the following
rule
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Question 800903: A circle has 29 points arranged in a clockwise manner numbered from 0 to 28, as shown
in the figure below. A bug moves clockwise around the circle according to the following
rule. If it is at a point i on the circle, it moves clockwise in 1 second by ( 1 + r ) places,
where r is the reminder ( possibly 0 ) when i is divided by 11. Thus if it is at position 5, it
moves clockwise in one second by ( 1 + 5 ) places to point 11. Similarly if it is at position
28 it moves ( 1 + 6 ) or 7 places to point 6 in one second.
If it starts at point 23, at what point will it be after 2012 seconds?
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
If the bug starts at point 23,
since ,
the bug will move places at 1 second,
and will end at point .
Since ,
the bug will move places at 2 seconds,
and since it will end at point 0.
Since ,
the bug will move place at 3 seconds,
and since it will end at point 1.
Since ,
the bug will move places at 4 seconds,
and since it will end at point 3.
Since ,
the bug will move places at 5 seconds,
and since it will end at point 7.
Since ,
the bug will move places at 6 seconds,
and since it will end at point 15.
Since ,
the bug will move places at 7 seconds,
and since it will end at point 20.
Since ,
the bug will move places at 8 seconds,
and since the bug will go past 0 will end at point 1.
From that point on, life will become very boring for the bug (and for anyone watching that bug), because the bug will go through endlessly repeated cycles visiting points 3, 7, 15, 20, and 1, in that order, again, and again, and again.
Each cycle will take 5 seconds, so after landing on point 1 at 3 seconds, the bug will keep cycling through points 3, 7, 15, 20, and 1, returning to point 1 every 5 seconds.
In the seconds after landing on point 1 for the first time,
the bug will return to point 1 401 more times (2009 divided by 5 gives a quotient of 401),
the last two times at and seconds.
In 5 more seconds, at 2013 seconds, the bug would return to point 1 once again,
but at 2012 seconds it had landed on point .
I believe that "after 2012 seconds" the bug will rest for a while at point 20,
maybe for half a second, before moving to point 1 again.
After all if the bug moved continuously, "after 2012 seconds" would have no meaning.
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