A circle has 29 points arranged in clockwise manner from 0 to 28. A bug moves clockwise on the circle according to the following rule. If it is at a point i on the circle, it moves clockwise in 1 sec by (1+r) places, where r is the remainder (possibly 0) when i is divided by 17. If starts from position 0, at what position will it be after 2012 sec?

• after 1 sec its position will be at point 1.
  after 2 sec its position will be at point 3.
  after 3 sec its position will be at point 7.
  after 4 sec its position will be at point 15. after 5 sec its position will be at again on 3 .
  now the cycle goes on. after 2012 sec its position will be on 15.
• 11 years agoHelpfull: Yes(7) No(10)
• points are from 0-28
  possibly in i sec 1 place so in 30 sec it will be at o position again.
  so 2012/30=2.
  hence it will be at position 2.
• 11 years agoHelpfull: Yes(5) No(13)
• ans: 20
  i r(i%11) l+r newposition timesec
  23 1 2 23+2=25 1
  25 3 4 25+4=0(as positions start from 0 - 28)
  
• 11 years agoHelpfull: Yes(4) No(3)
• all the on the circle from 0-28 means 29 points
  bugs moves clockwise and its position will change (1+r)
  eg from 1 sec =1;2 sec=2 ......at 17 sec=17 means i=17 and again position will 0(zero).so 2012/17=118 and reminder =6 so position =6
• 11 years agoHelpfull: Yes(2) No(8)
• ans is 15...
• 11 years agoHelpfull: Yes(1) No(9)
• In 30 sec bug will move (30*31)/2=465 steps or (465%29)=1. so after 2010 sec it will be at point 1 and then in next 1 sec it will move 2(1+1) point so it will reach at 3rd point and in next 1 sec it will move 4(3+1) point so it will be at 7th point.
• 11 years agoHelpfull: Yes(1) No(6)
• after 1 sec - at point 0
  after 2 sec - at point 1
  after 3 sec - at point 3
  after 4 sec - at point 7
  
  after 5 sec - at point 15
  after 6 sec - at point 2
  after 7 sec - at point 5
  after 8 sec - at point 11
  
  after 9 sec - at point 23
  after 10 sec - at point 2
  after 11 sec - at point 5
  after 12 sec - at point 11
  after 13 sec - at point 23
  
  The same cycle is repeated now, so for every multiple of 4, it will be at point 23..
  Since 2012 is also a multiple of 4, so the ans will be 23
• 11 years agoHelpfull: Yes(1) No(2)
• he will be at 12th point.:) i think so..if not please tell me right ans
• 11 years agoHelpfull: Yes(0) No(4)
• i think the ans is 3 bcz by substituting we get 1,3,7,15,3,7,15 and the cycle repeats for the multiples of 3(2010 is the multiple of 3)and @ 2010th sec its @15 and the 1st one 1 is remaining therefore @ 2011 th sec its @ 15 and by 2012 it will be at position 3
• 11 years agoHelpfull: Yes(0) No(2)