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?

• after 1st second : (1+23%11 = 1) = 2 places [25]
  after 2nd second : (1+25%11 = 3) = 4 places [0]
  after 3rd second : (1+0%11 = 0) = 1 place [1]
  after 4th second : (1+1%11 = 1) = 2 places [3]
  after 5th second : (1+3%11 = 3) = 4 places [7]
  after 6th second : (1+7%11 = 7) = 8 places [15]
  after 7th second : (1+15%11 = 4) = 5 places [20]
  after 8th second : (1+20%11 = 9) = 10 places [1]
  now,for the same pattern from 4th sec to 8th sec will repeat itself (5 sec intervals)..
  
  total time = 2012 secs
  first 3 secs out of pattern...so time left 2012 - 3 =2009 secs
  
  now no. of repetitions in the leftover time = 2009/5 = 401....remainder = 4
  for the next 4 iterations following the similar pattern
  
  the position will be 20..
• 11 years agoHelpfull: Yes(109) No(7)
• Jasveer....see mate..consider the pattern from 4secs-8secs.....these 5 values will kepp on repeating after the 3rd second..in the pattern : 1-[3] 2-[7] 3-[15] 4-[20] 5-[1]....so after 401 complete cycles of these 5 nos repetition...and another 4 iterations u come to " 4-[20]".....gotcha ??
• 11 years agoHelpfull: Yes(5) No(5)
• ans is 20
  
  
• 11 years agoHelpfull: Yes(2) No(8)
• as starting point is 23..in 1 sec it will move (1+1)=2 places..in each sec..so in 2012 sec..it will move 2012*2=4024 places..from 23 to 28..it moves 5 places.. now it has to move 4024-5=4019 places..from 0 to 28..total 29 places..when 4018 is dvded by 29 leaves rem 12..so 12 steps are left and now it is at 28th..so ans is 11
• 11 years agoHelpfull: Yes(2) No(17)
• After 1st second, it moves 1 + (23/11)r = 1 + 1 = 2, So 25th position
  After 2nd second, it moves 1 + 25/11 = 1 + 3 = 4, So 29th position = 0
  After 3rd second, it moves 1 + 0/11 = 1 + 0 = 1, So 1st position
  After 4th second, it moves 1 + 1 = 3rd position
  after 5th, 1 + 3/11 = 4 So 7th
  After 6th, 1 + 7/11 = 8 so 15th
  After 7th, 1 + 15/11 = 5 so 20th
  After 8th, 1 + 20/11 = 10th, So 30th = 1st
  So it is on 1st after every 3 + 5n seconds. So it is on 1st position after 2008 seconds (3 + 5 x 401) So on 20th after 2012 position.
• 7 years agoHelpfull: Yes(2) No(1)
• SARBARTHA i get all the soln but not 2 second last lines when 4 is remainder then how position 20 come plz tell me as soon as possible
  
• 11 years agoHelpfull: Yes(1) No(2)
• answer is 15
  
• 8 years agoHelpfull: Yes(1) No(4)
• present position is 23 after 1 sec it become 25 and then 0,8,13 and again 23.........so after 2012 sec position is 25
• 11 years agoHelpfull: Yes(0) No(12)
• could you explain laxmipraba???
  
• 8 years agoHelpfull: Yes(0) No(2)
• After 1st second : (1+23%11 = 1) = 2 places [25]
  after 2nd second : (1+25%11 = 3) = 4 places [0]
  after 3rd second : (1+0%11 = 0) = 1 place [1]
  after 4th second : (1+1%11 = 1) = 2 places [3]
  after 5th second : (1+3%11 = 3) = 4 places [7]
  after 6th second : (1+7%11 = 7) = 8 places [15]
  after 7th second : (1+15%11 = 4) = 5 places [20]
  after 8th second : (1+20%11 = 9) = 10 places [1]
  now,for the same pattern from 4th sec to 8th sec will repeat itself (5 sec intervals)..
  
  total time = 2012 secs
  first 3 secs out of pattern...so time left 2012 - 3 =2009 secs
  
  now no. of repetitions in the leftover time = 2009/5 = 401....remainder = 4
  for the next 4 iterations following the similar pattern
  
  the position will be 20..
• 5 years agoHelpfull: Yes(0) No(0)
• Solution- after 1st second : (1+23%11 = 1) = 2 places [25] after 2nd second : (1+25%11 = 3) = 4 places [0] after 3rd second : (1+0%11 = 0) = 1 place [1] after 4th second : (1+1%11 = 1) = 2 places [3] after 5th second : (1+3%11 = 3) = 4 places [7]
  after 6th second : (1+7%11 = 7) = 8 places [15] after 7th second : (1+15%11 = 4) = 5 places [20] after 8th second : (1+20%11 = 9) = 10 places [1] now,for the same pattern from 4th sec to 8th sec will repeat itself (5 sec intervals)..
  
  total time = 2012 secs first 3 secs out of pattern...so time left 2012 - 3 =2009 secs
  
  now no. of repetitions in the leftover time = 2009/5 = 401....remainder = 4 for the next 4 iterations following the similar pattern
  
  the position will be 20..
• 4 years agoHelpfull: Yes(0) No(0)