A CIRCLE HAS 29 POINTS ARRANGED IN A CIRCULAR FASHION CLOCKWISE FROM 0 TO 28.A bug moves clockwise on the circle by foolowing rule:
if it is point i on the circle it moves clockwise by i+r places, where r is the remainder when i is divided by 17. Thus if it is at point 5, then it moves in one second clockwise to further 1+5; i.e at 11 position. If it is at position 28 it moves to 1+11 i.e 11 position in one second.
If he starts at point 0; where will he be after 2012 seconds?

• 0sec-1+0%17=1 1st poss
  1sec 1+1%17=2 2+1=3rd poss
  2sec 1+3=4 4+3=7th poss
  3sec 1+7=8 7+8=15th po
  4 sec 1+14%17=15 15+15=30=2nd poss
  5sec 1+2%17=3 3+2=5th poss
  6sec 1+5%17=6 6+5=11th poss
  7th 1+11%17=12 12+11=23th poss
  8th 1+23%17=1+6=7 7+23=30=1st poss
  after 7th poss circle repeates...
  2012%8=4
  15th possition
• 11 years agoHelpfull: Yes(31) No(7)
• ans is 15
  at 1sec o%17+1=1
  at 2 sec 1+1%17+1=3
  at3 sec 3+3%17+1=7
  at 4sec 7+7%17+1=15
  at 5 sec 15+15%17+1=3
  it will repet after 3second
  hence 2012-2=2010
  2010/3
  =0
  hence it will be 15
  
• 11 years agoHelpfull: Yes(8) No(3)
• thnks to JAYASRI but the positions repeat for every 8 secs (or 8 pos)
  Starting from 0 pos,
  pos. after 1sec=0+(1+0%17)=0+(1+0)=1
  ,, ,, 2sec=1+(1+1%17)=1+(1+1)=3
  ,, ,, 3sec=3+(1+3%17)=3+(1+3)=7
  ,, ,, 4sec=7+(1+7%17)=7+(1+7)=15
  ,, ,, 5sec=15+(1+15%17)=15+(1+15)=31=2
  ,, ,, 6sec=2+(1+2%17)=2+(1+2)=5
  ,, ,, 7sec=5+(1+5%17)=5+(1+5)=11
  ,, ,, 8sec=11+(1+11%17)=11+(1+11)=23
  ,, ,, 9sec=23+(1+23%17)=23+(1+6)=30=1
  so poss repeat for every 8 secs
  2012%8=4
  ANS 15 POSITION
• 11 years agoHelpfull: Yes(6) No(0)
• Ans is:7
  
  position calculation value
  -------- ----------- -----
  0 1+0/17 1
  1 1+1/17 2
  3 1+3/17 3
  7 1+7/17 8
  15 1+15/17 16
  2 1+2/17 3
  5 1+5/17 6
  11 1+11/17 12
  23 1+23/17 7
  1 Cycle formed
  
  1-3-7-15-2-5-11-23-1 Every 8 second the bug return to position '1'
  2012-1 (number of digit doesn't involved in the cycle)=2011
  2011/8 = 3 (remainder)
  so the third position in the cycle is '7'.
  hope you've got it.
• 11 years agoHelpfull: Yes(5) No(7)
• the ans is 3..bcoz when he starts from position 0..
  here i=0 and i divided by 17 will also give zero.therefore i+r(0+0) is again zero and hence its first move is only zero and will not be able to move further as this continues for infinite seconds.
  therefore,ans is ZERO.
• 11 years agoHelpfull: Yes(4) No(4)
• ans is o..
• 11 years agoHelpfull: Yes(1) No(5)
• ans is ZERO
  
• 11 years agoHelpfull: Yes(1) No(5)
• thanks jayasri. I made a mistake. answer would be 15. sorry guys.
• 11 years agoHelpfull: Yes(1) No(0)
• jayshri a small mistake at the end of 4th sec u have written 14 in place of 15
• 11 years agoHelpfull: Yes(1) No(0)
• 0sec-1+0%17=1 1st poss
  1sec 1+1%17=2 2+1=3rd poss
  2sec 1+3=4 4+3=7th poss
  3sec 1+7=8 7+8=15th po
  4 sec 1+14%17=15 15+15=30=2nd poss
  5sec 1+2%17=3 3+2=5th poss
  6sec 1+5%17=6 6+5=11th poss
  7th 1+11%17=12 12+11=23th poss
  8th 1+23%17=1+6=7 7+23=30=1st poss
  after 7th poss circle repeates...
  2012%8=4
  15th possition
• 11 years agoHelpfull: Yes(1) No(0)
• the ans is 0..
  
  bcoz when he starts from position 0..
  
  here i=0 and i divided by 17 will also give zero.therefore i+r(0+0) is again zero and hence its first move is only zero and will not be able to move further as this continues for infinite seconds.
  
  therefore,ans is ZERO.
• 10 years agoHelpfull: Yes(0) No(1)