A circle has 29 points arranged in a clockwise manner numbered from 0 to 28 . 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 remainder (possibly 0) when i is divided by 17. Thus it is at position 5 , it moves clockwise in one second by (1+5) places to point 11. similarly ,if moves (1+11) or 12 places to point 11 in one second.

If it starts at point 0,at what point will it be after 2012 seconds?

• Ans is 15th place
  0 0
  1 (1+0)=1 1
  2 (1+1)=2 2+1=3
  3 (1+3)=4 4+3=7
  4 (1+7)=8 8+7=15
  5 (1+15)=16 16+15=31 2
  6 (1+2)=3 3+2=5
  7 (1+5)=6 6+5=11
  8 (1+11)=12 12+11=23
  9 (1+23%17=6)=7 7+23=30 | 1
  
  so after every 8 second it reaches to position 1st,
  2012/8 gives 4..
  2012+1 1
  2012+2 3
  2013+3 7
  2012+4 15
  
  
  
  
  
• 9 years agoHelpfull: Yes(20) No(0)
• here it starts from position 0
  so the table will be as follows:
  time present position next position
  0 0
  1 0 0+(0+1=1
  2 1 1+(1+1)=3
  3 3 3+(1+3)=7
  4 7 7+(1+7)=15
  5 15 15+(1+15)=31(2)
  6 2 2+(1+2)=5
  7 5 5+(1+5)=11
  8 11 11+(1+11)=23
  9 23 23+(1+6)=1
  10 1 1+(1+1)=3
  
  and the position will continue to repeat in the same manner.
  Hence after every 9 sec the bug will come to 1.
  
  so 2012/8 gives 4 as reminder.
  so after 5 seconds the bug will be at 15. answer!
• 9 years agoHelpfull: Yes(5) No(0)
• Time Position
  0 0
  1 (1+0)=1 1
  2 (1+1)=2 2+1=3
  3 (1+3)=4 4+3=7
  4 (1+7)=8 8+7=15
  5 (1+15)=16 16+15=31 2
  6 (1+2)=3 3+2=5
  7 (1+5)=6 6+5=11
  8 (1+11)=12 12+11=23
  9 (1+23)=24 24+23=47 18
  10 (1+10)=11 11+18=29 0
  After every 10 sec it comes back on 0
  Now, (2012/10) gives remainder 2
  And after 2 sec it reaches at 3rd position.
• 9 years agoHelpfull: Yes(4) No(5)
• starts from 0 position,
  rule is=1+(previous position/17)|r,r=remainder
  1.0 position
  2.1+(0/17)|r=1+0=1,position=0+1=1st
  3.1+(1/17)|r=1+1=2,pos=1st+2=3rd
  4.1+(3/17)|r=1+3=4,pos=3rd+4=7Th
  5.1+(7/17)|r=1+7=8,pos=7th+8=15th
  6.1+(15/17)|r=1+15=16,pos=15th+16=21st
  7.1+(21/17)|r=1+4=5,pos=21st+6=26th
  8.1+(26/17)|r=1+9=10,pos=26th+10=36th position=0th position
  so after every 8 sec it reaches to its 0th position bcoz it starts from 0
  2008 sec again reaches to 0 position(:2012/8|r=4)
  if 2008 sec=0th position
  2009 sec=1st
  2010 sec=3rd
  2011 sec=7th
  2012 sec=15th position
  
  
  ans-15th place
  
• 9 years agoHelpfull: Yes(3) No(0)
• 0 1 2 3 4 5 6 7
  0 1 3 7 11 16 22 0
  aftr every 7 sec it comes back on 0 ... (2012/7) = gives remainder 3
  now after 3 sec it reaches at 7 ..
  
• 9 years agoHelpfull: Yes(0) No(2)
• How did u take 7 sec ?..can u elaboratly explain me..
• 9 years agoHelpfull: Yes(0) No(0)
• submit the right answer....
  
• 9 years agoHelpfull: Yes(0) No(0)