4 thieves rob a bakery of the breadone after the other.each thief takes half of what is present ,& half
a
bread...if at the end 3 bread remains,what is the no of bread that was present initiallly?

• it will be 63
  
  procedure:
  
  let initially der are x breads
  
  1st thief takes=x/2 +1/2 ,so remaining=x-(x/2+1/2)=x/2-1/2
  
  2nd thief takes=1/2+1/2*(x/2-1/2)=1/2+x/4-1/4=x/4+1/4 so remain=x/2-1/2-x/4-1/4=x/4-3/4
  
  3rd thief takes=1/2 +1/2*(x/4-3/4)=1/2+x/8-3/8=x/8+1/8 so remain=x/4-3/4-x/8-1/8=x/8-7/8
  
  4th thief takes=1/2+1/2*(x/8-7/8)=1/2+x/16-7/16=x/16+1/16 so remain=x/8-7/8-x/16-1/16=x/16-15/16
  
  so x/16-15/16=3
  
  x=48+15=63(ans)
• 12 years agoHelpfull: Yes(22) No(0)
• If at the end,3 bread remains,
  The 4th thief must have taken 4breads(out of 7 breads, he has taken 3.5+0.5)
  The third thief have taken=8 breads
  The second thief have taken=16 breads
  The first thief have taken=32
  No. of breads initially there was=3+4+8+16+32=63
• 12 years agoHelpfull: Yes(6) No(2)