Q. There are five thieves, each loot a bakery one after the other such that the first one takes 1/2 of the total no. of the breads plus 1/2 of a bread. Similarly 2nd, 3rd,4th and 5th also did the same. After the fifth one no. of breads remained are 3. Initially how many breads were there?

• suppose total no. of breads =x
  for 1st thief:x/2+1/2=(x+1)/2
  remaining breads=x-(x+1)/2=(x-1)/2
  2nd thief:((x-1)/4)+1/2=(x+1)/4
  remaining bread=(x-1)/2-(x+1)/4=(x-3)/4
  3rd thief:((x-3)/8)+1/2=(x+1)/8
  remaining bread=((x-3)/4)-(x+1)/8=(x-7)/8
  4th thief:((x-7/16)+1/2)=x+1/16
  remainig bread:(x-7/8)-(x+1/16)=(x-15)/16
  5th thief:((x-15/32)+1/2)=(x+1)/32
  hence x+1(1/2+1/4+1/8+1/16+1/32)+3=x
  x=127
• 12 years agoHelpfull: Yes(21) No(2)
• suppose there were x breads 2 thiefs come and perfom the same action.
  first = x/2 + 1/2
  second=1/2{x-(x/2+1/2)}+1/2
  frist + second = x
  x/2 + 1/2 + 1/2{x-(x/2 + 1/2)} + 1/2 = x
  on solving we will get x = 3
  from here we can conclude that for three thiefs = 3 x 2 + 1 = 7
  and there were total of if 7 thives steel we will left no breads
  for four = 7 x 2 + 1 = 15
  for five = 15 x 2 + 1 = 31
  for six = 31 x 2 + 1 = 63
  for seven = 63 x 2 + 1 = 127
  
  so there was a toatal 127 breads initially
• 12 years agoHelpfull: Yes(8) No(12)
• 127
  if total no of breads is x then after 1st one the remaing portion is (x/2-1)+1/2
  after 2nd one (x-3)/4
  after 3rd one (x-7)/8
  after 4th one (x-15)/16
  after 5th one (x-31)/32
  then (x-31)/32=3
• 12 years agoHelpfull: Yes(4) No(3)
• suppose total breads before 1 theft : x
  remaining breads after 1 theft : y
  so, y = 1/2(x)-(1/2)
  => x = 2(y + 1/2)
  now, after the fifth one remaining bread : 3
  put 3 in the place of y
  x=7,put 7 in the place of y
  x=15,put 15 in the place of y
  x=31,put 31 in the place of y
  x=63,put 63 in the place of y
  x=127, initial bread
• 10 years agoHelpfull: Yes(2) No(1)
• ans should be 112
  before the 5th thief : 2*3+(1/2)=6 and 1/2
  " 4th thief : 2*(6 and half)+(1/2)=13 and 1/2
  .
  .
  before the 1st thief: 2*(55 and half)+ 1/2+1/2=112
  
• 12 years agoHelpfull: Yes(1) No(9)
• suppose total no. of breads =x
  for 1st thief:x/2+1/2=(x+1)/2
  remaining breads=x-(x+1)/2=(x-1)/2
  2nd thief:((x-1)/4)+1/2=(x+1)/4
  remaining bread=(x-1)/2-(x+1)/4=(x-3)/4
  3rd thief:((x-3)/8)+1/2=(x+1)/8
  remaining bread=((x-3)/4)-(x+1)/8=(x-7)/8
  4th thief:((x-7/16)+1/2)=x+1/16
  remainig bread:(x-7/8)-(x+1/16)=(x-15)/16
  5th thief:((x-15/32)+1/2)=(x+1)/32
  hence x+1(1/2+1/4+1/8+1/16+1/32)+3=x
  x=127
• 10 years agoHelpfull: Yes(1) No(0)
• i think it would be 176!!
• 12 years agoHelpfull: Yes(0) No(20)
• Let the total no. of bread=x
  After the 1st thief fled,total no. of bread=x-(x/2+1/2)=(x-1)/2
  After the 2nd thief fled,total no. of bread=(x-1)/2-((x-1)/4+1/2)=(x-3)/4
  After the 3rd thief fled,total no. of bread=(x-3)/4-((x-3)/8+1/2)=(x-7)/8
  After the 4th thief fled,total no. of bread=(x-7)/8-((x-7)/16+1/2)=(x-15)/16
  After the 5th thief fled,total no. of bread=(x-15)/16-((x-15)/32+1/2)=(x-31)/32
  By the problem,
  (x-31)/32=3
  or,x=127 (Ans.)
  
• 11 years agoHelpfull: Yes(0) No(0)
• answer is 127
• 9 years agoHelpfull: Yes(0) No(0)
• GAUTAM could you please explain why in your remaining bread after 1st thief left (x-1)/2 instead of (x+1)/2 ?
• 8 years agoHelpfull: Yes(0) No(0)