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 initially?

Mr.Ramki,
Assume,total no of breads,x=48
first person stole,x/2 = 48/2 = 24.
second person stole,x/4 =(x/2)/2 = 24/12 =12.
third person stole, x/8 = (x/4)/2 = 12/2 =6.
fourth person stole, x/16 = (x/8)/2 = 6/2 =3.
Total breads stolen= 24+12+6+3 =45.
Still at the end there are 3 breads remaining what has
mentioned in the given question.
Hence no of breads initially are,48.

If we assume no of breads initially,x=63.
first person=31
second person=15
third person=7
fourth person=3
Breads stolen=31+15+7+3=56
Remaining breads=63-56=7
But in the question at the end there must be 3 breads
remaining,but in your logic 7 breads remaining,so 63 is
wrong one.

Well i guess 63 is right answer coz @Surya u have calculated
and just made half the breads evertime but u havent
considered the half the totle and also "half bread" !

at the end,total no of breads=3.
Let the no of breads in the bakery before 4th thief robbed
the bakery and after 3rd thief robbed=x.
Then x-(x/2+1/2)=3
=> x=7
similarly,no of breads in the bakery before 3rd thief ,2nd
thief and 1st thief robbed are 15,31 and 63 respectively.
So, total no of breads initially=No of breads before 1st
thief robbed=63.

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