Discussion

Explanation:

Giving chocolates to 5^th child:
Let the man be left with x chocolates after 4^th child.
Now he gives half of x i.e., x/2 chocolates to 5^th child. Hence, he will be left with x/2 chocolates.
The man again gives 1 chocolate to 5th child and is now left with no chocolates.
∴ x/2 – 1 = 0
⇒ x = 2

Giving chocolates to 4^th child:
Now, let the man be left with y chocolates after 3^rd child.
Now he gives half of y i.e., y/2 chocolates to 4^th child. Hence, he will be left with y/2 chocolates.
The man again gives 1 chocolate to 4^th child and is now left with 2 chocolates.
∴ y/2 – 1 = 2
⇒ y = 6

Giving chocolates to 3^rd child:
Now, let the man be left with z chocolates after 2^nd child.
∴ z/2 – 1 = 6
⇒ y = 14

Giving chocolates to 2^nd child:
Now, let the man be left with a chocolates after 1^st child.
∴ a/2 – 1 = 14
⇒ y = 30

Giving chocolates to 1^st child:
Now, let the man have c chocolates in the beginning.
∴ c/2 – 1 = 30
⇒ y = 62

Hence, 62.

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