Question

A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were there in his stock initially?

Answer 1

We are given that a person distributes toffees to 5 students. To each student, he gives one more than half the number of toffees he has at that moment.

After distributing toffees to the fifth student, he is left with none.

Reverse Calculation Approach

Let's work backwards from the fifth student:

• At the 5th stage: He gave away all toffees and was left with 0.
• If he had not given the extra 1 toffee, he would have been left with 1. So: \[ \text{Toffees before giving to 5th student} = 2 \text{ (i.e., } 1 \times 2 \text{)} \]
• At the 4th stage: \[ \text{Toffees} = (2 + 1) \times 2 = 6 \]
• At the 3rd stage: \[ \text{Toffees} = (6 + 1) \times 2 = 14 \]
• At the 2nd stage: \[ \text{Toffees} = (14 + 1) \times 2 = 30 \]
• At the 1st stage: \[ \text{Toffees} = (30 + 1) \times 2 = \boxed{62} \]

Final Answer

Therefore, the person initially had \( \boxed{62} \) toffees.

Answer 2

Let the initial number of toffees be \( 64x \).

Step-by-step Distribution:

1. First child: receives \( 32x + 1 \) toffees.
   Remaining: \( 64x - (32x + 1) = 32x - 1 \)
2. Second child: receives \( 16x + \frac{1}{2} \) toffees.
   Remaining: \( 32x - 1 - (16x + \frac{1}{2}) = 16x - \frac{3}{2} \)
3. Third child: receives \( 8x + \frac{1}{4} \) toffees.
   Remaining: \( 16x - \frac{3}{2} - (8x + \frac{1}{4}) = 8x - \frac{7}{4} \)
4. Fourth child: receives \( 4x + \frac{1}{8} \) toffees.
   Remaining: \( 8x - \frac{7}{4} - (4x + \frac{1}{8}) = 4x - \frac{15}{8} \)
5. Fifth child: receives \( 2x + \frac{1}{16} \) toffees.
   Remaining: \( 4x - \frac{15}{8} - (2x + \frac{1}{16}) = 2x - \frac{31}{16} \)

Since all toffees are exhausted after giving to the fifth child, we are given: \[ 2x - \frac{31}{16} = 0 \] Solving: \[ 2x = \frac{31}{16} \Rightarrow x = \frac{31}{32} \]

Final Calculation:

Total initial toffees: \[ 64x = 64 \times \frac{31}{32} = \boxed{62} \]

Answer:

The person initially had 62 toffees.