Solving by Working Backward

This type of puzzle is most easily solved by starting from the final state and reversing the operations step-by-step to find the initial quantity.

1. The Rule of Sale and Its Reverse

• The Sale: Each time, the dealer sells half of his eggs plus half an egg. If he has 'x' eggs, he sells $(\frac{x}{2} + 0.5)$. The number of eggs remaining is $x - (\frac{x}{2} + 0.5) = \frac{x-1}{2}$.
• The Reverse Operation: If 'y' is the number of eggs remaining after a sale, the number of eggs 'x' before the sale can be found by reversing the operation: $x = 2y + 1$.

2. Step-by-Step Calculation

We'll apply the reverse operation three times, starting from the 7 eggs he had at the end.

1. Before the sale to Sivani (the 3rd person):
   • Eggs remaining at the end = 7.
   • Eggs before this sale = (7 × 2) + 1 = 14 + 1 = 15.
2. Before the sale to Anurag (the 2nd person):
   • Eggs remaining after this sale = 15.
   • Eggs before this sale = (15 × 2) + 1 = 30 + 1 = 31.
3. Before the sale to Deepak (the 1st person):
   • Eggs remaining after this sale = 31.
   • Eggs before this sale = (31 × 2) + 1 = 62 + 1 = 63.

3. Verification (Working Forward)

• Start: 63 eggs.
• Sells to Deepak: Sells (63/2 + 0.5) = 32. Remaining: 63 - 32 = 31.
• Sells to Anurag: Sells (31/2 + 0.5) = 16. Remaining: 31 - 16 = 15.
• Sells to Sivani: Sells (15/2 + 0.5) = 8. Remaining: 15 - 8 = 7.

The final amount matches the problem statement.

Conclusion

The dealer started with 63 eggs.