Problem Analysis

This is a classic logical puzzle that is best solved by working backward from the end result. We know 3 breads remained, and we need to reverse the actions of the three thieves to find the initial amount.

Each thief's action was to take half of the current amount plus half a bread. Let 'x' be the number of breads before a thief's turn. The number of breads taken is $(\frac{x}{2} + 0.5)$. The amount remaining is $x - (\frac{x}{2} + 0.5) = \frac{x}{2} - 0.5$.

To reverse this, if we know the amount 'y' that was remaining, the amount 'x' before the thief acted was $(y + 0.5) \times 2$.

Step-by-Step Calculation (Working Backward)

1. Before the 3rd Thief:
   • There were 3 breads remaining at the very end.
   • Applying the reverse formula: Amount = (3 + 0.5) × 2
   • Amount = 3.5 × 2 = 7
   • So, there were 7 breads before the 3rd thief arrived.
2. Before the 2nd Thief:
   • There were 7 breads before the 3rd thief, which means there were 7 breads remaining after the 2nd thief left.
   • Applying the reverse formula: Amount = (7 + 0.5) × 2
   • Amount = 7.5 × 2 = 15
   • So, there were 15 breads before the 2nd thief arrived.
3. Before the 1st Thief (The Initial Amount):
   • There were 15 breads before the 2nd thief, which means there were 15 breads remaining after the 1st thief left.
   • Applying the reverse formula: Amount = (15 + 0.5) × 2
   • Amount = 15.5 × 2 = 31
   • So, there were 31 breads initially.

Conclusion

The number of breads that were present initially was 31.