There are five thieves; each one robs a bakery, one after another. The first…
2024
There are five thieves; each one robs a bakery, one after another. The first thief takes half of the total number of breads plus half a bread. The second, third, fourth, and fifth thieves each do the same — taking half of the breads remaining at that point, plus half a bread. After the fifth thief has taken his share, 3 breads remain. Initially, how many breads were there?
- A.
125
- B.
126
- C.
127
- D.
128
Attempted by 2 students.
Show answer & explanation
Correct answer: C
Concept: When a quantity is repeatedly reduced by ‘half of the current amount plus half a unit’, work backward from the final remaining amount using the inverse relation: amount BEFORE a step = 2 × (amount AFTER that step) + 1. This follows directly from the forward rule after = before/2 − 1/2, i.e. before = 2×after + 1.
Application: Apply this inverse relation five times, starting from the final remainder of 3 breads (left after the fifth thief) and working backward to before the first thief took his share.
Before the fifth thief (i.e. the remainder after the fourth thief) = 2 × 3 + 1 = 7.
Before the fourth thief (remainder after the third thief) = 2 × 7 + 1 = 15.
Before the third thief (remainder after the second thief) = 2 × 15 + 1 = 31.
Before the second thief (remainder after the first thief) = 2 × 31 + 1 = 63.
Before the first thief — the initial number of breads — = 2 × 63 + 1 = 127.
Cross-check (forward simulation starting from 127):
Thief 1 takes 127/2 + 1/2 = 64; remaining = 127 − 64 = 63.
Thief 2 takes 63/2 + 1/2 = 32; remaining = 63 − 32 = 31.
Thief 3 takes 31/2 + 1/2 = 16; remaining = 31 − 16 = 15.
Thief 4 takes 15/2 + 1/2 = 8; remaining = 15 − 8 = 7.
Thief 5 takes 7/2 + 1/2 = 4; remaining = 7 − 4 = 3 — matching the final remainder given in the question.
Result: The bakery initially had 127 breads.