A shop had a certain number of diamonds. A thief stole half of the diamonds…

Concept:

When a quantity is repeatedly reduced by "half of it, plus a fixed extra", and the final leftover is known, the original quantity is recovered by reversing the operations one at a time from the end: the amount present just before a theft equals 2 times (the amount left after that theft, plus the fixed extra taken).

Step-by-step (working backward from the empty shop):

1. After the 4th theft, 0 diamonds were left. So before the 4th theft: 2 x (0 + 2) = 4 diamonds.

2. Before the 3rd theft: 2 x (4 + 2) = 12 diamonds.

3. Before the 2nd theft: 2 x (12 + 2) = 28 diamonds.

4. Before the 1st theft (the original count): 2 x (28 + 2) = 60 diamonds.

Cross-check (forward, from the recovered count):

1. 1st theft: half of 60 is 30, plus 2 more = 32 stolen; 60 - 32 = 28 left.

2. 2nd theft: half of 28 is 14, plus 2 more = 16 stolen; 28 - 16 = 12 left.

3. 3rd theft: half of 12 is 6, plus 2 more = 8 stolen; 12 - 8 = 4 left.

4. 4th theft: half of 4 is 2, plus 2 more = 4 stolen; 4 - 4 = 0 left, matching that nothing remained.

This forward check confirms the shop started with 60 diamonds. (Algebraically, if x is the starting count and each theft maps the remaining amount r to r/2 - 2, solving r4 = 0 backward through four stages gives the same value, x = 60.)