The owner of a local jewellery store hired 3 watchmen to guard his diamonds, but a thief still got in and stole some diamonds. On the wayout, the thief met each watchman, one at a time. To each he gave $$\frac{1}{2}$$ of the diamonds he had then, and 2 more besides. He escaped with one diamond. How manydid he steal originally?

We can solve this problem by backtracking. 

The thief left with one diamond. So when he met the final guards, he must have given him two more than his half. 
If he still had 1, adding 2 equals 3, which must be half of what he had. 

When reaching the third guard, the thief had six diamonds; he gave half and then two more, giving five diamonds to the third watchmen. 

We have to apply the same logic to the two earlier guards. 

When he left the second watchman, he had six diamonds, meaning when approaching the second watchman, he must have had 2(6+2) = 16 diamonds. 

When he left the first watchman, he had 16 diamonds, meaning when he approached the first watchman, he must have had 2(16+2) = 36 diamonds. 

Therefore, Option B is the correct answer.