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 way out, 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 many did he steal originally?

Explanation:-

$$\eqalign{ & {\text{At last thief is left with one diamond}}{\text{.}} \cr & {\text{Hence, the number of diamonds before }} \cr & {\text{he gave some diamonds to the third watchman}}, \cr & = x - \left( {\left( {\frac{x}{2}} \right) + 2} \right) = 1; \cr & Or,\,\frac{{\left( {x - 4} \right)}}{2} = 1 \cr & Or,\,x = 6 \cr & {\text{Hence,}}\,{\text{he}}\,{\text{had}}\,{\text{6}}\,{\text{diamonds}}\,{\text{before}}\,{\text{he}}\,{\text{gave}}\,{\text{5}}\,{\text{to}}\,{\text{the}}\,{\text{third}}\,{\text{watchman}}{\text{.}} \cr & {\text{Similarly number of diamonds before giving to second watchman,}} \cr & \frac{{\left( {x - 4} \right)}}{2} = 6; \cr & Or,\,x = 16. \cr & {\text{And}}\,{\text{number}}\,{\text{of}}\,{\text{diamonds}}\,{\text{before}}\,{\text{giving}}\,{\text{to}}\,{\text{the}}\,{\text{first}}\,{\text{watchman}}, \cr & \frac{{\left( {x - 4} \right)}}{2} = 16; \cr & Or,\,x = 36 \cr & \therefore {\text{The}}\,{\text{thief}}\,{\text{has}}\,{\text{stolen}}\,{\text{36}}\,{\text{diamonds}}\,{\text{originally}} \cr}$$