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?

Solution(By Examveda Team)

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