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?
a) 36
b) 25
c) 640
d) None of these
Answer: a
Explanation: At last thief is left with one diamond.
Hence, the number of diamonds before he gave some diamonds to the third watchman,
$$\eqalign{
& x – \left( { {\frac{x}{2}} + 2} \right) = 1 \cr
& {\kern 1pt} \frac{{ {x – 4} }}{2} = 1 \cr
& {\kern 1pt} x = 6 \cr} $$
Hence, he had 6 diamonds before he gave 5 to the third watchman.
Similarly number of diamonds before giving to second watchman,
$$\eqalign{
& \frac{{ {x – 4} }}{2} = 6 \cr
& {\kern 1pt} x = 16 \cr} $$
And number of diamonds before giving to the first watchman,
$$\eqalign{
& \frac{{ {x – 4} }}{2} = 16 \cr
& {\kern 1pt} x = 36 \cr} $$
The thief has stolen 36 diamonds originally
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