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Selling down a position

BrainteasersEasy

A trader holds some shares. On each of $3$ days she sells half of the shares she holds that morning and then $2$ more. After the $3$ rd day she holds $6$ shares. How many shares did she hold at the start?

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Hint

Reverse one day: add back the extra shares, then double.

Solution

If a day ends with $x$ shares, the morning holding $m$ satisfies $\frac{m}{2} - 2 = x$ , so $m = 2 (x + 2)$ .
Before the $3$ rd day she held $2 \times (6 + 2) = 16$ shares.
Before the $2$ nd day she held $2 \times (16 + 2) = 36$ shares.
At the start she held $2 \times (36 + 2) = 76$ shares.

Answer: $76$

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