A1. John had a box of candies. On the first day he ate exactly half of the candies and gave one to his little sister. On the second day he ate exactly half of the remaining candies and gave one to his little sister. On the third day he ate exactly half of the remaining candies and gave one to his little sister, at which point no candies remained. How many candies were in the box at the start?
Solution 1: We work backwards. At the start of the third day, John had 2 candies. At the start of the second day, he had $2 \times(2+1)=6$ candies. At the start of the first day, he had $2 \times(6+1)=14$ candies, so there were 14 candies in the box at the start.

Solution 2: Assume that there were $x$ candies in the box at the start. After the first day, there are $\frac{x}{2}-1$ candies. After the second day, there are
$$
\frac{\frac{x}{2}-1}{2}-1
$$
candies, and after the last day, there are
$$
\frac{\frac{\frac{x}{2}-1}{2}-1}{2}-1=0
$$
candies. Expanding the left-hand side, we see that this is equivalent to
$$
\frac{x}{8}-\frac{7}{4}=0,
$$
so we have $x=8 \times \frac{7}{4}=14$.
Answer: 14 .