Maria has just won a huge chocolate bar as an Easter gift. She decides to break it into pieces to eat it gradually. On the first day, she divides it into 10 pieces and eats only one of them. On the second day, she divides one of the remaining pieces from the previous day into 10 more pieces and eats only one of them. On the third day, she does the same, that is, she divides one of the remaining pieces from the previous day into 10 more and eats only one of them. She continues repeating this process until the next Easter.

a) How many pieces will she have at the end of the third day?

b) Is it possible for her to have exactly 2014 pieces on some day?
Solution

a) At the end of the first day, she will have $10-1=9$ pieces. At the end of the next day, she will have $9-1+10-1=17$ pieces. From a practical standpoint, it's as if she added $10-1-1=8$ new pieces, because one piece is always lost to the division into $10$ and another is always eaten. At the end of the third day, she adds eight more new pieces and ends up with 25.

b) Since the total always increases by 8 every day, after $n$ days from the initial day, she will have $9+8n$ pieces. If it is possible to have exactly 2014 pieces, we must have:

$$
\begin{aligned}
9+8n & =2014 \\
n & =\frac{2005}{8}
\end{aligned}
$$

Since $\frac{2005}{8}$ is not an integer, such a day will never occur.