, 2016

Solution:
Let $r_{i}$ denote the remainder when $2^{i}$ is divided by $25$. Note that because $2^{\phi(25)} \equiv 2^{20} \equiv 1 \pmod{25}$, $r$ is periodic with length $20$. In addition, we find that $20$ is the order of $2 \bmod 25$. Since $2^{i}$ is never a multiple of $5$, all possible integers from $1$ to $24$ are represented by $r_{1}, r_{2}, \ldots, r_{20}$ with the exceptions of $5, 10, 15$, and $20$. Hence,
$$\sum_{i=1}^{20} r_{i} = \sum_{i=1}^{24} i - (5+10+15+20) = 250.$$
We also have
$$\begin{aligned} \sum_{i=0}^{2015}\left\lfloor\frac{2^{i}}{25}\right\rfloor &= \sum_{i=0}^{2015} \frac{2^{i} - r_{i}}{25} \\ &= \sum_{i=0}^{2015} \frac{2^{i}}{25} - \sum_{i=0}^{2015} \frac{r_{i}}{25} \\ &= \frac{2^{2016} - 1}{25} - \sum_{i=0}^{1999} \frac{r_{i}}{25} - \sum_{i=0}^{15} \frac{r_{i}}{25} \\ &= \frac{2^{2016} - 1}{25} - 100\left(\frac{250}{25}\right) - \sum_{i=0}^{15} \frac{r_{i}}{25} \\ &\equiv \frac{2^{2016} - 1}{25} - \sum_{i=0}^{15} \frac{r_{i}}{25} \pmod{100} \end{aligned}$$
We can calculate $\sum_{i=0}^{15} r_{i} = 185$, so
$$\sum_{i=0}^{2015}\left\lfloor\frac{2^{i}}{25}\right\rfloor \equiv \frac{2^{2016} - 186}{25} \pmod{100}$$
Now $2^{\phi(625)} \equiv 2^{500} \equiv 1 \pmod{625}$, so $2^{2016} \equiv 2^{16} \equiv 536 \pmod{625}$. Hence $2^{2016} - 186 \equiv 350 \pmod{625}$, and $2^{2016} - 186 \equiv 2 \pmod{4}$. This implies that $2^{2016} - 186 \equiv 350 \pmod{2500}$, and so $\frac{2^{2016} - 186}{25} \equiv 14 \pmod{100}$.