Let $r_{i}$ denote the remainder when $2^{i}$ is divided by 25. Note that because $2^{\phi(25)} \equiv 2^{20} \equiv 1(\bmod 25)$, $r$ 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 $$\begin{aligned} \sum_{i=0}^{2015}\left\lfloor\frac{2^{i}}{25}\right\rfloor & =\sum_{i=0}^{2015} \frac{2^{i}-r_{i}}{25} \\ & =\sum_{i=0}^{2015} \frac{2^{i}}{25}-\sum_{i=0}^{2015} \frac{r_{i}}{25} \\ & =\frac{2^{2016}-1}{25}-\sum_{i=0}^{1999} \frac{r_{i}}{25}-\sum_{i=0}^{15} \frac{r_{i}}{25} \\ & =\frac{2^{2016}-1}{25}-100\left(\frac{250}{25}\right)-\sum_{i=0}^{15} \frac{r_{i}}{25} \\ & \equiv \frac{2^{2016}-1}{25}-\sum_{i=0}^{15} \frac{r_{i}}{25}(\bmod 100) \end{aligned}$$ We can calculate $\sum_{i=0}^{15} r_{i}=185$, so $$\sum_{i=0}^{2015}\left\lfloor\frac{2^{i}}{25}\right\rfloor \equiv \frac{2^{2016}-186}{25}(\bmod 100)$$ Now $2^{\phi(625)} \equiv 2^{500} \equiv 1(\bmod 625)$, so $2^{2016} \equiv 2^{16} \equiv 536(\bmod 625)$. Hence $2^{2016}-186 \equiv 350$ $(\bmod 625)$, and $2^{2016}-186 \equiv 2(\bmod 4)$. This implies that $2^{2016}-186 \equiv 350(\bmod 2500)$, and so $\frac{2^{2016}-186}{25} \equiv 14(\bmod 100)$