Number Theory Gone mad!

To simplify the problem, consider that by definition

$n=\lfloor n\rfloor + \{n\}$

Hence

$S=\displaystyle\sum_{i=0}^{2015} \left\lfloor\dfrac{2^i}{25}\right\rfloor=\underbrace{\displaystyle\sum_{i=0}^{2015} \dfrac{2^i}{25}}_{S_1}-\underbrace{\displaystyle\sum_{i=0}^{2015} \left\{\dfrac{2^i}{25}\right\}}_{S_2}$

From the formula for a geometric series we get

$S_1=\dfrac{2^{2016}-1}{25}$

For $S_2$ we'll see how $2^i$ behaves when considered $\mod 25$

$\begin{array}{|c|c|} \hline i & 2^i \mod 25\\ \hline 0&1\\ 1&2\\ 2&4\\ 3&8\\ 4&16\\ 5&7\\ 6&14\\ 7&3\\ 8&6\\ 9&12\\ 10&24\\ \hline \end{array}$

$ord_{25}(2)|\phi(25)$ , or $ord_{25}(2)|20$ and we can see that $ord_{25}(2)>10$ so it must be $20$ . Hence $2$ is a primitive root modulo $25$ and it generates $(\mathbb{Z}/25\mathbb{Z})^*$ .

With this we can calculate $S_2$ , since if

$2^j\equiv k \mod 25\qquad 0<k<25$

Then

$\left\{\dfrac{2^j}{25}\right\}=\left\{\dfrac{25m+k}{25}\right\}=\dfrac{k}{25}$

So $S_2$ can be subdivided in groups of $20$ addends whose sum is

$\dfrac{\displaystyle\sum_{x\in(\mathbb{Z}/25\mathbb{Z})^*} x}{25}=10$

There are $\left\lfloor\dfrac{2016}{20}\right\rfloor=100$ such groups

The last $16$ addends add up to $\dfrac{185}{25}$

Hence

$S_2=10\cdot 1000+\dfrac{185}{25}=\dfrac{25185}{25}$

Consequently

$S=S_1-S_2=\dfrac{2^{2016}-25186}{25}$

We now need $S \mod 100$ , considering $\mod 4$ and $\mod 25$ separately

$S\equiv 2^{2016}-25186 \equiv 2\mod 4$

Consider the numerator of the fraction $\mod 625$

$\phi(625)=500\Longrightarrow 2^{2016}\equiv 2^{16}\mod 625$

$2^{16}-25186\equiv 65536-186\equiv 536-186\equiv 350\mod 625$

This is equivalent to

$25S=625n+350\Longrightarrow S=25n+14$

So

$\begin{cases} S\equiv 2 \mod 4\\ S\equiv 14\mod 25\\ \end{cases}$

And from CRT this implies that $S\equiv \boxed{14}\mod 100$