Large Exponents

We will calculate $Z_i$ one by one for the required values, using the same trick to reduce the problem each time.

$1^{2013} + 2^{2013} + 3^{2013} + \dots + 2014^{2013} \equiv Z_{2013} \pmod{2013}$

Grouping terms:

$(1^{2013} + 2012^{2013}) + (2^{2013} + 2011^{2013}) + \dots + 2013^{2013} + 2014^{2013}$

Which factors as,

$(1 + 2012)(\dots) + (2+2011)(\dots) + (3 + 2010)(\dots) + \dots + 2013^{2013} + 2014^{2013}$

Every pair is divisible by $2013$ and thus is $\equiv 0 \pmod{2013}$ , and we are left with $2013^{2013} + 2014^{2013} \equiv 0^{2013} + 1^{2013} = 1 \pmod{2013}$ , so $Z_{2013} = 1$ .

The same trick reduces $Z_{2014}$ to $1007^{2013} + 2014^{2013} \equiv 1007^{2013} + 0^{2013} = 1007^{2013} \pmod{2014}$ , and then we calculate $1007^{2013} \textrm{mod}\:2014$ .

$2 \cdot 1007^{2013} \equiv 0 \pmod{2014} \:\:\Rightarrow\:\: 1007^{2013} \equiv 0 \:\textrm{or}\: 1007 \!\pmod{2014}$

Clearly, $2014$ does not divide $1007$ , so $1007^{2013} \equiv 1007 \pmod{2014}$ , and thus $Z_{2014} = 1007$ .

Continuing the same ideas, we find $Z_{2015} = 0$ , and $Z_{2016} = 1$ .

$2Z_{2013} + Z_{2014} + 5Z_{2015} - 100Z_{2016} = 2 + 1007 + 0 - 100 = \boxed{909}$