Euler $\phi$ function

Let $N=2^{2017}$ . We need to find $N \bmod 1000$ . Since $2, 1000$ are not coprime integers, we have to consider the factors of 1000 that are 8 and 125 separately.

Factor 8: $N \equiv 2^{2017} \equiv 0 \text{ (mod 8)}$

Factor 125:

$\begin{aligned} N & \equiv 2^{\color{#3D99F6}2017 \bmod \phi(125)} \text{ (mod 125)} & \small \color{#3D99F6} \text{Since }\gcd(2,125) =1 \text{, Euler's theorem applies.} \\ & \equiv 2^{\color{#3D99F6}2017 \bmod 100} \text{ (mod 125)} & \small \color{#3D99F6} \text{Euler's totient function }\phi(125) = 100 \\ & \equiv 2^{17} \text{ (mod 125)} \\ & \equiv 2^{10}\times 2^7 \text{ (mod 125)} \\ & \equiv 1024\times 128 \text{ (mod 125)} \\ & \equiv 24\times 3 \text{ (mod 125)} \\ & \equiv 72 \text{ (mod 125)} \end{aligned}$

Since $72 \equiv 0 \text{ (mod 8)} \implies N \equiv \boxed{72} \text{ (mod 1000)}$ .