Good Bye 2017

We need to find $2017!! \bmod 100$ . Since $\gcd(2017!!, 100) \ne 1$ , we have to consider the factors 4 and 25 of 100 separately using the Chinese remainder theorem.

Consider factor 4:

$\begin{aligned} 2017!! & \equiv 1 \times 3 \times 1 \times 3 \times 1 \cdots \times 3 \times 1 \text{ (mod 4)} \\ & \equiv 3^{504 \bmod \phi(4)} \text{ (mod 4)} & \small \color{#3D99F6} \text{Since }\gcd (3,4) = 1\text{, Euler's theorem applies.} \\ & \equiv 3^{504 \bmod 2} \text{ (mod 4)} & \small \color{#3D99F6} \text{Euler's function }\phi(4) = 2 \\ & \equiv 1 \text{ (mod 4)} \end{aligned}$

$\begin{aligned} \implies 2017!! & \equiv 4n + 1 & \small \color{#3D99F6} \text{where }n \in \mathbb N \end{aligned}$

Consider factor 25:

$\begin{aligned} 2017!! & \equiv 0 \text{ (mod 25)} \\ 4n + 1 & \equiv 0 \text{ (mod 25)} \\ \implies n & = 6 \end{aligned}$

$\implies 2017!! \equiv 4(6) + 1 \equiv \boxed{25} \text{ (mod 100)}$

$\Large \color{#69047E} \textit{Wishing Everyone A Happy 2018.}$