Does it cycle? Maybe...

We notice that the last two digits of $2016^{n}$ , $2017^{n}$ and $2018^{n}$ each have their respective cycles:

• Last two digits of $2016^{n}$ : $\boxed{16,56,96,36,76,16,56......}$
  Notice that the last two digits $2016^{n}$ has a fixed five-number cycle $\boxed{16,56,96,36,76}$ before repeating itself again. Therefore, we can deduce that the last two digits of $2016^{2016}$ is 16, which is the first number in the cycle, because $2016\equiv 1(mod5)$ .
  

Similarly for $2017^{2017}$ and $2018^{2018}$ , we can use the same method we used in $2016^{2016}$ .

• Last two digits of $2017^{n}$ : $\boxed{17,89,13,21,57,69,73,41,97,49,33,61,37,29,93,81,77,09,53,01,17,......}$ The last two digits of $2017^{n}$ is has a fixed 20-number cycle $\boxed{17,89,13,21,57,69,73,41,97,49,33,61,37,29,93,81,77,09,53,01}$ , so we know that the last two digits of $2017^{n}$ is the 17th number in the cycle, which is 77 because $2016\equiv 17(mod20)$ .

• Last two digits of $2018^{n}$ : $\boxed{18,24,32,76,68,24,32,......}$ The last two digits of $2018^{n}$ is has a fixed 4-number cycle $\boxed{24,32,76,68}$ after the first term 18, so we know that the last two digits of $2017^{n}$ is the 1st number in the cycle, which is 24 because $(2018-1)\equiv 1(mod4)$ .

Since we have the last two digits of $2016^{n}$ , $2017^{n}$ and $2018^{n}$ which are 16,77 and 24 respectively, so we can find the last two digits of their product, which is the last two digits of $16\times77\times24$ , which is $\boxed{68}$ .

Let the number given be $N$ . We need to find $N \bmod 100$ .

$\begin{aligned} N & \equiv 2016^{2016} \times 2017^{2017} \times 2018^{2018} \text{ (mod 100)} \\ & \equiv 16^{2016} \times 17^{2017} \times 18^{2018} \text{ (mod 100)} \end{aligned}$

Since $\gcd (16, 18, 100) \ne 1$ , we have to consider the factors 4 and 25 of 100 separately using Chinese remainder theorem.

For factor 4 : $16^{2016} \equiv 0 \text{ (mod 4)}$ and $18^{2018} \equiv (16+2)^{2018} \equiv 0 \text{ (mod 4)}$

For factor 25 :

$\begin{aligned} 16^{2016} & \equiv 16^{2016 \bmod \color{#3D99F6} \phi (25)} \text{ (mod 25)} & \small \color{#3D99F6} \text{Since }\gcd(16,25) = 1 \text{, Euler theorem applies.} \\ & \equiv 16^{2016 \bmod \color{#3D99F6} 20} \text{ (mod 25)} & \small \color{#3D99F6} \text{Euler totient function }\phi (25) = 20 \\ & \equiv 16^{16} \text{ (mod 25)} \\ & \equiv 2^{64 \bmod 20} \text{ (mod 25)} & \small \color{#3D99F6} \text{Apply Euler theorem again.} \\ & \equiv 2^4 \equiv 16 \text{ (mod 25)} & \small \color{#3D99F6} \text{Since }16^{2016} \equiv 0 \text{ (mod 4)} \\ & \equiv 16 \text{ (mod 100)} \end{aligned}$

$\begin{aligned} 18^{2018} & \equiv 18^{2018 \bmod \color{#3D99F6} \phi (25)} \text{ (mod 25)} & \small \color{#3D99F6} \text{Since }\gcd(18,25) = 1 \text{, Euler theorem applies.} \\ & \equiv 18^{2018 \bmod \color{#3D99F6} 20} \text{ (mod 25)} & \small \color{#3D99F6} \text{Euler totient function }\phi (25) = 20 \\ & \equiv 18^{18} \text{ (mod 25)} \\ & \equiv 2^{18}3^{36 \bmod 20} \text{ (mod 25)} & \small \color{#3D99F6} \text{Apply Euler theorem again.} \\ & \equiv 2^{10}2^83^{16} \text{ (mod 25)} \\ & \equiv 24 \times 6 \times 3 \times (25+2)^5 \text{ (mod 25)} \\ & \equiv (-1) \times 6 \times 3 \times 7 \text{ (mod 25)} \\ & \equiv -1 \equiv 24 \text{ (mod 25)} & \small \color{#3D99F6} \text{Since }18^{2018} \equiv 0 \text{ (mod 4)} \\ & \equiv 24 \text{ (mod 100)} \end{aligned}$

Now consider

$\begin{aligned} 17^{2017} & \equiv 17^{2017 \bmod \color{#3D99F6} \phi (100)} \text{ (mod 100)} & \small \color{#3D99F6} \text{Since }\gcd(17,100) = 1 \text{, Euler theorem applies.} \\ & \equiv 17^{2017 \bmod \color{#3D99F6} 40} \text{ (mod 100)} & \small \color{#3D99F6} \text{Euler totient function }\phi (100) = 40 \\ & \equiv 17^{17} \text{ (mod 100)} \\ & \equiv (20-3)^{15} 17^2 \text{ (mod 100)} \\ & \equiv -3^{15} 89 \text{ (mod 100)} \\ & \equiv 3 (10-1)^7 11 \text{ (mod 100)} \\ & \equiv 3\times 69 \times 11 \text{ (mod 100)} \\ & \equiv 77 \text{ (mod 100)} \end{aligned}$

Therefore, $N \equiv 16 \times 77 \times 24 \equiv 32 \times 24 \equiv \boxed{68} \text{ (mod 100)}$ .