Last Two Digits?

The last two digits of $2018^{2018}$ will be the same as the last two digits of $18^{2018}$ . Consider the last two digits of the first few powers of $18$ : $\begin{aligned} 18^2 &\equiv 24 \mod 100 \\ 18^3 &\equiv 32 \mod 100 \\ 18^4 &\equiv 76 \mod 100 \\ 18^5 &\equiv 68 \mod 100 \\ 18^6 &\equiv 24 \mod 100 \end{aligned}$ Thus the last two digits of $18^n$ repeat with a period of $4$ . In other words, $18^m \equiv 18^n \bmod 100$ if and only if $m\equiv n \bmod 4$ (for $n,m\geq 2$ ). Since $2018\equiv 2\bmod 4$ , the last two digits of $18^{2018}$ , and therefore $2018^{2018}$ will be $\boxed{24}$ .

Let $N=2018^{2018}$ . We need to find $N \bmod 100$ . Since $\gcd(2018, 100) \ne 1$ , we consider the factors 4 and 25 of 100 separately using Chinese remainder theorem .

Factor 4: $N \equiv 0 \text{ (mod 4)}$

Factor 25: Since $\gcd(2018, 25) = 1$ , we can apply Euler's theorem and note that Euler's totient function $\phi(25) = 20$ .

$\begin{aligned} N & \equiv 2018^{2018 \bmod \phi(25)} \text{ (mod 25)} \\ & \equiv 2018^{2018 \bmod 20} \text{ (mod 25)} \\ & \equiv 2018^{18} \text{ (mod 25)} \\ & \equiv (2000+18)^{18} \text{ (mod 25)} \\ & \equiv 18^{18} \text{ (mod 25)} \end{aligned}$

Now consider,

$\begin{aligned} 18^{20} & \equiv 18^{20 \bmod \phi(25)} \equiv 18^{20 \bmod 20} \equiv 18^0 \equiv 1 \text{ (mod 25)} \\ 18^2\cdot 18^{18} & \equiv 1 \text{ (mod 25)} \\ 4 \cdot 81\cdot 18^{18} & \equiv 1 \text{ (mod 25)} \\ 4 \cdot 6 \cdot 18^{18} & \equiv 1 \text{ (mod 25)} \\ 24 \cdot 18^{18} & \equiv 1 \text{ (mod 25)} \\ -18^{18} & \equiv 1 \text{ (mod 25)} \\ 18^{18} & \equiv -1 \equiv 24 \text{ (mod 25)} \end{aligned}$

Since $24 \equiv 0 \text{ (mod 4)}$ , implying that $N \equiv \boxed{24} \text{ (mod 100)}$ .