Last two digits??

Let us consider the four $f(x) \bmod 100$ separately. Since $\gcd(17, 100) = 1$ , we can apply Euler's theorem and use Carmichael's lambda function $\lambda (\cdot)$ . Note that $\lambda (100) = 20$ and $\lambda (20) = 4$ . Then

$\begin{aligned} f(17) & \equiv 17^{17^{17^{17} \bmod 4} \bmod 20} \text{ (mod 100)} \\ & \equiv 17^{17^{(16+1)^{17} \bmod 4} \bmod 20} \text{ (mod 100)} \\ & \equiv 17^{17^1 \bmod 20} \text{ (mod 100)} \\ & \equiv 17^{17} \text{ (mod 100)} \\ & \equiv 17(289)^8 \text{ (mod 100)} \\ & \equiv 17(90-1)^8 \text{ (mod 100)} \\ & \equiv 17(-720+1) \text{ (mod 100)} \\ & \equiv 17(-19) \text{ (mod 100)} \\ & \equiv -23 \text{ (mod 100)} \\ & \equiv 77 \text{ (mod 100)} \end{aligned}$

Since $\gcd(18, 100) \ne 1$ , we have to use Chinese remainder theorem and consider $f(18) \bmod 4$ and $f(18) \bmod 25$ separately.

• First $f(18) \equiv 0 \text{ (mod 4)}$ . Then, since $\gcd(18, 25)$ , Euler's theorem applies and Euler's totient function $\phi (25) = 20$ . Then $f(18) \equiv 18^{18^{18^{18}} \bmod 20} \text{ (mod 25)}$ . But $\gcd(18,20) \ne 1$ , we have to consider $g(18) = 18^{18^{18}} \bmod 20$ using Chinese remainder theorem.
• Again $g(18) \equiv 18^{18^{18}} \equiv 0 \text{ (mod 4)}$ . And $g(18) \equiv 18^{18^{18}} \equiv 3^{18^{18} \mod \phi (5)} \equiv 3^{18^{18} \mod 4} \equiv 3^{2^{18} \mod 4} \equiv 3^0 \equiv 1 \text{ (mod 5)}$ . Then $g(18) \equiv 5n + 1 \equiv 0 \text{ (mod 4)} \implies n \equiv 3$ and $g(18) \equiv 16$ .
• Then $f(18) \equiv 18^{16} \equiv (-7)^{16} \equiv 7^{16} \equiv (50-1)^8 \equiv 1 \text{ (mod 25)}$ .
• Hence, $f(18) \equiv 25n + 1 \equiv 0 \text{ (mod 4)} \implies n \equiv 3$ and $f(18) \equiv 76 \text{ (mod 100)}$ .

$\begin{aligned} f(19) & \equiv 19^{19^{19^{19}}\bmod \lambda(100)} \equiv 19^{(20-1)^{19^{19}}\bmod 20} \equiv 19^{19} \text{ (mod 100)} \\ & \equiv (20-1)^{19} \equiv (380 - 1) \equiv 79 \text{ (mod 100)} \end{aligned}$

And $f(20) \equiv 20^{20^{20^{20}}} \equiv 0 \text{ (mod 100)}$

Therefore, $f(17) + f(18) + f(19) + f(20) = 77 + 76 + 79+ 0 \equiv \boxed{32} \text{ (mod 100)}$ .

This is my solution, I have definitely used Euler's Theorem