New Year's Countdown Day 14: Yay for 2017!

Let $N = 2017^{2017^{2017}}$ . We need to find $N \bmod 10000$ . Since 2017 is prime, we can use Euler's theorem and the respective Carmichael lambda values are $\lambda (10000) = 500$ and $\lambda (500) = 100$ . Then, we have:

$\large \begin{aligned} N & \equiv 2017^{2017^{2017 \bmod 100}\bmod 500} \text{ (mod 10000)} \\ & \equiv 2017^{2017^{17}\bmod 500} \text{ (mod 10000)} \\ & \equiv 2017^{17^{17}\bmod 500} \text{ (mod 10000)} \\ & \equiv 2017^{(10+7)^{15}(17^2) \bmod 500} \text{ (mod 10000)} \\ & \equiv 2017^{(150+7^{15})(289) \bmod 500} \text{ (mod 10000)} \\ & \equiv 2017^{(150+(50-1)^7(7))(289) \bmod 500} \text{ (mod 10000)} \\ & \equiv 2017^{(150+(349)(7))(289) \bmod 500} \text{ (mod 10000)} \\ & \equiv 2017^{(93)(289) \bmod 500} \text{ (mod 10000)} \\ & \equiv 2017^{377} \text{ (mod 10000)} \\ & \equiv (2000+17)^{375}(2000+17)^2 \text{ (mod 10000)} \\ & \equiv 17^{375}(4000+289) \text{ (mod 10000)} \\ & \equiv (20-3)^{375}(4289) \text{ (mod 10000)} \\ & \equiv (7500-{\color{#3D99F6}3^{375}})(4289) \text{ (mod 10000)} & \small \color{#3D99F6} \text{By modular inverse (see note)} \\ & \equiv 7500-{\color{#3D99F6}3307}(4289) \text{ (mod 10000)} \\ & \equiv 3777 \text{ (mod 10000)} \end{aligned}$

Therefore, the sum of the last four digits is $3+7+7+7 = \boxed{24}$ .

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Note:

$\begin{aligned} 3^{380} & \equiv (10-1)^{190} \text{ (mod 10000)} \\ & \equiv \frac {190\times 189 \times 10^2}2 - 190\times 10 + 1 \text{ (mod 10000)} \\ & \equiv 3601 \text{ (mod 10000)} \\ \implies 3^5\times 3^{375} & \equiv 3601 \text{ (mod 10000)} \\ 243 \times 3^{375} & \equiv 3601 \text{ (mod 10000)} \\ 243 \times 7 & \equiv 1701 \text{ (mod 10000)} \\ 243 \times 307 & \equiv 4601 \text{ (mod 10000)} \\ 243 \times 3307 & \equiv 1601 \text{ (mod 10000)} \\ \implies 3^{375} & \equiv 3307 \text{ (mod 10000)} \end{aligned}$