Enormous number

Let $N=2012^{2014^{2016}}$ . We need to find $\lfloor N \bmod 100 \div 10\rfloor$ . Since $\gcd(2012, 10) \ne 1$ , we have to consider the factors of 100, that is 4 and 25 separately using Chinese remainder theorem (CRT) .

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

Factor 25 :

$\begin{aligned} N & \equiv (2000 + 12)^{2014^{2016}} \text{ (mod 25)} \\ & \equiv 12^{\color{#3D99F6}2014^{2016}\bmod \phi (25)} \text{ (mod 25)} & \small \color{#3D99F6} \text{Since }\gcd(12,25) = 1\text{, Euler's theorem applies.} \\ & \equiv 12^{\color{#3D99F6}2014^{2016}\bmod 20} \text{ (mod 25)} & \small \color{#3D99F6} \text{Euler's totient function }\phi(25) = 20 \\ & \equiv 12^{\color{#3D99F6}14^{2016}\bmod 20} \text{ (mod 25)} & \small \color{#3D99F6} \text{Since }\gcd(14,20) \ne 1\text{, applies CRT.} \\ & \equiv 12^{\color{#3D99F6}16} \text{ (mod 25)} & \small \color{#3D99F6} \text{See note.} \\ & \equiv (10+2)^{15}(12) \text{ (mod 25)} \\ & \equiv 2^{10}(2^5)(12) \text{ (mod 25)} \\ & \equiv 24(7)(12) \text{ (mod 25)} \\ & \equiv (-1)(7)(12) \text{ (mod 25)} \\ & \equiv -84 \equiv -9 \equiv 16 \text{ (mod 25)} \end{aligned}$

This implies that $N \equiv 25n + 16 \equiv 0 \text{ (mod 4)}$ , where $n$ s an integer, $\implies n=0$ and $N \equiv 16 \text{ (mod 100)}$ and the tens digit of $N$ is $\boxed{1}$ .

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Note: Let $M = 14^{2016}$ and we need to find $M \bmod 20$ using CRT. We note that $M = 14^{2016} \equiv 0 \text{ (mod 4)}$ and $M = 14^{2016}$ $\equiv 4^{2016}$ $\equiv 4^{2016 \bmod \phi (5)}$ $\equiv 4^{2016 \bmod 4}$ $\equiv 4^0 \equiv 1 \text{ (mod 5)}$ . Then $M \equiv 5m+1 \equiv 0 \text{ (mod 4)}$ $\implies m \equiv 3$ and $M \equiv 16 \text{ (mod 20)}$ .

As you see, that's an enormous number. If written in standard notation, its digits will form a row many miles long. By the way, the last five digits are 21216. Thank you for your interest. E.G.