Last Digit Problem in 2018

Let $N=2018^{2018}$ . We need to find $N \bmod 10$ . Since $\gcd(2018,10) \ne 1$ , we have to consider the factors of 10 that are 2 and 5 separately using Chinese remainder theorem .

Factor 2 : $N = 2018^{2018} \equiv 0 \text{ (mod 2)}$

Factor 5 :

$\begin{aligned} N & \equiv 2018^{2018} \text{ (mod 5)} \\ & \equiv 3^{2018} \text{ (mod 5)} \\ & \equiv 9^{1009} \text{ (mod 5)} \\ & \equiv (10-1)^{1009} \text{ (mod 5)} \\ & \equiv (-1)^{1009} \text{ (mod 5)} \\ & \equiv -1 \text{ (mod 5)} \\ & \equiv 4 \text{ (mod 5)} \end{aligned}$

Implies that $N \equiv 5n+4$ , where $n$ is a positive integer. And we have $N\equiv 5n+4 \equiv n \equiv 0 \text{ (mod 2)}$ , implying $n = 0$ and $N \equiv \boxed{4} \text{ (mod 10)}$ .

We need to find $2018^{2018}\pmod{10}$ .

We know: $2018^{2018} \pmod{10} \equiv 2^{6054} \pmod{10}$

$2^1 \pmod{10} \equiv 2 \pmod{10}$ $2^2 \pmod{10} \equiv 4 \pmod{10}$ $2^3 \pmod{10} \equiv 8 \pmod{10}$ $2^4 \pmod{10} \equiv 6 \pmod{10}$ $2^5 \pmod{10} \equiv 2 \pmod{10}$ $2^6 \pmod{10} \equiv 4 \pmod{10}$

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So, we have: $2018^{2018} \pmod{10} \equiv 2^{6054} \pmod{10} \equiv 2^2 \pmod{10} \equiv \boxed{4} \pmod{10}$