If $A = 4^{0} + 4^{1} + 4^{2} + 4^{3} + ... + 4^{2017} + 4^{2018}$ , find the last digit of $A$ .

5
9
1
7

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We need to find $A \bmod 10$ .

$\begin{aligned} A & \equiv 4^0 + 4^1 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6 + \cdots + 4^{2017} + 4^{2018} \text{ (mod 10)} \\ & \equiv 1 + (4^1 + 4^2) + 4^2(4^1 + 4^2) + 4^4(4^1 + 4^2) + \cdots + 4^{2016}(4^1 + 4^2) \text{ (mod 10)} \\ & \equiv 1 + 20 + 4^2(20) + 4^4(20) + \cdots + 4^{2016}(20) \text{ (mod 10)} \\ & \equiv \boxed{1} \text{ (mod 10)} \end{aligned}$