Division and stuff.

$\begin{aligned} 1^{2018}+2^{2018}+3^{2018}+4^{2018} & \equiv 1 + 2^{\color{#3D99F6} 2018 \bmod \phi (5)} + 3^{\color{#3D99F6} 2018 \bmod \phi (5)} + 4^{\color{#3D99F6} 2018 \bmod \phi (5)} \text{ (mod 5)} & \small \color{#3D99F6} \text{Since }\gcd(2,3,5) = \gcd(4,5)=1 \text{, Euler's theorem applies.} \\ & \equiv 1 + 2^{\color{#3D99F6} 2018 \bmod 4} + 3^{\color{#3D99F6} 2018 \bmod 4} + 4^{\color{#3D99F6} 2018 \bmod 4} \text{ (mod 5)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (5) = 4 \\ & \equiv 1 + 2^2 + 3^2 + 4^2 \text{ (mod 5)} \\ & \equiv 1 + 4 + 4 + 1 \text{ (mod 5)} \\ & \equiv \boxed 0 \text{ (mod 5)} \end{aligned}$

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References:

• Euler's theorem
• Euler's totient function

We add the last digits of each number.We get 1^2018 has 1.Now consider 4^n.If n=2k+1(k is whole) then the last digit of 4^n is 4.Otherwise it is 6.So now we have 1 and 6.3^n has the pattern 3,9,7,1.As n= 2018 is at place 2,3^2018's last digit is 9.2^n has the pattern 2,4,8,6.As n=2018 is at place 2,the last digit of 2^2018 is 4.1+6+9+4=20 whose last digit is 0.By common sense we can identify that a number whose last digit is 0 is divisible 5 with r=0.