$73\mid 3^{2018} + K$

Relevant wiki: Fermat's Little Theorem

By Fermat's Little Theorem, if $p$ is prime and $a$ is not divisible by $p$ , $a^{p-1}\equiv 1\bmod p$ . Thus, since $73$ is prime, $3^{72} \equiv 1\bmod 73$ . Since $2018\equiv 2\bmod 72$ , we can conclude that $3^{2018} \equiv 3^2 \bmod 73$ .

We know know that the remainder when $3^{2018}$ is divided by $73$ is $3^2=9$ . So the smallest positive value of $K$ is $73-9=\boxed{64}$ .

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As a note, any integer of the form $K=73k+64$ will be a solution.

Relevant wiki: Euler's Theorem

$\begin{aligned} 3^{2018} & \equiv 3^{\color{#3D99F6}2018 \bmod \phi (73)} \text{ (mod 73)} & \small \color{#3D99F6} \text{Since }\gcd (3,73) = 1 \text{, Euler's theorem applies.} \\ & \equiv 3^{\color{#3D99F6}2018 \bmod 72} \text{ (mod 73)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (73) = 72 \\ & \equiv 3^2 \equiv 9 \text{ (mod 73)} \end{aligned}$

Therefore, for $3^{2018} + K$ to be divisible by 73, the smallest $K = 73-9 = \boxed{64}$ .

lets -> 3^2018 + k = 0 (mod 73) (3^72)^28 (3)^2 + k = 0 (mod 73) 1 (3^2) + k = 0 (mod 73) 9 + k = 0 (mod 73)

so the smallest number of k is 64