If $17^{200}$ is divided by $18$, the remainder is:

We can rewrite $17$ relative to the divisor $18$ using negative remainders:
$$17\equiv -1(\mathrm{mod}18)$$

Now, raise both sides to the power of $200$:

$$17^{200}\equiv (-1{)}^{200}(\mathrm{mod}18)$$

Since $200$ is an even number, $(-1{)}^{200}=1$:

$$17^{200}\equiv 1(\mathrm{mod}18)$$

Thus, the remainder is $1$.