Let's suppose a set of identically uniformly distributed random variables $A,B,C,D,E$ and assume that we want to find the distribution $X=\min{\{A,B,C,D,E\}}$.
Then the probability $P(X≤x)=P(\min{\{A,B,C,D,E\}}≤x)$ implies that one can find at least one of the elements of $\{A,B,C,D,E\}$ smaller than $x$.
Or one can say that the probability that at least one of them is smaller than $x$ equals to:
$$\begin{eqnarray}P(\min{\{A,B,C,D,E\}}≤x)&=&1-P(A>x,B>x,C>x,D>x,E>x)=\\&=&1-P(A>x)P(B>x)P(C>x)P(D>x)P(E>x)
=\\&=&1-\left(P(Y>x)\right)^5
\end{eqnarray}$$
where $Y$ is any of the variables $\{A,B,C,D,E\}$ and $F_Y(y)$ its cdf (since they are identically distributed.)
Then one can conclude that the cdf of $X$ would look like:$F_X(x)=1-(1-F_Y(x))^5.$
Since you have $Y\sim \text{Uniform} (0,18)$, then the cdf $F_Y=\frac{x}{18}$. Then $F_X=1-(1-\frac{x}{18})^5$.