= Solution
The degree $[K:F]$ is the dimension of $K$ as a vector space over $F$. The tower theorem states that for intermediate fields $F\subseteq E\subseteq K$ in a finite extension,
$$
[K:F]=[K:E][E:F].
$$
If the degree is prime, one factor must be one, forcing $E=K$ or $E=F$; no strictly intermediate field exists.
Let $K$ have 32 elements. Its prime subfield is $\mathbb F_2$ and $[K:\mathbb F_2]=5$. Every element outside that subfield has <minimal polynomial> of degree five, by the tower theorem. Also every member of $K$ satisfies $x^{32}=x$: nonzero elements form a <group> of order 31. Hence $x^{32}-x$ splits over $K$, with distinct roots because its derivative is $-1=1$ in characteristic two.
Every irreducible degree-five polynomial over $\mathbb F_2$ divides $x^{32}-x$. Indeed its quotient field has 32 elements, and its <residue> class of $x$ satisfies that equation. Thus every such polynomial splits into five distinct roots in $K$. Conversely each of the 30 elements outside $\mathbb F_2$ has one such <minimal polynomial>, and different minimal polynomials have disjoint root sets. Consequently the number is
$$
\boxed{30/5=6}.
$$
There is no extra leading-coefficient factor, since every nonzero leading coefficient over $\mathbb F_2$ is one.
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