= Solution
{c}
Suppose there were only finitely many <prime numbers> congruent to $5$ modulo $6$, and list them as $p_1,\ldots,p_k$. The list is nonempty since it contains $5$. Let $N=6p_1\cdots p_k-1$. It is greater than one and divisible by neither $2$ nor $3$. Every prime factor therefore has residue $1$ or $-1$ modulo $6$. Since $N\equiv-1\pmod6$, at least one prime factor $q$ must have residue $-1$. But $N\equiv-1\pmod{p_i}$ for every listed prime, so $q$ is not on the list. This contradiction proves \b[there are infinitely many primes congruent to $5$ modulo $6$].
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