If $n > 1$ and all $n$ positive integers $a, a + k, \cdots , a+ (n - 1)k$ are odd primes, show every prime $<n \mid k$.

Assume that $p\nmid k$ for some prime $p<n.$ Then $kx\equiv-a\pmod p$ for some integer $0\leqslant x<p$ so $a+kx\equiv0\pmod p$ and since each of the numbers $a+ik,$ with $0\leqslant i<n,$ is a prime then $a+kx=p.$ Now notice that for each $1\leqslant i<n$ we have $\gcd(a,i)=1$ for otherwise the number $a+ik=\gcd(a,i)\left(\frac{a}{\gcd(a,i)}+\frac{i}{\gcd(a,i)}k\right)$ would not be a prime. Thus $a\geqslant n.$ Therefore we have $n\leqslant a\leqslant a+kx=p<n,$ which is a contradiction. Thus the primorial $(n-1)\#$ divides $k.$