Problem 3

If $b<3$, then $b=2$ and $n=2a>2p_k$. The prime $a$ is therefore greater than $p_k$; by maximality of $p_k$ below $k$, $a\ge k$, so $n-k=2a-k\ge a$. The factors $2$ and $a$ occur in the factorial. If $b=a$, then $n=a^2$ and $a\ge6$ because $n>26$; hence $n-k>n/3=a^2/3\ge2a$, so $a$ and $2a$ occur and their product is $2a^2$, in particular supplying the two copies of $a$ needed for $a^2\mid(n-k)!$.