Simple. Note that $24\mid k(k+2)$ iff $k$ is even and $k\not\equiv 2\pmod{3}$. We have \[ p:=13\cdots 3 = \frac{4\cdot 10^{k-1}-1}{3}. \]Suppose $k$ is odd. Then, $k-1=2u\ge 2$ is even, thus the given number factorizes as $(2\cdot 10^u-1)(2\cdot 10^u+1)/3$, and since $2\cdot 10^u-1>3$, $p$ cannot be a prime. So, $k$ is even. Likewise, if $k\equiv 2\pmod{3}$, then it's easy to see $13\mid p$, while $p>13$, so a contradiction.