, 2023

Solution:

The largest $a$ such that $7^{a}$ divides $\binom{2k}{k}$ is equal to the number of carries when you add $k + k$ in base $7$, by Kummer's Theorem. Thus, we need two carries, so $2k$ must have at least $3$ digits in base $7$. Hence, $2k \geq 49$, so $k \geq 25$. We know $k = 25$ works because $25 + 25 = 34_{7} + 34_{7} = 101_{7}$ has two carries.