Number of $5$s dividing $\prod_{n=1}^{50} n! = \sum_{n=1}^{50} v_5(n!)$ where $v_5(n!) = \lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + \cdots$.

For $n \le 50$, $v_5(n!) = \lfloor n/5\rfloor + \lfloor n/25\rfloor$ (higher terms vanish).

$\sum_{n=1}^{50} \lfloor n/5 \rfloor$: $\lfloor n/5 \rfloor = k$ for $n \in \{5k, 5k+1, 5k+2, 5k+3, 5k+4\}$, contributing $5k$.

• $n = 1\text{--}4$: $0$, sum $0$.
• $n = 5\text{--}9$: $1$, sum $5$.
• $n = 10\text{--}14$: $2$, sum $10$.
• $n = 15\text{--}19$: $3$, sum $15$.
• $n = 20\text{--}24$: $4$, sum $20$.
• $n = 25\text{--}29$: $5$, sum $25$.
• $n = 30\text{--}34$: $6$, sum $30$.
• $n = 35\text{--}39$: $7$, sum $35$.
• $n = 40\text{--}44$: $8$, sum $40$.
• $n = 45\text{--}49$: $9$, sum $45$.
• $n = 50$: $10$, sum $10$.

Total: $0 + 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 10 = 235$.

$\sum_{n=1}^{50} \lfloor n/25 \rfloor$: $0$ for $n \le 24$, $1$ for $n \in [25, 49]$ ($25$ values), $2$ for $n = 50$. Sum $= 25 + 2 = 27$.

Total $k = 235 + 27 = 262$.

Verified answer: $262$.

Why this is CAT-level: Double summation $\sum_n v_5(n!)$ instead of $v_5(\text{single factorial})$; the structural switch is non-obvious and brute-force expansion is infeasible.