A number theory problem by Md Zuhair

Let the number of trailing zeros of a integer $N$ be denoted by $n_z(N)$ . Then we have:

$\begin{aligned} n_z \left(\prod_{r=1}^{1000} r! \right) & = \sum_{r=1}^{1000} n_z(r!) \quad \quad \quad \quad \small \color{#3D99F6} \text{Note that } n_z(n!) = \sum_{k=1}^\infty \left \lfloor \frac n{5^k} \right \rfloor \\ & = \sum_{r=1}^{1000} \sum_{k=1}^\infty \left \lfloor \frac r{5^k} \right \rfloor \\ & = {\color{#3D99F6}\sum_{r=1}^{1000} \left \lfloor \frac r{5} \right \rfloor} + {\color{#D61F06} \sum_{r=1}^{1000} \left \lfloor \frac r{25} \right \rfloor} + {\color{#3D99F6}\sum_{r=1}^{1000} \left \lfloor \frac r{125} \right \rfloor} + \color{#D61F06}\sum_{r=1}^{1000} \left \lfloor \frac r{625} \right \rfloor \\ & = {\color{#3D99F6} 5 \sum_{r=1}^{199} r + 200} + {\color{#D61F06}25 \sum_{r=1}^{39} r + 40} + {\color{#3D99F6}125 \sum_{r=1}^7 r + 8} + {\color{#D61F06} \sum_{r=625}^{1000} 1} \\ & = 5 \times \frac {199 \times 200}2 + 200 + 25 \times \frac {39 \times 40}2 + 40 + 125 \times \frac {7 \times 8}2 + 8 + 1000 - 624 \\ & = 99700 + 19540 + 3508 + 376 \\ & = \boxed{123124} \end{aligned}$