Child Labor #2

The number of trailing zeros of $k!$ is given by: $z_k =\displaystyle \sum_{i=1}^\infty \left \lfloor \frac{k}{5^i} \right \rfloor$ , where $\lfloor x \rfloor$ is the greatest integer function. The number of trailing zeros of a product of factorial is the sum of $z_k$ as follows:

$\begin{aligned} Z_n & = \sum_{k=1}^n z_k = \sum_{k=1}^n \sum_{i=1}^\infty \left \lfloor \frac{k}{5^i} \right \rfloor \\ & = 0 + 0 + 0 + 0 + 1 + 1 + 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + ... \end{aligned}$

\(\begin{array} {} \Rightarrow & Z_9 & = 5 & \color{red}{<} 2\times 9 \\ & Z_{14} & = 10 + 5 = 15 & \color{red}{<} 2\times 14 \\ & Z_{19} & = 15+15 = 30 & \color {red} {<} 2\times 19 \\ & Z_{24} & = 20 +30 = 50 & \color {red} {>} 2\times 24 \end{array} \)

$\begin{aligned} \Rightarrow 19 < n < 24 \quad \Rightarrow 4(n - 19) + 30 & = 2n \\ 4n - 76 + 30 & = 2n \\ 2n & = 46 \\ n & = \boxed{23} \end{aligned}$