Inverse question of zeroes

The number of trailing zeros $Z$ of $x!$ is given by:

$Z = \sum _{ i=1 }^{ N }{ \left\lfloor \frac { x! }{ { 5 }^{ i } } \right\rfloor }$ , where $\left\lfloor \right\rfloor$ is the greatest integer function.

Therefore,

$Z = \left\lfloor \frac { 5n }{ 5 } \right\rfloor + \left\lfloor \frac { 5n }{ 25 } \right\rfloor + \left\lfloor \frac { 5n }{ 125 } \right\rfloor + ...$

$\quad = n + \left\lfloor \frac { n }{ 5 } \right\rfloor + \left\lfloor \frac { n }{ 25 } \right\rfloor + \left\lfloor \frac { n }{ 125 } \right\rfloor + ...$

Since $5^5 = 3125 > Z$ , it implies that:

$2014 = n + \left\lfloor \frac { n }{ 5 } \right\rfloor + \left\lfloor \frac { n }{ 25 } \right\rfloor + \left\lfloor \frac { n }{ 125 } \right\rfloor + \left\lfloor \frac { n }{ 625 } \right\rfloor$

$\Rightarrow n (1 + \frac{1}{5} + \frac{1}{25} + \frac{1}{125} \frac{1}{625}) \approx 2014$

$\quad 1.2486n \approx 2014$

$\Rightarrow n \approx 1612$

When $n = 1612$ , $Z = 2012$ , therefore, $n = 1612+2 = \boxed{1614}$ , then $Z = 2014$ .