Think!

Using the formula for the number of trailing zeroes:

$f\left( n \right) =\sum _{ i=1 }^{ \left\lfloor \log _{ 5 }{ n } \right\rfloor }{ \left\lfloor \frac { n }{ { 5 }^{ i } } \right\rfloor }$

we can see that

$f\left( 5k-1 \right) <f\left( 5k \right) =f\left( 5k+1 \right) =f\left( 5k+2 \right) =f\left( 5k+3 \right) =f\left( 5k+4 \right) <f\left( 5k+5 \right)$ .

So there are $5$ numbers $n$ for which $f\left( n \right) =x$ or in this case $20152015$ .

To get a zero, you need a factor of 5, and a factor of 2. THERE ARE ENOUGH FACTORS OF 2. There are 5 distinct numbers that have x as their greatest multiple of 5 in their factorial. Sorry for bad english.