Happy X'mas

Number of zeroes in $25!$ ,

$=\left \lfloor \frac{25}{5^1} \right \rfloor + \left \lfloor \frac{25}{5^2} \right \rfloor$

$=5+1=6$

Number of zeroes in $12!$ ,

$=\left \lfloor \frac{12}{5^1} \right \rfloor$

$=2$

Number of zeroes in $2015!$

$=\left \lfloor \frac{2015}{5^1} \right \rfloor + \left \lfloor \frac{2015}{5^2} \right \rfloor + \left \lfloor \frac{2015}{5^3} \right \rfloor + \left \lfloor \frac{2015}{5^4} \right \rfloor$

$=403+80+16+3=502$

Therefore,

Number of zeroes in $25!×12!×2015!$ ,

$=10^6×10^2×10^{502}$

$=10^{510} \Rightarrow \boxed{510}$

Take 2015 and divide it by 5 succeivly and add all quotients you get 403+80+16+3=502. Similarly divide 25 we get 6 also divide 12 we get 2 . so 506+12+2=510.so it ends with 510 zeros.