How many trailing zeros 75! have ?

$\lfloor\frac{75}5\rfloor+\lfloor\frac{75}{25}\rfloor=18$

$\dfrac{75}{5}+\dfrac{75}{5^2}=15+3=18$

The answer is $18$ .

Note that after dividing $75$ by $5^3$ , the answer is less than $1$ , so we don't need to divide $75$ by $5^4$ and so on.

75 in 75! is the multiple of 5. which means there are 75/5 = 15 fives in 75. again 75 is the multiple of 25 too. so total fives in 75 will be 15+3 = 18. 18 fives will produce 10 tens which will result in the trailing zeros. so there will be 18 trailing zeros.