A number theory problem by Paola Ramírez

Note that $\lfloor{n}\rfloor=n$ if $n\in\mathbb{Z}$

$\because 10=2\cdot 5$ , we need a $2$ and a $5$ to give a trailing zero. Also, since the multiplication of the first 100 numbers contains more $2$ 's than $5$ 's, we shall just compute the number of $5$ 's in the multiplication.

There are $20$ multiples of $5$ and $4$ multiples of $25$ in the multiplication $\Rightarrow 20+4= \boxed{24}$ trailing zeroes.

Trailing zeros in $100!$ are $\lfloor \frac{100}{5} \rfloor+\lfloor \frac{100}{5^2} \rfloor=20+4=\boxed{\large{24}}$

100/5 + 100/25 = 20 + 4 = 24

$number~of~trailing~zeros=\dfrac{100}{5}+\dfrac{100}{25}=20+4=$ $\boxed{24}$

We just have to find the no. of 5's as the no. 10=2*5. And as explained by @Tay Yong Qiang that no. of 2's will be greater so we just need to find no. of 5's . To find that we can use floor function. Like [100/5] + [100/5^2] + [100/5^3] but the last part is zero so we don't have to continue this but if it wasnt we had To. But this we get no. of 5's which will be equal to no. of trailing zeros.

[.] : Floor function