1. ## N!

For a positive integer, $N$, we define N! (read N factorial) to
be the product of all the integers from 1 to N. Thus:
$N! = 1 . 2 . 3 .... . N$
It is clear that N! will end in a zero if N is larger or equal to
5. For big values of N, N! will end in many, many zeros.
How many zeros will $75!$ end in?

2. This is known as the trailing zeros problem.
This function will answer the question for any $N<5^{10}$,
$Z(N) = \sum\limits_{k = 1}^{10} {\left\lfloor {\frac{N}
{{5^k }}} \right\rfloor }$

3. Hello, perash!

For a positive integer, $N$, we define N! (read N factorial)
to be the product of all the integers from 1 to N.
Thus: . $N! = 1\cdot2\cdot3 \cdots N$

It is clear that $N!$ will end in a zero if $N \geq 5$.
For large values of $N,\;N!$ will end in many, many zeros.

How many zeros will $75!$ end in?

Plato is absolutely correct . . . Here's the basis of that formula . . .

Every factor of 5 (paired with an even factor) will produce a final zero.

The question becomes: how many 5's are in the prime factorization of $75!$ ?

Every fifth number has a factor of 5.
. . Hence, there are: . $\frac{75}{5} \:=\:15$ factors of 5.

But every twenty-fifth number has a factor of $5^2 = 25.$
. . Each of them contributes one more 5 to the total.
And there are: . $\frac{75}{25} \:=\:3$ of them.

Hence, $75!$ contains: $15 + 3 \:=\: 18$ factors of 5

Therefore, $75!$ ends in eighteen zeros.