So many zeroes!

If a number has one trailing zero, then it must be a multiple of $10$ or $5\times2$ . So, for a number to have $n$ trailing zeroes it should have exactly $n$ pairs of $5$ & $2$ in it's factor.

For example, $10!=1\times2\times3\times4\times5\times6\times7\times8\times 9\times 10$ $=1\times2\times3\times4\times5\times6\times7\times8\times 9\times2\times 5$ $=3628800$ Which has $2$ pairs of $5~\& ~2$ hence, 2 trailing zeroes.

Now, we have to find how many pairs are there in $1000!$ $1000!=1\times2\times3\times4.....\times 1000$

Let's find the number of fives first. We can find them in every multiple of five. [e. g. $5, 10, 15, ...995, 1000$ ] . So there are $\boldsymbol {\frac {1000}{5}=200}$ multiples of five in $1000!$

But some of them has more than one five in their factor. $\boldsymbol {[}e.g. 25=5\times 5, 125= 5\times5 \times5, 625=5\times5\times5 \times5$ $\boldsymbol {]}$

There are $\boldsymbol {\frac{1000}{25}=40}$ multiples of 25, $\boldsymbol {\frac{1000}{125} =8}$ multiples of $125$ & $1$ multiple of $625$ in $1000!$

Hence, the total number of $5$ in the prime factorization of $1000!$ is, $\boldsymbol {200+40+8+1=\color{#69047E} {\boxed{249}}}$

There are $500$ even numbers between $1$ to $1000$ . So, it's very easy to understand that there will be far more number of $2$ than $5$ in the prime factorization of $1000!$

So, the number of $5$ & $2$ pairs will be $\color{#69047E} {\boxed{249}}$ .

We can do this directly by a formula,

$=[\frac{n}{k}]+[\frac{n}{k^{2}}]+[\frac{n}{k^{3}}]+[\frac{n}{k^{4}}]+.......$

We will keep doing this till $n>k^{r}$ .

Here,

$n$ = of $g!$

$k$ = prime integer

$[...]$ = highest integral value

From this formula we can calculate the power of a prime number $[k]$ which can divide $[g!]$ .

Hence the one having less power will be the limiting number .

Therefore ,to calculate the number of zeroes ,we'll take $k=5$ and $n=1000$ .

$=[\frac{1000}{5}]+[\frac{1000}{5^{2}}]+[\frac{1000}{5^{3}}]+[\frac{1000}{5^{4}}]$

$=[200]+[40]+[8]+[1.6]$

$=200+40+8+1=\color{#3D99F6}{\boxed{249}}$

Let's make it easier.

1000 ÷ 5 = 200

200 ÷ 5 = 40

40 ÷ 5 = 8

8 ÷ 5 = 1.6

200 + 40 + 8 + 1 = 249

For example, 12345!, do it this way.

12345 ÷ 5 = 2469

2469 ÷ 5 = 493.8

493 ÷ 5 = 98.6

98 ÷ 5 = 19.6

19 ÷ 5 = 3.8

2469 + 493 + 98 + 19 + 3 = 3082