# 1*2*3*4*5*6.....upto 1000 Find the number of zeroes at the end?

Apr 24, 2016

$249$

#### Explanation:

This product is commonly known as the factorial of $1000$, written 1000!

The number of zeros is determined by how many times $10 = 2 \times 5$ occurs in the prime factorisation of 1000!.

There are plenty of factors of $2$ in it, so the number of zeros is limited by the number of factors of $5$ in it.

These numbers have at least one factor $5$:

$5 , 10 , 15 , 20 , 25 , \ldots , 1000$ which is $\frac{1000}{5} = 200$ numbers.

These numbers have at least two factors $5$:

$25 , 50 , 75 , 100 , \ldots , 1000$ which is $\frac{1000}{25} = 40$ numbers.

These numbers have at least three factors $5$:

$125 , 250 , 375 , 500 , \ldots , 1000$ which is $\frac{1000}{125} = 8$ numbers

This number has four factors $5$:

$625$ which is $1$ number.

So the total number of factors $5$ in 1000! is:

$200 + 40 + 8 + 1 = 249$

Hence there are $249$ zeros at the end of 1000!