123*456.....upto 1000 Find the number of zeroes at the end?

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Answer 1 · 2016-04-24T23:08:05

$249$ $249$

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

The number of zeros is determined by how many times $10 = 2 \times 5$ $10=2xx5$ occurs in the prime factorisation of $1000!$ $1000!$ .

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

These numbers have at least one factor $5$ $5$ :

$5, 10, 15, 20, 25,..., 1000$ which is $1000/5 = 200$ numbers.

These numbers have at least two factors $5$ :

$25, 50, 75, 100,..., 1000$ which is $1000/25 = 40$ numbers.

These numbers have at least three factors $5$ :

$125, 250, 375, 500,..., 1000$ which is $1000/125 = 8$ 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 $1000!$

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