Question

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

Answer 1

`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=2xx5` 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,..., 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!`