Question

Explain why there is no factorial number that ends with exactly five 0's?

Answer 1

See explanation...

Explanation:

Factorial is defined recursively by the following rules:

• `0! = 1`

• `(n+1)! = (n+1) n!`

As `n` increases, the factorial of `n` only acquires more factors.

The factors that result in trailing `0`'s are `2`'s and `5`'s. There are plenty of `2`'s to go around, so the number of `0`'s is really determined by how many `5`'s you have.

So as `n` increases, the number of zeros increases each time you encounter a multiple of `5`.

So we find:

• `5! = 120` is the first factorial with a trailing `0`

• `10! = 3628800` is the first factorial with `2` trailing `0`'s

• `15!` adds another trailing zero to make `3`

• `20!` adds another trailing zero to make `4`

• `25!` adds two trailing zeros, since `25 = 5^2`, to make `6`

Thereafter, all factorials will have `6` or more trailing zeros.

Answer 2

See below.

Explanation:

Because for each `10` we need a `5` and the `5` appearing in a factorial into the factors

`5,10,15,20,25,cdots` and the associated number of `5` is

`1,1,1,1,2, cdots` and the corresponding trailing zeros,

`0,00,000,0000,000000,cdots` so no `00000` trailing zeros.