Standing in the hall of fame

Let $n!$ be one of the number having $23$ trailing zeroes.

So an approximation for trailing zeroes in $n!$ ,

$\begin{aligned} 23 & = \left \lfloor \frac{n}{5^1} \right \rfloor + \left \lfloor \frac{n}{5^2} \right \rfloor + \left \lfloor \frac{n}{5^3} \right \rfloor + \ldots \\ 23 & = \frac{\frac{n}{5}}{1-\frac{1}{5}} \\ 23 & = \frac{n}{4} \\ 92 & = n \end{aligned}$

Now, we can see that $92!$ has $21$ trailing zeroes.

So $95!$ would have $22$ trailing zeroes.

Then $100!$ would have $24$ trailing zeroes.

For any number $n≥100$ , $n!$ would have trailing zeroes $≥24$ .

Therefore, the number $n!$ having $23$ trailing zeroes should be $95<n<100$ .

But no such number really exist.

$\therefore n!$ can never end in $23$ trailing zeroes.

nice problem

95! Has 22 trailing zero while 100! have 24 trivial zero:

So if there is any number n then 95<n<100:

If you want you can try each case or you can simply assert that there must not be any number n because there is no multiple of 5 in between 95 and 100

We know that 5! has 1 trailing zero because it is the first integer factorial with a 5 x 2 within it. As we increase out number 5 to 10! then 15! then 20! we notice it increases by one. (10! = 2, 15! = 3, 20! = 4) but when we hit 25! it is equal to 6 because 25 is a integer power of 5 so it increases by another 2. Using this logic one can conclude that 100! must have 24 trailing zeros because 100 = 25 *4 so we multiply 25! trailing zeros by 4. Notice that this only works because 100 is less than 125 which is important because 125 is 5 cubed means it would have 31 trailing zeros instead of 30. Now with the knowledge that 100! is 24 we would like to know how many trailing zeros 95! has. We can conclude that since 100 is multiple of 25 we must subtract two from 100! trailing zeros to get the trailing zero's of 95!. This brings us back to the whole idea of how 25! has 6 trailing zeros and 20! has 4 because 25 is an integer power of 5 multiplied by another integer. Using this we now have that 100! has 24 trailing zeros and 95! has 22, this means that no integer factorial has 23 trailing zeros since 96, 97, 98, or 99 factorial wouldn't add another zero because they are not multiples of 5.