Factorial divisors

By counter-example:

Call the number of trailing zeros in 99! "z".

100! must have z+2 trailing zeros.

Since multiplying will never decrease the amount of trailing zeros, we will never have z+1 trailing zeros, and therefore the statement is false.

The number of trailing 0's is determined by the number of factors of 5 the final factorial has. This is found by the infinite series trailing 0's =floor(n/5)+floor(n/5^2)+floor(n/5^3)+... which does not cover all integers

I.e. there is no solution for n=5 (24! has 4 0's 25! has 6)

When a multiple of 25 is added, the number of zeroes becomes n+2 ,where n is previous number of zeroes. Thus 1 number is skipped. So its false

The fact that for every integer $n$ there exists another integer $m$ where $m!$ will have $n$ number of trailing zeroes is not true because when we try to find out the trailing number of zeroes we look for number of 5 in the expression. Because $5×2=10$ and thus a $0$ is produced.

For example, take $23!$ , it will have $[\dfrac{23}{5}]=4$ , where $[]$ denotes greatest integer, so here $n=4$

But let's take the case of $25!$ , it will have $[\dfrac{25}{5}]=5$ and $[\dfrac{25}{25}]=1$ , here I divided 25 also as $25=5×5$ and has two 5, while dividing 25 by 5, I only count one of the 5. So, total number of 5 in $25!$ are 6. And so $n=5$ never exists. And thus our the statement above is not true.

Integers can be negative?