Problem: $1995!$ has a number of divisors that are powers of $10$, i.e. $10^n$. Find the greatest possible $n$, $n\in Z$
My solution: I reckon that the maximum $n$ equals to the number of trailing zeroes of $1995!$ I think there are two sources of zeroes: numbers from $1$ to $1995$ that are divisible by $10$ and numbers that are divisible by $5$ but not by $10$.
1) The number of numbers that are divisible by $5$ but not divisible by $10$ is $200$. Thus, they give at least $200$ zeros. However, some numbers are divisible by $25$, which means they contain two $5$'s in their prime factorization, which means that they give one more zero each. There are $40$ such numbers. Similarly, there are numbers that are divisible by $125$, they give one more zero each, there are $8$ such numbers. Finally, there are $2$ numbers that are divisible by $625$, they give one more zero each. In total, all these numbers give $250$ zeros.
2) There are $199$ numbers that are divisible by $10$. Among these numbers there are 19 numbers that are also divisible by $100$ and $1$ number that is divisible by $1000$. Thus, all these numbers give $219$.
Totally, we have $250+219 = 469$ trailing zeros, which gives the answer.
However, the answer key says I'm wrong and the answer is $496$.
I'd like to know where my reasoning is wrong and how I can reach the correct answer.
I appreciate any help.
The answer given is correct. To find the number of zeros, we have to coubt the number of $5$ that are contanied in $1995!$. This is given by: $$\sum_{i=1}^{4}\left\lfloor \frac{1995}{5^i}\right\rfloor$$ Note that $4$ is the integer part of $\log_5(1995)$. This sum gives as result $399+79+15+3=496$, so there are $496$ trailing zeros in $1995!$. You can find some references here: https://www.purplemath.com/modules/factzero.htm and https://www.handakafunda.com/number-of-trailing-zeros/.