ending zeros in 100! I'm working through Hammack's Book of Proof.  Section 3.2 has an weird question, and unfortunately it's even-numbered, so there is no answer key.
"There are two 0's at the ned of 10! = 3,628,800.  Using only pencil and paper, determine how many 0's are at the end of the number 100!."
I used the special case De Polignac's formula for factorials to get tz(100!) = 24.  I believe this is the right answer. 
But the thing is,  this question (and this solution) is totally unlike everything else I've seen in the book.   This chapter is about 'Counting' (factorials, unique lists, etc).  I'm wondering if there is some other way to get the answer using tools I'd seen in text so far, and not this kind of weird excursion into number theory.  Like some intuitive way to think through the question.
 A: The task is a counting one!
The question asks how many times you can divide $100!$ by $10$, and if you write $100!=2^m\cdot 5^n \cdot x$ where $x$ is not divisible by $2$ or $5$, you can see the question really asks what $\min\{m,n\}$ is.
So, the question is asking how many twos and how many fives you have in the product $1\cdot 2\cdots 100$.

The solution:
So let's find $n$.
There are $20$ multiples of $5$ in $100!$, and of them $4$ are multiples of $25$, so the answer is $20 + 4$ - each multiple of $5$ gives you one factor of $5$ in the product, and $25, 50, 75, 100$ give you one aditional factor of $5$.
As for $m$, it's enough to see that $m>n$, which is clear since there are at least $50$ divisors of $2$ in the product $100!$, so $m>50$, meaning that
$$\min\{m,n\} = 24.$$
(in fact, $m = 50 + 25 + 12 + 6 + 3 + 1 = 98$)
A: I often use the old-fashioned $[x]$ for the largest integer not exceeding $x$.
To show, combinatorially, for prime $p,$ that the largest $m$ for which $p^m|n!$ is $M(p,n)=\sum_{j=1}^{\infty}[n p^{-j}].$ 
For positive integer $x\leq n$  let $v(x)$  be the largest $m$ such that $p^m|x. $  As $p$ is prime, we have $M(p,n)=\sum_{x=1}^n v(x).$ 
Write, on a separate line for each $x\leq n ,$ a sequence of $1$'s with the number of $1$'s being equal to $v(x).$ The line is left blank if $ v(x)=0.$ Now if you read down the columns, the number of $1$'s in the  1st column is $[np^{-1}],$  the number of $1$'s in the second column is $[np^{-2}],$  et cetera. And the total number of $1$'s is $M(p,n).$ 
A: I believe the best way is what you did! Indeed the simple formula applied to that gives you
$$n = \text{int}\left(\frac{100}{5^1}\right) + \text{int}\left(\frac{100}{5^2}\right) = 20 + 4 = 24$$
A: The exponent of $2$ is equal to
 $$\sum_{n\ge1}\left[\frac{100}{2^n}\right]=\left[\frac{100}{2}\right]+\left[\frac{100}{2^2}\right]=75$$ and the exponent of $5$ is
$$\sum_{n\ge1}\left[\frac{100}{5^n}\right]=\left[\frac{100}{5}\right]+\left[\frac{100}{5^2}\right]=24$$ Hence the exponent of $10$ is equal to $24$. One has $$100!=10^{24}M\text{ where } (M,10)=1$$
NOTE.- It is clear that the exponent of $2$ being surely greater than the exponent of $5$ it would be enough to calculate the exponent of $5$. However I wanted to expose the general procedure without the quite particular case of $10$.
A: To get a number ending with 0 in decimal system, we need a 5 and an even number to be multiplied.
The first 0 appears in 5! because we have 5 times an even number (I.e. 2 or 4). It is when we get to 10! That we encounter with the next 5. And in 10! we also have an even number to times it with 5 and make it a number ending with 0. Thus 10! will give a number ending with two 0s, 15! three 0s, 20! four 0s. But 25! will give a number ending with six 0s and not five 0s. And that's because we have 5 getting multiplied twice in 25 and each time it can combine with an even number to get a 0.
Thus 50! will end with twelve 0s, 75! with eighteen 0s and 100! with twenty four 0s 
