Find the number of trailing zeros in 50! My attempt:  
50! = 50 * 49 *48 ....  
Even * even = even number  
Even * odd = even number  
odd * odd = odd number  
25 evens and 25 odds  
Atleast 26 of the numbers will lead to an even multiple (24 evens + 1 even * 1 odd) so at most 26 trailing zeros. 
50 is divisible by 5: 10 times. Atleast 10 trailing zeros. 
What is the answer? 
 A: Every factor of 5 will mate with a factor of 2 and produce a trailing 0.  How do you count the factors of 5 in the product?
A: You need to find the highest power of $10$ that divides $50!$, which is same as the highest power of $5$ that divides $50!$, since $10 = 5 \times 2$, and there are fewer multiples of $5$ in the first $50$ numbers. It's easy to count the 10 multiples of $5$, namely $5, 10, 15, \cdots, 50$, but also add two more powers of $5$ as $25$ and $50$ have $5^2$ as the highest power of $5$ as factor. 
The answer is $12$.
A: There are $\lfloor \frac{50}{5}\rfloor$ numbers  between $1$ and $50$ that are divisible by $5$. Similiarily $\lfloor \frac{50}{5^2}\rfloor$ and $\lfloor \frac{50}{5^3}\rfloor$ numbers divisible by $5^2$ and $5^3$ respectively.
Thus, the highest power of $5$ dividing $50!$ is
$$\lfloor \frac{50}{5}\rfloor + \lfloor \frac{50}{5^2}\rfloor + \lfloor \frac{50}{5^3}\rfloor = 10 + 2 + 0 = 12$$
With similiar arguments, one can show the greatest power of $2$ dividing $50!$ is greater then $12$. Thus, the greatest power of $10 = 2 \cdot 5$ dividing $50!$ is $12$.
