<s>[INST] 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? 
 [/INST] 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 $12$.
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