Does $18^{247}$ divide (500!)? I wanted to find the highest power of 18 that divides 500!  
I did this :  
$18=6*3=(2*3)*3$  
For 2, highest power is 494  
For 3, highest power is 247,  
Therefore highest power of 6 is min{494, 247}=247  
And hence highest power of 18 is min{247, 247}=247  
But I think I have done something wrong and the answer should be 123, the highest power of 9.  
Now, I just want to know how can I check if $18^{247}$ divides 500! or not.
 A: The power of $3$ in the prime factorization of $500!$ is
$$
\lfloor \frac{500}{3} \rfloor + \lfloor \frac{500}{9} \rfloor + \lfloor \frac{500}{27} \rfloor + \lfloor \frac{500}{81} \rfloor + \lfloor \frac{500}{243} \rfloor = 166 + 55 + 18 + 6 + 2 = 247 < 494
$$
So it is not possible that $18^{247}$ divides $500!$.
A: You want twice as many '$3$'s, not '$2$'s. In other words, you started correctly but you need two '$3$'s to get one '$18$', so you only have at most $123$ of them, and it's the 'limiting factor' since you have more than enough '$2$'s.
A: The decomposition of $500!$ is, as you said, : $2^{494} \cdot 3^{247} \cdot 5^{?} ... $
$18 = 2 \cdot 3^2$ so you will be able to divides $500!$ $494$ times by $2$ and be able to divide $500!$ 123 times by $3^2$. Taking the min of the two, you'll be able to divide $500!$ 123 times by $18$.
A: You are backwards between $2$ and $3$.  Because $18=2^1 \cdot 3^2, 18^{247}=(2^1\cdot3^2)^{247}=2^{247}3^{494}$ Now you can check whether these powers each divide $500!$ as shown in this question.  You should find that $2^{247}$ works, but $3^{494}$ does not.
