If we start off with the first digit, we know that it can\textquotesingle t be $9$ since $9$ is not a factor of $120$ . We go down to the digit $8$ , which does work since it is a factor of $120$ . Now, we have to know what digits will take up the remaining four spots. To find this result, just divide $\frac{120}{8}=15$. The next place can be $5$ , as it is the largest factor, aside from $15$ . Consequently, our next three values will be $3,1$ and $1$ if we use the same logic. Therefore, our five-digit number is $85311$ , so the sum is $8+5+3+1+1=18 \Longrightarrow($ D) 18 . 
Let $N$ be the greatest five-digit number whose digits have a product of $120$ . What is the sum of the digits of $N$ ? (2018 AMC 8 Problem, Question \#14) 

a) $$15$$ 
b) $$16$$ 
c) $$17$$ 
d) $$18$$ 
e) $$20$$ 