To make the largest possible five$$-$$digit $$N$$, you should make the number in the biggest digit as large as possible. $$120=8\times5\times3\times1\times1$$. So, the sum of digits is $$8+5+3+1+1=18$$. 
Let $$N$$ be the greatest five$$-$$digit number whose digits have a product of $$120$$. What is the sum of the digits of $$N$$? 
Options:
A. $$15$$ 
B. $$16$$ 
C. $$17$$ 
D. $$18$$ 
E. $$20$$ 