Let $$N$$ be the greatest five$$-$$digit number whose digits have a product of $$120$$. What is the sum of the digits of $$N$$? 
Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization
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 $$8+5+3+1+1=18$$. 