14. If $N$ is the largest five-digit number whose digits multiply to 120, then the sum of the digits of $N$ is ( ).
(A) 15
(B) 16
(C) 17
(D) 18
(E) 20
14. D.

Since 9 is not a factor of 120, while 8 is a factor of 120, we know that the ten-thousands digit of $N$ is 8.
We only need to determine the last four digits.
Since $\frac{120}{8}=15$, the thousands digit is 5, the hundreds digit is 3, and the tens and units digits are both 1.
Thus, $N=85311$.
Therefore, the sum of the digits is $8+5+3+1+1=18$.