# Difference between revisions of "2018 AMC 8 Problems/Problem 14"

## Problem 14

Let $N$ be the greatest five-digit number whose digits have a product of $120$. What is the sum of the digits of $N$?

## Solution

If we start off with the first digit, we know that it can't by $9$ since $9$ is not a factor of $120$. We scale 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 $8+5+3+1+1=\boxed{18}, \textbf{(D)}$ -mathmaster010

$\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20$

 2018 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions