# 2018 AMC 8 Problems/Problem 14

## Problem

Let $N$ be the greatest five-digit number whose digits have a product of $120$. What is the sum of the digits of $N$? $\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20$

## Solution

If we start off with the first digit, we know that it can'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 $8+5+3+1+1=18\implies \boxed{\textbf{(D) }18}$.

## Solution(factorial)

120 is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multple numbers.

(5)(4)(3)(2)(1) = 120 make the greatest integer

(5)(4 times 2)(3)(2 divided by 2)(1)

= (5)(8)(3)(1)(1) =120


8 is the largest value and will go in the front so we can express it as 5,8,3,1,1

We don't even need the number just add

5+8+3+1+1 = 18

## Video Solution (CREATIVE ANALYSIS!!!)

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## Video Solutions

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