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 $8+5+3+1+1=18\implies \boxed{\textbf{(D)}}$ .

Video Solution

https://youtu.be/IAKhC_A0kok

https://youtu.be/7an5wU9Q5hk?t=13

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2018 AMC 8 Problems/Problem 14

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Video Solution

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