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

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==Problem 14==
 
==Problem 14==
 
Let <math>N</math> be the greatest five-digit number whose digits have a product of <math>120</math>. What is the sum of the digits of <math>N</math>?
 
Let <math>N</math> be the greatest five-digit number whose digits have a product of <math>120</math>. What is the sum of the digits of <math>N</math>?
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If we start off with the first digit, we know that it can't by <math>9</math> since <math>9</math> is not a factor of <math>120</math>. We scale down to the digit <math>8</math>, which does work since it is a factor of <math>120</math>. Now, we have to know what digits will take up the remaining four spots. To find this result, just divide <math>\frac{120}{8}=15</math>. The next place can be <math>5</math>, as it is the largest factor, aside from <math>15</math>. Consequently, our next three values will be <math>3,1</math> and <math>1</math> if we use the same logic! Therefore, our five-digit number is <math>85311</math>, so the sum is <math>8+5+3+1+1=\boxed{18}, \textbf{(D)}</math>
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<math>\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20</math>
 
<math>\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20</math>
  
 
{{AMC8 box|year=2018|num-b=13|num-a=15}}
 
{{AMC8 box|year=2018|num-b=13|num-a=15}}

Revision as of 12:47, 21 November 2018

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 $N$?


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


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

2018 AMC 8 (Problems • Answer Key • Resources)
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Problem 13 		Followed by
Problem 15
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All AJHSME/AMC 8 Problems and Solutions
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