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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>?
<math>\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20</math>


== Solution ==  
== Solution ==  
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> -mathmaster010
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{\textbf{D }18}</math> -mathmaster010


==See Also==
{{AMC8 box|year=2018|num-b=13|num-a=15}}


 
{{MAA Notice}}
<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}}

Revision as of 18:47, 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 $N$?

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

Solution

If we start off with the first digit, we know that it can't 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 $8+5+3+1+1=\boxed{\textbf{D }18}$ -mathmaster010

See Also

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

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