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

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== Solution ==  
 
== Solution ==  
If we start off with the first digit, we know that it can't be <math>9</math> since <math>9</math> is not a factor of <math>120</math>. We go 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=18\implies \boxed{\textbf{(D)}}</math>.
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If we start off with the first digit, we know that it can't be <math>9</math> since <math>9</math> is not a factor of <math>120</math>. We go 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=18\implies \boxed{\textbf{(D) 18}}</math>.
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 14:02, 15 June 2021

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

Video Solution

https://youtu.be/IAKhC_A0kok

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

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
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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