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

Line 2: Line 2:
 
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>?
  
 
+
== 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{18}, \textbf{(D)}</math> -mathmaster010
  

Revision as of 17:51, 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 $N$?

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 $8+5+3+1+1=\boxed{18}, \textbf{(D)}$ -mathmaster010


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

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
Invalid username
Login to AoPS