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

(Solution(factorial))
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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>
  
== Solution ==  
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== Solution 1==  
 
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>.
 
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>.
  
 
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== Solution 2 (Factorial) ==
 
 
== Solution(factorial) ==
 
  
 
120 is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multiple numbers.
 
120 is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multiple numbers.
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8 is the largest value and will go in the front.  
 
8 is the largest value and will go in the front.  
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We can express the number as <math>85311</math>.
 
We can express the number as <math>85311</math>.
<math>8+5+3+1+1=18</math>.
+
 
 +
<math>8+5+3+1+1=\boxed{\textbf{(D) }18}</math>.
  
 
==Video Solution (CREATIVE ANALYSIS!!!)==
 
==Video Solution (CREATIVE ANALYSIS!!!)==

Revision as of 21:10, 24 November 2023

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 1

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

Solution 2 (Factorial)

120 is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multiple numbers.

$(5)(4)(3)(2)(1) = 120$ Making the greatest integer,

$(5)(4 \cdot 2)(3)\left(\frac{2}{2}\right)(1)$ $= (5)(8)(3)(1)(1) =120$

8 is the largest value and will go in the front.

We can express the number as $85311$.

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

Video Solution (CREATIVE ANALYSIS!!!)

https://youtu.be/zxEO6amczPU

~Education, the Study of Everything

Video Solutions

https://youtu.be/IAKhC_A0kok

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

https://youtu.be/Q5YrDW62VDU

~savannahsolver

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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