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

(Solution(factorial))
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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 20:09, 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

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(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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