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2018 AMC 8 Problems/Problem 14: Difference between revisions

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== Solution(factorial) ==
== Solution(factorial) ==


120 is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multple numbers.
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
<math>(5)(4)(3)(2)(1) = 120</math>
make the greatest integer  
make the greatest integer  


(5)(4 times 2)(3)(2 divided by 2)(1)
<math>(5)(4 * 2)(3)(2 / 2)(1)</math>
= (5)(8)(3)(1)(1) =120
<math> = (5)(8)(3)(1)(1) =120</math>


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

Revision as of 20:05, 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.

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

$(5)(4 * 2)(3)(2 / 2)(1)$ $= (5)(8)(3)(1)(1) =120$

8 is the largest value and will go in the front so we can express it as 8,5,3,1,1

We don't even need the number just add

8+5+3+1+1 = 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 (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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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