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

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(→‎Solution(factorial))
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<math>(5)(4)(3)(2)(1) = 120</math>
 
<math>(5)(4)(3)(2)(1) = 120</math>
−
make the greatest integer  
+
Making the greatest integer,
  
−
<math>(5)(4 * 2)(3)(2 / 2)(1)</math>
+
<math>(5)(4 \cdot 2)(3)(2 / 2)(1)</math>
 
<math> = (5)(8)(3)(1)(1) =120</math>
 
<math> = (5)(8)(3)(1)(1) =120</math>
  
Line 24: Line 24:
 
We don't even need the number just add
 
We don't even need the number just add
  
−
8+5+3+1+1 = 18
+
<math>8+5+3+1+1=18</math>
  
 
==Video Solution (CREATIVE ANALYSIS!!!)==
 
==Video Solution (CREATIVE ANALYSIS!!!)==

Revision as of 20:06, 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,

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

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

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