Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 2 (Factorial))
 
(27 intermediate revisions by 13 users not shown)
Line 1: Line 1:
−
==Problem 14==
+
==Problem==
 
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>?
  
 
<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 ==  
+
== 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 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{\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>.
 +
 
 +
== Solution 2 (Factorial) ==
 +
 
 +
<math>120</math> is <math>5!</math>, so we have <math>(5)(4)(3)(2)(1) = 120</math>. (Alternatively, you could identify the prime factors <math>(5)(3)(2)(2)(2) = 120</math>.) Now look for the largest digit you can create by combining these factors.
 +
 
 +
<math>8=4 \cdot 2</math>
 +
 
 +
Use this largest digit for the ten-thousands place: <math>8</math>_ , _ _ _
 +
 
 +
Next you use the <math>5</math> and the <math>3</math> for the next places: <math>85,3</math> _ _ (You can't use <math>3 \cdot 2=6</math> because the <math>2</math> was used to make <math>8</math>.)
 +
 
 +
Fill the remaining places with 1: <math>85,311</math>
 +
 
 +
<math> = (5)(8)(3)(1)(1) =120</math>
 +
 
 +
<math>8+5+3+1+1=\boxed{\textbf{(D) }18}</math>.
 +
 
 +
==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==
 
==See Also==

Latest revision as of 09:01, 10 April 2024

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 $(5)(4)(3)(2)(1) = 120$. (Alternatively, you could identify the prime factors $(5)(3)(2)(2)(2) = 120$.) Now look for the largest digit you can create by combining these factors.

$8=4 \cdot 2$

Use this largest digit for the ten-thousands place: $8$_ , _ _ _

Next you use the $5$ and the $3$ for the next places: $85,3$ _ _ (You can't use $3 \cdot 2=6$ because the $2$ was used to make $8$.)

Fill the remaining places with 1: $85,311$

$= (5)(8)(3)(1)(1) =120$

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

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_14&oldid=218057"