Art of Problem Solving
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

2016 AMC 8 Problems/Problem 9: Difference between revisions

← Older edit
Hashtagmath (talk | contribs)
editor
496 edits
No edit summary
Aoum (talk | contribs)
editor
4,951 edits
Categorized problem
 
(11 intermediate revisions by 7 users not shown)
Line 11: Line 11:
The prime factorization is <math>2016=2^5\times3^2\times7</math>.  Since the problem is only asking us for the distinct prime factors, we have <math>2,3,7</math>.  Their desired sum is then <math>\boxed{\textbf{(B) }12}</math>.
The prime factorization is <math>2016=2^5\times3^2\times7</math>.  Since the problem is only asking us for the distinct prime factors, we have <math>2,3,7</math>.  Their desired sum is then <math>\boxed{\textbf{(B) }12}</math>.


===Solution 2===
==Video Solution==


We notice that <math>9 \mid 2016</math>, since <math>2+0+1+6 = 9</math>, and <math>9 \mid 9</math>. We can divide <math>2016</math> by <math>9</math> to get <math>224</math>. This is divisible by <math>4</math>, as <math>4 \mid 24</math>. Dividing <math>224</math> by <math>4</math>, we have <math>56</math>. This is clearly divisible by <math>7</math>, leaving <math>8</math>. We have <math>2016 = 9\cdot 4\cdot 7\cdot 8</math>. We know that <math>4</math> and <math>8</math> are both multiples of <math>2</math>, <math>9</math> is <math>3^2</math>, and <math>7</math> is prime. This means that the distinct prime factors are <math>2,3,</math> and <math>7</math>. Their sum is <math>\boxed{\textbf{(B) }12}</math>.
https://youtu.be/I_jevKp3Kyg?si=qKDYbUmjFkioeIvE


~Elijahman~
==Video Solution (CREATIVE THINKING!!!)==
https://youtu.be/GNFnta9MF9E
~Education, the Study of Everything
==Video Solution==
https://youtu.be/1KN7OTG3k-0
~savannahsolver


==See Also==
==See Also==
{{AMC8 box|year=2016|num-b=8|num-a=10}}
{{AMC8 box|year=2016|num-b=8|num-a=10}}
{{MAA Notice}}
{{MAA Notice}}
[[Category:Introductory Number Theory Problems]]

Latest revision as of 17:04, 25 June 2025

Problem

What is the sum of the distinct prime integer divisors of $2016$?

$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$

Solutions

Solution 1

The prime factorization is $2016=2^5\times3^2\times7$. Since the problem is only asking us for the distinct prime factors, we have $2,3,7$. Their desired sum is then $\boxed{\textbf{(B) }12}$.

Video Solution

https://youtu.be/I_jevKp3Kyg?si=qKDYbUmjFkioeIvE

~Elijahman~

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/GNFnta9MF9E

~Education, the Study of Everything

Video Solution

https://youtu.be/1KN7OTG3k-0

~savannahsolver

See Also

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_9&oldid=252322"