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Difference between revisions of "2016 AMC 8 Problems/Problem 9"
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<math>\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63</math>
 
<math>\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63</math>
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[[2016 AMC 8 Problems/Problem 9|Solution
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==Solutions==
 
==Solutions==
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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>.
  
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===Solution 2===
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==Video Solution==
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https://youtu.be/I_jevKp3Kyg?si=qKDYbUmjFkioeIvE
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A solution so simple that a 12-year-old made it!
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 +
~Elijahman~
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 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/GNFnta9MF9E
 +
 
 +
~Education, the Study of Everything
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 +
==Video Solution==
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https://youtu.be/1KN7OTG3k-0
  
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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>.
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~savannahsolver
  
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==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}}

Latest revision as of 10:51, 20 June 2024

Problem

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

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

Solution

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

A solution so simple that a 12-year-old made it!

~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

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