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Difference between revisions of "2016 AMC 8 Problems/Problem 9"
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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>.
 
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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==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}}

Revision as of 16:31, 21 April 2021

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 $\boxed{\textbf{(B) }12}$.

Solution 2

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


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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