2016 AMC 8 Problems/Problem 9: Difference between revisions

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

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: Unable to save thumbnail to destination

@@ Line 5: / Line 5: @@
 <math>\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63</math>
-==Solution 1==
+==Solutions==
+===Solution 1===
 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==
+===Solution 2===
 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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