Difference between revisions of "2016 AMC 8 Problems/Problem 9"
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+ | == Problem == | ||
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What is the sum of the distinct prime integer divisors of <math>2016</math>? | What is the sum of the distinct prime integer divisors of <math>2016</math>? | ||
<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> | ||
− | ==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>. | 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>. | 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>. | ||
+ | |||
+ | ==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 17:31, 21 April 2021
Problem
What is the sum of the distinct prime integer divisors of ?
Solutions
Solution 1
The prime factorization is . Since the problem is only asking us for the distinct prime factors, we have . Their desired sum is then .
Solution 2
We notice that , since , and . We can divide by to get . This is divisible by , as . Dividing by , we have . This is clearly divisible by , leaving . We have . We know that and are both multiples of , is , and is prime. This means that the distinct prime factors are and . Their sum is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.