Difference between revisions of "2016 AMC 8 Problems/Problem 9"

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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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{{AMC8 box|year=2016|num-b=8|num-a=10}}
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{{MAA Notice}}

Revision as of 09:46, 23 November 2016

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

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

2016 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AJHSME/AMC 8 Problems and Solutions

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