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

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(Solution 1)
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===Solution 1===
 
===Solution 1===
  
The prime factorization is <math>735=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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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>.
  
 
==Video Solution (CREATIVE THINKING!!!)==
 
==Video Solution (CREATIVE THINKING!!!)==

Revision as of 20:23, 4 June 2024

Problem

What is the sum of the distinct prime integer divisors of 735 ?

$\textbf{(A) }9\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$

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

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 (ProblemsAnswer KeyResources)
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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