Difference between revisions of "2010 AMC 8 Problems/Problem 14"

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==Solution==
 
==Solution==
First, we must find the prime factorization of <math>2010</math>. <math>2010=2\cdot 3 \cdot 5 \cdot 67</math>. We add the factors up to get <math>\boxed{\textbf{(C)}\ 77}</math>
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First, we must find the prime factorization of 2010 2010=2*3*5*67 We add the factors up to get C:77
  
 
==Video Solution==
 
==Video Solution==

Revision as of 13:55, 15 August 2021

Problem

What is the sum of the prime factors of $2010$?

$\textbf{(A)}\ 67\qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 77\qquad\textbf{(D)}\ 201\qquad\textbf{(E)}\ 210$

Solution

First, we must find the prime factorization of 2010 2010=2*3*5*67 We add the factors up to get C:77

Video Solution

https://youtu.be/6xNkyDgIhEE?t=1236

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AJHSME/AMC 8 Problems and Solutions

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