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 <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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==Video Solution by OmegaLearn==
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https://youtu.be/6xNkyDgIhEE?t=1236
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==Video by MathTalks==
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https://www.youtube.com/watch?v=6hRHZxSieKc
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==See Also==
 
==See Also==
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{{AMC8 box|year=2010|num-b=13|num-a=15}}
 
{{AMC8 box|year=2010|num-b=13|num-a=15}}
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{{MAA Notice}}

Latest revision as of 19:55, 22 January 2023

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 $2010$. $2010=2\cdot 3 \cdot 5 \cdot 67$. We add the factors up to get $\boxed{\textbf{(C)}\ 77}$

Video Solution by OmegaLearn

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

Video by MathTalks

https://www.youtube.com/watch?v=6hRHZxSieKc




See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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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