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

Latest revision as of 20:55, 22 January 2023

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

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

First, we must find the prime factorization of $2010$ . $2010=2\cdot 3 \cdot 5 \cdot 67$ . We add the factors up to get $\boxed{\textbf{(C)}\ 77}$

2010 AMC 8 (Problems • Answer Key • Resources)
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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-What is the sum of the prime factors of 2010?
+==Problem==
-Well to answer this question let's take the prime factorization of 2010. Which is 2,3,5, and 67. So all you do is add 2+3+5+67=77
+What is the sum of the prime factors of <math>2010</math>?
+<math> \textbf{(A)}\ 67\qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 77\qquad\textbf{(D)}\ 201\qquad\textbf{(E)}\ 210 </math>
+==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>
+==Video Solution by OmegaLearn==
+https://youtu.be/6xNkyDgIhEE?t=1236
+==Video by MathTalks==
+https://www.youtube.com/watch?v=6hRHZxSieKc
+==See Also==
+{{AMC8 box|year=2010|num-b=13|num-a=15}}
+{{MAA Notice}}