Difference between revisions of "2007 AMC 8 Problems/Problem 3"

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<math>\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 5 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 12</math>
 
<math>\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 5 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 12</math>
  
[[2007 AMC 8 Problems/Problem 3|Solution]]
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==Video Solution by OmegaLearn==
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https://youtu.be/7an5wU9Q5hk?t=272
  
 
== Solution ==
 
== Solution ==
  
We prime factor 250.  
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The prime factorization of <math>250</math> is <math>2 \cdot 5^3</math>. The smallest two are <math>2</math> and <math>5</math>. <math>2+5 = \boxed{\text{(C) }7}</math>.
  
We get <math>2 * 5 * 5 * 5</math>
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==See Also==
 
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{{AMC8 box|year=2007|num-b=2|num-a=4}}
The two smallest are <math>2</math> and <math>5</math>.
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{{MAA Notice}}
 
So, <math>2 + 5 = 7</math>
 
 
 
The answer is <math>\boxed{C}</math>
 

Latest revision as of 03:59, 29 December 2022

Problem

What is the sum of the two smallest prime factors of $250$?

$\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 5 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 12$

Video Solution by OmegaLearn

https://youtu.be/7an5wU9Q5hk?t=272

Solution

The prime factorization of $250$ is $2 \cdot 5^3$. The smallest two are $2$ and $5$. $2+5 = \boxed{\text{(C) }7}$.

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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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