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

m (Reverted edits by Raina0708 (talk) to last revision by Pi is 3.14)
(Tag: Rollback)
(Video Solution)
Line 5: Line 5:
 
<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>
  
==Video Solution==
+
==Video Solution by OmegaLearn==
 
https://youtu.be/7an5wU9Q5hk?t=272
 
https://youtu.be/7an5wU9Q5hk?t=272
  

Revision as of 02:59, 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png