2007 AMC 8 Problems/Problem 3: Difference between revisions

Latest revision as of 16:30, 28 October 2024

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

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

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

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AJHSME/AMC 8 Problems and Solutions

@@ Line 4: / Line 4: @@
 <math>\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 5 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 12</math>
+==Video Solution by OmegaLearn==
+https://youtu.be/7an5wU9Q5hk?t=272
 == Solution ==
-We prime factor 250.
+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>
+==Video Solution by SpreadTheMathLove==
+https://www.youtube.com/watch?v=omFpSGMWhFc
-The two smallest are <math>2</math> and <math>5</math>.
+==Video Solution by WhyMath==
+https://youtu.be/_8lDRd1FEd4
-So, <math>2 + 5 = 7</math>
-The answer is <math>\boxed{C}</math>
 ==See Also==
-{{AMC8 box|year=2007|before=First<br />Question|num-a=2}}
+{{AMC8 box|year=2007|num-b=2|num-a=4}}
+{{MAA Notice}}