Difference between revisions of "2007 AMC 12A Problems/Problem 3"

Revision as of 14:59, 4 January 2012

The larger of two consecutive odd integers is three times the smaller. What is their sum?

$\displaystyle \mathrm{(A)}\ 4\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 16\qquad \mathrm{(E)}\ 20$

Solution 1 Let $n$ be the smaller term. Then $n+2=3n \Longrightarrow 2n = 2 \Longrightarrow n=1$

Solution 2

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AMC 12 Problems and Solutions

@@ Line 6: / Line 6: @@
 == Solution ==
 '''Solution 1'''
-Let <math>n</math> be the middle term. Then <math>n+1=3(n-1) \Longrightarrow 2n = 4 \Longrightarrow n=2</math>
+Let <math>n</math> be the smaller term. Then <math>n+2=3n \Longrightarrow 2n = 2 \Longrightarrow n=1</math>
-*Thus, the answer is <math>(2-1)+(2+1)=4 \mathrm{(A)}</math>
+*Thus, the answer is <math>1+(1+2)=4 \mathrm{(A)}</math>
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