2007 AMC 12A Problems/Problem 3: Difference between revisions

Revision as of 10:44, 9 September 2007

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

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

Solution 1 Let $n$ be the middle term. Then $n+1=3(n-1) \Longrightarrow 2n = 4 \Longrightarrow n=2$

Solution 2

2007 AMC 12A (Problems • Answer Key • Resources)
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@@ Line 2: / Line 2: @@
 The larger of two consecutive odd integers is three times the smaller. What is their sum?
+<math>\mathrm{(A)}\ 4\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 16\qquad \mathrm{(E)}\ 20</math>
 == Solution ==
-Solution a.
+'''Solution 1'''
-* n+1=3n-3
+Let <math>n</math> be the middle term. Then <math>n+1=3(n-1) \Longrightarrow 2n = 4 \Longrightarrow n=2</math>
-* 4=2n
+*Thus, the answer is <math>(2-1)+(2+1)=4 \mathrm{(A)}</math>
-* n=2
-*(2-1)+(2+1)=4
+----
+'''Solution 2'''
+* By trial and error, 1 and 3 work. 1+3=4.
-Solution b.
+== See also ==
-* By inspection, 1 and 3 work. 1+3=4.
+{{AMC12 box|year=2007|ab=A|num-b=2|num-a=4}}
-== See also ==
+[[Category:Introductory Algebra Problems]]
-* [[2007 AMC 12A Problems/Problem 2 | Previous problem]]
-* [[2007 AMC 12A Problems/Problem 4 | Next problem]]
-* [[2007 AMC 12A Problems]]