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

Latest revision as of 02:09, 16 February 2021

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

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

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

2007 AMC 12A (Problems • Answer Key • Resources)
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 AMC 12 Problems and Solutions

@@ Line 3: / Line 3: @@
 <math>\mathrm{(A)}\ 4\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 16\qquad \mathrm{(E)}\ 20</math>
-== 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>
-*Thus, the answer is <math>(2-1)+(2+1)=4 \mathrm{(A)}</math>
-----
+==Solution 1==
-'''Solution 2'''
+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>1+(1+2)=4 \mathrm{(A)}</math>
+==Solution 2==
 * By trial and error, 1 and 3 work. 1+3=4.
@@ Line 16: / Line 16: @@
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}