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

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

Revision as of 10:44, 9 September 2007

Problem

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

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

  • Thus, the answer is $(2-1)+(2+1)=4 \mathrm{(A)}$

Solution 2

  • By trial and error, 1 and 3 work. 1+3=4.

See also

2007 AMC 12A (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 AMC 12 Problems and Solutions