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

 
 
(4 intermediate revisions by 3 users not shown)
Line 2: Line 2:
 
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?
  
== Solution ==
+
<math>\mathrm{(A)}\ 4\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 16\qquad \mathrm{(E)}\ 20</math>
Solution a.
 
* n+1=3n-3
 
* 4=2n
 
* n=2
 
*(2-1)+(2+1)=4
 
  
 +
==Solution 1==
 +
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 b.
+
 
* By inspection, 1 and 3 work. 1+3=4.
+
==Solution 2==
 +
* By trial and error, 1 and 3 work. 1+3=4.
  
 
== See also ==
 
== See also ==
* [[2007 AMC 12A Problems/Problem 2 | Previous problem]]
+
{{AMC12 box|year=2007|ab=A|num-b=2|num-a=4}}
* [[2007 AMC 12A Problems/Problem 4 | Next problem]]
+
 
* [[2007 AMC 12A Problems]]
+
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 01:09, 16 February 2021

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 1

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

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


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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png