Difference between revisions of "2007 AMC 10A Problems/Problem 4"

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== Solution ==
 
== Solution ==
Let the two consecutive odd integers be <math>a</math>, <math>a+2</math>. Then <math>a+2 = 3a</math>, so <math>a = 1</math> and their sum is <math>4\ \mathrm{(B)}</math>.
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Let the two consecutive odd integers be <math>a</math>, <math>a+2</math>. Then <math>a+2 = 3a</math>, so <math>a = 1</math> and their sum is <math>4\ \mathrm{(A)}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 14:26, 3 February 2008

Problem

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

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

Solution

Let the two consecutive odd integers be $a$, $a+2$. Then $a+2 = 3a$, so $a = 1$ and their sum is $4\ \mathrm{(A)}$.

See also

2007 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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All AMC 10 Problems and Solutions
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