2007 AMC 12A Problems/Problem 3

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
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All AMC 12 Problems and Solutions

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