2007 AMC 12A Problems/Problem 3

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Problem

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

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

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