2002 AMC 10A Problems/Problem 2

Revision as of 18:05, 26 December 2008 by Xpmath (talk | contribs) (New page: ==Problem== Given that a, b, and c are non-zero real numbers, define <math>(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}</math>, find <math>(2, 12, 9)</math>. <math>\text{(A)}\ 4 \q...)
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Problem

Given that a, b, and c are non-zero real numbers, define $(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$, find $(2, 12, 9)$.

$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution

$(2, 12, 9)=\frac{2}{12}+\frac{12}{9}+\frac{9}{2}=\frac{1}{6}+\frac{4}{3}+\frac{9}{2}=\frac{1}{6}+\frac{8}{6}+\frac{27}{6}=\frac{36}{6}=\boxed{6}$. Our answer is then $\text{(C)}\ 6 \qquad$.