2002 AMC 10B Problems/Problem 2

Revision as of 02:26, 27 December 2008 by Lifeisacircle (talk | contribs) (created page)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

For the nonzero numbers a, b, and c, define

$(a,b,c)=\frac{abc}{a+b+c}$

Find $(2,4,6)$.

$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 24$

Solution

$\frac{2\cdot 4\cdot 6}{2+4+6}=\frac{48}{12}=4\Longrightarrow\mathrm{ (C) \ }$