Difference between revisions of "2002 AMC 10B Problems/Problem 2"

Revision as of 12:57, 27 December 2008

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

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

Find $(2,4,6)$ .

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

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

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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 10 Problems and Solutions

@@ Line 12: / Line 12: @@
 <math>\frac{2\cdot 4\cdot 6}{2+4+6}=\frac{48}{12}=4\Longrightarrow\mathrm{ (C) \ }</math>
+==See Also==
+{{AMC10 box|year=2002|ab=B|num-b=1|num-a=3}}
+[[Category:Introductory Algebra Problems]]