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

(Problem)
 
Line 3: Line 3:
 
For the nonzero numbers a, b, and c, define
 
For the nonzero numbers a, b, and c, define
  
<math>(a,b,c)=\frac{abc}{a+b+c}</math>
+
<math>D(a,b,c)=\frac{abc}{a+b+c}</math>
  
Find <math>(2,4,6)</math>.
+
Find <math>D(2,4,6)</math>.
  
 
<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 24 </math>
 
<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 24 </math>

Latest revision as of 11:06, 1 October 2018

Problem

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

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

Find $D(2,4,6)$.

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

Solution

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

See Also

2002 AMC 10B (ProblemsAnswer KeyResources)
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

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