Difference between revisions of "2005 PMWC Problems/Problem T2"

(See also)
(Solution)
 
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== Solution ==
 
== Solution ==
 
<math>\dfrac{abc}{2*3*5}=64=\dfrac{a^3}{8}=\dfrac{b^3}{27}=\dfrac{c^3}{125}</math>
 
<math>\dfrac{abc}{2*3*5}=64=\dfrac{a^3}{8}=\dfrac{b^3}{27}=\dfrac{c^3}{125}</math>
 +
  
 
<math>a=8</math>
 
<math>a=8</math>
 +
  
 
<math>b=12</math>
 
<math>b=12</math>
 +
  
 
<math>c=20</math>
 
<math>c=20</math>
 +
  
 
<math>a+b+c=\boxed{40}</math>
 
<math>a+b+c=\boxed{40}</math>

Latest revision as of 07:13, 6 October 2007

Problem

Compute the sum of $a$, $b$, and $c$ given that $\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}$ and the product of $a$, $b$, and $c$ is $1920$.

Solution

$\dfrac{abc}{2*3*5}=64=\dfrac{a^3}{8}=\dfrac{b^3}{27}=\dfrac{c^3}{125}$


$a=8$


$b=12$


$c=20$


$a+b+c=\boxed{40}$

See also

2005 PMWC (Problems)
Preceded by
Problem T1
Followed by
Problem T3
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10