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

Revision as of 07:13, 6 October 2007

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 $1920$ .

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

$a=8$

$b=12$

$c=20$

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

Revision as of 07:12, 6 October 2007 (view source) 1=2 (talk \| contribs) (New page: == Problem == Compute the sum of <math>a</math>, <math>b</math>, and <math>c</math> given that <math>\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}</math> and the product of <math>a</math>, <math>...)		Revision as of 07:13, 6 October 2007 (view source) 1=2 (talk \| contribs) (→‎See also) Newer edit →
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	== See also ==		== See also ==
−	{{PMWC box\|year=2005\|num-b=~~I14~~\|num-a=T1}}	+	{{PMWC box\|year=2005\|num-b=T1\|num-a=T3}}