Difference between revisions of "2005 PMWC Problems/Problem T2"
(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>...) |
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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> | ||
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<math>a=8</math> | <math>a=8</math> | ||
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<math>b=12</math> | <math>b=12</math> | ||
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<math>c=20</math> | <math>c=20</math> | ||
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<math>a+b+c=\boxed{40}</math> | <math>a+b+c=\boxed{40}</math> | ||
== See also == | == See also == | ||
| − | {{PMWC box|year=2005|num-b= | + | {{PMWC box|year=2005|num-b=T1|num-a=T3}} |
, 
, and 
given that 
and the product of 
.
