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> | ||
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| + | ==Solution 2== | ||
| + | You can multiply the whole equation by the lcd of the three fractions (30). You are now left with the equation(s) <math>15a=10b=6c</math>. If we now set <math>a</math> as our desired "testing" variable and multiply the equation by <math>bc</math> then we will get: | ||
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| + | <math>15abc=10b^2c=6bc^2</math> | ||
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| + | <math>abc=1920\implies 28800=10b^2c=6bc^2</math> | ||
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| + | Now we should make this a system of equations. | ||
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| + | <math>28800=10b^2c</math> | ||
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| + | <math>28800=6bc^2</math> | ||
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| + | <math>6bc^2=10b^2c \implies 3c=5b \implies c=\frac{5b}{3}</math> | ||
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| + | Plugging in to the first equation, | ||
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| + | <math>28800=\frac{50b^3}{3}</math> | ||
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| + | <math>576=\frac{b^3}{3}</math> | ||
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| + | <math>1728=b^3</math> | ||
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| + | <math>b=12</math> | ||
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| + | Using our third equation, if <math>b=12</math> then <math>c=20</math> which means <math>a</math> has to be <math>8</math>. | ||
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| + | <math>8+12+20=\boxed{\textbf{40}}</math> | ||
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| + | [https://aops.com/wiki/index.php/User:Am24 ~AM24] | ||
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| + | Note: I have never done this compitition, so if you are practicing this, I am unsure about the size of the teams (if there even is a team round) and if you get a calculator or not. I am assuming it is a team of four with calculator but you don't need a calculator for this (assuming based off of Mathcounts). I am writing this to say only use a rigorous solution like this if you have a big and/or a calculator. | ||
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== See also == | == See also == | ||
| − | {{PMWC box|year=2005|num-b= | + | {{PMWC box|year=2005|num-b=T1|num-a=T3}} |
Latest revision as of 22:06, 20 September 2025
Contents
Problem
Compute the sum of 
, 
, and 
given that 
and the product of , , and is 
.
Solution
Solution 2
You can multiply the whole equation by the lcd of the three fractions (30). You are now left with the equation(s) 
. If we now set as our desired "testing" variable and multiply the equation by 
then we will get:
Now we should make this a system of equations.
Plugging in to the first equation,
Using our third equation, if then which means has to be 
.
Note: I have never done this compitition, so if you are practicing this, I am unsure about the size of the teams (if there even is a team round) and if you get a calculator or not. I am assuming it is a team of four with calculator but you don't need a calculator for this (assuming based off of Mathcounts). I am writing this to say only use a rigorous solution like this if you have a big and/or a calculator.
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 | ||
