2005 PMWC Problems/Problem T2

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

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

$a=8$

$b=12$

$c=20$

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

Solution 2

You can multiply the whole equation by the lcd of the three fractions (30). You are now left with the equation(s) $15a=10b=6c$ . If we now set $a$ as our desired "testing" variable and multiply the equation by $bc$ then we will get:

$15abc=10b^2c=6bc^2$

$abc=1920\implies 28800=10b^2c=6bc^2$

Now we should make this a system of equations.

$28800=10b^2c$

$28800=6bc^2$

$6bc^2=10b^2c \implies 3c=5b \implies c=\frac{5b}{3}$

Plugging in to the first equation,

$28800=\frac{50b^3}{3}$

$576=\frac{b^3}{3}$

$1728=b^3$

$b=12$

Using our third equation, if $b=12$ then $c=20$ which means $a$ has to be $8$ .

$8+12+20=\boxed{\textbf{40}}$

~AM24

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.

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

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2005 PMWC Problems/Problem T2

Contents

Problem

Solution

Solution 2

See also