Find ab given its relationship to the number 432

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If a,b are positive integers and (a1/2b1/3)6 = 432, then what is the value of ab?
 
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fk378 said:
If a,b are positive integers and (a1/2b1/3)6 = 432, then what is the value of ab?

Is this a problem from the SAT?

First bring the 6 inside the a and b term to get a^6/2 * b^6/3 = 432
 
jedishrfu said:
Is this a problem from the SAT?

First bring the 6 inside the a and b term to get a^6/2 * b^6/3 = 432

Yes, I did that. Don't know where to go from here. Seems like I just go in circles when I try to make two equations to solve for the two unknowns.
 
Is the only way to do this just to get a3b2=432, then find the factors of 432? I tried this and then got 16*27=432, so then a=3, b=2. But I feel like there must be a different way to do this problem...
 
Hmmmm... It would be good if there was another way. It seems a bit too easy.
 
fk378 said:
Is the only way to do this just to get a3b2=432, then find the factors of 432? I tried this and then got 16*27=432, so then a=3, b=2. But I feel like there must be a different way to do this problem...

You mean b=4 :smile:

I'm not aware of another way if there is one, and I'd imagine if there were, it'd be fairly more complicated.
 
fk378 said:
If a,b are positive integers and (a1/2b1/3)6 = 432, then what is the value of ab?

We are given
##(a^{1/2} b^{1/3})^6 = 432##
So
##a^3 b^2 = a(ab)^2 = 432##
##(ab)^2 = \frac{432}{a}##
LHS is a square, so test different a.
##a = 2 \implies \frac{432}{a} = 216## not a square
##a = 3 \implies \frac{432}{a} = 144##
144 is a square, so ab = 12.
 
pwsnafu said:
We are given
##(a^{1/2} b^{1/3})^6 = 432##
So
##a^3 b^2 = a(ab)^2 = 432##
##(ab)^2 = \frac{432}{a}##
LHS is a square, so test different a.
##a = 2 \implies \frac{432}{a} = 216## not a square
##a = 3 \implies \frac{432}{a} = 144##
144 is a square, so ab = 12.

Nice!
 
Notice that the condition "a,b are positive integers" is crucial here. If a and b were allowed to be negative, there would be more solutions. If a and b were allowed to be any real numbers there would be an infinite number of solutions.
 
HallsofIvy said:
Notice that the condition "a,b are positive integers" is crucial here. If a and b were allowed to be negative, there would be more solutions.

If a and b were allowed to be negative and we are allowed to use complex algebra, then yes.
Otherwise a1/2 is undefined.