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edit: NO, it is not a^2 * b^3 = 432, see photo proof attached, but I did missread the question :)

This SAT test question has me stuck:

If a and b are positive integers and ${({a^{(1/2)}} \cdot {b^{(1/3)}})^6} = 432$ what is the value of $ab$?

(a) 6
(b) 12
(c) 18
(d) 24
(e) 36

The correct answer is (b), but why? Any suggestions as to how to solve such problems efficiently?

Thanks!

Photo of the question: enter image description here

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8  
If $a$ and $b$ are positive integers, $\left( \frac{1}{a^2} \frac{1}{b^3} \right)^6$ can not be integer, unless $a=b=1$. –  Sasha Nov 1 '11 at 19:27
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I think it should be $a^2\cdot b^3 = 432$. Which is consistent with answer b) . –  Raskolnikov Nov 1 '11 at 19:38
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@all: If you are downvoting the post, please explain your reason in a comment, so that the OP gets a chance to improve the question. –  Srivatsan Nov 1 '11 at 19:51
    
See edits with the picture, problem is still open (sorry for the typo at first!) –  Robin Nov 1 '11 at 19:59
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4 Answers

up vote 11 down vote accepted

EDIT: Ah, we now have the correct question. $$\left(a^{1/2} \cdot b^{1/3}\right)^{6} = 432$$

Note the following two properties of exponentiation: $$a^{bc}=(a^b)^c\qquad (ab)^c=a^cb^c.$$ Thus $$\left(a^{1/2} \cdot b^{1/3}\right)^{6} =(a^{1/2})^{6}(b^{1/3})^{6}=a^{(1/2)(6)}b^{(1/3)(6)}=a^3b^2$$

Now consider $432$'s prime factorization to find the answer:

$$432=2^4\cdot 3^3=2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 3$$

You want to find two pieces of the factorization such that the first piece occurs 3 times, the second piece occurs 2 times, and put together, those repetitions form the entire factorization. Thus the only possible answer is $$432=2^4\cdot 3^3=\underbrace{2\cdot 2}_{b}\cdot \underbrace{2\cdot 2}_{b}\cdot \underbrace{3}_{a}\cdot \underbrace{3}_{a}\cdot \underbrace{3}_{a}$$ Hence $a=3$ and $b=4$, hence $ab=12$.

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See edit with the image and the actual problem restated –  Robin Nov 1 '11 at 19:58
    
@Robin: Ah, thanks, I've updated my answer. –  Zev Chonoles Nov 1 '11 at 20:05
    
Thanks Zev, now it makes sense :) –  Robin Nov 1 '11 at 20:15
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Here's a less exponent intensive method. Since $a$ and $b$ are positive integers, it follows that $a^2$ is a perfect square and $b^3$ is a perfect cube. Thus, we want to write $432$ as a perfect square times a perfect cube. Having done this, we then multiply the square root of the perfect square (i.e. $a$) by the cube root of the perfect cube (i.e. $b$).

Perfect squares (omitting $1$) are $4,$ $9,$ $16,$ $25,$ etc. If you check to see if $432$ is divisible by $4$ (motivated because $432$ is even; assured by the 4 divisibility rule), you'll find it is, with $432 = 4 \cdot 108.$

Since $108$ is not a perfect cube ($108$ isn't one of $8,$ $27,$ $64,$ $125,$ etc.), there must be a larger perfect square factor of $432$ than $4,$ or equivalently, there must be a perfect square factor of $108.$ Checking for divisibility by $4$ (motivated because $108$ is even; assured by the 4 divisibility rule), we find that $108 = 4 \cdot 27.$ Therefore, from $432 = 4 \cdot 108$ and $108 = 4 \cdot 27,$ we get:

$$432 \;= \;4 \cdot (4 \cdot 27) \;= \;16 \cdot 27$$

Now that we have $432$ written as a perfect square times a perfect cube, it's easy to see that $a = 4$ and $b = 3,$ and so $ab = 4 \cdot 3 = 12.$ [Note to others: In some of the statements above I've assumed the item is sound.]

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Just pick numbers from the answer. For instance, $12=4\times 3$. if you use 3 as (a) and 4 as (b), $3^{(1/2)(6)}$--or simply $3^3$--and $4^{(1/3)(6)}$--or $4^2$-- you multiply to get $432$. You then know that 4 and 3 are the correct integers. since $4\times 3=12$ [remember the SAT wants you to finish the answer]

Back-solving is your best bet: trial and error is key.

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$(a^\frac{1}{2}b^\frac{1}{3})^6=432$ would require that you distribute outside the exponent, thus resulting in $a^3b^2=432$. The other answers can then take you the rest of the way.

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The other answers can take you here, then the rest of the way! –  The Chaz 2.0 Nov 1 '11 at 20:20
    
@World: You mean $a^3b^2$, not $a^2b^3$. –  Zev Chonoles Nov 1 '11 at 20:24
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