Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


Photo of the question: enter image description here

share|cite|improve this question
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
I think it should be $a^2\cdot b^3 = 432$. Which is consistent with answer b) . – Raskolnikov Nov 1 '11 at 19:38
@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
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 $ab=12$.

share|cite|improve this answer
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

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.]

share|cite|improve this answer

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.

share|cite|improve this answer

$(a^\frac{1}{2}b^\frac{1}{3})^6=432$ 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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.