vinny380

I was taking one of the blue book's SAT tests when I came across this "hard" SAT math question. I am pretty good at math and usually get all the hard questions right, but this one blew me away. I know there is probably a very simple solution, but I cant seem to find it. Anyway, here it is. Thanks for helping!
BTW - PLease give me an explanation on how you got the answer.

If a and b are positive intergers and (a^1/2 * b^1/3) = 432, what is the value of ab?
A- 6
B-12
C-18
D-24
E-36

........sigh

yoshi503

You miswrote the question: It should actually be 432^(1/6). That may be why you were having a hard time. If not, then I know how to solve it, and I'll gladly help. A quick hint: the exponent on the 432 should be your guide to solving the problem (i.e. try to get RID of it!!!).

vinny380

Oops... I did write the question wrong. The real questions is ....
If a and b are positive intergers and (a^1/2 * b^1/3)^6 = 432 what is the value of ab?
A- 6
B-12
C-18
D-24
E-36

Yoshi... I am kinda confused by your post. The question does not say 432^1/6. Are you looking in the Blue Book, test 4 (thats where I got it from). Sorry Yoshi .. still confused =(

yoshi503

OK, you're right, Vinny, and thankfully that only makes the question easier. This is how you solve it, step by step:



[ a^(1/2) * b^(1/3) ] ^(6) = 432 <----- given

a^3 * b^2 = 432 <----- simplify by multiplying the exponents

Now here is the slightly tricky part...

You know that one number cubed times another one squared equals 432. So, to find what the numbers are, simply factor 432:

______________432
_____________/____\
____________2_____216
_________________/____\
________________2_____108
______________________/___\
_____________________2____54
__________________________/__\
_________________________2____27
_____________________________/__\
____________________________3____9
________________________________/_\
_______________________________3___3

Note in particular how 432 = 2^4 times 3^3, or:

432 = 2 * 2 * 2 * 2 * 3 * 3 * 3

Your number that is cubed (a) must be 3. But what about the number that is squared (b)? Simply group two pairs of twos together to find that number, which is 4:

432 = (2 * 2) * (2 * 2) * (3 * 3 * 3)

432 = (2 * 2)^2 * (3 * 3 * 3)

432 = (4)^2 * (3)^3

Therefore, a = 3 and b = 4, and ab = (3)(4), or 12.

optimizerdad

Vinny:
If (a^1/2 * b^1/3) = 432^1/6, raise both RHS & LHS to the power of 6; you'd then have

(a^1/2 * b^1/3)^6 = 432

*Cough* A case of six of one, half-a-dozen of the other...

panic

I think what Yoshi means is:
"(a^1/2 * b^1/3)^6 = 432"
can be changed so
"(a^1/2 * b^1/3) = 432^(1/6)"

yoshi503

had to edit my second post...the prime factorization tree wasn't working with spaces...I put underscores in tho

harry3734

Choice (B) is correct. Simplifying the exponential part of the expression gives (a^(6/2)*b^(6/3)) =a^3*b^2=432. The prime factorization of 432 is 3^3*2^4 so a^3*b^2=3^3*2^4. Since a & b must be positive integers, it follows that a^=3^3 and b^2=2^4=(2^2)^2 This yields a=3 and b=4 The question asks for the product of a and b which is 3*4=12



ps: from the CB...i missed it too

Quix

Harry how do you know when to use prime factorization? i know thats a stupid question but i factored it in 2s and it got really messy.

gcf101

^Quix - it's unlikely you'll get an answer from Harry: his last post was on 09-17-2007.
I'd like to help you with this, but I don't quite understand your question.

===============
A while ago somebody posted a very efficient SAT-ish solution to this question, but somehow it never got into the Consolidated List.
Recently a student of mine came up with the same one, and I find it's worthwhile to post it here.
(a^1/2 * b^1/3)^6 = 432
a^3*b^2 = 432
(ab)^2 * b = 432
Let's examine given answers for ab:
A- 6
B-12
C-18
D-24
E-36
C is too big (18^2 = 324), so are D and E; A is too small; that leaves B.
If you are pressed for time you don't even need to find a and b.

Just in case:
ab = 12
12^2*a = 432
a = 432/144
a = 3
b = 4

Co11apse

gcf101, Thanks from 2012. That solution is really the most efficient