Hi Mike,

This is a great question. Thanks for posting it. I'm curious how other people solved it. I looked at it as a combinations question:


Possibilities for the slots in team A

1 or 2----1 or 3---1, 5, or 5^2

2^3--------3^4----------5^7

so 2 * 2 * 3 = 12 possibilities

A^3 creates the limitations so it is necessary to figure out what those limitations are. The logic is that B has to cover for every factor that is not covered by A. So the number of possibilities for A dictates the number of possibilities for B.

HG.
_________________

"It is a curious property of research activity that after the problem has been solved the solution seems obvious. This is true not only for those who have not previously been acquainted with the problem, but also for those who have worked over it for years." -Dr. Edwin Land

"Learning hurts but enlightenment keeps you company along the way" -HerrGrau

If you found my post useful KUDOS are much appreciated.

Let's assume2^3, 3^3 and 5^3 as three different entities.

The entity5^3is there twice.

Thus, out of these three, we can take none, 1, two or all three. This gives a total of :

3C0+3C1+3C2+3C3 = 1+3+3+1 = 8 ways. Now the other5^3 will not make a difference when being selected in the mentioned way. It will only make a difference when it is either selected with5^3 only(1 case), or when both of them together are considered with either 2^3(1 case) or 3^3(1 case). Finally, when all the 4 are considered together, we have one final case. Thus a total of 8+3+1 = 12 cases.

E.

Really nice question , it took my grey matter spin for a while .

My Logic .

Try to write both LHS and RHS in common form

LHS can be written as (2^3)*(3^3)*(5^3 )*(1*3*5)*(5^3)
------------------------- -----

Ways to arrange them in RHS way ie ... a^3 * b

calculating LHS ways to arrange in RHS ways = 4! / 2! = 12 , because 5^3 is coming twice . Anagram method.

-eski
_________________

Your Kudos will motivate me :)

Dear vaj18psu,

I'm sorry to tell you --- you happened to get very lucky here, but your approach is a complete invalid way of thinking about the problem that actually involves some harmful misconceptions. Perhaps the biggest is --- when you have N ways something could happen in one case, and M ways in other, and P ways in another, etc. then the way to get to total number of ways is to use addition. That is one of the most poisonous misconceptions on the entire GMAT Quant section. The correct view involves the Fundamental Counting Principle, about which you can read here:
http://magoosh.com/gmat/2012/gmat-quant-how-to-count/

Does all this make sense?

Mike
_________________

Mike McGarry
Magoosh Test Prep

Yes, those are the twelve possible values. Rajkiranmareedu, first of all, please don't use the word "permutations", which means something very specific, for "counting methods". This problem has zero to do with permutations, but it is all about counting methods. Using those two words interchangeably will permanently confused you about this already challenging topic. I strongly suggest that you take a look at the blogs listed in my first post at the head of this thread. Among other things, those blogs will make clear the distinctions such as "permutations" vs. "counting methods". You see, listing, as you did here, can be a good secondary strategy for building intuition, but if the problem had involved dozens or hundreds of possibilities, you simply would not be able to list them all. In order to get everything correct that the GMAT will ask, you need to understand how to approach questions like this more methodically, more symbolically.
Does all this make sense?
Mike
_________________

Mike McGarry
Magoosh Test Prep

really good question - fooled me, but it makes great sense now. Thank you for teaching me something

Yes, that's exactly how I would do it and exactly what I get.
Mike
_________________

Mike McGarry
Magoosh Test Prep

Dear smartyman,
First of all, understand that this is a very hard question, perhaps an 800-level question. It involves several sophisticated concepts, including
(a) prime factorization
http://magoosh.com/gmat/2012/gmat-math-factors/
(b) counting & the FCP
http://magoosh.com/gmat/2012/gmat-quant-how-to-count/

Insight #1 --- the question asks for number of possible values of b, but it's much much easier to count the number of possible values for (a^3), and every value of (a^3) will be paired with a unique value of b, so we will count the (a^3)'s as a way to get the answer.

Insight #2 --- a^3 is a perfect cube, and every prime factor in a perfect cube must appear either three times or some number of times that is a multiple of three. Thus, (a^3) simply could be 1, and all the factors could be in b. If (a^3) has any factors of 2, it only could have 2^3, because that's all the factors of two available. If (a^3) has any factors of 3, it only could have 3^3, because the available powers of 3 don't go up as high as any other multiple of 3. If (a^3) has any factors of 5, it could have either (5^3) or (5^6) --- we have enough factors of 5 to construct either one of those, so we have both of them as options.

Insight #3 --- how many combinations in (a^3)?
For the factors of 2, we have two choices ---- 2^0 = 1 or 2^3
For the factors of 3, we have two choices ---- 3^0 = 1 or 3^3
For the factors of 5, we have three choices ---- 5^0 = 1 or 5^3 or 5^6
Any option in any one category could be matched with any option from any other category, so this is a case in which we can employ the FCP:
total number of combinations = 2*2*3 = 12

There are 12 possibilities for (a^3), which means there are 12 possibilities for b.

Because there are only 12, I will list everything for clarity, to demonstrate that the FCP works. Notice, in the (a^3) term, all prime factors always have powers divisible by 3.

(1) a = 1, b = (2^3)(3^4)(5^7)
(2) a = (2^3), b = (3^4)(5^7)
(3) a = (3^3), b = (2^3)(3^1)(5^7)
(4) a = (5^3), b = (2^3)(3^4)(5^4)
(5) a = (5^6), b = (2^3)(3^4)(5^1)
(6) a = (2^3)(3^3), b = (3^1)(5^7)
(7) a = (2^3)(5^3), b = (3^4)(5^4)
(8) a = (3^3)(5^3), b = (2^3)(3^1)(5^4)
(9) a = (2^3)(3^3)(5^3), b = (3^1)(5^4)
(10) a = (2^3)(5^6), b = (3^4)(5^1)
(11) a = (3^3)(5^6), b = (2^3)(3^1)(5^1)
(12) a = (2^3)(3^3)(5^6), b = (3^1)(5^1)

Does all this make sense?
Mike
_________________

Mike McGarry
Magoosh Test Prep