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Hey Mike, I would like to see the full solution if you have time.

Based on the logic in the "easier" question, I see that a could be 1,2,3,5,6,10,15,30 but I'm missing the last four possibilities. Care to explain? Had 8 and 10 also been answer choices, I wouldn't have been able to guess correctly.

Thanks!!

Dear holidayhero

First of all, I would suggest thinking about it this way

a^3 = (slot #1)*(slot #2)*(slot #2)

In slot #1, we could put either 1 or 2^3

In slot #2, we could put either 1 or 3^3

In slot #3, we could put either 1 or 5^3 or 5^6 (which is [5^2]^2 = 25^3)

I believe you forget about that last possibility in the last slot. All together, that would give 2 possibilities in the first slot, 2 in the second, and 3 in the third, for a total of 2*2*3 = 12. For the GMAT, it's usually better to use the FCP instead of just trying to make an exhaustive list --- this same question could be repeated with much higher exponent (i.e. (2^23)*(3^34)(5*45) = (a^3)*b), and in that case, listing out all the factors would be a prohibitive approach.

The last two possibilities for a that you missed are a = 50 and a = 75, the two that involve factors of 5^2 = 25.

BIG idea set of ideas: If n is any even number, than Q^n is a perfect square If n is any number divisible by 3, then Q^n is a perfect cube If n is any number divisible by 4, then Q^n is a perfect fourth power and a perfect square If n is any number divisible by 5, then Q^n is a perfect fifth power If n is any number divisible by 6, then Q^n is a perfect sixth power and a perfect cube and a perfect square etc. etc.

Does all this make sense?

Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

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

Dear mike, I still have problem understanding the following question, could you please further elaborate? Thanks Question:If a and b are positive integers, and (2^3)(3^4)(5^7) = (a^3)*b, how many different possible values of b are there?

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

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Re: If a and b are positive integers, and [#permalink]

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27 Feb 2013, 00:53

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Let's assume\(2^3, 3^3 and 5^3\) as three different entities.

The entity\(5^3\)is 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 other\(5^3\) will not make a difference when being selected in the mentioned way. It will only make a difference when it is either selected with\(5^3\) 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.

Re: If a and b are positive integers, and [#permalink]

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26 Feb 2013, 19:10

Hey Mike, I would like to see the full solution if you have time.

Based on the logic in the "easier" question, I see that a could be 1,2,3,5,6,10,15,30 but I'm missing the last four possibilities. Care to explain? Had 8 and 10 also been answer choices, I wouldn't have been able to guess correctly.

Not sure if I got lucky or this method works, but I did it a different way: Because the LHS is prime factors, the total number of factors on the LHS is 17 ((3+1)+(4+1)+(7+1)) The number of factors on the RHS is ((3+1)+(1+1)) = 5 17 - 5 = 12 thus, b can be 12 different numbers. Lucky guess or is this a valid method?

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

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

I am not good at permutations so i tried to find all the possible values A. In my first attempt i could find only 9.

The possible values of a are: 1. 1 2. 2 3. 3 4. 5 6. 25 7. (2*3) 8. (3*25) 9. (5*3) 10. (25*2) 11. (2*3*5) 12. (2*3*25)

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

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Re: If a and b are positive integers, and [#permalink]

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13 May 2013, 13:06

mikemcgarry wrote:

Rajkiranmareedu wrote:

I am not good at permutations so i tried to find all the possible values A. In my first attempt i could find only 9.

The possible values of a are: 1. 1 2. 2 3. 3 4. 5 6. 25 7. (2*3) 8. (3*25) 9. (5*3) 10. (25*2) 11. (2*3*5) 12. (2*3*25)

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

Thank you for pointing. Counting methods is always a difficult topic for me, but at present I don't have enough time to prepare.

Re: If a and b are positive integers, and [#permalink]

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16 May 2013, 16:41

Dear mike, I still have problem understanding the following question, could you please further elaborate? Thanks Question:If a and b are positive integers, and (2^3)(3^4)(5^7) = (a^3)*b, how many different possible values of b are there?

Re: If a and b are positive integers, and [#permalink]

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29 Dec 2013, 18:24

mikemcgarry wrote:

smartyman wrote:

Dear mike, I still have problem understanding the following question, could you please further elaborate? Thanks Question:If a and b are positive integers, and (2^3)(3^4)(5^7) = (a^3)*b, how many different possible values of b are there?

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

Hey Mike, whats up?

Is there a way to find out all these choices more systematically rather than just counting, I have tried looking at other posts but unlucky so far

The way I was trying to do it was getting the cubes together, that is 5^3 and 2^3 as one entity 10^3 which left me with 15^4 = b

Then I guess that B could only one value, but I may be understanding the question in a different way

Will you shed some light on what did you intended to test with this question and an alternate approach?

Is there a way to find out all these choices more systematically rather than just counting, I have tried looking at other posts but unlucky so far

The way I was trying to do it was getting the cubes together, that is 5^3 and 2^3 as one entity 10^3 which left me with 15^4 = b

Then I guess that B could only one value, but I may be understanding the question in a different way

Will you shed some light on what did you intended to test with this question and an alternate approach?

Thank you! Cheers! J

Dear jlgdr, I was listing the terms out for clarity, to respond to someone else who did so --- that's not how I would solve the problem myself. The problem can be done very efficiently with the Fundamental Counting Principle. See this post if that vital idea is unfamiliar: http://magoosh.com/gmat/2012/gmat-quant-how-to-count/

Here's the expression: (2^3)(3^4)(5^7) = (a^3)*b The cube term, a^3, can contain: 1) either 2^0 or 2^3, two possibilities 2) either 3^0 or 3^3, two possibilities 3) either 5^0, 5^3, or 5^6, three possibilities By the FCP, 2*2*3 = 12, so there are 12 possible values for (a^3), and hence, 12 possible ways to distribute the factors.

Does this make sense? Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)