Author 
Message 
TAGS:

Hide Tags

Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4130

If a and b are positive integers, and [#permalink]
Show Tags
26 Feb 2013, 11:32
10
This post received KUDOS
Expert's post
22
This post was BOOKMARKED
Question Stats:
46% (02:33) correct
54% (01:17) wrong based on 457 sessions
HideShow timer Statistics
I was inspired by another, easier question to create this question, just for fun. 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? (A) 2 (B) 3 (C) 4 (D) 6 (E) 12You may find this blog on prime factors helpful in thinking about this problem: http://magoosh.com/gmat/2012/gmatmathfactors/You may also find this blog on counting helpful. http://magoosh.com/gmat/2012/gmatquanthowtocount/I will post a full solution if there's interest.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Mike McGarry Magoosh Test Prep





Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4130

Re: If a and b are positive integers, and [#permalink]
Show Tags
27 Feb 2013, 12:53
6
This post received KUDOS
Expert's post
3
This post was BOOKMARKED
holidayhero wrote: 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 holidayheroFirst 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 squareetc. etc. Does all this make sense? Mike
_________________
Mike McGarry Magoosh Test Prep



Manager
Joined: 27 Jan 2013
Posts: 230

Re: If a and b are positive integers, and [#permalink]
Show Tags
26 Feb 2013, 15:15
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 possibilitiesA^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
GMAT vs GRE Comparison
If you found my post useful KUDOS are much appreciated.
IMPROVE YOUR READING COMPREHENSION with the ECONOMIST READING COMPREHENSION CHALLENGE:
Here is the first set along with some strategies for approaching this work: http://gmatclub.com/forum/theeconomistreadingcomprehensionchallenge151479.html



Intern
Joined: 09 Sep 2012
Posts: 26

Re: If a and b are positive integers, and [#permalink]
Show Tags
18 Mar 2013, 06:51
2
This post received KUDOS
1
This post was BOOKMARKED
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 :)



Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4130

Re: If a and b are positive integers, and [#permalink]
Show Tags
17 May 2013, 12:38
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? Dear smartyman, First of all, understand that this is a very hard question, perhaps an 800level question. It involves several sophisticated concepts, including (a) prime factorization http://magoosh.com/gmat/2012/gmatmathfactors/(b) counting & the FCP http://magoosh.com/gmat/2012/gmatquanthowtocount/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



Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 629

Re: If a and b are positive integers, and [#permalink]
Show Tags
27 Feb 2013, 00:53
1
This post received KUDOS
1
This post was BOOKMARKED
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. E.
_________________
All that is equal and notDeep Dive Inequality
Hit and Trial for Integral Solutions



Manager
Joined: 06 Jun 2012
Posts: 142

Re: If a and b are positive integers, and [#permalink]
Show Tags
14 May 2013, 05:13
1
This post received KUDOS
1
This post was BOOKMARKED
Wow this question really did get me!! Just to make sure i have got it right, i tried solving (2^23)*(3^34)(5*45) = (a^3)*b. Let me know if i did something wrong a^3 = (slot 1) (slot 2) (slot 3) Slot 1 = 1 or 2^3 or 2^6 or 2^9 ...2^21 = 8 possibilities Slot 2 = 1 or 3^3 or 3^6 or 3^9....3^33 = 12 possibilities Slot 3 = 1 or 5^3 or 5^6 or 5^9....5^45 = 16 possibilities So 1536 possibilities??
_________________
Please give Kudos if you like the post



Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4130

Re: If a and b are positive integers, and [#permalink]
Show Tags
14 May 2013, 10:56
summer101 wrote: Wow this question really did get me!!
Just to make sure i have got it right, i tried solving (2^23)*(3^34)(5*45) = (a^3)*b. Let me know if i did something wrong
a^3 = (slot 1) (slot 2) (slot 3) Slot 1 = 1 or 2^3 or 2^6 or 2^9 ...2^21 = 8 possibilities Slot 2 = 1 or 3^3 or 3^6 or 3^9....3^33 = 12 possibilities Slot 3 = 1 or 5^3 or 5^6 or 5^9....5^45 = 16 possibilities
So 1536 possibilities?? Yes, that's exactly how I would do it and exactly what I get. Mike
_________________
Mike McGarry Magoosh Test Prep



Senior Manager
Joined: 13 Oct 2016
Posts: 366
GPA: 3.98

Re: If a and b are positive integers, and [#permalink]
Show Tags
14 Nov 2016, 11:19
1
This post received KUDOS
1
This post was BOOKMARKED
mikemcgarry wrote: I was inspired by another, easier question to create this question, just for fun. 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? (A) 2 (B) 3 (C) 4 (D) 6 (E) 12You may find this blog on prime factors helpful in thinking about this problem: http://magoosh.com/gmat/2012/gmatmathfactors/You may also find this blog on counting helpful. http://magoosh.com/gmat/2012/gmatquanthowtocount/I will post a full solution if there's interest. Nice question and great job! It really encompass different fields of math. I’m too late but here is my view for the sake of diversity. I prefer to solve it for \(a^3\) rather than \(b\) as \(a^3\) is the most restrictive part. Let’s create different cases for different perfect cubes. 1. \(a^3\) consists of a single prime – \(p^c\), where \(c\) multiple of 3. We have \(2^3\), \(3^3\), \(5^3\) and \(5^6\) \(3C1 + 1 = 4\) 2. \(a^3\) is composed from two distinct primes – \(p^cq^d\), where \(c,d\) – multiples of 3. \(2^33^3, 2^35^3, 3^35^3\) and \(2^35^6, 3^35^6\) \(3C2 + 1*2C1 = 5\) 3. \(a^3\) – 3 distinct primes – \(p^cq^dr^e\), where \(c,d,e\) – multiples of 3. \(2^33^35^3\) and \(2^33^35^6\) together \(= 2\) 4. and final case when all primes will be absorbed by \(b\) – \(1\) (also a perfect cube) \(p^0q^0r^0\) = 1 Total # = \(4 + 5 + 2 + 1 = 12\)



Intern
Joined: 08 Feb 2011
Posts: 11

Re: If a and b are positive integers, and [#permalink]
Show Tags
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.
Thanks!!



Current Student
Joined: 29 Dec 2012
Posts: 23
GMAT 1: 680 Q45 V38 GMAT 2: 690 Q47 V38
GPA: 3.69

Re: If a and b are positive integers, and [#permalink]
Show Tags
09 Apr 2013, 19:33
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?



Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4130

Re: If a and b are positive integers, and [#permalink]
Show Tags
10 Apr 2013, 10:51
vaj18psu wrote: 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/gmatquanthowtocount/Does all this make sense? Mike
_________________
Mike McGarry Magoosh Test Prep



Intern
Joined: 20 Apr 2013
Posts: 24
Concentration: Finance, Finance
GMAT Date: 06032013
GPA: 3.3
WE: Accounting (Accounting)

Re: If a and b are positive integers, and [#permalink]
Show Tags
13 May 2013, 05:33
I am not good at permutations so i tied 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*2) 9. (5*3) 10. (25*2) 11. (2*3*5) 12. (2*3*25)



Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4130

Re: If a and b are positive integers, and [#permalink]
Show Tags
13 May 2013, 10:37
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
_________________
Mike McGarry Magoosh Test Prep



Intern
Joined: 20 Apr 2013
Posts: 24
Concentration: Finance, Finance
GMAT Date: 06032013
GPA: 3.3
WE: Accounting (Accounting)

Re: If a and b are positive integers, and [#permalink]
Show Tags
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. I will keep ur advice for future. Regards Ratnakar



Manager
Joined: 09 Apr 2013
Posts: 209
Location: United States
Concentration: Finance, Economics
GMAT 1: 710 Q44 V44 GMAT 2: 740 Q48 V44
GPA: 3.1
WE: Sales (Mutual Funds and Brokerage)

Re: If a and b are positive integers, and [#permalink]
Show Tags
13 May 2013, 15:49
really good question  fooled me, but it makes great sense now. Thank you for teaching me something



Manager
Joined: 27 Jul 2011
Posts: 61

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



Current Student
Joined: 06 Sep 2013
Posts: 1997
Concentration: Finance

Re: If a and b are positive integers, and [#permalink]
Show Tags
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? Dear smartyman, First of all, understand that this is a very hard question, perhaps an 800level question. It involves several sophisticated concepts, including (a) prime factorization http://magoosh.com/gmat/2012/gmatmathfactors/(b) counting & the FCP http://magoosh.com/gmat/2012/gmatquanthowtocount/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? Thank you! Cheers! J



Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4130

Re: If a and b are positive integers, and [#permalink]
Show Tags
30 Dec 2013, 07:20
jlgdr wrote: 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? 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/gmatquanthowtocount/Here's the expression: (2^3)(3^4)(5^7) = (a^3)*bThe 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



Current Student
Joined: 06 Sep 2013
Posts: 1997
Concentration: Finance

Re: If a and b are positive integers, and [#permalink]
Show Tags
30 Dec 2013, 07:52
Yes awesome, makes perfect sense! Thanks! Cheers J Posted from my mobile device




Re: If a and b are positive integers, and
[#permalink]
30 Dec 2013, 07:52



Go to page
1 2
Next
[ 26 posts ]




