Find all School-related info fast with the new School-Specific MBA Forum

It is currently 25 Jul 2014, 01:26

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If a and b are positive integers, and

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Expert Post
9 KUDOS received
Magoosh GMAT Instructor
User avatar
Joined: 28 Dec 2011
Posts: 1939
Followers: 466

Kudos [?]: 1846 [9] , given: 29

If a and b are positive integers, and [#permalink] New post 26 Feb 2013, 10:32
9
This post received
KUDOS
Expert's post
2
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  55% (medium)

Question Stats:

44% (02:27) correct 56% (01:18) wrong based on 182 sessions
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) 12


You may find this blog on prime factors helpful in thinking about this problem:
http://magoosh.com/gmat/2012/gmat-math-factors/
You may also find this blog on counting helpful.
http://magoosh.com/gmat/2012/gmat-quant-how-to-count/

I will post a full solution if there's interest.
[Reveal] Spoiler: OA

_________________

Mike McGarry
Magoosh Test Prep

Image

Image

Kaplan GMAT Prep Discount CodesKnewton GMAT Discount CodesManhattan GMAT Discount Codes
Expert Post
3 KUDOS received
Manager
Manager
User avatar
Joined: 27 Jan 2013
Posts: 183
GMAT 1: 770 Q49 V46
Followers: 26

Kudos [?]: 103 [3] , given: 11

CAT Tests
Re: If a and b are positive integers, and [#permalink] New post 26 Feb 2013, 14:15
3
This post received
KUDOS
Expert's post
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

GMAT Blog and Tutoring: http://AtlanticGMAT.com

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 and some strategies for approaching this work: http://gmatclub.com/forum/the-economist-reading-comprehension-challenge-151479.html

Intern
Intern
User avatar
Joined: 08 Feb 2011
Posts: 11
Followers: 1

Kudos [?]: 6 [0], given: 9

Re: If a and b are positive integers, and [#permalink] New post 26 Feb 2013, 18:10
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!!
Expert Post
Verbal Forum Moderator
Verbal Forum Moderator
User avatar
Joined: 10 Oct 2012
Posts: 626
Followers: 40

Kudos [?]: 535 [0], given: 135

Premium Member
Re: If a and b are positive integers, and [#permalink] New post 26 Feb 2013, 23:53
Expert's post
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.
_________________

All that is equal and not-Deep Dive In-equality

Hit and Trial for Integral Solutions

Expert Post
5 KUDOS received
Magoosh GMAT Instructor
User avatar
Joined: 28 Dec 2011
Posts: 1939
Followers: 466

Kudos [?]: 1846 [5] , given: 29

Re: If a and b are positive integers, and [#permalink] New post 27 Feb 2013, 11:53
5
This post received
KUDOS
Expert's post
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 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

Image

Image

1 KUDOS received
Intern
Intern
avatar
Joined: 09 Sep 2012
Posts: 29
Schools: LBS '14, IMD '16
Followers: 0

Kudos [?]: 10 [1] , given: 25

Re: If a and b are positive integers, and [#permalink] New post 18 Mar 2013, 05:51
1
This post received
KUDOS
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 :)

Intern
Intern
avatar
Joined: 29 Dec 2012
Posts: 15
GMAT 1: 680 Q45 V38
GMAT 2: 690 Q47 V38
GPA: 3.69
Followers: 0

Kudos [?]: 0 [0], given: 2

CAT Tests
Re: If a and b are positive integers, and [#permalink] New post 09 Apr 2013, 18:33
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?
Expert Post
Magoosh GMAT Instructor
User avatar
Joined: 28 Dec 2011
Posts: 1939
Followers: 466

Kudos [?]: 1846 [0], given: 29

Re: If a and b are positive integers, and [#permalink] New post 10 Apr 2013, 09:51
Expert's post
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/gmat-quant-how-to-count/

Does all this make sense?

Mike :-)
_________________

Mike McGarry
Magoosh Test Prep

Image

Image

Intern
Intern
avatar
Joined: 20 Apr 2013
Posts: 24
Concentration: Finance, Finance
GMAT Date: 06-03-2013
GPA: 3.3
WE: Accounting (Accounting)
Followers: 0

Kudos [?]: 1 [0], given: 99

Re: If a and b are positive integers, and [#permalink] New post 13 May 2013, 04:33
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)
Expert Post
Magoosh GMAT Instructor
User avatar
Joined: 28 Dec 2011
Posts: 1939
Followers: 466

Kudos [?]: 1846 [0], given: 29

Re: If a and b are positive integers, and [#permalink] New post 13 May 2013, 09:37
Expert's post
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

Image

Image

Intern
Intern
avatar
Joined: 20 Apr 2013
Posts: 24
Concentration: Finance, Finance
GMAT Date: 06-03-2013
GPA: 3.3
WE: Accounting (Accounting)
Followers: 0

Kudos [?]: 1 [0], given: 99

Re: If a and b are positive integers, and [#permalink] New post 13 May 2013, 12:06
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
Manager
avatar
Joined: 09 Apr 2013
Posts: 214
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)
Followers: 4

Kudos [?]: 56 [0], given: 40

Re: If a and b are positive integers, and [#permalink] New post 13 May 2013, 14:49
really good question - fooled me, but it makes great sense now. Thank you for teaching me something :)
Manager
Manager
avatar
Joined: 06 Jun 2012
Posts: 150
Followers: 0

Kudos [?]: 26 [0], given: 37

GMAT ToolKit User
Re: If a and b are positive integers, and [#permalink] New post 14 May 2013, 04:13
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

Expert Post
Magoosh GMAT Instructor
User avatar
Joined: 28 Dec 2011
Posts: 1939
Followers: 466

Kudos [?]: 1846 [0], given: 29

Re: If a and b are positive integers, and [#permalink] New post 14 May 2013, 09:56
Expert's post
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

Image

Image

Intern
Intern
avatar
Joined: 27 Jul 2011
Posts: 47
Followers: 1

Kudos [?]: 12 [0], given: 2

Re: If a and b are positive integers, and [#permalink] New post 16 May 2013, 15:41
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?
Expert Post
Magoosh GMAT Instructor
User avatar
Joined: 28 Dec 2011
Posts: 1939
Followers: 466

Kudos [?]: 1846 [0], given: 29

Re: If a and b are positive integers, and [#permalink] New post 17 May 2013, 11:38
Expert's post
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 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

Image

Image

SVP
SVP
User avatar
Joined: 06 Sep 2013
Posts: 1629
Location: United States
Concentration: Finance
GMAT 1: 710 Q48 V39
WE: Corporate Finance (Investment Banking)
Followers: 10

Kudos [?]: 132 [0], given: 254

GMAT ToolKit User
Re: If a and b are positive integers, and [#permalink] New post 29 Dec 2013, 17:24
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 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 :-)


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 :)
Expert Post
Magoosh GMAT Instructor
User avatar
Joined: 28 Dec 2011
Posts: 1939
Followers: 466

Kudos [?]: 1846 [0], given: 29

Re: If a and b are positive integers, and [#permalink] New post 30 Dec 2013, 06:20
Expert's post
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/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

Image

Image

SVP
SVP
User avatar
Joined: 06 Sep 2013
Posts: 1629
Location: United States
Concentration: Finance
GMAT 1: 710 Q48 V39
WE: Corporate Finance (Investment Banking)
Followers: 10

Kudos [?]: 132 [0], given: 254

GMAT ToolKit User
Re: If a and b are positive integers, and [#permalink] New post 30 Dec 2013, 06:52
Yes awesome, makes perfect sense!
Thanks!

Cheers
J :)

Posted from my mobile device Image
Re: If a and b are positive integers, and   [#permalink] 30 Dec 2013, 06:52
    Similar topics Author Replies Last post
Similar
Topics:
5 Experts publish their posts in the topic If a and b are positive integers, is a!/b! an integer? josemarioamaya 2 20 Sep 2013, 13:01
Experts publish their posts in the topic If a and b are positive integers Shawshank 2 13 Oct 2012, 17:18
4 Experts publish their posts in the topic When positive integer A is divided by positive integer B prasadrg 6 22 Dec 2009, 18:25
If a and b are positive integers such that a-b and a/b andy1979 6 04 Jul 2005, 02:34
If a and b are positive integers such that a b and are marine 1 05 Sep 2004, 06:08
Display posts from previous: Sort by

If a and b are positive integers, and

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.