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

It is currently 18 Apr 2014, 07:56

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 the product of all the unique positive divisors of n, a p

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
4 KUDOS received
Senior Manager
Senior Manager
Status: ready to boMBArd
Joined: 31 Oct 2010
Posts: 493
Location: India
Concentration: Entrepreneurship, Strategy
GMAT 1: 710 Q48 V40
WE: Project Management (Manufacturing)
Followers: 13

Kudos [?]: 31 [4] , given: 71

If the product of all the unique positive divisors of n, a p [#permalink] New post 02 Feb 2011, 11:03
4
This post received
KUDOS
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

21% (03:05) correct 78% (01:52) wrong based on 540 sessions
If the product of all the unique positive divisors of n, a positive integer which is not a perfect cube, is n^2, then the product of all the unique positive divisors of n^2 is

(A) n^3
(B) n^4
(C) n^6
(D) n^8
(E) n^9
[Reveal] Spoiler: OA

_________________

My GMAT debrief: from-620-to-710-my-gmat-journey-114437.html


Last edited by gmatpapa on 03 Feb 2011, 01:05, edited 1 time in total.
7 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 17307
Followers: 2873

Kudos [?]: 18374 [7] , given: 2348

GMAT Tests User CAT Tests
Re: Divisibility Question [#permalink] New post 02 Feb 2011, 11:46
7
This post received
KUDOS
Expert's post
gmatpapa wrote:
If the product of all the unique positive divisors of n, a positive integer which is not a perfect cube, is n^2, then the product of all the unique positive divisors of n^2 is
(A) n^3
(B) n^4
(C) n^6
(D) n^8
(E) n^9


All positive integers n which equal to n=p_1*p_2, where p_1 and p_2 are distinct primes satisfy the condition in the stem. Because the factors of n in this case would be: 1, p_1, p_2, and n itself, so the product of the factors will be 1*(p_1*p_2)*n=n^2.

(Note that if n=p^3 where p is a prime number also satisfies this condition as the factors of n in this case would be 1, p, p^2 and n itself, so the product of the factors will be 1*(p*p^2)*n=p^3*n=n^2, but we are told that n is not a perfect cube, so this case is out, as well as the case n=1.)

For example if n=6=2*3 --> the product of all the unique positive divisors of 6 will be: 1*2*3*6=6^2;
Or if n=10=2*5 --> the product of all the unique positive divisors of 10 will be: 1*2*5*10=10^2;

Now, take n=10 --> n^2=100 --> the product of all the unique positive divisors of 100 is: 1*2*4*5*10*20*25*50*100=(2*50)*(4*25)*(5*20)*10*100=10^2*10^2*10^2*10*10^2=10^9 (you can do this with formula to get (p_1)^9*(p_2)^9=n^9 but think this way is quicker).

Answer: E.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

11 KUDOS received
Manager
Manager
Joined: 11 Sep 2009
Posts: 129
Followers: 4

Kudos [?]: 142 [11] , given: 6

GMAT Tests User
Re: Divisibility Question [#permalink] New post 02 Feb 2011, 19:57
11
This post received
KUDOS
I agree with E: n^9.

Another way to look at it (using the same methodology outlined by Bunuel in his post):

We know that the product of all the divisors of n is equivalent to n^2. Therefore, n = x * y, where x and y are both prime positive and distinct integers (as outlined by the conditions set in the original question, and explained in the previous post).

We need to determine, the product of all the divisors of n^2, which we know to be equivalent to:

n^2 = n * n

n^2 = (x * y)(x * y)

Since both x and y are known to be prime numbers, the divisors of n^2 are therefore:

1, x, y, xy, x^2, y^2, x^2y, xy^2, and x^2y^2

The product of these divisors is therefore:

(1)(x)(y)(xy)(x^2)(y^2)(x^2y)(xy^2)(x^2y^2)

= x^9y^9

= n^9
2 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4177
Location: Pune, India
Followers: 894

Kudos [?]: 3787 [2] , given: 148

Re: Divisibility Question [#permalink] New post 02 Feb 2011, 20:09
2
This post received
KUDOS
Expert's post
gmatpapa wrote:
If the product of all the unique positive divisors of n, a positive integer which is not a perfect cube, is n^2, then the product of all the unique positive divisors of n^2 is
(A) n^3
(B) n^4
(C) n^6
(D) n^8
(E) n^9


My take on it is following:
If I can find one example such that 'product of all the unique positive divisors of n is n^2' (n not a perfect cube), I will just find the product of all unique positive divisors of n^2 and get the answer.

So what do the factors of n listed out look like?
1 .. .. .. .. n

1 and n are definitely factors of n so out of the required product,n^2, one n is already accounted for. I need only one more n and hence I must have 2 factors in the middle.
e.g.
1, 2, 3, 6 (6 is not a perfect cube but product of all factors of 6 is 36. So I got the one example I was looking for.)
36 = 2^2*3^2
So it will have 9 factors. Four pairs of factors equidistant from the beg and end will give the product 36 (i.e. n^2). Factor in the middle will be 6 i.e. n.
Hence product of all factors of 36 will be n^8*n = n^9.

If there are doubts in this theory, check out http://www.veritasprep.com/blog/2010/12/quarter-wit-quarter-wisdom-factors-of-perfect-squares/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Senior Manager
Senior Manager
Status: ready to boMBArd
Joined: 31 Oct 2010
Posts: 493
Location: India
Concentration: Entrepreneurship, Strategy
GMAT 1: 710 Q48 V40
WE: Project Management (Manufacturing)
Followers: 13

Kudos [?]: 31 [0], given: 71

Re: Divisibility Question [#permalink] New post 03 Feb 2011, 01:04
I think I misunderstood the question..

I picked the example of n=10=2^1*5^1. Unique divisors: 2, 5, 10 and their product is 100 which is n^2. Unique divisors of 100 are 1,2,4,5,10,20,50,100, the product of these equal 2^9*5^9. The mistake I committed was i divided this product by n, landing at (2^9*5^9)/(2^1*5^1)= 2^8*5^8= n^8, and then losing my sleep over the wrong answer lol GMAT mode, what can I say!

Thanks for your explanation guys!
_________________

My GMAT debrief: from-620-to-710-my-gmat-journey-114437.html

Moderator
Moderator
User avatar
Joined: 01 Sep 2010
Posts: 2175
Followers: 172

Kudos [?]: 1517 [0], given: 610

GMAT Tests User Premium Member
Re: Divisibility Question [#permalink] New post 11 Mar 2012, 03:46
Expert's post
Bunuel wrote:
gmatpapa wrote:
If the product of all the unique positive divisors of n, a positive integer which is not a perfect cube, is n^2, then the product of all the unique positive divisors of n^2 is
(A) n^3
(B) n^4
(C) n^6
(D) n^8
(E) n^9


All positive integers n which equal to n=p_1*p_2, where p_1 and p_2 are distinct primes satisfy the condition in the stem. Because the factors of n in this case would be: 1, p_1, p_2, and n itself, so the product of the factors will be 1*(p_1*p_2)*n=n^2.

(Note that if n=p^3 where p is a prime number also satisfies this condition as the factors of n in this case would be 1, p, p^2 and n itself, so the product of the factors will be 1*(p*p^2)*n=p^3*n=n^2, but we are told that n is not a perfect cube, so this case is out, as well as the case n=1.)

For example if n=6=2*3 --> the product of all the unique positive divisors of 6 will be: 1*2*3*6=6^2;
Or if n=10=2*5 --> the product of all the unique positive divisors of 10 will be: 1*2*5*10=10^2;

Now, take n=10 --> n^2=100 --> the product of all the unique positive divisors of 100 is: 1*2*4*5*10*20*25*50*100=(2*50)*(4*25)*(5*20)*10*100=10^2*10^2*10^2*10*10^2=10^9 (you can do this with formula to get (p_1)^9*(p_2)^9=n^9 but think this way is quicker).

Answer: E.


For me when we deal with this kind of problems is important to find a clue and here is that the number n is NOT a perfect cube.

So, a NOT perfect CUBE is a number when the exponents of our number are not 3 or multiple of 3 (the same reasoning is for perfect square contains only even powers of primes, otherwise is not a perfect square).

In this scenario at glance I purge B C and D. From A and E are the only answer where the LENGTH of the total number of primes in the prime box of integer N, so n^3 is not possible. The answer is E

This is correct?? I 'd like an opinion o explenation about that. Is unclear for me if this process is totally correct or no...

Thanks.
_________________

KUDOS is the good manner to help the entire community.

1 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4177
Location: Pune, India
Followers: 894

Kudos [?]: 3787 [1] , given: 148

Re: Divisibility Question [#permalink] New post 11 Mar 2012, 20:06
1
This post received
KUDOS
Expert's post
carcass wrote:
For me when we deal with this kind of problems is important to find a clue and here is that the number n is NOT a perfect cube.

So, a NOT perfect CUBE is a number when the exponents of our number are not 3 or multiple of 3 (the same reasoning is for perfect square contains only even powers of primes, otherwise is not a perfect square).

In this scenario at glance I purge B C and D. From A and E are the only answer where the LENGTH of the total number of primes in the prime box of integer N, so n^3 is not possible. The answer is E

This is correct?? I 'd like an opinion o explenation about that. Is unclear for me if this process is totally correct or no...

Thanks.


I am unable to understand your reasoning for rejecting (B), (C) and (D). Since the product of all factors is n^2 and factors equidistant from the beginning and end multiply to give n, we know we have 2 pairs of factors. 1, a, b, n (1*n gives n and a*b gives n)
The reason they gave that n is not a perfect cube is to reject the scenario where b is just the square of a e.g. to reject this scenario: 1, 2, 4, 8
So basically, the question is telling you that a and b are primes and n is the product of a and b e.g. 1, 2, 5, 10
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Moderator
Moderator
User avatar
Joined: 01 Sep 2010
Posts: 2175
Followers: 172

Kudos [?]: 1517 [0], given: 610

GMAT Tests User Premium Member
Re: Divisibility Question [#permalink] New post 12 Mar 2012, 04:21
Expert's post
VeritasPrepKarishma wrote:
carcass wrote:
For me when we deal with this kind of problems is important to find a clue and here is that the number n is NOT a perfect cube.

So, a NOT perfect CUBE is a number when the exponents of our number are not 3 or multiple of 3 (the same reasoning is for perfect square contains only even powers of primes, otherwise is not a perfect square).

In this scenario at glance I purge B C and D. From A and E are the only answer where the LENGTH of the total number of primes in the prime box of integer N, so n^3 is not possible. The answer is E

This is correct?? I 'd like an opinion o explenation about that. Is unclear for me if this process is totally correct or no...

Thanks.


I am unable to understand your reasoning for rejecting (B), (C) and (D). Since the product of all factors is n^2 and factors equidistant from the beginning and end multiply to give n, we know we have 2 pairs of factors. 1, a, b, n (1*n gives n and a*b gives n)
The reason they gave that n is not a perfect cube is to reject the scenario where b is just the square of a e.g. to reject this scenario: 1, 2, 4, 8
So basically, the question is telling you that a and b are primes and n is the product of a and b e.g. 1, 2, 5, 10


Thanks :) Infact I said that is unclear if my reasoning was right or wrong.

Now is more clear thanks to you .
_________________

KUDOS is the good manner to help the entire community.

1 KUDOS received
Intern
Intern
User avatar
Joined: 02 May 2013
Posts: 26
Concentration: International Business, Technology
WE: Engineering (Aerospace and Defense)
Followers: 1

Kudos [?]: 15 [1] , given: 16

Re: Divisibility Question [#permalink] New post 16 May 2013, 21:00
1
This post received
KUDOS
VeritasPrepKarishma wrote:
gmatpapa wrote:
If the product of all the unique positive divisors of n, a positive integer which is not a perfect cube, is n^2, then the product of all the unique positive divisors of n^2 is
(A) n^3
(B) n^4
(C) n^6
(D) n^8
(E) n^9


My take on it is following:
If I can find one example such that 'product of all the unique positive divisors of n is n^2' (n not a perfect cube), I will just find the product of all unique positive divisors of n^2 and get the answer.

So what do the factors of n listed out look like?
1 .. .. .. .. n

1 and n are definitely factors of n so out of the required product,n^2, one n is already accounted for. I need only one more n and hence I must have 2 factors in the middle.
e.g.
1, 2, 3, 6 (6 is not a perfect cube but product of all factors of 6 is 36. So I got the one example I was looking for.)
36 = 2^2*3^2
So it will have 9 factors. Four pairs of factors equidistant from the beg and end will give the product 36 (i.e. n^2). Factor in the middle will be 6 i.e. n.
Hence product of all factors of 36 will be n^8*n = n^9.

If there are doubts in this theory, check out http://www.veritasprep.com/blog/2010/12/quarter-wit-quarter-wisdom-factors-of-perfect-squares/





Generalized approach

product of factors of n = n^2 implies n^(f/2) = n^2 where f is no. of factors
here f= 4. So obviously along with 1 and n, we do have 2 more factors. let say x,y (prime factors)
i.e, 1*x*y*n =n^2
==> x*y=n
n^2 = x^2*y^2
Now the product of all factors of n^2 = (n^2)^(g/2) where g = no. of factors of n^2
we know g=(2+1)(2+1) =9
As n^2 is a perfect square , product of factors of n^2 will be n^9
Intern
Intern
Joined: 12 May 2013
Posts: 4
Followers: 0

Kudos [?]: 1 [0], given: 3

Re: Divisibility Question [#permalink] New post 08 Jun 2013, 10:58
VeritasPrepKarishma wrote:
gmatpapa wrote:
1 and n are definitely factors of n so out of the required product,n^2, one n is already accounted for. I need only one more n and hence I must have 2 factors in the middle.
e.g.
1, 2, 3, 6 (6 is not a perfect cube but product of all factors of 6 is 36. So I got the one example I was looking for.)
36 = 2^2*3^2


Can you please explain how you reached to the conclusion that you must have 2 factors in the middle ??
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4177
Location: Pune, India
Followers: 894

Kudos [?]: 3787 [0], given: 148

Re: Divisibility Question [#permalink] New post 09 Jun 2013, 21:55
Expert's post
atilarora wrote:
VeritasPrepKarishma wrote:
gmatpapa wrote:
1 and n are definitely factors of n so out of the required product,n^2, one n is already accounted for. I need only one more n and hence I must have 2 factors in the middle.
e.g.
1, 2, 3, 6 (6 is not a perfect cube but product of all factors of 6 is 36. So I got the one example I was looking for.)
36 = 2^2*3^2


Can you please explain how you reached to the conclusion that you must have 2 factors in the middle ??


When you write down all factors of n in increasing order, you notice that factors equidistant from the two ends have the product n.

e.g. n = 10
1, 2, 5, 10
1*10 = 10
2*5 = 10

n = 24
1, 2, 3, 4, 6, 8, 12, 24
1*24 = 24
2*12 = 24
3*8 = 24
4*6 = 24

This theory is also discussed here: http://www.veritasprep.com/blog/2010/12 ... ly-number/

So if the product of all factors is n^2, you will get one n from 1*n. Then there must be two factors in the middle which multiply to give another n. (Note that this is not valid for perfect squares. Check out this post: http://www.veritasprep.com/blog/2010/12 ... t-squares/)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Founder
Founder
User avatar
Status: In Mexico.... NO PM's :-)
Affiliations: UA-1K, SPG-G, HH-D
Joined: 04 Dec 2002
Posts: 11818
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
WE: Information Technology (Hospitality and Tourism)
Followers: 1983

Kudos [?]: 6411 [0], given: 3565

GMAT ToolKit User Premium Member CAT Tests
Re: If the product of all the unique positive divisors of n, a p [#permalink] New post 31 Jul 2013, 05:56
Expert's post
This is the Hard and Tricky question for today, which means it is one of the top 100 hardest PS Questions on GMAT Club.
_________________

Founder of GMAT Club

Just starting out with GMAT? Start here... | Want to know your GMAT Score? Try GMAT Score Estimator
Need GMAT Book Recommendations? Best GMAT Books

Co-author of the GMAT Club tests

Have a blog? Feature it on GMAT Club!

Manager
Manager
Joined: 30 May 2013
Posts: 153
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82
Followers: 0

Kudos [?]: 20 [0], given: 65

GMAT ToolKit User
Re: Divisibility Question [#permalink] New post 20 Aug 2013, 23:26
ButwhY wrote:
VeritasPrepKarishma wrote:
gmatpapa wrote:
If the product of all the unique positive divisors of n, a positive integer which is not a perfect cube, is n^2, then the product of all the unique positive divisors of n^2 is
(A) n^3
(B) n^4
(C) n^6
(D) n^8
(E) n^9


My take on it is following:
If I can find one example such that 'product of all the unique positive divisors of n is n^2' (n not a perfect cube), I will just find the product of all unique positive divisors of n^2 and get the answer.

So what do the factors of n listed out look like?
1 .. .. .. .. n

1 and n are definitely factors of n so out of the required product,n^2, one n is already accounted for. I need only one more n and hence I must have 2 factors in the middle.
e.g.
1, 2, 3, 6 (6 is not a perfect cube but product of all factors of 6 is 36. So I got the one example I was looking for.)
36 = 2^2*3^2
So it will have 9 factors. Four pairs of factors equidistant from the beg and end will give the product 36 (i.e. n^2). Factor in the middle will be 6 i.e. n.
Hence product of all factors of 36 will be n^8*n = n^9.

If there are doubts in this theory, check out http://www.veritasprep.com/blog/2010/12/quarter-wit-quarter-wisdom-factors-of-perfect-squares/





Generalized approach

product of factors of n = n^2 implies n^(f/2) = n^2 where f is no. of factors
here f= 4. So obviously along with 1 and n, we do have 2 more factors. let say x,y (prime factors)
i.e, 1*x*y*n =n^2
==> x*y=n
n^2 = x^2*y^2
Now the product of all factors of n^2 = (n^2)^(g/2) where g = no. of factors of n^2
we know g=(2+1)(2+1) =9
As n^2 is a perfect square , product of factors of n^2 will be n^9

Hi,

i have a doubt in this
product of all factors of n^2 = (n^2)^(g/2) where g = no. of factors of n^2

g=9
Prod of factors of n^2 = (n^2)^3=n^6

how come it is n^9

please explain.

Regards,
Rrsnathan.
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4177
Location: Pune, India
Followers: 894

Kudos [?]: 3787 [0], given: 148

Re: Divisibility Question [#permalink] New post 21 Aug 2013, 02:39
Expert's post
rrsnathan wrote:
g=9
Prod of factors of n^2 = (n^2)^3=n^6

how come it is n^9

please explain.

Regards,
Rrsnathan.


Prod of factors of n^2 = (n^2)^(9/2) (Your error - 9/2 is not 3)
= n^9
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Manager
Manager
Joined: 30 May 2013
Posts: 153
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82
Followers: 0

Kudos [?]: 20 [0], given: 65

GMAT ToolKit User
Re: Divisibility Question [#permalink] New post 21 Aug 2013, 02:46
VeritasPrepKarishma wrote:
rrsnathan wrote:
g=9
Prod of factors of n^2 = (n^2)^3=n^6

how come it is n^9

please explain.

Regards,
Rrsnathan.


Prod of factors of n^2 = (n^2)^(9/2) (Your error - 9/2 is not 3)
= n^9


Hi,

Thanks Karishma.

:oops: Sorry for mis calculation.

Regards,
Rrsnathan.
Senior Manager
Senior Manager
Joined: 17 Apr 2013
Posts: 414
Concentration: Entrepreneurship, Leadership
Schools: HBS '16
GMAT Date: 11-30-2013
GPA: 3.3
Followers: 1

Kudos [?]: 3 [0], given: 173

CAT Tests
Re: Divisibility Question [#permalink] New post 14 Sep 2013, 02:04
VeritasPrepKarishma wrote:
gmatpapa wrote:
If the product of all the unique positive divisors of n, a positive integer which is not a perfect cube, is n^2, then the product of all the unique positive divisors of n^2 is
(A) n^3
(B) n^4
(C) n^6
(D) n^8
(E) n^9


My take on it is following:
If I can find one example such that 'product of all the unique positive divisors of n is n^2' (n not a perfect cube), I will just find the product of all unique positive divisors of n^2 and get the answer.

So what do the factors of n listed out look like?
1 .. .. .. .. n

1 and n are definitely factors of n so out of the required product,n^2, one n is already accounted for. I need only one more n and hence I must have 2 factors in the middle.
e.g.
1, 2, 3, 6 (6 is not a perfect cube but product of all factors of 6 is 36. So I got the one example I was looking for.)
36 = 2^2*3^2
So it will have 9 factors. Four pairs of factors equidistant from the beg and end will give the product 36 (i.e. n^2). Factor in the middle will be 6 i.e. n.
Hence product of all factors of 36 will be n^8*n = n^9.

If there are doubts in this theory, check out http://www.veritasprep.com/blog/2010/12/quarter-wit-quarter-wisdom-factors-of-perfect-squares/



Karishma You are the best.

I have no doubt in theory, still couldn't understand this-
"So it will have 9 factors. Four pairs of factors equidistant from the beg and end will give the product 36 (i.e. n^2). Factor in the middle will be 6 i.e. n.
Hence product of all factors of 36 will be n^8*n = n^9.
"
_________________

Like my post Send me a Kudos :) It is a Good manner.

Intern
Intern
Joined: 15 May 2013
Posts: 8
Concentration: Entrepreneurship, Technology
GMAT Date: 08-12-2013
Followers: 0

Kudos [?]: 0 [0], given: 13

Re: If the product of all the unique positive divisors of n, a p [#permalink] New post 05 Nov 2013, 07:39
gmatpapa wrote:
If the product of all the unique positive divisors of n, a positive integer which is not a perfect cube, is n^2, then the product of all the unique positive divisors of n^2 is

(A) n^3
(B) n^4
(C) n^6
(D) n^8
(E) n^9


Product of factors of n= n^(f/2) (f=number of factors)
=>n^(f/2) =n^2
=>f=4
=>n=p1^1*p2^1 (this is the way we can get f=4. If n were a cube then n could have been p1^3 which could have resulted f=4 as well.)

Now question asks product of divisors of n^2.
=>n^2=p1^2*p2^2
=>total factors for n^2=(2+1)(2+1)=9
=>Product of divisors of n^2 = (n^2)(f/2) =(n^2)(9/2) =n^9

Do correct me if I missed something.
Intern
Intern
Joined: 15 Jan 2014
Posts: 3
Followers: 0

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

Re: If the product of all the unique positive divisors of n, a p [#permalink] New post 15 Jan 2014, 07:22
the product of factors of a number 'n' would be (n)^(no.of factors/2).given the product of all the unique divisors is n^2.so we can conclude no.of factors for the given number is 4.such numbers would be of the form (p1)^1*(p2)^1 where p1 and p2 are prime numbers.
n=(p1)^1*(p2)^1
n^2=(p1)^2*(p2)^2 .
product of all the factors of n^2 would be (n^2)^(no.of factors of n^2/2)=(n^2)^(3*3/2)=n^9
_________________

TWICKKKK

Re: If the product of all the unique positive divisors of n, a p   [#permalink] 15 Jan 2014, 07:22
    Similar topics Author Replies Last post
Similar
Topics:
Popular new posts If n is a positive integer and the product of all the Antmavel 19 30 Jan 2005, 17:29
New posts If n is a positive integer and the product of all the TOUGH GUY 1 10 Dec 2005, 15:16
New posts If n is a positive integer and the product of all the yach 3 06 Mar 2006, 15:11
New posts Experts publish their posts in the topic If n is a positive integer and the product of all the bsjames2 5 05 Jan 2008, 13:30
New posts If n is a positive integer and the product of all the tarek99 3 31 Mar 2008, 05:43
Display posts from previous: Sort by

If the product of all the unique positive divisors of n, a p

  Question banks Downloads My Bookmarks Reviews Important topics  


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®.