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

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If the product of all the unique positive divisors of n, a p  [#permalink]

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Updated on: 03 Feb 2011, 02:05
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23% (02:51) correct 77% (02:14) wrong based on 3004 sessions

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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) $$n^9$$

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Originally posted by gmatpapa on 02 Feb 2011, 12:03.
Last edited by gmatpapa on 03 Feb 2011, 02:05, edited 1 time in total.
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Re: Divisibility Question  [#permalink]

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02 Feb 2011, 12:46
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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).

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Re: Divisibility Question  [#permalink]

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02 Feb 2011, 20:57
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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:

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

$$= x^9y^9$$

$$= n^9$$
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Re: Divisibility Question  [#permalink]

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02 Feb 2011, 21:09
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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/
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Re: Divisibility Question  [#permalink]

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03 Feb 2011, 02:04
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!
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Re: Divisibility Question  [#permalink]

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11 Mar 2012, 04:46
1
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).

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.
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Re: Divisibility Question  [#permalink]

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11 Mar 2012, 21:06
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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
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Re: Divisibility Question  [#permalink]

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12 Mar 2012, 05:21
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 .
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Re: Divisibility Question  [#permalink]

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16 May 2013, 22:00
1
1
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
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Re: Divisibility Question  [#permalink]

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08 Jun 2013, 11:58
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 ??
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Re: Divisibility Question  [#permalink]

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09 Jun 2013, 22:55
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/)
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Re: If the product of all the unique positive divisors of n, a p  [#permalink]

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31 Jul 2013, 06:56
This is the Hard and Tricky question for today, which means it is one of the top 100 hardest PS Questions on GMAT Club.
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Re: Divisibility Question  [#permalink]

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21 Aug 2013, 00:26
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

Regards,
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Re: Divisibility Question  [#permalink]

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21 Aug 2013, 03:39
rrsnathan wrote:
g=9
Prod of factors of n^2 = (n^2)^3=n^6

how come it is n^9

Regards,
Rrsnathan.

Prod of factors of n^2 = (n^2)^(9/2) (Your error - 9/2 is not 3)
= n^9
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Re: Divisibility Question  [#permalink]

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21 Aug 2013, 03:46
VeritasPrepKarishma wrote:
rrsnathan wrote:
g=9
Prod of factors of n^2 = (n^2)^3=n^6

how come it is n^9

Regards,
Rrsnathan.

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

Hi,

Thanks Karishma.

Sorry for mis calculation.

Regards,
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Re: Divisibility Question  [#permalink]

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14 Sep 2013, 03:04
1
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.
"
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Re: If the product of all the unique positive divisors of n, a p  [#permalink]

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05 Nov 2013, 08:39
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.
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Re: If the product of all the unique positive divisors of n, a p  [#permalink]

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15 Jan 2014, 08:22
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
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Re: If the product of all the unique positive divisors of n, a p  [#permalink]

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29 Apr 2014, 23:25
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) $$n^9$$

Could solve this in under 2min by below approach
Easy one, if you assume a proper number .Lets suppose the number is n= 6. The number is not a perfect cube.
n^2 = 36
Find the divisors of 36 :- 2,3,4,6,9,12,18,36 :- Now multiply them all [You don't need to multiply them. Just re-arrange them in multiples of 6] .
2*3*4*6*9*12*18*36 = 6*6*6*6*6*6*6*6*6 = 6^9 or n^9 .
"E"
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Re: If the product of all the unique positive divisors of n, a p  [#permalink]

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01 Aug 2014, 19:00
AKProdigy87 wrote:
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:

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

$$= x^9y^9$$

$$= n^9$$

It saves more time than trying a real number ( in my opinion)
your exp is so helpful. + 1 kudos
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Re: If the product of all the unique positive divisors of n, a p &nbs [#permalink] 01 Aug 2014, 19:00

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

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