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

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Manager
Joined: 27 Nov 2014
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Re: If the product of all the unique positive divisors of n, a p [#permalink]

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30 Nov 2014, 02:57
hi karishma VeritasPrepKarishma

a shortcut i have learned for counting factors is if n^2 = 36 then factors are 2^2 and 3^2 so total unique divisors or factors ( i can say divisors as factors ??) are (power of the factors +1 and product of them) i.e powers and product (2+1)(2+1) = 9 this is how i reach the answer.

IT is applicable if n= 100 also

but it doesn't apply when i take n^2 = 49 ??

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

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30 Nov 2014, 22:17
smartyguy wrote:
hi karishma VeritasPrepKarishma

a shortcut i have learned for counting factors is if n^2 = 36 then factors are 2^2 and 3^2 so total unique divisors or factors ( i can say divisors as factors ??) are (power of the factors +1 and product of them) i.e powers and product (2+1)(2+1) = 9 this is how i reach the answer.

IT is applicable if n= 100 also

but it doesn't apply when i take n^2 = 49 ??

Regards
SG

You can use it for finding the total number of factors of any positive integer.

n^2 = 49

n^2 = 7^2
The total number of factors of n^2 is (2+1) = 3
The factors are 1, 7 and 49.

Of course, if you want to instead find the total number of factors of n, you will do
n = 7 (the positive value of n)
n = 7^1
Total number of factors = (1+1) = 2
The factors are 1 and 7.

Note here in this question, you need the product of all unique divisors, not the number of unique divisors.

When n = 6 = 2*3, the total number of factors is (1+1)(1+1) = 4
The factors are 1, 2, 3, 6 and the product of all factors = 1*2*3*6 = 6^2 = n^2
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 27 Nov 2014 Posts: 61 Re: If the product of all the unique positive divisors of n, a p [#permalink] Show Tags 01 Dec 2014, 03:57 ohh now i got it here if we take n = 7 than the product of factors is not constituting n^2 i.e 1*7 is not equal to 49 = N^2 gosh its a tricky question thanks Karishma Regards SG Manager Joined: 14 Oct 2014 Posts: 53 Re: If the product of all the unique positive divisors of n, a p [#permalink] Show Tags 18 Mar 2015, 17:57 Take n=6, factors are 12,3,6. n^2= 36. factors of 36 are 1,2,3,4,6,9,12,18,36. and 1*2*3*4*6*9*12*18*36 is divisible by 6^9. so n^9 is answer. Manager Joined: 22 Aug 2014 Posts: 176 Re: If the product of all the unique positive divisors of n, a p [#permalink] Show Tags 03 May 2015, 01:22 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: $$(1)(x)(y)(xy)(x^2)(y^2)(x^2y)(xy^2)(x^2y^2)$$ $$= x^9y^9$$ $$= n^9$$ great explanation.no doubt at all thank you Intern Joined: 30 Dec 2014 Posts: 1 Re: If the product of all the unique positive divisors of n, a p [#permalink] Show Tags 10 May 2015, 00:06 gmatpapa wrote: 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! wht if I take n =12, then uniqe divisor will be 2,6,3,4,12 = 12^3 ,,how to take this Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8058 Location: Pune, India Re: If the product of all the unique positive divisors of n, a p [#permalink] Show Tags 10 May 2015, 21:42 1 This post received KUDOS Expert's post vishthree wrote: gmatpapa wrote: 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! wht if I take n =12, then uniqe divisor will be 2,6,3,4,12 = 12^3 ,,how to take this You are given that the product of all unique factors of n must be n^2. So n cannot be 12 since the product of all unique factors of 12 is n^3 (12^3), not n^2. The product of all factors of a number will be n^2 when it has exactly 4 factors e.g. n = 6 or 10 or 21 (whenever n is a product of two prime numbers). I suggest you to check out these two posts: http://www.veritasprep.com/blog/2010/12 ... ly-number/ http://www.veritasprep.com/blog/2010/12 ... t-squares/ The logic will make a lot more sense then. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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27 Jun 2015, 11:11
Quote:
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/)

How do you conclude that n is not a perfect square? The question only states that n is not a perfect cube...
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Re: If the product of all the unique positive divisors of n, a p [#permalink]

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28 Jun 2015, 22:40
avgroh wrote:
Quote:
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/)

How do you conclude that n is not a perfect square? The question only states that n is not a perfect cube...

Because factors of perfect squares could be of the form:

1, sqrt(n), n

e.g.
Factors of 9: 1, 3, 9
so they may not have two factors in the middle.

Suggest you to check out the two posts I have mentioned above.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 23 Nov 2014 Posts: 58 Location: India GMAT 1: 730 Q49 V40 GPA: 3.14 WE: Sales (Consumer Products) Re: If the product of all the unique positive divisors of n, a p [#permalink] Show Tags 29 Jun 2015, 09:21 VeritasPrepKarishma wrote: avgroh wrote: Quote: 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/) How do you conclude that n is not a perfect square? The question only states that n is not a perfect cube... Because factors of perfect squares could be of the form: 1, sqrt(n), n e.g. Factors of 9: 1, 3, 9 so they may not have two factors in the middle. Suggest you to check out the two posts I have mentioned above. Hey, Thanks, I get it now. I wasn't thinking straight. Manager Joined: 25 Nov 2014 Posts: 160 WE: Engineering (Manufacturing) Re: If the product of all the unique positive divisors of n, a p [#permalink] Show Tags 08 Aug 2015, 02:53 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 Interesting Math Fact: For all perfect cubes with the given product = n^2, the asked product = n^7 Eg. take n = 8 Manager Joined: 25 Nov 2014 Posts: 160 WE: Engineering (Manufacturing) Re: If the product of all the unique positive divisors of n, a p [#permalink] Show Tags 08 Aug 2015, 02:56 one doubt though: while solving this ques i also plugged in n = 10 directly. i wanna ask the great Bunuel as to how you reached the generalization you showed in your solution. i wanna know because this is my greatest prob in Number Properties questions - Generalizations of constrained num sets. Please please please explain Intern Joined: 02 Dec 2015 Posts: 12 If the product of all the unique positive divisors of n, a p [#permalink] Show Tags 30 Jun 2016, 09:23 Hi Why cant it be solved as follows It is given that product of unique divisors of n is $$n^2$$ And we got to do the same for n^2 So $$n^2$$ =$$n * n$$ and since we have the result of n = $$n^2$$ hence for getting the result of $$n^2$$ = $$n^2$$*$$n^2$$ = $$n^4$$ I know its not true but i couldnt figure why Regards 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. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8058 Location: Pune, India Re: If the product of all the unique positive divisors of n, a p [#permalink] Show Tags 30 Jun 2016, 22:46 2 This post received KUDOS Expert's post ppb1487 wrote: Hi Why cant it be solved as follows It is given that product of unique divisors of n is $$n^2$$ And we got to do the same for n^2 So $$n^2$$ =$$n * n$$ and since we have the result of n = $$n^2$$ hence for getting the result of $$n^2$$ = $$n^2$$*$$n^2$$ = $$n^4$$ I know its not true but i couldnt figure why The product of factors depends on the number of factors. The number of factors depends on the exponent of prime factors. When the exponent of prime factors is multiplied by 2, the number of factors doesn't multiply by 2. It increase by much more because: Number of factors = (a+1)(b+1)... etc If a =1 , b= 1, Number of factors = (a+1)(b+1) = 2*2 = 4 If a = 2, b = 2, Number of factors = (a+1)(b+1) = 3*3 = 9 Whenever you feel stuck with such concept issues, try out some numbers to get clarity. n = 6 All factors of n: 1, 2, 3, 6 Product of all factors: 1*2*3*6 = 36 = 6^2 = n^2 36 = 2^2 * 3^2 How many factors will it have? (2+1)*(2+1) = 9 All factors of 36: 1, 2, 3, 4, 6, ... , 36 When you find their product, each pair equidistant from the extremes will give 36 (n^2). There will be 4 such pairs to get n^8 and an n so it all, you will get n^9. Check out this post: http://www.veritasprep.com/blog/2015/08 ... questions/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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15 Jul 2016, 23:51
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).

Can someone please explain why we nave taken +ve integer n of the form n= p1 x p2?

I'm unable to comprehend as to which constraint given in the question bounds us to imply that n is a product of only 2 prime numbers (without any power)..
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Re: If the product of all the unique positive divisors of n, a p [#permalink]

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17 Jul 2016, 22:16
1
KUDOS
Expert's post
sudhirgupta93 wrote:
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).

Can someone please explain why we nave taken +ve integer n of the form n= p1 x p2?

I'm unable to comprehend as to which constraint given in the question bounds us to imply that n is a product of only 2 prime numbers (without any power)..

"the product of all the unique positive divisors of n, a positive integer which is not a perfect cube, is n^2"

Had n been a prime number, it would have only 2 factors: 1 and n
Product of all factors = 1*n = n

If product of all factors is instead n^2, it means that other than 1 and n, it has 2 more factors which also multiply to give n.
All factors: 1, p1, p2, n (where p1 and p2 are two prime numbers)
Product of all factors = 1*p1*p2*n = n^2

What if p1 and p2 were not prime? e.g. 1, 4, 6, 24
Here, there are other factors too: 2, 3, 8 etc. So the product of all factors would be much more than n^2.

If there have to be only 4 factors, they must be 1, p1, p2 and n where p1 and p2 must be prime factors.
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If the product of all the unique positive divisors of n, a p [#permalink]

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03 Nov 2016, 07:11
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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) $$n^9$$

Let’s use some abstract approach:
We know that product of all factors of a number n (not a perfect square) can be calculated as
$$n^\frac{x}{2}$$. Where$$x$$ is the total number of factors of n.

Here we have$$\frac{x}{2} =2$$. =>$$x=4$$.

Our number has 4 divisors in total. That can be achieved in 2 cases: $$n=p^3$$or $$n=p*q$$, where p and q are prime numbers. The question says that n is not a perfect cube, so we can discard the first choice.

Now we square our n
$$n^2=p^2*q^2$$
Total number of factors (2+1)*(2+1)=9

Now let’s find the product of all unique divisors for a perfect square.
$$Q^\frac{(N-1)}{2}*\sqrt{Q}$$ , where Q is our new number$$(n^2)$$, N – total number of factors of Q

Plugging in:
($$(p*q)^2)^\frac{(9-1)}{2}*p*q=(p*q)^8*p*q=(p*q)^9=n^9$$
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