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If the product of all the unique positive divisors of n, a positive in
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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 Bunuel on 11 Jan 2019, 02:14, edited 2 times in total.
Renamed the topic and edited the question.




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Re: If the product of all the unique positive divisors of n, a positive in
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02 Feb 2011, 12:46
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.
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Re: If the product of all the unique positive divisors of n, a positive in
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02 Feb 2011, 20:57
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: If the product of all the unique positive divisors of n, a positive in
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02 Feb 2011, 21:09
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/quarterwitquarterwisdomfactorsofperfectsquares/
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Re: If the product of all the unique positive divisors of n, a positive in
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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: If the product of all the unique positive divisors of n, a positive in
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11 Mar 2012, 04:46
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.
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Re: If the product of all the unique positive divisors of n, a positive in
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11 Mar 2012, 21:06
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: If the product of all the unique positive divisors of n, a positive in
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16 May 2013, 22:00
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/quarterwitquarterwisdomfactorsofperfectsquares/ 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: If the product of all the unique positive divisors of n, a positive in
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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: If the product of all the unique positive divisors of n, a positive in
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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 ... lynumber/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 ... tsquares/)
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Re: If the product of all the unique positive divisors of n, a positive in
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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/quarterwitquarterwisdomfactorsofperfectsquares/ 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^9Hi, 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.



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Re: If the product of all the unique positive divisors of n, a positive in
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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
please explain.
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: If the product of all the unique positive divisors of n, a positive in
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30 Nov 2014, 02:57
hi karishma VeritasPrepKarishmai was going through your explanation. it was really helpful 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 ??can you please help what am i missing ? Regards SG



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Re: If the product of all the unique positive divisors of n, a positive in
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30 Nov 2014, 22:17
smartyguy wrote: hi karishma VeritasPrepKarishmai was going through your explanation. it was really helpful 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 ??can you please help what am i missing ? 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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Re: If the product of all the unique positive divisors of n, a positive in
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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



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Re: If the product of all the unique positive divisors of n, a positive in
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10 May 2015, 21:42
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 ... lynumber/http://www.veritasprep.com/blog/2010/12 ... tsquares/The logic will make a lot more sense then.
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Re: If the product of all the unique positive divisors of n, a positive in
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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 ... lynumber/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 ... tsquares/) 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 positive in
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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 ... lynumber/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 ... tsquares/) 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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Re: If the product of all the unique positive divisors of n, a positive in
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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.



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Re: If the product of all the unique positive divisors of n, a positive in
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30 Jun 2016, 22:46
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/
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Re: If the product of all the unique positive divisors of n, a positive in
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30 Jun 2016, 22:46



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