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

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

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15 Jul 2016, 22:51
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 positive in  [#permalink]

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17 Jul 2016, 21:16
1
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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Re: If the product of all the unique positive divisors of n, a positive in  [#permalink]

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03 Nov 2016, 06:11
1
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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Re: If the product of all the unique positive divisors of n, a positive in  [#permalink]

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17 Aug 2018, 02:26
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:

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

$$= x^9y^9$$

$$= n^9$$

why is it assumed that that there are ONLY 2 prime factors?
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Re: If the product of all the unique positive divisors of n, a positive in  [#permalink]

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11 Jan 2019, 01:06
If we assume more than three prime factors for n, then the original condition of the product of all positive factors on n is n^2 does not hold.
Let n = a*b*c, where a,b,c are unique primes

# of factors = 2*2*2 = 8

And the product of all factors is definitely larger than n^2. Hence it must be the case that # of unique prime factors must be less than 3.

It cannot be 1 also, as then the product of all factors would equal n and not n^2.

So it has to be the case that n = a * b

where a & b are unique primes.

Mansoor50 wrote:

why is it assumed that that there are ONLY 2 prime factors?

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Re: If the product of all the unique positive divisors of n, a positive in &nbs [#permalink] 11 Jan 2019, 01:06

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