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

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sudhirgupta93

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15 Jul 2016, 23:51

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Bunuel

gmatpapa

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.

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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sudhirgupta93

Bunuel

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"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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03 Nov 2016, 07:11

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gmatpapa

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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17 Aug 2018, 03:26

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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:

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

$= x^9y^9$

$= n^9$

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

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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.

Hope this answer is still relevant to your query

Mansoor50

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

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19 Jun 2020, 04:55

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Why can't the answer be n^8 since the question says unique divisors.

So surely in case of n^2 you should not count n (n will not be a unique divisor to n^2 ? )

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The factors of a number always exist in a pair except for perfect square in a sense that every pair from extreme multiply to a number. Now the product of factors of n is n^2, so if we take n as a product of two prime numbers like a and b then n=a×b and its factors are 1,a,b, a×b and product of its factor are a^2×b^2.
Now, n^2= a^2×b^2
So total number of factors of n^2 is 3×3=9 and four pair will multiply to a^2×b^2 and one factor will be a×b so product of all pairs is a^9×b^9

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I hate that n counts as a divisor of n. It's just "meh". Not taking this into account made it unnecessarily hard for me to find n. However, I finally found 28, but of course ended up with the wrong answer, again not counting n as a divisor of n.

28 = 2, 4, 7, 14

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What a tough question.

N is made up prime factors.

When we multiple all the Factors of N (which are themselves composed of all the different arrangements of prime factors) ———> the Product result = (N)^2

I started from the perfect square and worked backwards.

Using 4, 9, or 25 ———> the only unique factors that make up the perfect square are the Prime Number that is the square root and the Number itself

Thus, for (N)^2 = 25, for example, the unique factors are: 1 , 5, and 25

The corresponding value for N would be 5: factors of 5 are only 1 and 5.

1 * 5 does not equal 25 ——I.e., the Product of the factors of N do not multiply to (N)^2 if N is a Prime Number.

Thus we are looking for a Perfect Square that is NOT a Prime Number Squared

Let’s try to make N with the 2 smallest prime numbers: 2 and 3

If N = (2)(3) = 6

All the unique factors of N are: 1, 2, 3, and 6

1 * 2 * 3 * 6 = 36 = (N)^2

Works!

So we are looking for a Number N whose Prime Factorization is of the form:

N = p * k

Where p and k are 2 different prime factors.

The Divisors of N are all the different arrangements of the Prime Factors

Multiplying these arrangements ——> 1 * (p) * (k) * (pk) = (p)^2 * (k)^2 ———-> which is (N)^2

Now let’s take all the factors of (N)^2 and see what the result is:

Using 36 as an example again:

All the Unique Divisors of any Number will just be all the different arrangements we can make with the Prime Factors that make up the number’s prime factorization.

The prime factors are like “building blocks” and we use them to put together each Unique Divisor.

(2)^0 (3)^0 * (2)^0 (3)^1 * (2)^0 (3)^2

* (2)^1 (3)^0 * (2)^1 (3)^1 * (2)^1 (3)^2

* (2)^2 (3)^0 * (2)^2 (3)^1 * (2)^2 (3)^2

= (2)^9 (3)^9

= (N)^9

(E)

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Hi,

I believe this to be a simple and in-depth solution to this problem which doesn't require the memorization of any special rules, except the common prime factorization principle:

Any number n has n as factor, meaning the product of the remaining unique factors must equal n. But we know that their product already yielded n, meaning that there can be no other factors except ones such that the remaining factors equal n -> we have a number which consists of exactly 2 prime numbers to the power of 1.

Here is an example:

n=10 -> Factors are 2,5,10. We said that any number contains itself as factor, and the remaining factors must equal n=10. Removing the 10, we are left with 2 and 5, which yield 10! Now we can directly see that this is possible if we have a number n which consists of 2 unique primes to the power of 1. Why? Try it with 3 unique primes to the power of one, you will get too many factors.

We don't have to proof anything further, since we found a group of numbers (as defined above) which fulfill the requirements. Take n=6 and test the formula. We get 2,3,6,9,12,18,36 as unique factors of n^2=36, and in total, we can write that product as 6^9.

Hence (E)

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The formula for finding the product of all factor of a Number n is = $n^{Total Number of factors/2}$
Total number of factors of a number n with prime factors $X,Y$

Prime factors $n = X^{a}*Y^{b}$
Total number of factors including prime and compound factors = $(a+1)*(b+1)$

with the above formulas in place let us a number with only prime factors as the main factors to satisfy the condition in the question stem

$6=2*3$
Product of all factors = $1*2*3*6 = 6^{2}$
So 6 satisfies the question

now we have to find the product of the factors of $6^{2}$

which can be written as product of primes $= 2^{2}*3^{2}$
Total factors = $(2+1)*(2+1) = 9$

From the formula above the product of all factors is $n^{Total Number of factors/2}$
or
$36^{\frac9 2}$
$(6^{2})^{\frac9 2}$
$6^{9}$
so the answer is $n^{9}$ or $E$

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If the product of all the unique positive divisors of n, a positive integer that is not a perfect cube, is$ n^2$, then the product of all the unique positive divisors of $n^2$ is

For every divisor of a number n if x is a divisor there exists another divisor in the form such that: $\left(\frac{n}{x}\right)$ . Unless the number is a perfect square.

Hence we have ordered pairs of divisors such that the product is equal to n.

Since it is given that the product of the divisors is equal to $n^2$.

This is possible if the divisors are in the form : $\left(1,\ a,\ b,\ n\right)$ where n = ab.

Hence all the possible unique divisors for $\left(a^2b^2\right)$ $\left(1,\ a,\ b,\ a^2,b^2,a^2b^2,\ a^2b,\ ab^2,ab\right)$

The product is given by : $\left(ab\right)^9$

Hence this is equal to $n^9$

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gmatpapa

Let n = a^x * b^y * c^z ... where a, b, c, ... are primes

Product of the unique positive divisors (factors) of a number n equals n^(F/2), where F is the number of factors of n
Number of factors of n = (x+1)(y+1)(z+1) ... = F

Thus, n^(F/2) = n^2
=> F/2 = 2
=> F = 4

Thus, n has 4 factors
=> n = a^1 * b^1 [since number of factors would be (1+1)(1+1) = 4]
Or n = a^3 [since number of factors would be (3+1) = 4]

However, we know that n is NOT a cube.
Hence, n = a * b
=> n^2 = a^2 * b^2
Number of factors of n^2 = (2+1)(2+1) = 9
Thus, product of the factors = (n^2)^(9/2) = n^9

Answer E

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Say, n can be written as: 1*n, a*b, c*d (you can further expand this list)
Now, factors of n^2: 1*n^2, n*n, a*(ab^2),b*(a^2b),c*(cd^2),d*(c^2d)
Now, if a,b,c and d are unique, product of factors: 1*n*n^2*(a*(ab^2))*(b*(a^2b))*(c*(cd^2))*(d*(c^2d))
Upon simplification: (n^3)(n^2)(n^2)... and so on, that is, after the basic n^3, the exponent increases by 2, thus, the exponent has to be odd. It cannot be n^3, therefore, n^9 is the answer.

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Easy approach, think Analytically !!

Guess the number whose factors when multiplied result in square of that number...... a typical behaviour of any number ending with 0.....so let's take 10.

factors of 10 -> 2,5,10 . Multiplied : 100 which is 10^2
Factors of 100 when multiplied : 1&100 will be n^2, 2&50, 4&25, 5&20, left will be one 10, that is n

Therefore n^9.

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