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

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

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

great.... u made it look so easy

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 ? )

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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How can we conclusively say that there must be only 2 factors in the middle? Please help.

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

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)

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 \(E\)

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 \(n^9\)

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