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- anonymous

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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- anonymous

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- Kainui

Seems to me that your new "n^2" term behaves just like the other "n" so you can "plug it in" and follow through with the same logic and see that
(n) => (n)^2
(n^2) => (n^2)^2

- Kainui

Reason through this and give "n" an actual number that works, like 10.
10's factors are 1, 10, 2, and 5 right?
1*10*2*5=10*10=10^2
Now let's see about 100
1, 100, 2, 50, 25, 4, 5, 20
1*100*2*50*25*4...
Well that doesn't equal 10^4 so my logic is flawed. Maybe this helps though in some way?

- anonymous

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

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- shubhamsrg

since you ,without loss of generality,correctly let p_1 * p_2 = n
we see,
n^2 = (p_1)^2 * (p_2)^2
so all factors of n^2 would be
1,
p_1 ,
p_2 ,
p_1* p_2,
p_1 ^2 ,
p_2 ^2,
(p_1)^2 p_2 ,
(p_2)^2 p_1 ,
(p_1)^2 (p_2)^2 ..
multiplying all , you should get n^9

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