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suneja
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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
 one year ago
 one year ago
suneja Group Title
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
 one year ago
 one year ago

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Kainui Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Kainui Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

suneja Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago
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