anonymous
  • 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
Mathematics
schrodinger
  • schrodinger
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Kainui
  • 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
  • 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
  • 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
  • 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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