Question

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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i got E. is that right?

Answer 1

k thanks

Answer 2

Can someone explain how to do this problem? Really curious as of why that is the answer.

Answer 3

?

Answer 4

weird question

Answer 5

n^2= n*n
or
n^2=n^2 *1

Answer 6

so hehe since n^2 * n^2 * (n^2)^2 *1=n^8
how did u got 9 ?

Answer 7

*

Answer 8

ok lol i forget one factor xD
factors of n^2 are
1
n
n^2
factors from n there multi = n^2
factors from n^2 with multi =(n^2 )^2=n^4

so we have 1*n*n^2 * n^2 *n^4 =n^9
cool ^^

Answer 9

nice :)

Answer 10

no one trusts me :(

Answer 11

lol why :o