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# If p = 12k^3, where k is a prime number greater than 3 how many differ

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Re: If p = 12k^3, where k is a prime number greater than 3 how many differ [#permalink]
Since K is a prime number>3, K will take values as 3,5,7,11.....
Note:- Here K will always be odd and K^3 will have 4 factors.

Now 12*(k^3)=(2^2)*(3)*(K^3)

We need positive EVEN divisors, so there can be only 2 cases out of 3 in 2^2.

One when we have scenario as 2^1*3*(K^3)
Number of factors in this case:- 2*4=8

OR when we have scenario as 2^2*3*(K^3)
Number of factors in this case:- 2*4=8

Total Number of Even positive divisors=16

Therefore B is correct
Re: If p = 12k^3, where k is a prime number greater than 3 how many differ [#permalink]
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