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If n=8p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 8


since p is a prime number greater than 2, it's always an odd number. since there are 6 even divisors (2,4,8,2p,4p,8p) the answer is D.



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GMATGuruNY wrote:

koby_gen wrote:If n=8p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 8

We can plug in a value for p.
Let p = 3.
Then n = 8*3 = 24.
Even factors of 24 are 2, 4, 6, 8, 12, 24 = 6 even factors.

The correct answer is D.

can somebody explain to me why everybody is always only considering the first prime number bigger than two which indeed is 3 so the result would be 24.

But when I add the next prime numbers greater than 2 I get the following number line:

8*3 = 24
8*5 = 40
8*7 = 56
8*11 = 88
8* 13 = 104

Assuming that this line is representative for all the higher prime numbers to come I want to find out the common positive factors of 24, 40, 56, 88 and 104 which actually are 4 in total (factor 2, 4, 8 and n itself).

Can somebody explain why this approach is incorrect?

Best regards,
Max