sanaa.rizwan wrote:If n=4p 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

general approach here is

for any no. N = 2^x * 3^y * 5^z ...so on, where x, y and z are their power.
then total no. of factors of N will be (x + 1) * ( y + 1) * (z + 1) ... A
then total no. of odd factors will be ( y + 1) * (z + 1) ... B
then total no. of even factors will be A - B
for example

n = 6 = 2 * 3
total no. of factors = (1 + 1) * (1 + 1) = 4
total no. of odd factors = 2
total no. of even factors = 4-2 = 2

This approach is good when you need to find the even or odd factor of larger no.