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If n = 4p, where p is a prime number greater than 2, how man [#permalink]
chacinluis wrote:
n=4p
Since p is a prime number greater than 2, then p is odd.
Also since p is prime the only factors of p are p and 1.

factors of 4p are: 1,2,4,p,2p,4p
Therefore
2,4,2p,4p are all even factors
The fact that p is not 2 and not 1, tells us 2p is not 4 and 4p is not 1 correspondingly. So they are all distinct.

Final Answer: 4


Can someone explain the reason behind the highlighted portion above?
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink]
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I suggest you to read our Quant book: the section how to find the factor of a number.

Finding the Number of Factors of an Integer

https://gre.myprepclub.com/forum/gre-quant ... tml#p51913

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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink]
n=4p, where p is some prime greater than 2 (p=3,5,7,11,13...)
Lets just assume that p=5. We know from number properties, that because p will not be raised to any power or such, the answer will not change if p is 5 or 17 or 11 etc.
n=20=2*2*5
n has a total of (2+1)(1+1)=6 factors
Of these factors the only odds one are: 1 and 5
the rest must be even divisors. So it has 2 odd divisors and 4 even divisors.

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