Divisibility/Primes: If n=4p, where p is prime number greate

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If n=4p, where p is 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

Interested in learning the DIFFERENT approaches to solve this problem. Ultimately in the best and quickest approach.

My approach:
whenever I see the following words or terms in a question: "prime" and "divisors", I automatically start to think about breaking a number down into its prime factors which are the foundation for all the numbers.
So n = 4p.
Ok ... so 4p has the following prime factors: 2, 2, p.
"p is prime number greater than 2" - this statement is telling us that p CANNOT be even since 2 is the only even prime number. I dont know why they placed this "greater than 2" statement into the question. Is this a trap ... were GMAT trying to trip us up ... when eventually asking us to fine the possible "even divisors" of n.

Ok ... so we know n=4p ... 4p has the following prime factors: 2, 2, p.
So 4p has the following factors: 2, 2p, 4, 4p.
"how many different positive EVEN divisors does n have ?" well we know anything multiplied by an even number will always be even ... so even though p is a prime number greater than 2 (in other words odd) ... 2p and 4p will always be even.
So we can say that n has 4 positive even factors (or divisors), including n (n=4p).

Any other approaches/strategies to tackle this question. Any spot any other traps/tricks here ?

Thanks.
Last edited by II on Mon May 05, 2008 1:36 am, edited 1 time in total.
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by Kaunteya1 » Tue Apr 01, 2008 2:42 pm
You started off exactly how I would in that:

n=4p

prime factor of 4p like you said is 2,2 and p. Now the reason they mentioned that p is greater than 2 is because had p been 2 then n=8

8 - prime factorized is 2, 2, and 2

The question asks how many DIFFERENT Positive EVEN divisors does n have including n.

Well if n=8 then the number of different positive even divisors the n would have are : 2, 2x2, 2x2x2 otherwise 3 even positive divisors.

BUT THE questions says that n is greater than 2 throwing in a odd number into the mix cause all prime greater than 2 are odd.

so now p is odd: n=4p
The way to get an even number out of an odd one is multiply by an even number, in this case 2 or 4.
The even divisors now are 2, 2x2, 2 x p and finally 2 x 2 x p

And yes p is a divisor of 4p but it is an ODD one so for this question don't count it.

So if n = 4p and p is a prime number greater than 2 then there are four even divisors.

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by II » Tue Apr 01, 2008 3:04 pm
Good points. Thanks.

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by Stuart@KaplanGMAT » Tue Apr 01, 2008 3:26 pm
You also could have just picked a number - a much quicker approach.

let p=3

n= 4*3 = 12

the even divisors of 12 are 2, 4, 6 and 12: choose (C).

We can be confident that no matter which number we pick we'll get the same answer, because only one of the 5 choices can possibly be correct.
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by II » Tue Apr 01, 2008 4:01 pm
Excellent point Stuart.

My objective is to have 2 approaches to solving problems.
Typically the 2 approaches tend to be 1) using theory - which is what my approach was based on, and 2) plugging numbers (as you have just shown)

In this case the plugging numbers would be much quicker.

Thanks.

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by gmat765 » Wed Apr 16, 2008 7:05 pm
thanks, Stuart

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by lunarpower » Wed Apr 16, 2008 11:56 pm
II wrote:Excellent point Stuart.

My objective is to have 2 approaches to solving problems.
Typically the 2 approaches tend to be 1) using theory - which is what my approach was based on, and 2) plugging numbers (as you have just shown)

In this case the plugging numbers would be much quicker.

Thanks.
yeah.

you should take the time and effort to develop an instinct for the problems on which plugging numbers is the BEST way. on this problem, plugging is a slam dunk: the answer choices are constants, so it's clear that any choice of p will yield the correct answer.

if you see this general template of problem - on which there's an undetermined variable, but answer choices that can never be the same - go for the plugging right away. it will make you a better person, and will enable you to win friends and influence people.
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by vkb16 » Sun May 03, 2009 5:52 am
replyin to the post after a LOOONG time, but here goes

I got the apporach, however, the qs says how many factors....., INCLUDING n

shouldnt the answer be 5? since we're including n as well?

pardon me if this a dumb comment.. just ramping up my doubts.. :roll:

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by lilu » Sun May 03, 2009 10:16 am
vkb16 wrote:replyin to the post after a LOOONG time, but here goes

I got the apporach, however, the qs says how many factors....., INCLUDING n

shouldnt the answer be 5? since we're including n as well?

pardon me if this a dumb comment.. just ramping up my doubts.. :roll:
4p IS n,so it's already counted
The more you look, the more you see.

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by vkb16 » Sun May 03, 2009 8:53 pm
oh ok.. my bad!

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by Baten80 » Thu Sep 16, 2010 7:12 am
Stuart Kovinsky wrote:You also could have just picked a number - a much quicker approach.

let p=3

n= 4*3 = 12

the even divisors of 12 are 2, 4, 6 and 12: choose (C).

We can be confident that no matter which number we pick we'll get the same answer, because only one of the 5 choices can possibly be correct.
thanks stuart

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by minkathebest » Mon Oct 01, 2012 8:02 pm
Hello all,

In the plug-in approach, one minor clarification, why don't we consider 20, 28, etc as the divisor possibilities? I.e., 4(5) = 20, 4(7) = 28. Thank you all in advance.

I may have answered my own question, there are four possibilities in each combination:
4(7) = 28, Even divisors = 2, 4, 14, 28
4(5) = 20, Even divisors = 2, 4, 10, 20