Since we cannot have two correct answers just pick a prime greater than 2, and see how many different positive even divisors will 4p have.

p = 3 --> 4p = 12--> 12 has 4 even divisors: 2, 4, 6, and 12.

Answer: C.

Or this way: since p is prime greater than 2, then p=odd, thus 4p=even, which means that it has 4 even divisors: 2, 4, 2p, and 4p.

Similar question to practice: if-n-is-a-prime-number-greater-than-3-what-is-the-remainder-137869.html

Hope it helps.
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n= 4p = 2^2*p^1
so total no. of factors: (2+1)*(1+1)= 6
total no. of odd factors, since p is odd as it is a prime>2: p^1 and p^0 = 2
Total no. of even factors: 6 - 2 = 4

Now if n was n=4pq where p and q are both prime no.s greater than 2 then:
total no. of factors: (2+1)*(1+1)*(1+1)= 12
total no. of odd factors, since p is odd as it is a prime>2: (1+1)*(1+1)= 4
Total no. of even factors: 12 - 4 = 8

Hi Bunuel, could you validate my logic pls?
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Is one not a divisor?
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It is but its' not even (how many different positive even divisors does n have, including n).
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p can get any prime number greater than 2 so there should be unlimited different divisor for the n because p can be 11,13,17,19....

What is wrong with this ?

No matter which prime p is, 4p will have only four EVEN divisors: 2, 4, 2p, and 4p. Try to check it with any prime greater than 2.

Does this make sense?
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4 has two even divisors >> 2 & 4

Any prime no p divisors will have 2p & 4p (from above)

So total = 4

n=4*p

And p(prime) > 2

n= 2*2*p

Positive even divisors 'n' can have including 'n':

2
2p
4
4p which is also equal to 'n'

Hence 4 positive even divisors.

Yes, your approach is correct.
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if we apply theory of finding number of divisor
Let,s plug in p= 3
then n= 12
Factor of 12 are 3, 2, 2
p>2, so 2s are out
Now the formula (p+1)+(q+1)+(r+1) = 3+1 = 4
Ans C
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@p=3, n = 4*3 = 12, Positive Even divisor of n = {2, 4, 6, 12} i.e. 4 Divisors
@p=5, n = 4*5 = 20, Positive Even divisor of n = {2, 4, 10, 20} i.e. 4 Divisors
@p=7, n = 4*7 = 28, Positive Even divisor of n = {2, 4, 14, 28} i.e. 4 Divisors
@p=11, n = 4*11 = 44, Positive Even divisor of n = {2, 4, 22, 44} i.e. 4 Divisors
@p=13, n = 4*13 = 52, Positive Even divisor of n = {2, 4, 26, 52} i.e. 4 Divisors
and so on...

Answer: option C
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Hi All,

This question can be solved rather easily by TESTing VALUES:

We're told that N = 4P and that P is a prime number greater than 2. Let's TEST P = 3; so N = 12

The question now asks how many DIFFERENT positive EVEN divisors does 12 have, including 12?

12:
1,12
2,6
3,4

How many of these divisors are EVEN? 2, 4, 6, 12 …..4 even divisors.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Hi All,

This question can be solved rather easily by TESTing VALUES:

We're told that N = 4P and that P is a prime number greater than 2. Let's TEST P = 3; so N = 12

The question now asks how many DIFFERENT positive EVEN divisors does 12 have, including 12?

12:
1,12
2,6
3,4

How many of these divisors are EVEN? 2, 4, 6, and 12 …..that's a total of 4 even divisors.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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n = 4p
p prime no. \(> 2\)

We have to find the no. of even divisors which means even factors of n

P must be a odd no. because 2 is the only even prime no.

Let p be \(3\)

\(n= 4 * 3\)

= \(2^{2} * 3\)

Now , No of factors of P = \(2^{2+1} * 3^{1+1}\) = \(2^{3} * 3^{2}\)

= \(3 * 2 = 6\)

Even factors = Total factors - odd factors

To find Odd factors we take all the prime apart from 2. so here we are left with only 3

Odd factors = \(3^{1+1} = 3^{2}= 2\)

Total factors - Odd factors = Even factors
\(6-2 = 4\)

hence answer is 4

MATH REVOLUTION VIDEO SOLUTION:


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Since p is primes and is > 2, then p must be an odd number since 2 is the only even prime number.

I solved by simply plugging in:
p = 5 (a prime) ---> n = 20
Factors of 20 include:
20 1
10 2
5 4

In this factor tree there is a total of 4 even factors means C is the correct answer.

If you aren't convinced after this you should have time to plug in another prime number for p. I finished within 50 seconds using just p =5.

This is an interesting question because we are immediately given the option to insert any prime number we wish for p. Since this is a problem-solving question, and there can only be one correct answer, we can select any value for p, as long as it is a prime number greater than 2. We always want to work with small numbers, so we should select 3 for p. Thus, we have:

n = 4 x 3

n = 12

Next we have to determine all the factors, or divisors, of P. Remember the term factor is synonymous with the term divisor.

1, 12, 6, 2, 4, 3

From this we see that we have 4 even divisors: 12, 6, 2, and 4.

If you are concerned that trying just one value of p might not substantiate the answer, try another value for p. Let’s say p = 5, so

n = 4 x 5

n = 20

The divisors of 20 are: 1, 20, 2, 10, 4, 5. Of these, 4 are even: 20, 2, 10 and 4. As we can see, again we have 4 even divisors.

No matter what the value of p, as long as it is a prime number greater than 2, n will always have 4 even divisors.

The answer is C.
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One of those Quality Questions from the Official Guide.
here is my solution =>
Method 1->
n=2^2*p
as p is a odd prime (all primes >2 are odd)
number of even factors => 2*2-> four
Four factors are -> 2,4,2p,4p
Alternatively let p=3
so n=12
factors => 1,2,3,4,6,12
even factors => 2,4,6,12
hence four

Hence C
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n=4p
Let;s factorize n
n= 2^2 * p^a (a is some positive power of p and we know that p is a prime number greater than 2 i.e. it is odd as the only even prime number is 2. So we don't have to worry about the power of p as it would yield odd integer and we have to just find even factors.)

Thus the answer is 2, 4(2^2), 2p (even), and 4p (even).