If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight

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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It is but its' not even (how many different positive even divisors does n have, including n).
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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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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.

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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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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MATH REVOLUTION VIDEO SOLUTION:


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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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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).
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