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Let p=3 and q=5, with the result that n = 2*3*5 = 30.lheiannie07 wrote:If n = 2pq, where p and q are distinct prime numbers greater than 2, how many different positive even divisors does n have, including n ?
(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight
Mitch's approach is definitely the best.lheiannie07 wrote:If n = 2pq, where p and q are distinct prime numbers greater than 2, how many different positive even divisors does n have, including n ?
(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight
We can let p = 3 and q = 5. Thus, the product of 2pq is 2 x 3 x 5 = 30. The factors of 30 are:lheiannie07 wrote:If n = 2pq, where p and q are distinct prime numbers greater than 2, how many different positive even divisors does n have, including n ?
(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight
Thanks a lot!Brent@GMATPrepNow wrote:Mitch's approach is definitely the best.lheiannie07 wrote:If n = 2pq, where p and q are distinct prime numbers greater than 2, how many different positive even divisors does n have, including n ?
(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight
However, if you didn't come up with that approach, here's another.
If p and q are distinct prime numbers greater than 2, then p and q are each odd
So, if n = 2pq, then the positive EVEN divisors of n are: 2, 2p, 2q and 2pq (4 in total)
Answer: C
Cheers,
Brent
Thanks a lot!Scott@TargetTestPrep wrote:We can let p = 3 and q = 5. Thus, the product of 2pq is 2 x 3 x 5 = 30. The factors of 30 are:lheiannie07 wrote:If n = 2pq, where p and q are distinct prime numbers greater than 2, how many different positive even divisors does n have, including n ?
(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight
1, 30, 2, 15, 3, 10, 5, 6
Since 30 has 4 even factors, n has 4 even factors.
Alternatively, we can solve the problem algebraically. Keep in mind that p and q will be odd primes since they are greater than 2.
The factors of n are:
1, 2pq, 2, pq, p, 2q, q, 2p
We see that the even factors of n are 2pq, 2, 2q, and 2p, so there are 4 even factors.
Answer: C
