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605-655 (Medium)|   Number Properties|                     
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dhruva09
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Hi iurequi ,

For simplicity rearrange the question as: n/4 = p
Where p is prime number greater than 2,
Now let's say p = 3
so n becomes 12
Even factors of 12: 2,4,6 & 12 =>4

Now let's say p = 5
so n becomes 20
Even factors of 20: 2,4,10 & 20 =>4

Hope I was able to help.

So the answer is 4
Option C

iurequi
The main issue I had to tackle was the scope of the question, we had P=3 as our one solution, but i was not sure about other prime numbers, that if we put them, would we still get 4 even divisors.

Anyone has any kind of tip that how to determine the scope, or extent to which we put put values to find out the solution?

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Bunuel,
I did, n=2^2p, and since p is prime greater than 2, it won't have even factors. So I calculated the total factors of even, as 2+1=3
Why is that wrong? isn't it how you calculate the no of factors?
Walkabout
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
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SwethaReddyL
Bunuel,
I did, n=2^2p, and since p is prime greater than 2, it won't have even factors. So I calculated the total factors of even, as 2+1=3
Why is that wrong? isn't it how you calculate the no of factors?


The mistake is that p being odd does not mean it cannot appear in an even divisor.

An even divisor just needs at least one factor of 2. Since n = 2^2 * p, the exponent of 2 can be 1 or 2, and the exponent of p can be 0 or 1.

So the even divisors are 2, 4, 2p, and 4p.

There are 4 even divisors.

Answer: C.
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