If n = 4p, where p is a prime number greater than 2, how : PS Archive
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# If n = 4p, where p is a prime number greater than 2, how

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If n = 4p, where p is a prime number greater than 2, how [#permalink]

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02 Jul 2008, 15:28
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If n = 4p, where p is a 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
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02 Jul 2008, 15:40
vksunder wrote:
If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n?

You could list all of the even divisors:

n = 2^2 * p
so the even divisors are: 2, 2^2, 2p and 2^2 * p. Four divisors. If it's easier, there's nothing wrong with choosing a number for p here, provided you choose an odd prime.

Or you might know directly how to find the number of even divisors by looking at the prime factorization. You just need to look at the exponents: add one to each power except the power on the 2, and multiply. Here we have 2*(1+1) = 4. That's to find the number of even divisors only; if you want the total number of divisors (odd and even), add one to every power in the prime factorization and multiply the numbers you get.
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02 Jul 2008, 20:32
just put p = 3
n = 12
factors of 12 - 1,2,3,4,6,12
even factors - 2,4,6,12

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02 Jul 2008, 22:19
n =4p
Take multiple values of p
p = 3,5,7,11,13 ...
n = 12,20,28,44,52 ....
different positive even divisors of n
{2,4,12},{2,4,20}, {2,4,28}, {2,4,44}, {2,4,52}...

Max 3 positive even divisors.

Ans B.
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02 Jul 2008, 22:24
vksunder wrote:
If n = 4p, where p is a 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

Since all the positive prime numbers greater/other than 2 are odd,

Out of the total 6 (i.e (2+1) * (1+1) ) divisiors of n, (2+1) + 1 (n itself) = 4 will be even.
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02 Jul 2008, 22:33
[quote="kapilnegi"]
different positive even divisors of n
{2,4,6,12},{2,4,10,20}, {2,4,14,28},

quote]
Re: PS: Number properties   [#permalink] 02 Jul 2008, 22:33
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