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Official Guide 11th. PS #207

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Official Guide 11th. PS #207 [#permalink] New post 17 Feb 2009, 18:22
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207. 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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Re: Official Guide 11th. PS #207 [#permalink] New post 17 Feb 2009, 19:47
emily0506 wrote:
207. 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


N= 4P = 2*2*P

even divisors : 2,4,2p,4p

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Re: Official Guide 11th. PS #207 [#permalink] New post 17 Feb 2009, 22:27
Another approach:

Since we know that p>2, you can simply take any prime value for p and find n. Once you have the value of n, you can find the different positive even divisors.

Ex. For p=3, n = 4*3 = 12

So, the no. of different positive even divisors are four i.e., 2, 4, 6, 12.
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Re: Official Guide 11th. PS #207 [#permalink] New post 06 Mar 2009, 23:07
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And in general, if you have a prime factorization:

n = 2^x 3^a 5^b

-the number of positive divisors of n will be (x+1)*(a+1)*(b+1)
-the number of odd positive divisors of n will be (a+1)*(b+1)
-the number of even positive divisors of n will be x*(a+1)*(b+1)

This will be true no matter what primes you have in your factorization - the odd primes don't need to be 3 and 5 - and regardless of how many primes you have in the prime factorization (I used two odd primes just for illustration).
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Re: Official Guide 11th. PS #207 [#permalink] New post 06 Mar 2009, 23:14
Nice tip to remember. Thank you for letting us know.

GMAT is all about solving smart rather than solving hard. :)
Re: Official Guide 11th. PS #207   [#permalink] 06 Mar 2009, 23:14
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Official Guide 11th. PS #207

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