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Is integer n odd ? 1. n is divisible by 3 2. 2n is divisible

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Senior Manager
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Is integer n odd ? 1. n is divisible by 3 2. 2n is divisible [#permalink]

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New post 27 Oct 2008, 23:30
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A
B
C
D
E

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Is integer n odd ?

1. n is divisible by 3
2. 2n is divisible by twice as many +ve integers as n

Please explain yur approach
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Thanks
rampuria

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Re: DS [#permalink]

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New post 27 Oct 2008, 23:44
Is the QA b ?

statement 1 is not sufficient , as n can be 12 or 15 both.

statement 2 , take n as 5, 6,8, 15 , different even and odd numbers and verify.
n=5 -> divisible by 1, 5 2n=10 -> divisible by 1,2,5,10 double
n=6 -> divisible by 1, 2,3,6 2n=12 -> divisible by 1,2,3,4,6,12 not double
n=8 -> divisible by 1, 2,4,8 2n=16 -> divisible by 1,2,4,8,16 not double
n=15 -> divisible by 1,3,5,15 2n=30 -> divisible by 1,2,3,5,6,10,15,30 double

So looks like if n is odd, 2n is divisible by twice num of +ive integers as n is.

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Re: DS [#permalink]

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New post 28 Oct 2008, 00:00
I would say B too.
Stat 1 insufficient

Stat 2 if n was even, it would share a common factor i.e 2 therefore it cannot be that 2n is divisible by twice as many factors as n. Therefore n must be odd.

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Re: DS [#permalink]

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New post 28 Oct 2008, 16:07
I just simply used numbers and realized that is not possible for 2n to have twice as many factors as n if n is even. When n is even, 2n will have only one factor more than n For example (4,8) (8,16) (16,32) (32,64).

B is sufficient

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Re: DS   [#permalink] 28 Oct 2008, 16:07
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Is integer n odd ? 1. n is divisible by 3 2. 2n is divisible

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