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Director
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Is the integer N odd? (1) N is divisible by 3 (2)2N is [#permalink]
 03 Oct 2004, 12:51
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Is the integer N odd?
(1) N is divisible by 3
(2)2N is divisible by twice as many positive integer as N.
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Senior Manager
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I pick C.
Seems like only N=3 or (-3) works...
This is the only prime that is divisible by 3... And YES, it is odd.
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Manager
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I would go with a B
2N has as many positive divisors as N.
Implies N is an odd multiple of 3.
Ok check for
2 : 1,2 ; 4 : 1,2,4 ; 8 : 1,2,4,8
3 : 1,3 ; 6 : 1,2,3,6 ; 12 : 1,2,3,4,6,12
9 : 1,3,9 ; 18: 1,2,3,6,9,18 ; 36 : 1,2,3,4,6,9,18,36
Clearly the pattern is seen
_________________
Franky
http://franky4gmat.blogspot.com
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Senior Manager
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Agree.
B it is.
It also has nothing to do with primes.. 
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Manager
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It is odd by B alone, but not necessarily mutliple of 3. Say N = 5. Multiples = 1,5. 2N = 10. Multiples = 1, 2, 5, 10.
...thinking of a different way rather than trial and error.
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Director
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Statement 2:
Either N is odd (=2k+1) then 2N has obviously twice as much divisors as N
or N is even (=2k) and then 2N has twice as much factors as N.
B is the answer
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Director
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myfrankenstein wrote: I would go with a B
2N has as many positive divisors as N.
Implies N is an odd multiple of 3.
Ok check for
2 : 1,2 ; 4 : 1,2,4 ; 8 : 1,2,4,8
3 : 1,3 ; 6 : 1,2,3,6 ; 12 : 1,2,3,4,6,12
9 : 1,3,9 ; 18: 1,2,3,6,9,18 ; 36 : 1,2,3,4,6,9,18,36
Clearly the pattern is seen
I am nto sure if N is 4 and 2N is 8, then this does not hold true.
So, restricitng to be factor of 3 along with statement B, should give C as answer.
S
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Intern
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Well....
lets try n=6
n is even and 2(6) has twice the factors of 6
try n=3
n is odd and 2(3) has twice the factors of 3
so i think the ans.. should be E.
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