n odd/

[g.shankaran]

is the integer n odd?

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

[Frankenstein]

Hi,
From(1): n is divisible by 3.
n can be 3,6,9,...
Insufficient

From(2): My understanding of st(2) is if n has p factors, 2n has 2p factors..is it right?
If so, my solution:
if n = (a^i).(b^j).(c^k)... where a,b,c are all distinct primes.
So, p = (i+1)(j+1)..
Now 2n = 2^1.(a^i).(b^j).(c^k)...
Number of factors of 2n will be (1+1).(i+1)(j+1)... For this to happen, a,b,c... should be distinct from 2. So, n is product of powers of primes other than 2 i.e. product of odd numbers. So, n is odd number.
Sufficient

Hence, B

[HSPA]

A is certianly No. Because we have both 3 and 6 divisisble by 3.

A Not sufficient

B) 2n has twice as many factors as N

N =3 , factors 2
N =6 , factors 4

N= 8, factors 4
N= 16, factors 5

N= 15,factors 3
N= 30, factors 6

Seems okay... B it is for me

[Brent@GMATPrepNow]

Removed my erroneous post

Cheers,
Brent

[Anurag@Gurome]

is the integer n odd?

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

Statement 1: n is divisible by 3.
If n = 3 --> Odd
If n = 6 --> Even

Not sufficient

Statement 2: 2n divisible by twice as many positive integers as n
Hence, number of factors of 2n is double as of n.

Say, n is even.
Then factors of n are 1, 2,..., a,... and n
Factors of 2n will include all the factors of n and double of them, i.e. factors of 2n are 1, 2,..., a,..., 2a,... and 2n

Note that 2 is common in both of them. Hence number of factors of 2n must be less than twice the number of factors of n. Hence, n is not even. Then n must be odd. (Similarly, we can show that n is in fact odd.)

Sufficient.

The correct answer is B.

Big point: Notice what happens when we find the number of divisors of 2(400)
Since 800 = (2^5)(5^2), the total number of positive divisors of 800 equals (5+1)(2+1)=30

Hi Brent, the bold part has a calculation error. It would evaluate to (5 + 1)*(2 + 1) = 6*3 = 18

[Brent@GMATPrepNow]

Argh . . . you're totally right Anurag!

And now back to bed

Cheers,
Brent

[Brent@GMATPrepNow]

Wow, I spent a very long time "proving" something that is incorrect.
For my next trick . . .

Cheers,
Brent