n odd/

This topic has expert replies
Source: — Data Sufficiency |

Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Sun Jun 12, 2011 3:02 am
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
Cheers!

Things are not what they appear to be... nor are they otherwise

User avatar
Legendary Member
Posts: 1101
Joined: Fri Jan 28, 2011 7:26 am
Thanked: 47 times
Followed by:13 members
GMAT Score:640

by HSPA » Sun Jun 12, 2011 3:03 am
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
First take: 640 (50M, 27V) - RC needs 300% improvement
Second take: coming soon..
Regards,
HSPA.

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sun Jun 12, 2011 6:39 am
Removed my erroneous post

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Tue May 14, 2013 7:28 am, edited 1 time in total.
Brent Hanneson - Creator of GMATPrepNow.com
Image

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Sun Jun 12, 2011 7:22 am
g.shankaran wrote: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.
Brent@GMATPrepNow wrote: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
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sun Jun 12, 2011 7:27 am
Argh . . . you're totally right Anurag!

And now back to bed :-)

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sun Jun 12, 2011 7:31 am
Wow, I spent a very long time "proving" something that is incorrect.
For my next trick . . .

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image