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Is the integer n odd?

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Is the integer n odd?

by mitzwillrockgmat » Mon May 24, 2010 3:22 am
Is the integer n odd?

(1) n is divisible by 3.

(2) 2n is divisible by twice as many positive integers as n.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

stat 1 is easy to kick out but can someone explain stat 2. i spent 5 minutes in a test on this question & still got it wrong!

basically stat 2 means that 2n has twice as many factors than n right?

for values of n from 1 to 10 i determined which numbers have double the no. of factors for 2n.

i.e n = 2 ..2 factors 1,2 while 2n=4 has 3 factors (1,2,4 )

basically for values of n between 1 & 10.... 3 , 4 , 5 , 7 & 9 fit the bill.

if i combined this result with stat 1 then only 3 & 9 fit the bill as they are both divisible by 3. hence, n will be odd.

is this correct? and is there a shorter way??
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by thephoenix » Mon May 24, 2010 3:44 am
imo B is sufficient to say that n is odd
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by gmatmachoman » Mon May 24, 2010 3:45 am
mitzwillrockgmat wrote:Is the integer n odd?

(1) n is divisible by 3.

(2) 2n is divisible by twice as many positive integers as n.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

stat 1 is easy to kick out but can someone explain stat 2. i spent 5 minutes in a test on this question & still got it wrong!

basically stat 2 means that 2n has twice as many factors than n right?

for values of n from 1 to 10 i determined which numbers have double the no. of factors for 2n.

i.e n = 2 ..2 factors 1,2 while 2n=4 has 3 factors (1,2,4 )

basically for values of n between 1 & 10.... 3 , 4 , 5 , 7 & 9 fit the bill.

if i combined this result with stat 1 then only 3 & 9 fit the bill as they are both divisible by 3. hence, n will be odd.

is this correct? and is there a shorter way??
Pick B!!

St 1:

case 1: n = 9 . Divisible by 3. Odd YES
case 2: n=12 Divisible by 3 Not ODD... so NO

Insufficient

St 2 :

2n is divisible by twice as many positive integers as n.

It means that : 2n will have twice the Number of factors of n

Case 1 :

n= 9
expressed as multiples of prime factors = 3^2
No of factors : (2 +1) = 3

2n =18
expressed as multiples of prime factors = 2* 3^2
No of factors : 2*3
: 6
YES

case 2:

n=12
expressed as multiples of prime factors: 2^2 *3
No of factors :3*2
: 6
2n =24
expressed as multiples of prime factors: 2^3 * 3
No of factors : 4 * 2
: 8

even Number Cannot be expressed as "2n is divisible by twice as many positive integers as n".

Pick B
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by gmatmachoman » Mon May 24, 2010 3:47 am
thephoenix wrote:imo B is sufficient to say that n is odd
Bhai, i tried harder keying all the concepts..u lead me by a minute.... Anyways..good wrk bro!!

It will be B indeed..Now let me wait for LR to pitch in....
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by albatross86 » Mon May 24, 2010 3:48 am
1. Clearly insufficient.

2. If 2n has twice as many factors as n, this should immediately alert you to realize that n is ODD.

(On a sidenote, if n is even, 2n would have 1.5 times as many factors)

Why? Because odd numbers have only ODD Factors. So 2n has all those factors + each of them multiplied by 2.

So Statement 2 is sufficient.
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by mitzwillrockgmat » Mon May 24, 2010 3:56 am
albatross86 wrote:1. Clearly insufficient.

2. If 2n has twice as many factors as n, this should immediately alert you to realize that n is ODD.

(On a sidenote, if n is even, 2n would have 1.5 times as many factors)

Why? Because odd numbers have only ODD Factors. So 2n has all those factors + each of them multiplied by 2.

So Statement 2 is sufficient.
hi you made a mistake! odd numbers DO NOT have have odd no. of factors i.e. 3, factors are 1 & 3......5, factors are 1 & 5.....15 factors are 1, 3, 5, 15... etc etc
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by mitzwillrockgmat » Mon May 24, 2010 3:58 am
gmatmachoman wrote:
mitzwillrockgmat wrote:Is the integer n odd?

(1) n is divisible by 3.

(2) 2n is divisible by twice as many positive integers as n.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

stat 1 is easy to kick out but can someone explain stat 2. i spent 5 minutes in a test on this question & still got it wrong!

basically stat 2 means that 2n has twice as many factors than n right?

for values of n from 1 to 10 i determined which numbers have double the no. of factors for 2n.

i.e n = 2 ..2 factors 1,2 while 2n=4 has 3 factors (1,2,4 )

basically for values of n between 1 & 10.... 3 , 4 , 5 , 7 & 9 fit the bill.

if i combined this result with stat 1 then only 3 & 9 fit the bill as they are both divisible by 3. hence, n will be odd.

is this correct? and is there a shorter way??
Pick B!!

St 1:

case 1: n = 9 . Divisible by 3. Odd YES
case 2: n=12 Divisible by 3 Not ODD... so NO

Insufficient

St 2 :

2n is divisible by twice as many positive integers as n.

It means that : 2n will have twice the Number of factors of n

Case 1 :

n= 9
expressed as multiples of prime factors = 3^2
No of factors : (2 +1) = 3

2n =18
expressed as multiples of prime factors = 2* 3^2
No of factors : 2*3
: 6
YES

case 2:

n=12
expressed as multiples of prime factors: 2^2 *3
No of factors :3*2
: 6
2n =24
expressed as multiples of prime factors: 2^3 * 3
No of factors : 4 * 2
: 8

even Number Cannot be expressed as "2n is divisible by twice as many positive integers as n".

Pick B
thanks i got it! basically i was on the right path but mistakenly included 4 . thanks! :)
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by albatross86 » Mon May 24, 2010 3:58 am
I didn't say ODD numbers have an odd number of factors, I said they have ODD factors.
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by mitzwillrockgmat » Mon May 24, 2010 4:00 am
albatross86 wrote:I didn't say ODD numbers have an odd number of factors, I said they have ODD factors.
aahhhhhh ok! now it makes sense! this is a good way of looking at it saves so much time! thanks alot!
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by albatross86 » Mon May 24, 2010 4:02 am
mitzwillrockgmat wrote:
albatross86 wrote:I didn't say ODD numbers have an odd number of factors, I said they have ODD factors.
aahhhhhh ok! now it makes sense! this is a good way of looking at it saves so much time! thanks alot!
No problem :)

Some of these number properties are really useful to know.
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by mitzwillrockgmat » Mon May 24, 2010 4:14 am
albatross86 wrote:
mitzwillrockgmat wrote:
albatross86 wrote:I didn't say ODD numbers have an odd number of factors, I said they have ODD factors.
aahhhhhh ok! now it makes sense! this is a good way of looking at it saves so much time! thanks alot!
No problem :)

Some of these number properties are really useful to know.
hey a random addition but u said:

odd numbers have only ODD Factors. So 2n has all those factors + each of them multiplied by 2.

so n =9 factors 1,3,9

so 2n=18 will have 1,3,9, & 1, 6, 18

but 18 has factors, 1, 2, 3, 6, 9, 18

so we should re phrase this as;

if n is odd, then 2n will have all of n's factors, the integer 2, and 2 times each of n's factors. Hence, 2n will always have twice the no. of factors of n.

So i just wanted to ask you...this property will hold through if 2n is 3n or 4n
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by albatross86 » Mon May 24, 2010 4:22 am
mitzwillrockgmat wrote:
albatross86 wrote:
mitzwillrockgmat wrote:
albatross86 wrote:I didn't say ODD numbers have an odd number of factors, I said they have ODD factors.
aahhhhhh ok! now it makes sense! this is a good way of looking at it saves so much time! thanks alot!
No problem :)

Some of these number properties are really useful to know.
hey a random addition but u said:

odd numbers have only ODD Factors. So 2n has all those factors + each of them multiplied by 2.

so n =9 factors 1,3,9

so 2n=18 will have 1,3,9, & 1, 6, 18

but 18 has factors, 1, 2, 3, 6, 9, 18

so we should re phrase this as;

if n is odd, then 2n will have all of n's factors, the integer 2, and 2 times each of n's factors. Hence, 2n will always have twice the no. of factors of n.

So i just wanted to ask you...this property will hold through if 2n is 3n or 4n
The better way to visualize this is:

n = 9 : Factors 1 , 3 and 9

2n = 18: Factors 1, 1*2, 3, 3*2, 9 and 9*2 i.e. each factor of n is a factor of 2n and so is twice that factor. This is why there are twice as many factors, since each factor of n contributes to 2 factors of 2n, WHEN N IS ODD.

This will get complicated with 3n and 4n, and would result in double counting, so I wouldn't advise taking on blanket rules for them.

However, if a statement says 3n has twice the number of factors as n, you can safely conclude that none of the factors of n are multiples of 3. Similarly for 4, and so on.

That is actually how you came to the conclusion that n is odd. Since 2n has twice as many factors, the factors of n are NOT multiples of 2, and are hence odd => n is odd.

Hope this is correct!
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by liferocks » Mon May 24, 2010 4:32 am
Some real good solutions here.Just wanted to add my 2 cents.

From 1
n=3k..k can be even or odd so not sufficient

From 2

2n has twice as many factors as n
now for any n=p^a*q^b and so on ..where p,q etc are prime numbers
number of factors =(a+1)(b+1)...

if for any Pn this number of factors has become double , P must be a prime which is not present in n i.e if number of factors of n is (a+1)(b+1)... ,number of factors of Pn=(1+1)(a+1)(b+1)...

clearly 2 is that prime for 2n..hence n is odd
if it said that 3n is that prime..then n is not divisible by 3
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by Patrick_GMATFix » Mon May 24, 2010 3:49 pm
If after all the explanations above you still have trouble, watch the video solution: This is GMATPrep question 1298. You can find similar questions by looking for topic='Exponents' and difficulty='500-600 AND 600-700' in the Solutions Engine

-Patrick
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by arora007 » Sat Jan 29, 2011 12:14 pm
great problem!!
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