yankee.musk wrote:This is a GMATPrep problem

Is the integer n odd?

(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n

For some reason the spoiler feature is not working for me. So, OA is the answer choice # 3+5-1+2-7

(1) is not enough since it could be either 3, 6, 9...
(2) gonna do some tests

3*2=6.

6 is divisible by 1,2,3,6,
3 is divisible by 1,3

63*2=126
63 is divisible by 1, 3, 7, 9, 21, 63
1,2,3,7,9,21,63,126

so it doesn't tell us whether or not it's odd. this should also be impossible to tell with positive numbers. We can tell though, that it will only double the number of multiples. The total number of multiple will be 4 probably, because by multiplying a number by 2, you're adding 2 extra multiples (2, if 2 wasn't already a multiple) and the number that equals 2n.

so let's take a look at the multiples of 3.

3, 6, 9, 12.

lets look at their multiples.

1,3
1,2,3,6
1,3,9
1,2,3,4,6,12

together, with (1) and (2), the multiples of 2n are

1,2,3,6
1,2,3,4,6,12
1,2,3,6,9,18
1,2,3,4,6,8,12,24

as you can see, the answer could be either 9 or 3, so we can tell that together, we can't solve the problem, so E would be the answer.

From stmt1,n can be 3,6,9,12.

So A insufficient.

From stmt2,2n is divisible by twice as many positive integers as n.

We need to check this with n as odd and n as even.

1)n is odd.

n=21

n is divisible by 1,3,7,21

2n=42 = 2*(21)=2*(3*7)

42 is divisible by 1,2,3,6,7,14,21,42.

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

2) n is even

n=30

n=2*3*5

30 is divisible by 1,2,3,5,6,10,15,30

2n=2*(30)=2*(2*3*5)


60 is divisible by 1,2,3,4,5,6,10,12,15,30,60.

When n is even 2n ,2n is not divisible by twice as many positive integers as n.

So n is odd.

B is sufficient.

Answer is B

Thank you all for the explanations

By the way, how do guys use spoiler feature in BTG. I'm having trouble with it.

Talkativetree wrote: 3*2=6.

6 is divisible by 1,2,3,6,
3 is divisible by 1,3

63*2=126
63 is divisible by 1, 3, 7, 9, 21, 63
1,2,3,7,9,21,63,126

Hey, how'd you happen to pick on 63 as the no.? I tried till 21 & stmt 2 seemed to be working till there

I'm wondering if we can do it like this?

A & B are out of picture based on what Talkative tree said

Considering both the statements means:

n=3k
2n=6k

for k=2:
n=6 which has factors 1,2,3,6 i.e. 4 factors
2n=12 which has factors 1,2,3,4,6 i.e. 6 factors which contradicts stmnt 2 - hence the 2 statements taken together is void.

therefore, answer is E

Jube,

U tried the stmt 2 with n as even.

with n as even,it contradicts stmt2.

Let consider n as odd.

n=9

9 is divisible by 1,3,9

2n=18=2*3*6

18 is divisible by 1,2,3,6,9,18

From this you can see that 2n is divisible by twice as many integers as n.

We dont need n to be a multiple of 3.

Take any odd and even number and you can check with stmt2

When n is odd only stmt2 can be true.

Hence B is sufficient.

selango wrote:Jube,

U tried the stmt 2 with n as even.

with n as even,it contradicts stmt2.

Let consider n as odd.

n=9

9 is divisible by 1,3,9

2n=18=2*3*6

18 is divisible by 1,2,3,6,9,18

From this you can see that 2n is divisible by twice as many integers as n.

We dont need n to be a multiple of 3.

Take any odd and even number and you can check with stmt2

When n is odd only stmt2 can be true.

Hence B is sufficient.

I see what you're saying. You look to be right! Thanks for pointing this out. Was just wondering if there's another way of solving this one other than plugging values because with values you don't know for sure that there's some value out there that you couldn't think of which would contradict this statement.

From 1 clearly n can be either even or odd.
From 2,
2n has twice as many positive factor as n ,so 2n must have one more prime factor than n
as for any number N=p^a*q^b the number of factors is(a+1)(b+1)

only prime factor that 2n can have other than all prime factors of n is 2..hence n is odd..sufficient

Ans option B

liferocks wrote: 2n has twice as many positive factor as n ,so 2n must have one more prime factor than n
as for any number N=p^a*q^b the number of factors is(a+1)(b+1)

Two questions:

1) I got the formula for calculating the no. of factors but why must there be only 1 more prime factor? Why not more than 1?
2) How do you know so much about no. properties? Is it something you've built up over the years or do you have a resource which mortals like me can use too?

jube wrote:

liferocks wrote: 2n has twice as many positive factor as n ,so 2n must have one more prime factor than n
as for any number N=p^a*q^b the number of factors is(a+1)(b+1)

Two questions:

1) I got the formula for calculating the no. of factors but why must there be only 1 more prime factor? Why not more than 1?
2) How do you know so much about no. properties? Is it something you've built up over the years or do you have a resource which mortals like me can use too?

for the question 1, let n=p^a*q^b so number of factors of n=(a+1)(b+1) and number of factors for 2n=2*(a+1)(b+1)
or 2n=(1+1)*(a+1)(b+1) so 2n has one more prime factor than n has

For question 2,I am not using any specific resource..just brushed up the basics..and different forum discussions really helps!

liferocks wrote:
for the question 1, let n=p^a*q^b so number of factors of n=(a+1)(b+1) and number of factors for 2n=2*(a+1)(b+1)
or 2n=(1+1)*(a+1)(b+1) so 2n has one more prime factor than n has

For question 2,I am not using any specific resource..just brushed up the basics..and different forum discussions really helps!

thanks! makes sense.

what other forums do you use?

jube wrote:

liferocks wrote:
for the question 1, let n=p^a*q^b so number of factors of n=(a+1)(b+1) and number of factors for 2n=2*(a+1)(b+1)
or 2n=(1+1)*(a+1)(b+1) so 2n has one more prime factor than n has

For question 2,I am not using any specific resource..just brushed up the basics..and different forum discussions really helps!

thanks! makes sense.

what other forums do you use?

As of now only btg..I should rephrase my sentence 'different forum discussions really helps' as 'different discussions in the forum really helps'

Is there a number properties rule that will help me think about this problem or is there an algebraic way to solve this. I mean, I really hate having to just pick numbers, but is that the only way to do this?