DS: Is the integer n odd?

[cesar.rodriguez.blanco]

Is the integer n odd?
1) n is divisible by 3
2) 2n is divisible by twice as many positive integers as n

I do not know why the correct answer is C.

[banakida]

Are you sure the correct answer is not B?

[RonPurewal]

Is the integer n odd?
1) n is divisible by 3
2) 2n is divisible by twice as many positive integers as n

I do not know why the correct answer is C.

this should be (b).

(1) is clearly insufficient. (3: yes; 6: no.)

as for (2), that's much more difficult.

there's only one way that 2n could be divisible by twice as many positive integers as n itself. here's how that would work:
for every factor of n, the multiplication by 2 would create a NEW factor of 2n (i.e., a number that is NOT already a factor of n).
this could only happen if NONE of the factors of n contains a '2' in its prime factorization.
if that's true, then n is odd.
therefore n must be odd.

the answer should be (b).

[naren_hn]

Ron,

As you have noted, it is easier to conclude that for a even number, multiplying it by 2 will not result in a new factor and for an odd number, multiplying it by 2 will result in a new factor. But, what does this sentence exactly mean - "is divisible by twice as many positive integers as n"?

[mschwrtz]

That means that if you list all the factors of n, and all the factors of 2n, the latter list will have twice as many members as will the former.

I can't really add anything to Ron's explanation except an example; I hope that that helps.

Could n be the odd number 35? Sure. 35 has four factors, (1, 5, 7, 35), while 70 has eight, (1, 2, 5, 7, 10, 14, 35, 70). Notice that each of the factors of 70 in bold is twice a corresponding factor of 35. For each of the original--all odd--factors of 35 there is a new corresponding even factor of 70.

But if n were even, then both 1 and 2 would already be among its factors, and doubling 1 would not add a new factor.

Could n be, say, the even number 30? 30 has eight even factors (1, 2, 3, 5, 6, 10, 15, 30), while 60, 2n, has twelve (1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60). Twelve ain't twice eight, so n could not be 30.

Notice these pairs of factors of 30: (1, 2), (3, 6), (5, 10), (15, 30). Doubling 30 would not add a new factor corresponding to 1, because 2 is already among the factors. Doubling 30 would not add a new factor corresponding to 3, because 6 is already among the factors, etc.

[mba012012]

you guys are simply superb ! Thanks for a thorough explanation !

[RonPurewal]

you guys are simply superb ! Thanks for a thorough explanation !

sure

[tsiria]

Another method would be to count factors using primes:

If n is odd: n = 15 = 3 * 5 = 2*2 = 4 factors
2n = 30 = 2*3*5 = 2*2*2 = 8 factors

If n is even: n= 10 = 2*5 = 2*2 = 4 factors
2n=20= (2^2)*5 = 3*2 = 6 factors

Since 2n can only have twice as many factors as n when n is odd, statement 2 is sufficient.

[RonPurewal]

tsiria, i'm not seeing how you get from here ...

Another method would be to count factors using primes:

If n is odd: n = 15 = 3 * 5 = 2*2 = 4 factors
2n = 30 = 2*3*5 = 2*2*2 = 8 factors

If n is even: n= 10 = 2*5 = 2*2 = 4 factors
2n=20= (2^2)*5 = 3*2 = 6 factors

... to here ...

[/quote]Since 2n can only have twice as many factors as n when n is odd[/quote]

are you implying (and just not saying) that you also tried a considerable number of other examples before coming to this conclusion?
because you definitely can't generalize from one example of each type.

[ven2]

Are you sure the correct answer is not B?

The answer IS b for this one

[jnelson0612]

Correct.

[mail2arifmd]

Generalisation of tisira explanation
Number N represented in factor form = a^p . b^q (where a, b are prime).
No.of factors of N = (p+1)(q+1)
Note that Odd numbers dont have "2" as a factor to them.

So when we take "2N" then factor form = 2^1 . a^p . b^q
No.of factors of 2N = (1+1)(p+1)(q+1) = 2[no.of factors of N]
Hence if N is odd then 2N will have twice the number of factors as N.

Consider N as Even. The Factors of which already contains "2" as factor. It could be 2 or 4 or 6 or 8 or any even factor of 2^r
N = a^p . b^q . 2^r
No.of Factors of N = (p+1)(q+1)(r+1)
Now 2N = a^p . b^q . 2^(r+1)
No.of Factors of 2N =(p+1)(q+1)(r+1+1) = (p+1)(q+1)(r+2)
The relation ship between factors of N and 2N is difficult to show here. It can be double in some cases and might not be double in some other cases.

This explanation is just a prove of concept and not a process for GMAT. The only thing we need to remember for GMAT is:

Rule1) The No.Of factors of 2N = twice the no.of factors of N, when N is Odd.

Rule2) Above rule doesn't fit for Even numbers. It may or may not be twice.

[tim]

your explanation is correct. i'm not sure this is a rule that one should memorize for the GMAT though, since it is so obscure and unlikely to show up on your test. instead, you should learn how to determine the number of factors of an integer and use that to inform your approach to this problem..