integer

[vietst]

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

[StaceyKoprince]

It's generally more useful if you tell us what you struggled with so that we can target our answer to your needs.

n is an integer
is n odd?
yes/no question, so I will try to prove it wrong (that is, get a yes and a no based upon the statements)

(1) n/3
n could be 6 (that is divisible by 3). Is n odd? No
n could be 9 (that is divisible by 3). Is n odd? Yes
Elim A and D

(2) 2n has twice as many factors as n
n could be 1, which has one factor; 2n would be 2, which has two factors; is n odd? Yes
n could be 2, which has two factors; 2n would be 4, which has three factors. Oops, can't use this combo of numbers (has to make statement 2 true, and this combo doesn't)

What's going on here?
general rule: 2n will be divisible by 2 and also by whatever number 2n is.
If I make n an even number, even numbers are already divisible by 2. So 2n will only be divisible by one new number, equal to 2n. That is, I add only one new factor for 2n. [editor: there's a mistake in this explanation - see below for a correction]
Any even number, by definition, has at least two factors - 1 and 2. So I would need to add at least two more factors to double the number of factors. But I can't - the setup of statement 2 only allows me to add one new factor if n is even. So I can never make statement 2 true using an even number for n.

Sufficient. Answer is B.

[vietst]

Stacey Koprince
Thanks

[StaceyKoprince]

no problem!

[goelmohit2002]

Hi Stacey,

Probably I am missing something here...but if you take the example number 52.

Here 52 is divisible by 1, 2, 4, 13, 26. [editor: 52 is also divisible by 52 itself.]

But 104 is divisible by 1, 2, 4, 8, 13, 26, 52, 104

Thus it is not that only one extra is added when multiplied by 2.....there can be more too...

Thanks
Mohit

[RonPurewal]

Hi Stacey,

Probably I am missing something here...but if you take the example number 52.

Here 52 is divisible by 1, 2, 4, 13, 26. [editor: 52 is also divisible by 52 itself.]

But 104 is divisible by 1, 2, 4, 8, 13, 26, 52, 104

Thus it is not that only one extra is added when multiplied by 2.....there can be more too...

Thanks
Mohit

very true. good catch.

here's a revised version:

let's say a number has "n" different factors.
when you multiply this number by 2, you POTENTIALLY create "n" MORE factors - by doubling each factor.
HOWEVER,
the only way that ALL of these factors can be NEW (i.e., not already listed in the original n factors) is if they are ALL ODD.
if there are ANY even factors to start with, then those factors will be repeated in the original list. (for instance, note that 2, 4, 26, and 52 all appear in both lists above.) therefore, if the number is even, then the number of factors will be less than doubled because of the repeat factors.

thus if statement (2) is true, then the number must be odd.

[goelmohit2002]

Hi Stacey,

Probably I am missing something here...but if you take the example number 52.

Here 52 is divisible by 1, 2, 4, 13, 26. [editor: 52 is also divisible by 52 itself.]

But 104 is divisible by 1, 2, 4, 8, 13, 26, 52, 104

Thus it is not that only one extra is added when multiplied by 2.....there can be more too...

Thanks
Mohit

very true. good catch.

here's a revised version:

let's say a number has "n" different factors.
when you multiply this number by 2, you POTENTIALLY create "n" MORE factors - by doubling each factor.
HOWEVER,
the only way that ALL of these factors can be NEW (i.e., not already listed in the original n factors) is if they are ALL ODD.
if there are ANY even factors to start with, then those factors will be repeated in the original list. (for instance, note that 2, 4, 26, and 52 all appear in both lists above.) therefore, if the number is even, then the number of factors will be less than doubled because of the repeat factors.

thus if statement (2) is true, then the number must be odd.

Thanks Ron !!!

[Ben Ku]

Glad it helped!

[gsingh058]

ist statement, If n=0, then n is divided by 3 but it is even. so answer is No. Can we consider n=0 for such division ?

2nd Statement : We can't consider n=0 becasue '2n=0" is divisble my all the positive integer (more than twice) but the statements says, n shall be divided by twice as many positive integer.

Is the above reasoning to choose n=0 in the 1st statement and not to choose n=0 in the 2nd statement is correct ?

Gagan

[RonPurewal]

ist statement, If n=0, then n is divided by 3 but it is even. so answer is No. Can we consider n=0 for such division ?

2nd Statement : We can't consider n=0 becasue '2n=0" is divisble my all the positive integer (more than twice) but the statements says, n shall be divided by twice as many positive integer.

Is the above reasoning to choose n=0 in the 1st statement and not to choose n=0 in the 2nd statement is correct ?

Gagan

technically, yes, but the gmat will definitely NOT test you on the divisibility properties of 0 or negative numbers. you will only be tested on divisibility for positive numbers.

so, these things are interesting curiosities, but don't dwell excessively on them.