Data Sufficiency

Is y an integer?

(1) y^3 is an integer.

(2) 3 y is an integer.

sanju09 wrote:Is y an integer?

(1) y^3 is an integer.

(2) 3 y is an integer.

1) y can be either of the type y=3^1/3,10^1/3.. or y= 2^3,3^3..

i.e. both integral as well as non integral values are possible. hence not sufficient.

2) similarly here y can be fraction, (y=1/3,) or an integer.
hence is not sufficient,.

combining together.. y^3-3y must be integer because integer-integer must be integer.
y(y^2-3) if y becomes fraction than y(y^2-3) will be a fraction.
therefore y should be a integer.

hence C

yes i agree should be C....same reasoning.

manpsingh87 wrote:combining together.. y^3-3y must be integer because integer-integer must be integer.
y(y^2-3) if y becomes fraction than y(y^2-3) will be a fraction.
therefore y should be a integer.

Hi there!

This is not good, for some reasons:

01. The negation of integer is non-integer (not "fraction") and this is important because of (2) below:

02. y non-integer does NOT imply y(y^2-3) non-integer!! Take y = sqrt(3) for instance!

Therefore there is still something to be done: y^3 and 3y integers are really enough to guarantee that y is integer?

The answer is yes, let us see why:

let y^3 = M, where M is an integer. That means that y = cubicroot(M) ;
from 3y = 3*cubicroot(M) = N, where N is integer, we get 3^3*M = N^3 , therefore 3^3*M must be a perfect cube, and that means that M must be a perfect cube (because 3^3 it is), and we finally proved that y = cubicroot(M) where M is a perfect cube, then y is an integer.

I hope you liked the arguments.

Regards,
Fabio.

fskilnik wrote:

manpsingh87 wrote:combining together.. y^3-3y must be integer because integer-integer must be integer.
y(y^2-3) if y becomes fraction than y(y^2-3) will be a fraction.
therefore y should be a integer.

Hi there!

This is not good, for some reasons:

01. The negation of integer is non-integer (not "fraction") and this is important because of (2) below:

02. y non-integer does NOT imply y(y^2-3) non-integer!! Take y = sqrt(3) for instance!

Therefore there is still something to be done: y^3 and 3y integers are really enough to guarantee that y is integer?

The answer is yes, let us see why:

let y^3 = M, where M is an integer. That means that y = cubicroot(M) ;
from 3y = 3*cubicroot(M) = N, where N is integer, we get 3^3*M = N^3 , therefore 3^3*M must be a perfect cube, and that means that M must be a perfect cube (because 3^3 it is), and we finally proved that y = cubicroot(M) where M is a perfect cube, then y is an integer.

I hope you liked the arguments.

Regards,
Fabio.

hi sir,

01. The negation of integer is non-integer (not "fraction") and this is important because of (2) below:

absolutely agree with you sir, considering the context of the question i choose the fractional values only, (because of statement 2 3y is an integer, if we consider y to sqrt(3) then it wouldn't become integer and same goes for statement 1!!

02. y non-integer does NOT imply y(y^2-3) non-integer!! Take y = sqrt(3) for instance!

now sir here we can't assume y to be sqrt(3), because as per the conditions 1 and 2 only possible cases are, fractions, integers, and cubic roots..!!! that's why i didn't consider it..!!!

sir please correct me if you find any error in my this post,and yes thanks a lot sir for alternate explanation...!!!!

manpsingh87 wrote:sir please correct me if you find any error in my this post,and yes thanks a lot sir for alternate explanation...!!!!

Hi manpsingh87,

First of all, congrats for your respectiful attitude. It´s a pleasure to make myself clearer!

01. You are right that fraction values are "good candidates" for counter-examples, because (for instance) sqrt(3) will not satisfy sttm(2)... but what I tried to say is that you should not look for counter-examples only where you BELIEVE they may be found, you should look for any POSSIBLE counter-examples. In other words, I was simply trying to argue that when looking for the negation of a certain stament that ends with "then y is an integer", you may start with "suppose y is non-integer..." but you shouldn´t start with "suppose y is fractional", because then your argument will not cover ALL POSSIBLE situations.

02. I know sqrt(3) will not obbey sttms (1) and (2) together, because I PROVED there is NO non-integer value of y that satisfies both sttms (1) and (2) together! In other words, you are telling me that sqrt(3) does not "do the magic", and I agree, but you were just "illuding yourself" into the believe that where YOU THOUGHT you (perhaps) could find a counter-example, you did not find one! In other words, you should not say that y is an integer because you could NOT find a counter-example... you should say that y is an integer because THERE IS NO counter-example! Do you see the difference?

(If you tell me the official answer is (E), I am pretty sure that the official answer is wrong. But with your "argument", you could always be wrong... do you get it?)

Regards,
Fabio.

fskilnik wrote:

manpsingh87 wrote:sir please correct me if you find any error in my this post,and yes thanks a lot sir for alternate explanation...!!!!

Hi manpsingh87,

First of all, congrats for your respectiful attitude. It´s a pleasure to make myself clearer!

01. You are right that fraction values are "good candidates" for counter-examples, because (for instance) sqrt(3) will not satisfy sttm(2)... but what I tried to say is that you should not look for counter-examples only where you BELIEVE they may be found, you should look for any POSSIBLE counter-examples. In other words, I was simply trying to argue that when looking for the negation of a certain stament that ends with "then y is an integer", you may start with "suppose y is non-integer..." but you shouldn´t start with "suppose y is fractional", because then your argument will not cover ALL POSSIBLE situations.

02. I know sqrt(3) will not obbey sttms (1) and (2) together, because I PROVED there is NO non-integer value of y that satisfies both sttms (1) and (2) together! In other words, you are telling me that sqrt(3) does not "do the magic", and I agree, but you were just "illuding yourself" into the believe that where YOU THOUGHT you (perhaps) could find a counter-example, you did not find one! In other words, you should not say that y is an integer because you could NOT find a counter-example... you should say that y is an integer because THERE IS NO counter-example! Do you see the difference?

(If you tell me the official answer is (E), I am pretty sure that the official answer is wrong. But with your "argument", you could always be wrong... do you get it?)

Regards,
Fabio.

i agree with you sir, and thanks a lot for your help..!!!!

manpsingh87 wrote:i agree with you sir, and thanks a lot for your help..!!!!

My pleasure! I´m glad you like it, manpsingh87.

People believe this sort of "care" is not GMAT-oriented but, from my experience, students aiming at least scaled score of 48 (in the quantitative section) must know what they are doing... and the most careful/deep you study, the better!

See you in other BTG posts.

Cheers,
Fabio.

sanju09 wrote:Is y an integer?

(1) y^3 is an integer.

(2) 3 y is an integer.

I don't understand why the answer is not D. May someone break this down step by step?

Thanks,

Lawri

lawri wrote:

sanju09 wrote:Is y an integer?

(1) y^3 is an integer.

(2) 3 y is an integer.

I don't understand why the answer is not D. May someone break this down step by step?

Thanks,

Lawri

Hi, Lawri!

(1) Insufficient:

Take y=0, then (y^3 = 0 is an integer and) y is an integer
Take y = cubicroot(3), then (y^3 = 3 is an integer and) y is NOT an integer

(2) Insufficient:

Take y = 0, then (3y = 0 is an integer and) y is an integer
Take y = 1/3, then (3y = 1 is an integer and) y is NOT an integer

(1+2) Sufficient (as I explained previously): please see the image attached.

I hope things got clearer now.

Regards,
Fabio.