Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Statements (1) and (2) TOGETHER are NOT sufficient Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and\(x^2 = 9\) which is divisible by 9?

Re: Is [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]

Show Tags

13 Sep 2010, 14:23

In fact, even B is sufficient.

since xy is divisible by 9, 18, 27, 36 . xy can have (1,9), (9,1), (3,3) , (2,9), (9,2), (6,3), (3,6) etc. all these pairs when substituted for x^2 y^4, are divisible by 9.

Re: Is [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]

Show Tags

13 Sep 2010, 14:32

4

This post received KUDOS

2

This post was BOOKMARKED

seekmba wrote:

Is \(x^2 y^4\) an integer divisible by 9 ?

1. x is an integer divisible by 3 2. xy is an integer divisible by 9

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and\(x^2 = 9\) which is divisible by 9?

the question ask if x^2y^4 is an INTEGER divisible by 9 but nowhere it said x or y has to be integer themselves!

1. Y could be non -integer so INSUFF 2. xy is divisible by 9. so it is reasonable to say x*x*y*y divisible by 9 right? so that we know x^2y^2 is divisible by 9. HOWEVER, we dont know what the extra y^2 is - it could be 9 or could be 16 what not so INSUFF

c. lets combine - we know x^2*y^2 div by 9 with y^2 unknown. also we know x is div by 3 BUT we can have some situations like: 3^2 * 1^2(the first part satisfies 2) *1^2 = 9 which is SUFF X=3, Y=1 OR 9^2 * (1/3)^2 (again satisfies 2) * (1/3)^2 = 1 which is INSUFF X=9 , Y = 1/3

E hope it helps...
_________________

If you like my answers please +1 kudos!

Last edited by shaselai on 13 Sep 2010, 14:34, edited 1 time in total.

1. x is an integer divisible by 3 2. xy is an integer divisible by 9

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and\(x^2 = 9\) which is divisible by 9?

Note that we are not told that \(x\) and \(y\) are integers.

(1) x is an integer divisible by 3 --> \(x=6\) and \(y=any \ integer\) then the answer would be YES but if \(x=6\) and \(y=\frac{3}{2}\) then the answer would be NO. Not sufficient.

(2) xy is an integer divisible by 9 --> the same example: \(x=6\) and \(y=any \ integer\) then the answer would be YES but if \(x=6\) and \(y=\frac{3}{2}\) then the answer would be NO. Not sufficient.

(1)+(2) We can use the above examples again:

If \(x=6\) (divisible by 3) and \(y=any \ integer\) (\(xy=multiple \ of \ 3\)) then the answer would be YES; If \(x=6\) (divisible by 3) and \(y=\frac{3}{2}\) (\(xy=9\), hence divisible by 9) then the answer would be NO --> \(x^2*y^4=(xy)^2*y^2=81*\frac{9}{4}\) not an integer hence not divisible by 9.

1. x is an integer divisible by 3 2. xy is an integer divisible by 9

One more thing about this question:

On GMAT when we are told that \(a\) is divisible by \(b\) (or which is the same: "\(a\) is multiple of \(b\)", or "\(b\) is a factor of \(a\)"), we can say that: 1. \(a\) is an integer; 2. \(b\) is an integer; 3. \(\frac{a}{b}=integer\).

So the terms "divisible", "multiple", "factor" ("divisor") are used only about integers (at least on GMAT).

Which means that we could omit words in red in the question. For example "is \(x^2 y^4\) an integer divisible by 9" is the same as "is \(x^2 y^4\) divisible by 9" as well as "xy is an integer divisible by 9" is the same as "xy is divisible by 9".

Re: Is x^2 y^4 an integer divisible by 9 ? 1. x is an integer [#permalink]

Show Tags

17 Aug 2012, 12:19

Bunuel, when you use y= any integer in your explanation, what if y was zero (in your example)? x^2=36 and y^2=0 (any integer) would not be divisible by 9, right?

Bunuel, when you use y= any integer in your explanation, what if y was zero (in your example)? x^2=36 and y^2=0 (any integer) would not be divisible by 9, right?

If y=0, then x^2*y^4=0={even integer}, because zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Statements (1) and (2) TOGETHER are NOT sufficient Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and\(x^2 = 9\) which is divisible by 9?

......... without any doubt the Answer is (E)......................... consider fractions, then you will find x^2*y^4 is divisible by 9 but still is a fraction due to y being a fraction.
_________________

Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]

Show Tags

09 Oct 2013, 22:23

but if we take y as a fraction then \(x^2y^4\) wouldn't be an integer. But it's stated in the question "is \(x^2y^4\) an integer divisible by 9?" So how can we take y as a fraction when the whole expression\(x^2y^4\) has to be an integer. I'm totally confused.

but if we take y as a fraction then \(x^2y^4\) wouldn't be an integer. But it's stated in the question "is \(x^2y^4\) an integer divisible by 9?" So how can we take y as a fraction when the whole expression\(x^2y^4\) has to be an integer. I'm totally confused.

The question asks IS x^2*y^4 an integer divisible by 9. So, we are not told that x^2*y^4 is an integer. It has to be an integer to be divisible by 9, but it's not given.
_________________

Is x^2*y^4 an integer divisible by 9 ? [#permalink]

Show Tags

28 Sep 2014, 11:43

Bunuel wrote:

seekmba wrote:

Is \(x^2 y^4\) an integer divisible by 9 ?

1. x is an integer divisible by 3 2. xy is an integer divisible by 9

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and\(x^2 = 9\) which is divisible by 9?

Note that we are not told that \(x\) and \(y\) are integers.

(1) x is an integer divisible by 3 --> \(x=6\) and \(y=any \ integer\) then the answer would be YES but if \(x=6\) and \(y=\frac{3}{2}\) then the answer would be NO. Not sufficient.

(2) xy is an integer divisible by 9 --> the same example: \(x=6\) and \(y=any \ integer\) then the answer would be YES but if \(x=6\) and \(y=\frac{3}{2}\) then the answer would be NO. Not sufficient.

(1)+(2) We can use the above examples again:

If \(x=6\) (divisible by 3) and \(y=any \ integer\) (\(xy=multiple \ of \ 3\)) then the answer would be YES; If \(x=6\) (divisible by 3) and \(y=\frac{3}{2}\) (\(xy=9\), hence divisible by 9) then the answer would be NO --> \(x^2*y^4=(xy)^2*y^2=81*\frac{9}{4}\) not an integer hence not divisible by 9.

Not sufficient.

Answer: E.

Hope it's clear.

Hi Bunuel,

The explanation is crystal clear here. However, i have a doubt with regard to the concept (Exponentiation does not produce primes) -> I will be referring to it as concept E for convenience.

So as per Concept E if x is divisible by 9 then x^2 will also be divisible by 9.

In the above question: x is an integer divisible by 3, then x^2 y^4 (assuming y is an integer) is divisible by 9 is via some other practical logic but not CONCEPT E right? If yes, then i am guessing it is done through basic plugging in of numbers.

Statements (1) and (2) TOGETHER are NOT sufficient Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and\(x^2 = 9\) which is divisible by 9?

Dear Bunuel, for statement 2 I chose x=1/2 and y = 18 and the result of x^2*y^4 was divisible by 9 how i can find

smart numbers as that you used under 2 minutes
_________________

Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]

Show Tags

24 Aug 2015, 06:21

Bunuel wrote:

seekmba wrote:

Is \(x^2 y^4\) an integer divisible by 9 ?

1. x is an integer divisible by 3 2. xy is an integer divisible by 9

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and\(x^2 = 9\) which is divisible by 9?

Note that we are not told that \(x\) and \(y\) are integers.

(1) x is an integer divisible by 3 --> \(x=6\) and \(y=any \ integer\) then the answer would be YES but if \(x=6\) and \(y=\frac{3}{2}\) then the answer would be NO. Not sufficient.

(2) xy is an integer divisible by 9 --> the same example: \(x=6\) and \(y=any \ integer\) then the answer would be YES but if \(x=6\) and \(y=\frac{3}{2}\) then the answer would be NO. Not sufficient.

(1)+(2) We can use the above examples again:

If \(x=6\) (divisible by 3) and \(y=any \ integer\) (\(xy=multiple \ of \ 3\)) then the answer would be YES; If \(x=6\) (divisible by 3) and \(y=\frac{3}{2}\) (\(xy=9\), hence divisible by 9) then the answer would be NO --> \(x^2*y^4=(xy)^2*y^2=81*\frac{9}{4}\) not an integer hence not divisible by 9.

Not sufficient.

Answer: E.

Hope it's clear.

i know in GMAT when they say x is divisible by 4 they assume x is an integer but here how did you come up with x=6 and y =3/2 ?

1. x is an integer divisible by 3 2. xy is an integer divisible by 9

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and\(x^2 = 9\) which is divisible by 9?

Note that we are not told that \(x\) and \(y\) are integers.

(1) x is an integer divisible by 3 --> \(x=6\) and \(y=any \ integer\) then the answer would be YES but if \(x=6\) and \(y=\frac{3}{2}\) then the answer would be NO. Not sufficient.

(2) xy is an integer divisible by 9 --> the same example: \(x=6\) and \(y=any \ integer\) then the answer would be YES but if \(x=6\) and \(y=\frac{3}{2}\) then the answer would be NO. Not sufficient.

(1)+(2) We can use the above examples again:

If \(x=6\) (divisible by 3) and \(y=any \ integer\) (\(xy=multiple \ of \ 3\)) then the answer would be YES; If \(x=6\) (divisible by 3) and \(y=\frac{3}{2}\) (\(xy=9\), hence divisible by 9) then the answer would be NO --> \(x^2*y^4=(xy)^2*y^2=81*\frac{9}{4}\) not an integer hence not divisible by 9.

Not sufficient.

Answer: E.

Hope it's clear.

i know in GMAT when they say x is divisible by 4 they assume x is an integer but here how did you come up with x=6 and y =3/2 ?

(2) says that xy is an integer divisible by 9. So, we are told that xy is an integer, not x and y separately. If \(x=6\) and \(y=\frac{3}{2}\), then \(xy=9=integer \ divisible \ by \ 9\).
_________________

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...