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:

If x and y are integers great than 1, is x a multiple of y?

(1) \(3y^2+7y=x\) --> \(y(3y+7)=x\) --> as \(3y+7=integer\), then \(y*integer=x\) --> \(x\) is a multiple of \(y\). Sufficient.

(2) \(x^2-x\) is a multiple of \(y\) --> \(x(x-1)\) is a multiple of \(y\) --> \(x\) can be multiple of \(y\) (\(x=2\) and \(y=2\)) OR \(x-1\) can be multiple of \(y\) (\(x=3\) and \(y=2\)) or their product can be multiple of \(y\) (\(x=3\) and \(y=6\)). Not sufficient.

Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]

Show Tags

05 Sep 2012, 08:00

Hi Bunuel, I have not understood the answer to the (2) statement, infact we have that x(x-1) = Y*F (because is a multiple), doesn't it depend on the number F if X is a multiple of y? For example: If x=2, then y could be either 2 or 1; so the answer would be yes; If x=3, then y could be either 2 or 3; so the answer would be no;

Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]

Show Tags

05 Sep 2012, 14:17

3

This post received KUDOS

3

This post was BOOKMARKED

for(2)

x(x-1) is multiple of y. so either x is divisible by y or x-1 is divisible by y. if x-1 is divisible by y, then x is not divisble by y.

plug in y=8, x=16--->16*15 is divisible by 8 --> yes x is a multiple of y. plug in y=8,x=17----> 17*16 is still divisible by 8 --> But x is not a multiple of y.

Re: Is x multiple of y? If: 3y^2 + 7y = x? If: x^2-x is mult? [#permalink]

Show Tags

31 Dec 2012, 21:27

15

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

Nadezda wrote:

x and y are integers > 1. Is x multiple of y?

(1): 3y^2 + 7y = x (2): x^2 - x is multiple of y

The question is asking whether \(x/y\) is an integer or not. Answer is A.

Statement 1 can be written as \(y(3y +7)=x\) or \(3y+7=x/y\). Since \(y\) is an integer, therefore 3y+7 must also be an integer. Hence \(x/y\) will be an integer or x is a multiple of y. Sufficient.

Statement 2 can be written as \(x(x-1)/y\) is an integer. Notice that we can't deduce that x is a multiple of y because it is quite possible that the product is a multiple of y, but not the individual entities. ex. 2*3/6 is a mltiple of 6 BUT neither 2 nor 3 is a multiple of 6. Insufficient.

+1A

Please do add the OA while posting questions. _________________

Re: Is x multiple of y? If: 3y^2 + 7y = x? If: x^2-x is mult? [#permalink]

Show Tags

31 Dec 2012, 22:41

Nadezda wrote:

x and y are integers > 1. Is x multiple of y?

(1): 3y^2 + 7y = x (2): x^2 - x is multiple of y

Hi Nadezda,

The Questions asks whether x/y is an integer or not.

from St 1 we have 3y^2 +7y=x ---> x/y= 3y+7 if y=2, x=13 and 13/2 is not an integer. If y=7, then x/y=4 which is an integer. So St1 alone not sufficient

From St 2 we have x^2-x is multiple of y ----> x(x-1)/y= Integer value Now we know x-1 and x are consecutive integers

X2-x is even because if x is odd then x2-x (odd-odd)=even and if x is even then x2-x is even. Now x can be odd or even and hence alone not sufficient.

Combining both statement,we get for x/y to be an integer y has to be 7 and x will be have to be multiple of 7. But we are not sure from above statement so ans should be E.

What is OA??

Thanks Mridul _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Re: Is x multiple of y? If: 3y^2 + 7y = x? If: x^2-x is mult? [#permalink]

Show Tags

31 Dec 2012, 22:59

Expert's post

mridulparashar1 wrote:

Nadezda wrote:

x and y are integers > 1. Is x multiple of y?

(1): 3y^2 + 7y = x (2): x^2 - x is multiple of y

Hi Nadezda,

The Questions asks whether x/y is an integer or not.

from St 1 we have 3y^2 +7y=x ---> x/y= 3y+7 if y=2, x=13 and 13/2 is not an integer. If y=7, then x/y=4 which is an integer. So St1 alone not sufficient

From St 2 we have x^2-x is multiple of y ----> x(x-1)/y= Integer value Now we know x-1 and x are consecutive integers

X2-x is even because if x is odd then x2-x (odd-odd)=even and if x is even then x2-x is even. Now x can be odd or even and hence alone not sufficient.

Combining both statement,we get for x/y to be an integer y has to be 7 and x will be have to be multiple of 7. But we are not sure from above statement so ans should be E.

What is OA??

Thanks Mridul

if x/y=3y+7, then taking y as 2 will yield x/y as 13 and not 13/2. _________________

Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]

Show Tags

02 Jan 2014, 00:27

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]

Show Tags

03 May 2014, 12:26

5

This post received KUDOS

3

This post was BOOKMARKED

FACTS ESTABLISHED BY STEMS

\(x\) is an integer

\(y\) is an integer

\(x > 1\)

\(y > 1\)

QUESTION TRANSLATED

Since both \(x\) and \(y\) are integers the question "is \(x\) a multiple of \(y\)?" really asks if

\(x = q \cdot y + r\)

where \(q\) is an integer (the quotient) and \(r\) is the remainder when you divide \(x\) by \(y\) so that \(r = 0\) which means that

\(x = q \cdot y + 0\)

which implies that

\(x = q \cdot y\)

So the question translates to:

is \(x = q \cdot y\) when \(x\), \(q\) and \(y\) are all integers, and when \(x>1\) and \(y > 1\)?

STATEMENT 1

\(3y^2 + 7y = x\)

can be rewritten as

\(y(3y + 7) = x\)

however, from the question stems we know that \(y\) is an integer, which implies \(3y\) is an integer and that \(3y+7\) is also an integer:

\(y \cdot integer = x\)

which can be rewritten as

\(x = y \cdot integer\) or better yet, as

\(x = integer \cdot y\)

which if you notice is an equation of the form \(x = q \cdot y\) since \(q\) is an integer which implies that \(x\) is indeed a multiple of \(y\).

Therefore, Statement A is sufficient.

STATEMENT 2

the statement "\(x^2 - x\) is a multiple of \(y\)" must be rewritten as "\(x (x - 1)\) is a multiple of \(y\)" which implies that

either \(x\) is a multiple of \(y\) or \(x-1\) is a multiple of \(y\)

the sub-statement \(x\) is a multiple of \(y\) implies that \(x = q \cdot y\) which would answer the question, but we still need to figure out if the second sub-statement answers the question with the same answer as the first sub-statement.

the second sub-statement \(x-1\) is a multiple of \(y\) implies that

\(x-1 = q \cdot y\)

which can be rewritten as

\(x = q \cdot y + 1\)

which is an equation of the form \(x = q \cdot y + r\) when \(r=1\) which implies that when \(x\) is divided by \(y\) we get a remainder of 1, meaning that \(x\) is NOT a multiple of \(y\).

Since both sub-statements are contradictory, this means that when \(x^2 - x\) is a multiple of \(y\), \(x\) is not always a multiple of \(y\). meaning that we cannot determine whether \(x\) is always a multiple of \(y\) according to Statement 2 alone.

Therefore, Statement 2 is not sufficient.

ANSWER

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]

Show Tags

18 Sep 2014, 18:36

Bunuel wrote:

If x and y are integers great than 1, is x a multiple of y?

(1) \(3y^2+7y=x\) --> \(y(3y+7)=x\) --> as \(3y+7=integer\), then \(y*integer=x\) --> \(x\) is a multiple of \(y\). Sufficient.

(2) \(x^2-x\) is a multiple of \(y\) --> \(x(x-1)\) is a multiple of \(y\) --> \(x\) can be multiple of \(y\) (\(x=2\) and \(y=2\)) OR \(x-1\) can be multiple of \(y\) (\(x=3\) and \(y=2\)) or their product can be multiple of \(y\) (\(x=3\) and \(y=6\)). Not sufficient.

Answer: A.

Hope it helps.

Hi Bunuel,

Two questions: 1) If x = 2 and y = 2, does that mean that x is a multiple of y? I always thought that x needed to be 1 multiple greater, meaning, x=4 and y=2 etc? 2) In situations like these, I usually stumble with either starting off with factoring(like you did above) or some other method. What do you suggest? 3) Lastly, the way I solved statement 1 was -- I saw it as (y)(multiple of y) = x, and the rule states that if i multiply two numbers that are multiples of y, then the result will be a multiple of y. Is that correct to assume?(I know that we can assume that with addition, right?)

Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]

Show Tags

19 Sep 2014, 02:31

Expert's post

russ9 wrote:

Bunuel wrote:

If x and y are integers great than 1, is x a multiple of y?

(1) \(3y^2+7y=x\) --> \(y(3y+7)=x\) --> as \(3y+7=integer\), then \(y*integer=x\) --> \(x\) is a multiple of \(y\). Sufficient.

(2) \(x^2-x\) is a multiple of \(y\) --> \(x(x-1)\) is a multiple of \(y\) --> \(x\) can be multiple of \(y\) (\(x=2\) and \(y=2\)) OR \(x-1\) can be multiple of \(y\) (\(x=3\) and \(y=2\)) or their product can be multiple of \(y\) (\(x=3\) and \(y=6\)). Not sufficient.

Answer: A.

Hope it helps.

Hi Bunuel,

Two questions: 1) If x = 2 and y = 2, does that mean that x is a multiple of y? I always thought that x needed to be 1 multiple greater, meaning, x=4 and y=2 etc? 2) In situations like these, I usually stumble with either starting off with factoring(like you did above) or some other method. What do you suggest? 3) Lastly, the way I solved statement 1 was -- I saw it as (y)(multiple of y) = x, and the rule states that if i multiply two numbers that are multiples of y, then the result will be a multiple of y. Is that correct to assume?(I know that we can assume that with addition, right?)

1. The least multiple of a positive integer is that integer itself. So, the least multiple of 2 is 2. 2. I suggest the way I used. 3. 3y+7 is not necessarily a multiple of y but it does not need to be. y*integer=x implies that x is a multiple of y. _________________

Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]

Show Tags

27 Sep 2015, 20:04

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

MBA Admission Calculator Officially Launched After 2 years of effort and over 1,000 hours of work, I have finally launched my MBA Admission Calculator . The calculator uses the...

Final decisions are in: Berkeley: Denied with interview Tepper: Waitlisted with interview Rotman: Admitted with scholarship (withdrawn) Random French School: Admitted to MSc in Management with scholarship (...

The London Business School Admits Weekend officially kicked off on Saturday morning with registrations and breakfast. We received a carry bag, name tags, schedules and an MBA2018 tee at...