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 r – s = 3p , is p an integer? (1) r is divisible by 735 [#permalink]

Show Tags

21 Jan 2012, 14:46

1

This post received KUDOS

11

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

72% (02:10) correct
28% (00:59) wrong based on 261 sessions

HideShow timer Statistics

If r – s = 3p , is p an integer?

(1) r is divisible by 735 (2) r + s is divisible by 3

OA is C. I am struggling to find how. This is how I approaching the question. Can someone please help?

Considering Questions Stem

We have to find whether r-s/3 as p is an integer?

Considering statement 1

r is a factor of 735. That means r is divisible by all factors of 735 and 3 is a factor of 735 [Because 7+3+5=15]. But as the statement doesn't mention anything about s, it's INSUFFICIENT to answer the question.

Considering statement 2

r+s is divisible by 3

Case 1

r=6 s = 3 then r+s and r-s both divisible by 3.

Case 2 r=7 and s =5 then r+s is divisible by 3 and r-s is NOT divisible by 3. Therefore this statement alone is INSUFFICIENT.

Now combining the two statements : I am struggling after this?

If r – s = 3p, is p an integer? (1) r is divisible by 735 (2) r + s is divisible by 3

If r – s = 3p , is p an integer?

Question basically asks whether r-s is a multiple of 3 (..., -6, -3, 0, 3, 6, ...), because if it is then p would be an integer.

(1) r is divisible by 735 --> r is a multiple of 3 (as the sum of the digits of 735 is divisible by 3), though this statement is insufficient as no info about s.

(2) r + s is divisible by 3 --> r + s is a multiple of 3. Now, if r=2 and s=1 then r-s=1 and the answer is NO but if r=s=0 then r-s=0 and the answer is YES. Not sufficient.

(1)+(2) From (1) r={multiple of 3} then as from (2) r+s={multiple of 3}+s={multiple of 3} then s is also a multiple of 3 --> r-s={multiple of 3}-{multiple of 3}={multiple of 3}. Sufficient.

Answer: C.

Below might help to understand this concept better.

If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)): Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

If r – s = 3p, is p an integer? (1) r is divisible by 735 (2) r + s is divisible by 3

If r – s = 3p , is p an integer?

Question basically asks whether r-s is a multiple of 3 (..., -6, -3, 0, 3, 6, ...), because if it is then p would be an integer.

(1) r is divisible by 735 --> r is a multiple of 3 (as the sum of the digits of 735 is divisible by 3), though this statement is insufficient as no info about s.

(2) r + s is divisible by 3 --> r + s is a multiple of 3. Now, if r=2 and s=1 then r-s=1 and the answer is NO but if r=s=0 then r-s=0 and the answer is YES. Not sufficient.

(1)+(2) From (1) r={multiple of 3} then as from (2) r+s={multiple of 3}+s={multiple of 3} then s is also a multiple of 3 --> r-s={multiple of 3}-{multiple of 3}={multiple of 3}. Sufficient.

Answer: C.

Below might help to understand this concept better.

If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)): Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

Hope it's clear.

Hi,

I think B should be sufficient. Here is how:

We are given r+s is divisible by 3.

r+s=3a (where a is any integer) -----(1) r-s+2s = 3a 3p + 2s = 3a (since we are given that r-s=3p) 2s = 3(a-p) since 2 and 3 are both primes, we can conclude that a-p is a multiple of 2 and s is a multiple of 3.

similarly if we replace r by (-r+2r), we will get that r is also divisible by 3.

hence, if both r and s are divisible by 3, therefore p is an integer.

If r – s = 3p, is p an integer? (1) r is divisible by 735 (2) r + s is divisible by 3

If r – s = 3p , is p an integer?

Question basically asks whether r-s is a multiple of 3 (..., -6, -3, 0, 3, 6, ...), because if it is then p would be an integer.

(1) r is divisible by 735 --> r is a multiple of 3 (as the sum of the digits of 735 is divisible by 3), though this statement is insufficient as no info about s.

(2) r + s is divisible by 3 --> r + s is a multiple of 3. Now, if r=2 and s=1 then r-s=1 and the answer is NO but if r=s=0 then r-s=0 and the answer is YES. Not sufficient.

(1)+(2) From (1) r={multiple of 3} then as from (2) r+s={multiple of 3}+s={multiple of 3} then s is also a multiple of 3 --> r-s={multiple of 3}-{multiple of 3}={multiple of 3}. Sufficient.

Answer: C.

Below might help to understand this concept better.

If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)): Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

Hope it's clear.

Hi,

I think B should be sufficient. Here is how:

We are given r+s is divisible by 3.

r+s=3a (where a is any integer) -----(1) r-s+2s = 3a 3p + 2s = 3a (since we are given that r-s=3p) 2s = 3(a-p) since 2 and 3 are both primes, we can conclude that a-p is a multiple of 2 and s is a multiple of 3.

similarly if we replace r by (-r+2r), we will get that r is also divisible by 3.

hence, if both r and s are divisible by 3, therefore p is an integer.

please correct me if i am wrong somewhere.

Thanks in advance.

OA for this question is C, not B. OA is given in the initial post under the spoiler.

Next, for the second statement there are two examples given in my post which give different answer to the question whether p is an integer.

(2) r + s is divisible by 3 --> r + s is a multiple of 3. Now, if r=2 and s=1 then r-s=1 and the answer is NO but if r=s=0 then r-s=0 and the answer is YES. Not sufficient.
_________________

The above mentioned solution is valid for integers .... What if s is a fraction say 3/4 ... The answer should be E in that case..

(1) says that r is divisible by 735, which implies that r is an integer. Next, (2) says that r + s is divisible by 3, which implies that r +s is an integer and since r is an integer then so is s. Thus, when we consider the two statements together we know that both r and s are integers.

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).

If r – s = 3p, is p an integer? (1) r is divisible by 735 (2) r + s is divisible by 3

If r – s = 3p , is p an integer?

Question basically asks whether r-s is a multiple of 3 (..., -6, -3, 0, 3, 6, ...), because if it is then p would be an integer.

(1) r is divisible by 735 --> r is a multiple of 3 (as the sum of the digits of 735 is divisible by 3), though this statement is insufficient as no info about s.

(2) r + s is divisible by 3 --> r + s is a multiple of 3. Now, if r=2 and s=1 then r-s=1 and the answer is NO but if r=s=0 then r-s=0 and the answer is YES. Not sufficient.

(1)+(2) From (1) r={multiple of 3} then as from (2) r+s={multiple of 3}+s={multiple of 3} then s is also a multiple of 3 --> r-s={multiple of 3}-{multiple of 3}={multiple of 3}. Sufficient.

Answer: C.

Below might help to understand this concept better.

If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)): Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

Hope it's clear.

Hi,

I think B should be sufficient. Here is how:

We are given r+s is divisible by 3.

r+s=3a (where a is any integer) -----(1) r-s+2s = 3a 3p + 2s = 3a (since we are given that r-s=3p) 2s = 3(a-p) since 2 and 3 are both primes, we can conclude that a-p is a multiple of 2 and s is a multiple of 3.

similarly if we replace r by (-r+2r), we will get that r is also divisible by 3.

hence, if both r and s are divisible by 3, therefore p is an integer.

please correct me if i am wrong somewhere.

Thanks in advance.

I did the same mistake as you and assumed statement 2 by itself would suffice . After pondering on it for a while , figured where I went wrong. To answer your question , consider your below explanation -

r+s=3a (where a is any integer) -----(1) r-s+2s = 3a 3p + 2s = 3a (since we are given that r-s=3p) 2s = 3(a-p) since 2 and 3 are both primes, we can conclude that a-p is a multiple of 2 and s is a multiple of 3.

Let me try to explain why statement 2 can be insufficient. 2s=3(a-p) . you got it right till here. But you cannot conclude with just statement 2 , that a-p is a multiple of 2 . Cos all we know untill this point is "a" is an integer. We do not know weather S or/and P is an integer . To illustrate what I mean above, lemme give you an example - consider 2s=3(a-p) ===> 2 (0.3) = 3(0.2) , where s=0.3 and a-p=0.2 , a is an integer , lets say 2 , in which case P would be 1.8 . Hence we can prove that p is not an integer. similarly , we can prove otherwise that P is an integer.

Now if you consider statement 1 , which says S is an integer , we can conclude from the equation 2S=3(a-p) , that P is an integer. cos 2*integer = 3 * integer , i.e a-p SHOULD be an integer , since a is an integer and S is an integer .

Re: If r – s = 3p , is p an integer? (1) r is divisible by 735 [#permalink]

Show Tags

22 Oct 2015, 04:38

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 r – s = 3p , is p an integer? (1) r is divisible by 735 [#permalink]

Show Tags

27 Oct 2016, 05:52

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.
_________________

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

[rss2posts title=The MBA Manual title_url=https://mbamanual.com/2016/11/22/mba-vs-mim-guest-post/ sub_title=MBA vs. MiM :3qa61fk6]Hey, guys! We have a great guest post by Abhyank Srinet of MiM-Essay . In a quick post and an...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...