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If r – s = 3p , is p an integer? (1) r is divisible by 735 [#permalink]
21 Jan 2012, 14:46

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

71% (02:08) correct
29% (00:52) wrong based on 150 sessions

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?

Re: Is p an integer? [#permalink]
21 Jan 2012, 15:01

6

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

enigma123 wrote:

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.

Re: Is p an integer? [#permalink]
25 May 2012, 00:18

Bunuel wrote:

enigma123 wrote:

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.

Re: Is p an integer? [#permalink]
25 May 2012, 00:32

Expert's post

kunalbh19 wrote:

Bunuel wrote:

enigma123 wrote:

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. _________________

Re: If r – s = 3p , is p an integer? (1) r is divisible by 735 [#permalink]
06 Nov 2012, 04:32

1

This post received KUDOS

Expert's post

himanshuhpr wrote:

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).

Re: Is p an integer? [#permalink]
06 May 2013, 12:32

kunalbh19 wrote:

Bunuel wrote:

enigma123 wrote:

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 .

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