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

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21 Jan 2012, 15:46

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A

B

C

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E

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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]

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22 Oct 2015, 05:38

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

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27 Oct 2016, 06:52

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

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13 Apr 2017, 01:47

1] r=735k (k is any integer) We can just say that r is an integer. Nothing known about s. So not sufficient.

2] r+s=3p Again, we can just say that r+s is an integer. nothing known independently of r and s. For example r=7/3, s=2/3, r+s=3 but r-s=5/3 So not sufficient.

Combined, r is an integer and s is also an integer. Sufficient. Ans C.

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

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07 Jun 2017, 12:57

This question can be rephrased: Is r – s divisible by 3? Or, are r and s each divisible by 3? Statement (1) tells us that r is divisible by 735. If r is divisible by 735, it is also divisible by all the factors of 735. 3 is a factor of 735. (To test whether 3 is a factor of a number, sum its digits; if the sum is divisible by 3, then 3 is a factor of the number.) However, statement (1) does not tell us anything about whether or not s is divisible by 3. Therefore it is insufficient. Statement (2) tells us that r + s is divisible by 3. This information alone is insufficient. Consider each of the following two cases: CASEONE:Ifr=9,ands=6,r+s=15whichisdivisibleby3,and r–s=3,whichisalsodivisibleby3. CASETWO:Ifr=7ands=5,r+s=12,whichisdivisibleby3,but r–s=2,whichisNOTdivisibleby3. Let's try both statements together. There is a mathematical rule that states that if two integers are each divisible by the integer x, then the sum or difference of those two integers is also divisible by x. We know from statement (1) that r is divisible by 3. We know from statement (2) that r + s is divisible by 3. Using the converse of the aforementioned rule, we can deduce that s is divisible by 3. Then, using the rule itself, we know that the difference r – s is also divisible by 3. : BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

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