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If r – s = 3p , is p an integer? (1) r is divisible by 735
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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 rs/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 rs both divisible by 3.
Case 2 r=7 and s =5 then r+s is divisible by 3 and rs is NOT divisible by 3. Therefore this statement alone is INSUFFICIENT.
Now combining the two statements : I am struggling after this?
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MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730
Originally posted by enigma123 on 21 Jan 2012, 15:46.
Last edited by HKD1710 on 07 Oct 2017, 06:56, edited 1 time in total.
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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735
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21 Jan 2012, 16:01
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 rs 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 rs=1 and the answer is NO but if r=s=0 then rs=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 > rs={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 \(ab=3\), 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 \(ab=1\), 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 \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), both are divisible by 4. Hope it's clear.
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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735
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21 Jan 2012, 16:15
Classic explanation. Thanks very much.
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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735
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25 May 2012, 01:18
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 rs 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 rs=1 and the answer is NO but if r=s=0 then rs=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 > rs={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 \(ab=3\), 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 \(ab=1\), 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 \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), 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) rs+2s = 3a 3p + 2s = 3a (since we are given that rs=3p) 2s = 3(ap) since 2 and 3 are both primes, we can conclude that ap 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.



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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735
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25 May 2012, 01:32
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 rs 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 rs=1 and the answer is NO but if r=s=0 then rs=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 > rs={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 \(ab=3\), 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 \(ab=1\), 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 \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), 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) rs+2s = 3a 3p + 2s = 3a (since we are given that rs=3p) 2s = 3(ap) since 2 and 3 are both primes, we can conclude that ap 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 rs=1 and the answer is NO but if r=s=0 then rs=0 and the answer is YES. Not sufficient.
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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735
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06 Nov 2012, 05:23
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..



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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735
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06 Nov 2012, 05:32
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). Hope it helps.
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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735
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06 Nov 2012, 05:36
Thank you Bunuel , it's a new and very important thing which I have come to know.



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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735
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06 May 2013, 13:32
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 rs 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 rs=1 and the answer is NO but if r=s=0 then rs=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 > rs={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 \(ab=3\), 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 \(ab=1\), 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 \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), 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) rs+2s = 3a 3p + 2s = 3a (since we are given that rs=3p) 2s = 3(ap) since 2 and 3 are both primes, we can conclude that ap 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) rs+2s = 3a 3p + 2s = 3a (since we are given that rs=3p) 2s = 3(ap) since 2 and 3 are both primes, we can conclude that ap 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(ap) . you got it right till here. But you cannot conclude with just statement 2 , that ap 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(ap) ===> 2 (0.3) = 3(0.2) , where s=0.3 and ap=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(ap) , that P is an integer. cos 2*integer = 3 * integer , i.e ap SHOULD be an integer , since a is an integer and S is an integer . Hope that helps Jyothi
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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735
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20 Feb 2014, 10:33
Terrific Bunuel, the way you have explained the solution is just amazing. +1 kudos for the excellent concept explained



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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735
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01 Jul 2014, 03:42
Clearly p=(rs)/3
Statement 1
r is divisible by 735,so p=(735xs)/3=245xs/3……..not sufficient
Statement 2
r+s=3y↪↪p=(3yss)/3=(3y2s)/3=y2s/3…….not sufficient
Combining (1)+(2) p=(735xs)/3
p=(735x3yr)/3
now r is a multiple of 3,so it is a multiple of 3
therefore,p=(735x3y3z)/3=integer…..sufficient
AnswerC



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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735
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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 rs=5/3 So not sufficient. Combined, r is an integer and s is also an integer. Sufficient. Ans C.



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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735
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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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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735
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07 Oct 2017, 04:45
I feel B is sufficient.
rs=3p If we add 2s on both sides: r+s=3p+2s Hence, 3p+2s is divisible by 3. Since 3p is separately divisible by 3, 2s has to be divisible by 3. For 2s to be divisible by 3, s has to be divisible by 3. We already know r+s is divisible by 3. Now we have s is divisible by 3. Thus r is divisible by 3.(same logic as above) Hence we get that both s and r are divisible by 3 from statement (ii).
Please do correct me if I'm wrong.
Thanks



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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735
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07 Oct 2017, 04:49
Srisk wrote: I feel B is sufficient.
rs=3p If we add 2s on both sides: r+s=3p+2s Hence, 3p+2s is divisible by 3. Since 3p is separately divisible by 3, 2s has to be divisible by 3. For 2s to be divisible by 3, s has to be divisible by 3. We already know r+s is divisible by 3. Now we have s is divisible by 3. Thus r is divisible by 3.(same logic as above) Hence we get that both s and r are divisible by 3 from statement (ii).
Please do correct me if I'm wrong.
Thanks 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 rs=1 and the answer is NO but if r=s=0 then rs=0 and the answer is YES. Not sufficient.
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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735 &nbs
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