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Is b/3 an integer?
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Updated on: 26 Nov 2012, 03:16
Question Stats:
45% (01:41) correct 55% (01:42) wrong based on 242 sessions
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Is b/3 an integer? (1) (b^2−9)/3 is an integer. (2) p and p + 1 are prime factors of b. All,
I am having trouble with even seemingly easy questions in DS and can't see my way out. Would you please advise? I picked A, but it is quite the opposite (B).
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Originally posted by valerjo79 on 26 Nov 2012, 03:11.
Last edited by Bunuel on 26 Nov 2012, 03:16, edited 1 time in total.
Renamed the topic and edited the question.




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Re: Is b/3 an integer?
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26 Nov 2012, 03:25
Is b/3 an integer?(1) (b^2−9)/3 is an integer. Given that \(\frac{b^29}{3}=integer\) > \(\frac{b^2}{3}3=integer\) > \(\frac{b^2}{3}=integer\). Now, if \(b=0\), then \(\frac{b}{3}=0=integer\) but if \(b=\sqrt{3}\), then \(\frac{b}{3}\neq{integer}\). Not sufficient. (2) p and p + 1 are prime factors of b. Since both \(p\) and \(p+1\) are primes, then \(p=2\) and \(p+1=3\). Thus we have that \(b\) is a multiple of 3. Sufficient. Answer: B. P.S. Please read carefully and follow: rulesforpostingpleasereadthisbeforeposting133935.html Pay attention to the rule #3: the name of the topic (subject field) MUST be the first 40 characters (~the first two sentences) of the question.
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Re: Is b/3 an integer?
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26 Nov 2012, 04:10
Thanks Bunuel! Apologies for not following the rules, will do so next time. M



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Is b/3 an integer?
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03 Nov 2014, 20:25
Bunuel wrote: Is b/3 an integer?(1) (b^2−9)/3 is an integer. Given that \(\frac{b^29}{3}=integer\) > \(\frac{b^2}{3}3=integer\) > \(\frac{b^2}{3}=integer\). Now, if \(b=0\), then \(\frac{b}{3}=0=integer\) but if \(b=\sqrt{3}\), then \(\frac{b}{3}\neq{integer}\). Not sufficient. (2) p and p + 1 are prime factors of b. Since both \(p\) and \(p+1\) are primes, then \(p=2\) and \(p+1=3\). Thus we have that \(b\) is a multiple of 3. Sufficient. Answer: B. P.S. Please read carefully and follow: rulesforpostingpleasereadthisbeforeposting133935.html Pay attention to the rule #3: the name of the topic (subject field) MUST be the first 40 characters (~the first two sentences) of the question. For (1) Can Someone or Bunuel explain why \(\frac{b^29}{3}=integer\) cannot > \(\frac{(b3)(b+3)}{3}=integer\) So that we can then assume that since b, b3 and b+3 share factors, (1) is sufficient



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Is b/3 an integer?
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04 Nov 2014, 04:06
Bunuel, can you please explain why cannot we factor b^2  9 to (b3)(b+3)? I marked D because I thought option A is sufficient. Please help! Ameya



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Re: Is b/3 an integer?
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04 Nov 2014, 04:58
Ameya85 wrote: Bunuel, can you please explain why cannot we factor b^2  9 to (b3)(b+3)? I marked D because I thought option A is sufficient. Please help! Ameya We can factor b^2  9 as (b  3)(b + 3) but how does this help to get sufficiency? Check examples in my post above.
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Re: Is b/3 an integer?
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04 Nov 2014, 07:45
valerjo79 wrote: Is b/3 an integer? (1) (b^2−9)/3 is an integer. (2) p and p + 1 are prime factors of b. All,
I am having trouble with even seemingly easy questions in DS and can't see my way out. Would you please advise? I picked A, but it is quite the opposite (B). I have a fundamental error here. The question asks Is b/3 an int? So basically it is asking if I can answer as Yes or No. So with (1) I get the answer as No with (2) I get as Yes So shouldn't the answer be D then because I can say yes or no in both (1) & (2)



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Re: Is b/3 an integer?
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04 Nov 2014, 07:51
anki2762 wrote: valerjo79 wrote: Is b/3 an integer? (1) (b^2−9)/3 is an integer. (2) p and p + 1 are prime factors of b. All,
I am having trouble with even seemingly easy questions in DS and can't see my way out. Would you please advise? I picked A, but it is quite the opposite (B). I have a fundamental error here. The question asks Is b/3 an int? So basically it is asking if I can answer as Yes or No. So with (1) I get the answer as No with (2) I get as Yes So shouldn't the answer be D then because I can say yes or no in both (1) & (2) In a Yes/No Data Sufficiency question, each statement is sufficient if the answer is “always yes” or “always no” while a statement is insufficient if the answer is "sometimes yes" and "sometimes no".For (1) the answer is "sometimes yes" and "sometimes no". If \(b = 0\), then the answer is YES but of \(b = \sqrt{3}\), the answer is NO. Also, on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other. So, we cannot have a NO answer from one statement and YES answer from another. Hope it's clear.
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Re: Is b/3 an integer?
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04 Nov 2014, 17:49
Bunuel wrote: anki2762 wrote: valerjo79 wrote: Is b/3 an integer? (1) (b^2−9)/3 is an integer. (2) p and p + 1 are prime factors of b. All,
I am having trouble with even seemingly easy questions in DS and can't see my way out. Would you please advise? I picked A, but it is quite the opposite (B). I have a fundamental error here. The question asks Is b/3 an int? So basically it is asking if I can answer as Yes or No. So with (1) I get the answer as No with (2) I get as Yes So shouldn't the answer be D then because I can say yes or no in both (1) & (2) In a Yes/No Data Sufficiency question, each statement is sufficient if the answer is “always yes” or “always no” while a statement is insufficient if the answer is "sometimes yes" and "sometimes no".For (1) the answer is "sometimes yes" and "sometimes no". If \(b = 0\), then the answer is YES but of \(b = \sqrt{3}\), the answer is NO. Also, on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other. So, we cannot have a NO answer from one statement and YES answer from another. Hope it's clear. I hate to keep coming back to this but for \(\frac{(b3)(b+3)}{3}\), if b is \(sqrt(3)\) the result is 2 which is an integer I really do not understand how we cannot conclude that if at least one of \(b3\) and \(b+3\) is a factor of 3, then both of them must be factors of 3 and consequently, \(b\) must also be a factor of 3. Can someone please show me where I'm wrong?



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Re: Is b/3 an integer?
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05 Nov 2014, 04:52
gooner wrote: Bunuel wrote: anki2762 wrote: In a Yes/No Data Sufficiency question, each statement is sufficient if the answer is “always yes” or “always no” while a statement is insufficient if the answer is "sometimes yes" and "sometimes no".
For (1) the answer is "sometimes yes" and "sometimes no". If \(b = 0\), then the answer is YES but of \(b = \sqrt{3}\), the answer is NO.
Also, on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other. So, we cannot have a NO answer from one statement and YES answer from another.
Hope it's clear.
I hate to keep coming back to this but for \(\frac{(b3)(b+3)}{3}\), if b is \(sqrt(3)\) the result is 2 which is an integer I really do not understand how we cannot conclude that if at least one of \(b3\) and \(b+3\) is a factor of 3, then both of them must be factors of 3 and consequently, \(b\) must also be a factor of 3. Can someone please show me where I'm wrong? Dear gooner, The question asks whether b/3 an integer. Now, for \(b = \sqrt{3}\) is \(\frac{\sqrt{3}}{3}\) an integer?
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Re: Is b/3 an integer?
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13 Jul 2016, 12:39
Hi Bunuel,
Can we say for a number such as [12][/3], the prime factors are 2 and 3. If yes, then Statement B doesn't give us a unique solution.
Thanks in advance.



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Re: Is b/3 an integer?
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13 Jul 2016, 12:45
adashis wrote: Hi Bunuel,
Can we say for a number such as [12][/3], the prime factors are 2 and 3. If yes, then Statement B doesn't give us a unique solution.
Thanks in advance. 12/3 = 4 and 4 has only one prime  2.
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Is b/3 an integer?
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13 Jul 2016, 12:51
Sorry a typo error, I meant for number 12*3^1/2



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Re: Is b/3 an integer?
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14 Jul 2016, 01:01
adashis wrote: Sorry a typo error, I meant for number 12*3^1/2 When we talk about prime factors, divisors, multiples, we always consider only integers. There is no sense talking about the factors of \(12\sqrt{3}\).
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Re: Is b/3 an integer?
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12 Sep 2017, 13:29
valerjo79 wrote: Is b/3 an integer?
(1) (b^2−9)/3 is an integer. (2) p and p + 1 are prime factors of b. ] Target question: Is b/3 an integer? Statement 1: (b² − 9)/3 is an integer There are several values of b that satisfy statement 1. Here are two: Case a: b = 6. Here, (b² − 9)/3 = (6² − 9)/3 = (36 − 9)/3 = 27/3 = 9, and 9 is an integer. In this case, b/3 = 6/3 = 2. So, b/3 IS an integerCase b: b = √18. Here, (b² − 9)/3 = (√18² − 9)/3 = (18 − 9)/3 = 9/3 = 3, and 3 is an integer. In this case, b/3 = (√18)/3 = (3√2)/3 = √2 (. So, b/3 is NOT an integerSince we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT Statement 2: p and p + 1 are prime factors of b Notice that p and p+1 are CONSECUTIVE integers Since p and p+1 are CONSECUTIVE integers, we know that one of the numbers must be odd and one must be even. Since 2 is the ONLY even integer, we can conclude that p = 2 and p+1 = 3 Since p and p+1 are factors of b, we can see that b is divisible by 2 and by 3. If b is divisible by 3, then b/3 must be an integer Since we can answer the target question with certainty, statement 2 is SUFFICIENT Answer: Cheers, Brent
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Is b/3 an integer?
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10 Jul 2018, 01:28
Bunuel wrote: Is b/3 an integer?(1) (b^2−9)/3 is an integer. Given that \(\frac{b^29}{3}=integer\) > \(\frac{b^2}{3}3=integer\) > \(\frac{b^2}{3}=integer\). Now, if \(b=0\), then \(\frac{b}{3}=0=integer\) but if \(b=\sqrt{3}\), then \(\frac{b}{3}\neq{integer}\). Not sufficient. (2) p and p + 1 are prime factors of b. Since both \(p\) and \(p+1\) are primes, then \(p=2\) and \(p+1=3\). Thus we have that \(b\) is a multiple of 3. Sufficient. Answer: B. P.S. Please read carefully and follow: http://gmatclub.com/forum/rulesforpos ... 33935.html Pay attention to the rule #3: the name of the topic (subject field) MUST be the first 40 characters (~the first two sentences) of the question. Hi Bunuel, Why can't we take b as 6 \(\sqrt{12}\) ?? we have 2 and 3 as prime factors here as well. Please explain. Thanks, Uma



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Is b/3 an integer?
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10 Jul 2018, 01:48
umabharatigudipalli wrote: Bunuel wrote: Is b/3 an integer?(1) (b^2−9)/3 is an integer. Given that \(\frac{b^29}{3}=integer\) > \(\frac{b^2}{3}3=integer\) > \(\frac{b^2}{3}=integer\). Now, if \(b=0\), then \(\frac{b}{3}=0=integer\) but if \(b=\sqrt{3}\), then \(\frac{b}{3}\neq{integer}\). Not sufficient. (2) p and p + 1 are prime factors of b. Since both \(p\) and \(p+1\) are primes, then \(p=2\) and \(p+1=3\). Thus we have that \(b\) is a multiple of 3. Sufficient. Answer: B. P.S. Please read carefully and follow: http://gmatclub.com/forum/rulesforpos ... 33935.html Pay attention to the rule #3: the name of the topic (subject field) MUST be the first 40 characters (~the first two sentences) of the question. Hi Bunuel, Why can't we take b as 6 \(\sqrt{12}\) ?? we have 2 and 3 as prime factors here as well. Please explain. Thanks, Uma Uma, when we talk of factors, ONLY integers can have factors \(6\sqrt{12}=6*\sqrt{4*3}=12\sqrt{3}=12*1.732...\) and this is NOT an integer.. you cannot take 6 in isolation as other part \(\sqrt{12}\) is not an integer.. example \(6*0.5\), this will not have 2 and 3 as prime factors because 6*0.5=3, so only 3 hope it helps
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Re: Is b/3 an integer?
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10 Jul 2018, 01:55
chetan2u wrote: umabharatigudipalli wrote: Bunuel wrote: Is b/3 an integer?(1) (b^2−9)/3 is an integer. Given that \(\frac{b^29}{3}=integer\) > \(\frac{b^2}{3}3=integer\) > \(\frac{b^2}{3}=integer\). Now, if \(b=0\), then \(\frac{b}{3}=0=integer\) but if \(b=\sqrt{3}\), then \(\frac{b}{3}\neq{integer}\). Not sufficient. (2) p and p + 1 are prime factors of b. Since both \(p\) and \(p+1\) are primes, then \(p=2\) and \(p+1=3\). Thus we have that \(b\) is a multiple of 3. Sufficient. Answer: B. P.S. Please read carefully and follow: http://gmatclub.com/forum/rulesforpos ... 33935.html Pay attention to the rule #3: the name of the topic (subject field) MUST be the first 40 characters (~the first two sentences) of the question. Hi Bunuel, Why can't we take b as 6 \(\sqrt{12}\) ?? we have 2 and 3 as prime factors here as well. Please explain. Thanks, Uma Uma, when we talk of factors, ONLY integers can have factors \(6\sqrt{12}=6*\sqrt{4*3}=12\sqrt{3}=12*1.732...\) and this is NOT an integer.. you cannot take 6 in isolation as other part \(\sqrt{12}\) is not an integer.. example \(6*0.5\), this will not have 2 and 3 as prime factors because 6*0.5=3, so only 3 hope it helps On the 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. \(\frac{a}{b}=integer\). Hope it helps.
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Re: Is b/3 an integer?
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08 Apr 2019, 02:48
valerjo79 wrote: Is b/3 an integer?
(1) (b^2−9)/3 is an integer. (2) p and p + 1 are prime factors of b.
Question : Is \(\frac{b}{3} = I\) ?
Or Is \(b = 3(I)\) , where I is an integer.
Statement 1 : \(\frac{{b^2  9}}{3} = I\) (integer)
\(b^2 = 3(I) + 9\)
\(b =\) \(+/ \sqrt{3(I) + 9}\)
So now, everything depends on the value of integer "I"
If I = 0, then b = +/ 3, and hence divisible by 3
If I = 1, then b will be decimal number, and hence not divisible by 3
INSUFFICIENT
Statement 2:
If b has prime factors as p and p+1, this means b = 6
Prime factors starts from 2, 3, 5, 7 etc.
Only 2 and 3 are of the form p and p+1.
(B) SUFFICIENT
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Re: Is b/3 an integer?
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