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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).

(1) (b^2−9)/3 is an integer. Given that \(\frac{b^2-9}{3}=integer\) --> \(\frac{b^2}{3}-3=integer\) --> \(\frac{b^2}{3}=integer\). 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: rules-for-posting-please-read-this-before-posting-133935.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. _________________

(1) (b^2−9)/3 is an integer. Given that \(\frac{b^2-9}{3}=integer\) --> \(\frac{b^2}{3}-3=integer\) --> \(\frac{b^2}{3}=integer\). 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: rules-for-posting-please-read-this-before-posting-133935.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 \(\frac{b^2-9}{3}=integer\) cannot --> \(\frac{(b-3)(b+3)}{3}=integer\) So that we can then assume that since b, b-3 and b+3 share factors, (1) is sufficient

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)

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.

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 \(\frac{(b-3)(b+3)}{3}\), 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 \(b-3\) 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.

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 \(\frac{(b-3)(b+3)}{3}\), 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 \(b-3\) 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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When we talk about prime factors, divisors, multiples, we always consider only integers. There is no sense talking about the factors of \(12\sqrt{3}\).
_________________

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).

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(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 integer Case 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 integer Since 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