Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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?
_________________

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

(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

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...