GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Oct 2019, 14:18 GMAT Club Daily Prep

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

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.  Is b/3 an integer?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Intern  Joined: 11 Oct 2012
Posts: 8
GMAT Date: 12-19-2012
WE: Analyst (Energy and Utilities)
Is b/3 an integer?  [#permalink]

Show Tags

7
18 00:00

Difficulty:   85% (hard)

Question Stats: 45% (01:40) correct 55% (01:40) wrong based on 258 sessions

HideShow timer Statistics

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).

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 58453
Re: Is b/3 an integer?  [#permalink]

Show Tags

3
7
Is b/3 an integer?

(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.

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.
_________________
General Discussion
Intern  Joined: 11 Oct 2012
Posts: 8
GMAT Date: 12-19-2012
WE: Analyst (Energy and Utilities)
Re: Is b/3 an integer?  [#permalink]

Show Tags

Thanks Bunuel! Apologies for not following the rules, will do so next time. M
Intern  Joined: 03 Oct 2011
Posts: 14
Is b/3 an integer?  [#permalink]

Show Tags

1
Bunuel wrote:
Is b/3 an integer?

(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.

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
Intern  Joined: 10 Apr 2008
Posts: 24
Location: India
GMAT 1: 560 Q39 V28 WE: Consulting (Computer Software)
Is b/3 an integer?  [#permalink]

Show Tags

1
Bunuel, can you please explain why cannot we factor b^2 - 9 to (b-3)(b+3)? I marked D because I thought option A is sufficient. Please help!

Ameya
Math Expert V
Joined: 02 Sep 2009
Posts: 58453
Re: Is b/3 an integer?  [#permalink]

Show Tags

1
Ameya85 wrote:
Bunuel, can you please explain why cannot we factor b^2 - 9 to (b-3)(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.
_________________
Intern  Joined: 10 Jan 2014
Posts: 7
GMAT 1: 600 Q40 V31 Re: Is b/3 an integer?  [#permalink]

Show Tags

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)
Math Expert V
Joined: 02 Sep 2009
Posts: 58453
Re: Is b/3 an integer?  [#permalink]

Show Tags

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.
_________________
Intern  Joined: 03 Oct 2011
Posts: 14
Re: Is b/3 an integer?  [#permalink]

Show Tags

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 $$\frac{(b-3)(b+3)}{3}$$, 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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58453
Re: Is b/3 an integer?  [#permalink]

Show Tags

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 $$\frac{(b-3)(b+3)}{3}$$, 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?
_________________
Intern  Joined: 14 Apr 2016
Posts: 4
Location: India
GMAT 1: 700 Q50 V34 GPA: 3.9
Re: Is b/3 an integer?  [#permalink]

Show Tags

Hi Bunuel,

Can we say for a number such as [/3], the prime factors are 2 and 3. If yes, then Statement B doesn't give us a unique solution.

Math Expert V
Joined: 02 Sep 2009
Posts: 58453
Re: Is b/3 an integer?  [#permalink]

Show Tags

Hi Bunuel,

Can we say for a number such as [/3], the prime factors are 2 and 3. If yes, then Statement B doesn't give us a unique solution.

12/3 = 4 and 4 has only one prime - 2.
_________________
Intern  Joined: 14 Apr 2016
Posts: 4
Location: India
GMAT 1: 700 Q50 V34 GPA: 3.9
Is b/3 an integer?  [#permalink]

Show Tags

Sorry a typo error, I meant for number 12*3^1/2
Math Expert V
Joined: 02 Sep 2009
Posts: 58453
Re: Is b/3 an integer?  [#permalink]

Show Tags

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 $$12\sqrt{3}$$.
_________________
GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4009
Re: Is b/3 an integer?  [#permalink]

Show Tags

Top Contributor
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 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

Cheers,
Brent
_________________
Intern  B
Joined: 21 Jan 2017
Posts: 31
Is b/3 an integer?  [#permalink]

Show Tags

Bunuel wrote:
Is b/3 an integer?

(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.

P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 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
Math Expert V
Joined: 02 Aug 2009
Posts: 7978
Is b/3 an integer?  [#permalink]

Show Tags

umabharatigudipalli wrote:
Bunuel wrote:
Is b/3 an integer?

(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.

P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 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
_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58453
Re: Is b/3 an integer?  [#permalink]

Show Tags

1
chetan2u wrote:
umabharatigudipalli wrote:
Bunuel wrote:
Is b/3 an integer?

(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.

P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 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.
_________________
SVP  V
Status: It's near - I can see.
Joined: 13 Apr 2013
Posts: 1684
Location: India
Concentration: International Business, Operations
Schools: INSEAD Jan '19
GPA: 3.01
WE: Engineering (Real Estate)
Re: Is b/3 an integer?  [#permalink]

Show Tags

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

_________________
"Do not watch clock; Do what it does. KEEP GOING." Re: Is b/3 an integer?   [#permalink] 08 Apr 2019, 02:48
Display posts from previous: Sort by

Is b/3 an integer?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  