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# Is b/3 an integer?

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Intern
Joined: 11 Oct 2012
Posts: 8
GMAT Date: 12-19-2012
WE: Analyst (Energy and Utilities)

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Updated on: 26 Nov 2012, 03:16
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45% (01:40) correct 55% (01:40) wrong based on 258 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).

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?  [#permalink]

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26 Nov 2012, 03:25
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.
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Re: Is b/3 an integer?  [#permalink]

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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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03 Nov 2014, 20:25
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
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04 Nov 2014, 04:06
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
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Re: Is b/3 an integer?  [#permalink]

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04 Nov 2014, 04:58
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.
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Re: Is b/3 an integer?  [#permalink]

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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?  [#permalink]

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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?  [#permalink]

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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 $$\frac{(b-3)(b+3)}{3}$$, 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?
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Re: Is b/3 an integer?  [#permalink]

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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 $$\frac{(b-3)(b+3)}{3}$$, 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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Re: Is b/3 an integer?  [#permalink]

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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.

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Re: Is b/3 an integer?  [#permalink]

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13 Jul 2016, 12:45
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.

12/3 = 4 and 4 has only one prime - 2.
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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?  [#permalink]

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14 Jul 2016, 01:01
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?  [#permalink]

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12 Sep 2017, 13:29
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
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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 $$\frac{b^2-9}{3}=integer$$ --> $$\frac{b^2}{3}-3=integer$$ --> $$\frac{b^2}{3}=integer$$. 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
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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 $$\frac{b^2-9}{3}=integer$$ --> $$\frac{b^2}{3}-3=integer$$ --> $$\frac{b^2}{3}=integer$$. 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
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Re: Is b/3 an integer?  [#permalink]

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10 Jul 2018, 01:55
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.
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Re: Is b/3 an integer?  [#permalink]

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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?   [#permalink] 08 Apr 2019, 02:48
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