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Is b/3 an integer?

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Is b/3 an integer? [#permalink]

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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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New post 26 Nov 2012, 03:25
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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.

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.
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Re: Is b/3 an integer? [#permalink]

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New post 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? [#permalink]

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New post 03 Nov 2014, 20:25
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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.

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
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Is b/3 an integer? [#permalink]

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New post 04 Nov 2014, 04:06
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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!


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Re: Is b/3 an integer? [#permalink]

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New post 04 Nov 2014, 04:58
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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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New post 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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New post 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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New post 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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New post 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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New post 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? [#permalink]

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New post 13 Jul 2016, 12:45
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Is b/3 an integer? [#permalink]

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New post 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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Re: Is b/3 an integer? [#permalink]

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New post 12 Sep 2017, 13:29
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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

Answer:

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Is b/3 an integer? [#permalink]

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New post 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.

Answer: B.

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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Is b/3 an integer? [#permalink]

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New post 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.

Answer: B.

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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New post 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 \(\frac{b^2-9}{3}=integer\) --> \(\frac{b^2}{3}-3=integer\) --> \(\frac{b^2}{3}=integer\). 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: 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] 10 Jul 2018, 01:55
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