Is b/3 an integer?

Question

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).

Bunuel · Accepted Answer

Is b/3 an integer?

(1) (b^2−9)/3 is an integer.

\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 . However, if  b=\sqrt{3} , then  \frac{b}{3}
eq{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.

valerjo79 · Answer

Thanks Bunuel! Apologies for not following the rules, will do so next time. M

gooner · Answer

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

Ameya85 · Answer

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

Bunuel · Answer

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.

anki2762 · Answer

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)

Bunuel · Answer

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.

gooner · Answer

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?

Bunuel · Answer

Dear gooner,

The question asks whether b/3 an integer. Now, for  b = \sqrt{3}  is  \frac{\sqrt{3}}{3}  an integer?

adashis · Answer

Hi Bunuel,

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

Thanks in advance.

Bunuel · Answer

12/3 = 4 and 4 has only one prime - 2.

adashis · Answer

Sorry a typo error, I meant for number 12*3^1/2

Bunuel · Answer

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} .

BrentGMATPrepNow · Answer

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:  B

Cheers,
Brent

umabharatigudipalli · Answer

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

chetan2u · Answer

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

Bunuel · Answer

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.

AkshdeepS · Answer

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

ScottTargetTestPrep · Answer

Solution:

Statement One Only:

(b^2−9)/3 is an integer.

Since (b^2−9)/3 is an integer and 9/3 is an integer, it must be true that b^2/3 is also an integer. However, since 3 is a prime, b must itself be divisible by 3 if b^2 is divisible by 3. So, b/3 must be an integer.

Statement Two Only:

p and p + 1 are prime factors of b.

We see that p and p + 1 are consecutive integers. If they are both prime factors of b, then they must be 2 and 3, respectively (because 2 and 3 are the only prime numbers that are consecutive integers). In other words, p = 2 and p + 1 = 3. Since 3 is a prime factor of b, b/3 must be an integer.

Answer: D

Is b/3 an integer?

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Is b/3 an integer?

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