Is b/3 an integer?

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

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.

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?

Dear gooner,

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

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.

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

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 \(12\sqrt{3}\).

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:

Cheers,
Brent

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.

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

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