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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]
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EgmatQuantExpert wrote:
If a, b and n are positive integers such that \(n = 3*a\) \(-b^3\), is \(n^2 + 3\) divisible by 2?

(1) \(a^2\) \(-4*b^{3}-5\) \(= 0\)

(2) \(3*\)\(b^3\) \(-a^2\) \(+ 6 = 0\)


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n^2+3 can only be divisible by 2 when n is odd.

This can only be possible when either a or b is odd. It follows that the other unknown variable should be even.

1. I assume there should be (a^2-4)(b^3-5)=0

Since b is an integer, this equation can only be true if a^2 is 4. So a is even.
But we do not know anything about b.--->Not Sufficient

2. 3*\(b^3\)-\(a^2\)=-6
OR 3*b-a=Even
If b is Even, then a is Even
If b is Odd, then a is Odd.

Answer is no. n^2+3 CANNOT be divisible by 2

Sufficient

Answer: B
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]
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Dear AmoyV

Thank you for posting your analysis! :)

The thing I was very happy to see in your solution was that you first analysed the question statement, determined what it is that we need to know to answer the question, and only then moved to St. 1 and 2. This is something that we tell our students to do in all DS Questions.

Further, you've answered the question correctly and your analysis of Statement 2 is spot on. But I'm afraid that your analysis of Statement 1 is flawed. Yes, you've arrived at the right conclusion, but the way you reached there was not right. Would you like to give another shot at why Statement 1 is not sufficient? :)

- Japinder


Originally posted by EgmatQuantExpert on 08 Apr 2015, 23:58.
Last edited by EgmatQuantExpert on 07 Aug 2018, 00:39, edited 1 time in total.
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]
EgmatQuantExpert wrote:
Detailed Solution

Step-I: Given Info:

We are given that \(a\), \(b\), \(n >0\) and are integers. Also \(n =3a – b^3\) and we are asked to find if \((n^2 + 3)\) is divisible by 2.

Step-II: Interpreting the Question Statement:

Let’s start from our expression i.e. \((n^2 + 3)\), this expression is divisible by 2 only if it’s even, since 3 is odd, for \(n^2 + 3\) to be even \(n^2\) has to be odd ( as odd + odd =even) and \(n^2\) can be odd only when \(n\) is odd.

Now, we know that \(n = 3a – b^3\) , for \(n\) to be odd, one of the \(3a\) or \(b^3\) has to be odd and other has to be even as the difference of an even and an odd number will always be odd. The even/odd nature of \(3a\) would depend on the even/odd nature of \(a\) and similarly the even odd nature of \(b^3\) would depend on the even/odd nature of \(b\). So, if we can establish that the even/odd nature of \(a\) and \(b\) are either similar or opposite, we will find our answer.

Step-III: Statement-I

Statement-I tells us that \(a^2 – 4b^3 = 5\), it tells us that difference of two numbers is odd. Since \(4b^3\) would always be even, for the difference of \(a^2\) and \(4b^3\) to be odd, \(a^2\) would have to be odd. For \(a^2\) to be odd, \(a\) has to be odd. But St-I does not tell us anything about the even/odd nature of \(b\).

So, Statement-I alone is insufficient.

Step-IV: Statement-II

Statement-II tells us that \(a^2 - 3b^3= 6\), it tells us that difference of two numbers is even. This is only possible in two cases:

a) When both \(a^2\) and \(3b^3\) are odd, for this to happen both \(a\) and \(b\) have to be odd or
b) When both \(a^2\) and \(3b^3\) are even, for this to happen both \(a\) and \(b\) have to be even

But, we know that for n to be odd, both \(a\) and \(b\) have to of opposite even/odd natures. We see that in St-II, in both the cases \(a\) and \(b\) are of the same nature, thus in both the cases, \(n\) would be even.

Hence, Statement-II is sufficient to answer our question.

Step-V: Combining Statements I & II

Since, we have a unique answer from Statement-II alone, we don’t need to combine the information from Statement-I and II.
Thus, the answer is Option B.

Key Takeaways

1. In even-odd questions, simplify complex expressions into simpler expressions using the properties of even-odd combinations.
2. Know the properties of even/odd combinations to save the time spent deriving them in the test
3. The even/odd nature of some expressions can be determined without knowing the exact even/odd nature of the variables of the expressions by using the even/odd combination property


Regards
Harsh


I read the expression a^2-4*b^3-5=0 as (a^2-4)*(b^3-5)=0
By this, a=2 and b^3=5. Subst. this is n=3a-b^3 which makes n= (3*2)-5=1
thus n^2+3 becomes, 4 which is divisible by 2!!!
thus making each statement alone sufficient (opt. D) as the answer!!

Pl use proper parenthesis..
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]
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vishnu23688 wrote:
EgmatQuantExpert wrote:
Detailed Solution

Step-I: Given Info:

We are given that \(a\), \(b\), \(n >0\) and are integers. Also \(n =3a – b^3\) and we are asked to find if \((n^2 + 3)\) is divisible by 2.

Step-II: Interpreting the Question Statement:

Let’s start from our expression i.e. \((n^2 + 3)\), this expression is divisible by 2 only if it’s even, since 3 is odd, for \(n^2 + 3\) to be even \(n^2\) has to be odd ( as odd + odd =even) and \(n^2\) can be odd only when \(n\) is odd.

Now, we know that \(n = 3a – b^3\) , for \(n\) to be odd, one of the \(3a\) or \(b^3\) has to be odd and other has to be even as the difference of an even and an odd number will always be odd. The even/odd nature of \(3a\) would depend on the even/odd nature of \(a\) and similarly the even odd nature of \(b^3\) would depend on the even/odd nature of \(b\). So, if we can establish that the even/odd nature of \(a\) and \(b\) are either similar or opposite, we will find our answer.

Step-III: Statement-I

Statement-I tells us that \(a^2 – 4b^3 = 5\), it tells us that difference of two numbers is odd. Since \(4b^3\) would always be even, for the difference of \(a^2\) and \(4b^3\) to be odd, \(a^2\) would have to be odd. For \(a^2\) to be odd, \(a\) has to be odd. But St-I does not tell us anything about the even/odd nature of \(b\).

So, Statement-I alone is insufficient.

Step-IV: Statement-II

Statement-II tells us that \(a^2 - 3b^3= 6\), it tells us that difference of two numbers is even. This is only possible in two cases:

a) When both \(a^2\) and \(3b^3\) are odd, for this to happen both \(a\) and \(b\) have to be odd or
b) When both \(a^2\) and \(3b^3\) are even, for this to happen both \(a\) and \(b\) have to be even

But, we know that for n to be odd, both \(a\) and \(b\) have to of opposite even/odd natures. We see that in St-II, in both the cases \(a\) and \(b\) are of the same nature, thus in both the cases, \(n\) would be even.

Hence, Statement-II is sufficient to answer our question.

Step-V: Combining Statements I & II

Since, we have a unique answer from Statement-II alone, we don’t need to combine the information from Statement-I and II.
Thus, the answer is Option B.

Key Takeaways

1. In even-odd questions, simplify complex expressions into simpler expressions using the properties of even-odd combinations.
2. Know the properties of even/odd combinations to save the time spent deriving them in the test
3. The even/odd nature of some expressions can be determined without knowing the exact even/odd nature of the variables of the expressions by using the even/odd combination property


Regards
Harsh


I read the expression a^2-4*b^3-5=0 as (a^2-4)*(b^3-5)=0
By this, a=2 and b^3=5. Subst. this is n=3a-b^3 which makes n= (3*2)-5=1
thus n^2+3 becomes, 4 which is divisible by 2!!!
thus making each statement alone sufficient (opt. D) as the answer!!

Pl use proper parenthesis..


No parenthesis are need there.

Mathematically \(a^2 - 4∗b^3-5 = 0\) can ONLY mean \((a^2) -(4∗b^3)-(5) = 0\) and nothing else.
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]
Can I get more questions like this, I need to practice more of these. I am facing difficulties in concept applications.

Thanks!
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]
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ujjwal80 wrote:
Can I get more questions like this, I need to practice more of these. I am facing difficulties in concept applications.

Thanks!


Check Our Questions' Banks


For more:
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Hope it helps.
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If a, b and n are positive integers such that n = 3a – b3 [#permalink]
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Hi All,

We have posted an article on the application of Process Skills in solving GMAT Quant questions. These process skills are critical to master GMAT Quant, they help you solve any quant question of any difficulty level.

You can go through the article from this Six Process Skills in Quant

To watch the video on the same, click on the image below

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Re: If a, b and n are positive integers such that n = 3a b3 [#permalink]
EgmatQuantExpert wrote:
Detailed Solution

Step-I: Given Info:

We are given that \(a\), \(b\), \(n >0\) and are integers. Also \(n =3a – b^3\) and we are asked to find if \((n^2 + 3)\) is divisible by 2.

Step-II: Interpreting the Question Statement:

Let’s start from our expression i.e. \((n^2 + 3)\), this expression is divisible by 2 only if it’s even, since 3 is odd, for \(n^2 + 3\) to be even \(n^2\) has to be odd ( as odd + odd =even) and \(n^2\) can be odd only when \(n\) is odd.

Now, we know that \(n = 3a – b^3\) , for \(n\) to be odd, one of the \(3a\) or \(b^3\) has to be odd and other has to be even as the difference of an even and an odd number will always be odd. The even/odd nature of \(3a\) would depend on the even/odd nature of \(a\) and similarly the even odd nature of \(b^3\) would depend on the even/odd nature of \(b\). So, if we can establish that the even/odd nature of \(a\) and \(b\) are either similar or opposite, we will find our answer.

Step-III: Statement-I

Statement-I tells us that \(a^2 – 4b^3 = 5\), it tells us that difference of two numbers is odd. Since \(4b^3\) would always be even, for the difference of \(a^2\) and \(4b^3\) to be odd, \(a^2\) would have to be odd. For \(a^2\) to be odd, \(a\) has to be odd. But St-I does not tell us anything about the even/odd nature of \(b\).

So, Statement-I alone is insufficient.

Step-IV: Statement-II

Statement-II tells us that \(a^2 - 3b^3= 6\), it tells us that difference of two numbers is even. This is only possible in two cases:

a) When both \(a^2\) and \(3b^3\) are odd, for this to happen both \(a\) and \(b\) have to be odd or
b) When both \(a^2\) and \(3b^3\) are even, for this to happen both \(a\) and \(b\) have to be even

But, we know that for n to be odd, both \(a\) and \(b\) have to of opposite even/odd natures. We see that in St-II, in both the cases \(a\) and \(b\) are of the same nature, thus in both the cases, \(n\) would be even.

Hence, Statement-II is sufficient to answer our question.

Step-V: Combining Statements I & II

Since, we have a unique answer from Statement-II alone, we don’t need to combine the information from Statement-I and II.
Thus, the answer is Option B.

Key Takeaways

1. In even-odd questions, simplify complex expressions into simpler expressions using the properties of even-odd combinations.
2. Know the properties of even/odd combinations to save the time spent deriving them in the test
3. The even/odd nature of some expressions can be determined without knowing the exact even/odd nature of the variables of the expressions by using the even/odd combination property


Regards
Harsh



Thanks EgmatQuantExpert for your detailed and great explanation. A bit confused in Step-III: Since \(4b^3\) would always be even, which mean b = even. Therefore not sure why the statement of St-I does not tell us anything about the even/odd nature of \(b\). Have I got my understanding wrong? Thanks for your time in advanced.
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Re: If a, b and n are positive integers such that n = 3a b3 [#permalink]
EgmatQuantExpert & Bunuel: I'm struggling a bit with this question especially Statement I. Looking at the equation, I don't see how a or b (with the power of 3) would be an integer . It is either a or b is an integer but not both. For instance, if b=4, a^2= 261, leading to a= 16.1 which is not an integer. The same logic could be applied in reverse, if a has a value of 7, for example and we solve for b. I think there's something problematic about the question (including the constraints mentioned) but please provide an explanation if I'm missing something. I'd also appreciate contributions from other forum members as well. Thanks!
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Re: If a, b and n are positive integers such that n = 3a b3 [#permalink]
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Re: If a, b and n are positive integers such that n = 3a b3 [#permalink]
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