GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 27 Jan 2020, 01:22 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If a, b and n are positive integers such that n = 3a – b3

Author Message
TAGS:

### Hide Tags

e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3230
If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

18
1
116 00:00

Difficulty:   95% (hard)

Question Stats: 35% (02:32) correct 65% (02:56) wrong based on 1272 sessions

### HideShow timer Statistics

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

This is

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts _________________

Originally posted by EgmatQuantExpert on 08 Apr 2015, 10:08.
Last edited by EgmatQuantExpert on 13 Aug 2018, 03:10, edited 5 times in total.
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3230
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

23
33
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

_________________

Originally posted by EgmatQuantExpert on 10 Apr 2015, 05:54.
Last edited by EgmatQuantExpert on 07 Aug 2018, 01:08, edited 2 times in total.
Retired Moderator Joined: 06 Jul 2014
Posts: 1214
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33
GMAT 2: 740 Q50 V40
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

10
6
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) $$3*$$$$b^3$$ $$-a^2$$ $$+ 6 = 0$$

This is

Register for our Free Session on Number Properties this Saturday to solve exciting 700+ Level Questions in a classroom environment under the guidance of our Experts!

From this question
$$n = 3*a$$ $$-b^3$$, is $$n^2 + 3$$ divisible by 2
we can remake this question in more convinient form:

firstly change N on $$n = 3*a$$ $$-b^3$$

$$(3*a -b^3)^2 + 3 = even?$$
As we know that A and B integers we can eliminate all powers because they don't have any influence on parity
and we have such question: $$3*a -b + 3 = even?$$

1) $$a^2 -4*b^{3}-5 = 0$$
remove powers: a-4b-5=0
0 is even
so we know that 4b even and 5 is odd
so a is odd but we need information about b and we don't have it because B can be odd and can be even, after multiplying on 4 it will even result
Insufficient

2) $$3*b^3-a^2 + 6 = 0$$
remove powers: 3b-a+6=0
and we see that a and b will be equal parity: both even or both odd
And it seems like insufficient but if we check question we will see that we don't need to know if b odd or even
we need to know about result of a-b+3
and from this statement we know that a-b equal even result and plus 3 will be odd result
Sufficient

_________________
##### General Discussion
Retired Moderator B
Status: On a mountain of skulls, in the castle of pain, I sit on a throne of blood.
Joined: 30 Jul 2013
Posts: 297
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

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

This is

Register for our Free Session on Number Properties this Saturday to solve exciting 700+ Level Questions in a classroom environment under the guidance of our Experts!

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

e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3230
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

2
1
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.
Manager  Joined: 17 Mar 2015
Posts: 113
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

To add to Amoy's analysis, if we take the #1 as it is written there:
4*B^3 = even regardless of B, 5 - odd, difference is odd, so A^2 has to be odd aswell which means A is odd.
Unfortunatelly the fact that the type of B is unknown we can't say if N = 3*A - B^3 is even or odd, coz its odd if B is even and even otherwise, that means we can't say if N^2+3 is divisible by 2 coz it is if N^2 is odd and it is not if otherwise.
Manager  Joined: 26 May 2013
Posts: 51
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

1
I'd suggest you use braces () instead of * or both to make the question more readable.

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

This is

Register for our Free Session on Number Properties this Saturday to solve exciting 700+ Level Questions in a classroom environment under the guidance of our Experts!
Manager  Joined: 29 Apr 2014
Posts: 116
Location: Viet Nam
Concentration: Finance, Technology
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q51 V27
GMAT 3: 680 Q50 V31
GMAT 4: 710 Q50 V35
GMAT 5: 760 Q50 V42
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

1
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) $$3*$$$$b^3$$ $$-a^2$$ $$+ 6 = 0$$

This is

Register for our Free Session on Number Properties this Saturday to solve exciting 700+ Level Questions in a classroom environment under the guidance of our Experts!

Determining whether n^2 +3 is divisible by 2 means that determine if n is odd or even.

(1) a^2-4*b^3-5=0
<=> (3*a - b^3) +a^2 - 3*a - 3*b^3 -5 = 0
<=> n + a*(a-3) -3*b^3 -5 = 0
a(a-3) is even, because one is odd, one is even;
5 is odd.
So it depends on 3*b^3 is odd or even, which in turn depends on b. (1) is NOT SUFFICIENT.

(2) 3*b^3 - a^2 +6 = 0
Do the same above method => (3*a - b^3) +4*b^3 - a^2 - 3*a + 6 = 0
<=> n +4*b^3 - a*(a+3) +6 = 0
4*b^3 is always even whatever b is
a*(a+3) is even, because one is odd, one is even
6 is even
=> n is even. (2) is SUFFICIENT => the correct answer is B
Intern  Joined: 11 May 2015
Posts: 2
Location: United States
GMAT 1: 720 Q50 V38
GPA: 3
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

By the following logic, statement-I should also be sufficient:

$$b^3 = (a^2 - 5)/4$$ => a has to be odd for the numerator to be divisible by 4

Now since a is odd I can represent it as a=2k-1,
So, $$b^3 = ((2k-1)^2 - 5)/4 => (4k^2-4k-4)/4 => (k^2 - k -1)$$ which is always odd. Hence b is also odd.
Intern  Joined: 07 Jun 2015
Posts: 7
WE: Information Technology (Computer Software)
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

Sol: n =3a - b^3
a>0, b>0 n>0 => 3a >b3
Is n^2 + 3 even?

1. a^2 - 4b^3 - 5 = 0 => a^2 = 4b^3 +5
4B^3 is always even hence a^2 should be odd => a is odd
b^3 can be either even or odd so n can be either even or odd, so we get Yes & No. hence A is insufficient
2. 3b^3 - a^2 +6 =0 => a^2 = 3b^3 +6
if b is odd then a will also be odd
and if b is even a will also be even.
This implies n = (even - even) + 3 => odd
or n = (odd-odd) +3 => odd
A definite NO.

Hence B is solution.
Intern  Joined: 05 Apr 2015
Posts: 9
GMAT 1: 480 Q36 V20
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

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..
Math Expert V
Joined: 02 Sep 2009
Posts: 60657
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

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.
_________________
Intern  Joined: 04 Dec 2014
Posts: 10
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

Just a different perspective. Not sure why i am wrong.

(1) a2 −4∗b3−5 =0 - this holds true when a=3 and b=1, considering both a and b are positive and integers

(2) 3∗b3 −a2 +6=0 - this holds true when a=3 and b=1, considering both a and b are positive and intergers

In both the above equations, none of the other values of a and b will satisfy since for any value of b>1, values of a will be fractions.

Given this, it is easy to see that n=8 and n^2+3 = 67, not divisible by 2.

Should it not be D. both statements are sufficient.
Current Student D
Joined: 12 Aug 2015
Posts: 2547
Schools: Boston U '20 (M)
GRE 1: Q169 V154 Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

Excellent Question egmat
Here is what i did
We need to check whether N^2+3 is divisible by 2 or not. i.e whether N^2+3=even or simply we need to check whether N is Even or Odd
Statement 1 => Here A^3-4B^3-5=0 hence A=even+even+odd=odd
So A is odd but we need the character of B to get to any conclusion on N
Hence This statement is insufficient because it gives us no clue as to what B's even/odd nature is.

Lets look at statement 2 now =>
here 3B^3-A^2+6=0 hence 3B^3-A^2=even
Now the sum of two integers is even when either both are even or both are odd
Also POWER does not effect the Even and odd nature of any expression
Hence A and B are either both odd or both even
Case 1 => if they are both odd => N=3A-B^3=> Odd-Odd=Even
Case 2 => if they are both even => N=3A-B^3 => Even -Even = Even
Hence N is always even
So the Answer to our Question is NO => N^2+3 will be Even +odd = odd and would never be divisible by 2

Hence Sufficient
SMASH that B
_________________
Intern  Joined: 12 Sep 2015
Posts: 10
Location: India
Concentration: Strategy, General Management
Schools: Duke '19
WE: Operations (Energy and Utilities)
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

The answer should be definitely: D

Statement 1: Only integer & positive solutions are 3,1 both are odd......statement is sufficient to answer the prompt

Statement 2: As above explanations reason and are correct

Any thoughts on this?
Director  G
Joined: 02 Sep 2016
Posts: 635
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

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

This is

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts  In option, I assumed (a^2-4)*(b^3-5)=0

But this is wrong in the explanation given by e-gmat.

My query is will the variables be put in bracket in the question stem if they form one unit ? Otherwise its little confusing.

Thanks.
Math Expert V
Joined: 02 Sep 2009
Posts: 60657
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

Shiv2016 wrote:
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) $$3*$$$$b^3$$ $$-a^2$$ $$+ 6 = 0$$

This is

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts  In option, I assumed (a^2-4)*(b^3-5)=0

But this is wrong in the explanation given by e-gmat.

My query is will the variables be put in bracket in the question stem if they form one unit ? Otherwise its little confusing.

Thanks.

The answer to your query is ABSOLUTELY. If it were (a^2-4)*(b^3-5)=0, then it would have been written this way, it's mathematically correct.
_________________
Manager  G
Joined: 20 Sep 2016
Posts: 103
GMAT 1: 680 Q49 V35
GPA: 3.99
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

My approach was little different, so I want to know if it is valid (i.e. can I use it safely for solving other problems)

Based on this article, when I see exponents or odd coefficients in E/O problems, I just ignore them (exponent or odd coefficient). i.e. x^5=x or 3x=x.

The question asks is n^2+3 (or n+3) even, so we have to know is n E or O. To understand this, we have to know is a-b even or odd. (3a-b^3 = a-b)

1.
a - Even = Odd gives us that A is odd. We know nothing about a-b => insufficient.

2. Ignore coefficient and exponent 3.
b-a = even. Sufficient.
Intern  B
Joined: 26 Mar 2017
Posts: 27
GMAT 1: 720 Q50 V38
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

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

Response:
I think it is D - either statement is sufficient alone.
Statement-1 is sufficient because:
a is odd for obvious reasons. Now rewrite the original equation $$a^2 - 4b^3 - 5 = 0$$ as $$a^2 - 1 = 4b^3 + 4$$
Then it becomes, $$(a-1)(a+1)/4 = b^3+1$$. Since a is odd, put a=2k+1, which means eq. is now: $$k(k+1) = b^3+1$$. LHS is always even, hence b has to be odd. Hence from statement 1, a and b both are odd. Thus n is even.

State-2 is also sufficient alone as a and b must be either both odd or both even. Thus n is even always.
Manager  B
Joined: 04 May 2014
Posts: 147
Location: India
WE: Sales (Mutual Funds and Brokerage)
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

### Show Tags

We are getting 2 different answers. Which one is right? Re: If a, b and n are positive integers such that n = 3a – b3   [#permalink] 02 Aug 2017, 03:37

Go to page    1   2    Next  [ 29 posts ]

Display posts from previous: Sort by

# If a, b and n are positive integers such that n = 3a – b3  