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

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

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Updated on: 13 Dec 2016, 06:42
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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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Originally posted by EgmatQuantExpert on 08 Apr 2015, 10:08.
Last edited by EgmatQuantExpert on 13 Dec 2016, 06:42, edited 3 times in total.
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]

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Updated on: 07 Dec 2016, 04:03
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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
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Originally posted by EgmatQuantExpert on 10 Apr 2015, 05:54.
Last edited by EgmatQuantExpert on 07 Dec 2016, 04:03, edited 1 time in total.
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]

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09 Apr 2015, 03:11
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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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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

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

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08 Apr 2015, 10:31
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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) $$3*$$$$b^3$$ $$-a^2$$ $$+ 6 = 0$$

This is

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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

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

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08 Apr 2015, 23:58
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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
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]

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09 Apr 2015, 03:14
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.
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]

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09 Apr 2015, 16:07
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

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

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10 Apr 2015, 01:25
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

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

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17 Jun 2015, 09:28
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.
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]

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21 Aug 2015, 00:10
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.
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]

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23 Aug 2015, 23:24
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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24 Aug 2015, 08:38
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]

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24 Sep 2015, 15:14
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.
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]

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22 Aug 2016, 07:05
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
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]

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03 Dec 2016, 09:21
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?
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]

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08 Jan 2017, 23:34
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

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

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08 Jan 2017, 23:44
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

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

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13 Apr 2017, 03:06
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.
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]

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23 Apr 2017, 23:24
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.
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Re: If a, b and n are positive integers such that n = 3a – b3 [#permalink]

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02 Aug 2017, 03:37
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

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