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

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

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New post 03 Sep 2017, 05:13
Bunuel, abhimahna, and @e-gmat

I got confused with the statements like others.

The question I interpreted was:

If a, b, and n are positive integers such that n= 3*a-b^3, is n^2+3 divisible by 2?

Given:
1) a,b,n >0
2) n= 3(a-b^3)

To find:
n^2+3 will be divisible by 2 when n^2+3 is even.
n^2+3 will be even when n is odd (odd+odd= even).


So the question becomes: Find the even- odd nature of n.

Statement 1:
(a^2-4)*(b^3-5)=0

a^2*b^3-5a^2-4b^3+20=0
a^2*b^3-5a^2-4b^3= -20
a^2*b^3-5a^2= Even+4b^3= Even

Possible cases:
1) Even-even= Even
Then a and b are even.
So n= even

2) Odd-Odd= Even
Then a and b are odd.
So n= 3(odd-odd)= even

Sufficient.

Statement 2: 3b^3-a^2+6= 0

3b^3-a^2= -6= Even

That means 3b^3-a^2 is even.
Therefore either a and b are even or odd.

Sufficient.


D


This is what I interpreted and marked D as the answer. I know this is not the question but if the question was framed like this, then would this be correct?
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Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

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New post 24 Jan 2018, 20:07
I solved this using the method posted by e-gmat. Only diff is that i ignored 3 from n=3a-\(b^{3}\) -> n =a-\(b{^3}\). Reason 3 is odd so doesn't play a role in even/odd of n. Eventually we just need to determine if a and b are a combination of even odd numbers for n to become an odd number so that its divisible by 2.
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Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

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New post 24 Jan 2018, 21:58
Shiv2016 wrote:
Bunuel, abhimahna, and @e-gmat

I got confused with the statements like others.

The question I interpreted was:

If a, b, and n are positive integers such that n= 3*a-b^3, is n^2+3 divisible by 2?

Given:
1) a,b,n >0
2) n= 3(a-b^3)

To find:
n^2+3 will be divisible by 2 when n^2+3 is even.
n^2+3 will be even when n is odd (odd+odd= even).


So the question becomes: Find the even- odd nature of n.

Statement 1:
(a^2-4)*(b^3-5)=0

a^2*b^3-5a^2-4b^3+20=0
a^2*b^3-5a^2-4b^3= -20
a^2*b^3-5a^2= Even+4b^3= Even

Possible cases:
1) Even-even= Even
Then a and b are even.
So n= even

2) Odd-Odd= Even
Then a and b are odd.
So n= 3(odd-odd)= even

Sufficient.

Statement 2: 3b^3-a^2+6= 0

3b^3-a^2= -6= Even

That means 3b^3-a^2 is even.
Therefore either a and b are even or odd.

Sufficient.


D


This is what I interpreted and marked D as the answer. I know this is not the question but if the question was framed like this, then would this be correct?


Hi Shiv

I think you have misinterpreted what was written in the question. n is NOT 3*(a - b^3) But n = (3*a) - (b^3)

Also the first statement is NOT (a^2-4)*(b^3-5)=0, Instead first statement is (a^2) - 4*(b^3) - 5 = 0.
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Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

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New post 31 Mar 2018, 07:22
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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New post 31 Mar 2018, 09:13
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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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Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

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New post 31 Mar 2018, 09:37
Bunuel wrote:
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:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.


Wow! Thanks for sharing such a useful resource Bunuel, this is great.
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Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

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New post 10 Nov 2018, 14:10
The trick of this question is understanding that exponent of a number N does not change its even/odd property. This means is N = Even, then N^2, N^3, N^x is Even. Same applies to Odd.

If a, b and n are positive integers such that n=3∗a-b^3, is n^2+3 divisible by 2?
The question is asking if n^2 +3 = Even -> is n+3 = Even -> since Odd+Odd=Even so is n=Odd?
Then we also know n= 3*a-b^3. This means we have following options:
if a = Even, b=Even then n=Even
if a=Odd, b=Odd then n=Even
if a = Even, b=Odd then n= Odd
if a=Odd, b=Even, then n = Odd
So if a and b have the same even/odd property then n is even, otherwise n is odd.

(1) a^3 - 4b^-5 =0 -> a^3-4b^3=5 (odd)
since 4b^3 is also even so we only care about a^3 and because exponent does not change a's even/odd property, only a matter. So a-even=odd -> this means a is odd, but we do not know the property of b so (1) is not sufficient

(2) 3b^3 - a^2+6=0 -> 3b^3-a^2 = -6 (even)
With exponent property: 3b-a = even
since 3b = even if b is even and odd if b is odd, only b matter in this case: b-a = even.
we have following options:
if b=even a=even then b-a = even
if b=even a=odd then b-a = odd
if b=odd a=even then b-a = odd
if b=odd a=odd then b-a = even.
b and a have same even/odd property. This is exactly what the question is looking for. So this is sufficient.

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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New post 12 Nov 2018, 01:54
Bunuel wrote:
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.


Can you please explain why statement 2 is sufficient?

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Re: If a, b and n are positive integers such that n = 3a – b3 &nbs [#permalink] 12 Nov 2018, 01:54

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