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

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Director
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G
Joined: 02 Sep 2016
Posts: 786
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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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Joined: 03 Oct 2016
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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.
Re: If a, b and n are positive integers such that n = 3a – b3   [#permalink] 24 Jan 2018, 21:58

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