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

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Joined: 02 Sep 2016
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Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

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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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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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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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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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31 Mar 2018, 09:13
1
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 ! ! !

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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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 ! ! !

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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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.

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Joined: 18 Jul 2018
Posts: 30
Re: If a, b and n are positive integers such that n = 3a – b3  [#permalink]

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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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