Bunuel,

abhimahna, and @

e-gmatI 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?

_________________

Help me make my explanation better by providing a logical feedback.

If you liked the post, HIT KUDOS !!

Don't quit.............Do it.