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Nice Question Here we can use the property that POWER does not effect the even or odd nature of any integer Hence statement 1=> 3n+odd=even => n is odd=> suff Statement 2=> -n-5=even=> n=even-odd=> odd => suff

Nice Question Here we can use the property that POWER does not effect the even or odd nature of any integer Hence statement 1=> 3n+odd=even => n is odd=> suff Statement 2=> -n-5=even=> n=even-odd=> odd => suff

Smash that D

stonecold - I am quite a bit lost on this question. Could you just explain a lil bit more how you get to "3n + odd = even" / "-n-5=even"?

Nice Question Here we can use the property that POWER does not effect the even or odd nature of any integer Hence statement 1=> 3n+odd=even => n is odd=> suff Statement 2=> -n-5=even=> n=even-odd=> odd => suff

Smash that D

stonecold - I am quite a bit lost on this question. Could you just explain a lil bit more how you get to "3n + odd = even" / "-n-5=even"?

Thanks in advance! p

Hi Here is what i would do in this question =>

Lets start from scratch -->

Given data => n is a positive integer. We are asked if it is odd or not.Hence we just need the even/odd nature of n.

Lets roll =>

Statement 1 => n^2+(n+1)^2+(n+2)^2 is odd. There are three ways to tackle this one. First way => Open the brackets using (a+b)^2=a^2+b^2+2ab n^2+(n+1)^2+(n+2)^2=> n^2+(n^2+1+2n)+(n^2+4+4n)=> 3n^2+6n+5

So 3n^2+6n+5 =even 3n^2 +even +odd =even 3n^2+odd=even 3n^2=odd-even 3n^2=odd As 3 is odd =>Hence n must be also to make 3n^2 even. Thus n is even

Hence this statement is sufficient.

Second way(The more reasonably very quick) => Use hit and trial. Since n is an integer => It can be either even or odd. If n is odd => n^2+(n+1)^2+(n+2)^2 => odd+(odd+odd)^2 +(odd+even)^2 => odd+even+odd => even Hence n can be odd

If n is even => n^2+(n+1)^2+(n+2)^2 => even+odd+even => odd REJECTED. Hence n must be always odd.

Sufficient

Third way=> Power does not affect the even/odd nature of any number. So if n is odd => n^2 is odd if n is even => n^2 is even too.

n^2+(n+1)^2+(n+2)^2 = even n+n+1+n+2=> even 3n+3=even 3n => even-odd=> odd Hence n must be odd.

The same way we can get Statement 2 => Sufficient.

I understood your explanation but can you suggest flaw in my approach:

I started with : Assuming n is even if st 1 is satisfied: If n is even, then n+1 is odd and n+2 is even so we have: LHS=e+o+e = o ; whereeas RHS is even. Where did I do wrong?
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thanks for chipping in. You mean to say that we are looking for unique property of n as even or odd. Since n is odd satisfies both hence n is odd Since n is even does not satisfy both hence n is even is not correct Since I got two answers for n as odd / even for st 1 i rejected it.
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