Is the positive integer n odd?

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.


Does this make sense ?



Regards
Stone Cold
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Is the positive integer n odd?


(1) \(n^2+(n+1)^2+(n+2)^2\) is even.

In n = odd, then \(n^2+(n+1)^2+(n+2)^2\) would be:

\(odd^2 + (odd+1)^2 + (odd+2)^2=odd + even^2 + odd^2=odd+even+odd=even\). Matches with the info given in the statement.

In n = even, then \(n^2+(n+1)^2+(n+2)^2\) would be:

\(even^2 + (even+1)^2 + (even+2)^2=even + odd^2 + even^2=even+odd+even=odd\). Does not match with the info given in the statement.

So, n = odd.

Sufficient.


(2) \(n^2-(n+1)^2-(n+2)^2\) is even

In n = odd, then \(n^2-(n+1)^2-(n+2)^2\) would be:

\(odd^2 - (odd+1)^2 - (odd+2)^2=odd - even^2 - odd^2=odd-even-odd=even\). Matches with the info given in the statement.

In n = even, then \(n^2-(n+1)^2-(n+2)^2\) would be:

\(even^2 - (even+1)^2 - (even+2)^2=even - odd^2 - even^2=even-odd-even=odd\). Does not match with the info given in the statement.

So, n = odd.

Sufficient.

Notice that (2) (n^2 - (n + 1)^2 - (n + 2)^2) is different from (1) (n^2 + (n + 1)^2 + (n + 2)^2) only by signs in two places, which won't affect the even/odd nature. Therefore, since (1) was sufficient, then (2) would also be sufficient and you could skip analyzing (2) altogether and say that it's sufficient as well.

Answer: D.
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D
given x is positive integer
only odd integer(n) satisfy both equations

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
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Correct answer is D.

Statement 1 :- N can only be odd.

as for the statement to be even either all the numbers have to even or there should be a pair of odd numbers i.e.

E+E+E = E
or O+E+O = E

Now as they are consecutive numbers first scenario is not possible. Hence second scenario is correct. Therefore N is odd.

Statement 2 :- Same explanation. Addition or subtraction dont make a difference in even and odd rule.

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 stonecold Bunuel

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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Stmnt 1)
If n is odd
O+E+O = E satisifies

If n is even
E+O+E = O does not satisfy
Hence N is odd

Stmnt 2)
If N is Odd
O-E-O = E satisfies

If N is even
E-O-E = O does not satisfy

Hence answr D

hi srikanth9502

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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Because powers don't affect whether the integer is odd or even.

it can be rewritten as n + n +1 + n + 2 = 3n + 3. Only way to get that as a even is if n is odd.

Sufficient.

Apply the same for statement 2

n - n - 1 -n - 2 = -n - 3 is even. the only way for it to be even is if n is odd.

Sufficient.

Answer choice D

Hey adkikani
What is "both" that u are referring to?

Question - Is n odd?
If we answer a YES or NO for that then it is sufficient.

Statement 1)
For N to satisfy statement one.
N has to be odd.
So it answers our main question. Yes N is odd.

Statement 2)
For N to satisfy statement two.
N has to be odd here too.
So again it answers our main question. Yes N is odd.

Note - You dont even have to check for statement 2 because all the EVEN and ODD properties for addition are same as that for subtraction.
Example:
2 + 1 = 3 ---------- EVEN + ODD = ODD
2 - 1 = 1 ---------- EVEN - ODD = ODD

Hope it helps.

(1) If \(n=1, 1+2^2+3^2=14=Even\), n is Odd.

if \(n=2, 2+3^2+4^2=4+9+16=25=Odd\), n is even and contradictory with the condition.

The answer n is ODD, Sufficient.

(2) If \(n=1, 1-2^2-3^2=1-4-9=-12=Even\) n is Odd.

if \(n=2, 2-3^2-4^2=4-9-16=-21=Odd\), n is even and contradictory with the condition.

The answer n is ODD, Sufficient.

The answer is D.
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Solution:

Question Stem Analysis:

We need to determine whether positive integer n is odd.

Let’s review a few even/odd rules:
If n is odd, then (n + 1) is even, and (n + 2) is odd
Odd x odd = odd, even x even = even, and even x odd = even
Odd + odd = even, even + even = even, and odd + even = odd

Statement One Alone:

If n is odd, then n^2 is odd, (n + 1)^2 is even, and (n + 2)^2 is odd. The sum n^2 + (n + 1)^2 + (n + 2)^2 = odd + even + odd = even.

If n is even, then n^2 is even, (n + 1)^2 is odd, and (n + 2)^2 is even. The sum n^2 + (n + 1)^2 + (n + 2)^2 = even + odd + even = odd.

From the above, we see that, in order for n^2 + (n + 1)^2 + (n + 2)^2 to be even,then n must be odd. Statement one alone is sufficient.

Statement Two Alone:

If n is odd, then n^2 is odd, (n + 1)^2 is even, and (n + 2)^2 is odd and n^2 - (n + 1)^2 - (n + 2)^2 = odd - even - odd = even.

If n is even, then n^2 is even, (n + 1)^2 is odd, and (n + 2)^2 is even and n^2 - (n + 1)^2 - (n + 2)^2 = even - odd - even = odd.

From the above, we see that in order for n^2 - (n + 1)^2 - (n + 2)^2 to be even, then n must be odd. Statement two alone is sufficient.

Answer: D
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