Is the positive integer n odd?

Question

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

OG Q 2017 New Question(Book Question: 194)

stonecold · Accepted Answer

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

Bunuel · Answer

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.

hsbinfy · Answer

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

stonecold · Answer

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

kamal108555 · Answer

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.

guenthermat · Answer

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

adkikani · Answer

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?

srikanth9502 · Answer

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

adkikani · Answer

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.

Salsanousi · Answer

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

blitzkriegxX · Answer

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.

MHIKER · Answer

(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.

ScottTargetTestPrep · Answer

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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Is the positive integer n odd?

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