Is k^2 odd?

How can u say K must be odd...even if its even k(k-1)/2 is an integer.

Ah, you are right. I have omitted one additional step.

Group the a's together and group the numbers together to get: ka + k(k-1)/2
Factor out a k.
k(a+ (k-1)/2)
For this to be divisible by k, (a + (k-1)/2) must be a integer. So therefore k must be odd, otherwise the (k-1)/2) term would be a fraction...

hey cellydan thanks alot for the explanation....btw when is your GMAT?

Question Is k^2 odd?

Stmt1: k-1 is divisible by 2.

(k-1)/2 = m where m is some integer
k-1 = 2m
k = 2m+1. Now this is the equation of any odd number

Hence k is odd. Therefore k^2 is odd.
Sufficient.

Stmt2: The sum of k consecutive integers is divisible by k
sum = k(k+1)/2
Sum/k = m where m is some integer
k(k+1)/2k = m
(k+1)/2 = m
k=2m-1. This again represents odd number.

Hence k is odd. Therefore k^2 is odd.
Sufficient.

OA D

jamifahad,

I would like to point out a small possible improvement in your analysis, which is otherwise on track.

The sum of k consecutive integers is not k(k+1)/2. This is the expression for the sum of the first k natural numbers.
Example: The sum of three consecutive integers 50,51,and 52 is not 3(3+1)/2.

The expression for the sum of k consecutive integers is ak + k(k-1)/2, where a is the first number and k is the number of terms. In this question, you may have assumed that k consecutive integers can also be n consecutive natural numbers and so used the expression k(k+1)/2.

The OA is (D) as explained by cellydan

Hi, What if I assume K is a fraction/decimal?

Question asks is k odd?

1. Clearly Suff, k-1 --> even so k is odd
2. Choose a case or two to test. 3,4,5 yes div by 3
3,4 not div by 2
1,2,3,4,5 - yes div by 5

So, click on D & move on

Here we need to get i k is even/odd
Statement 1
(k-1)/2=integer
hence k-1=2m for some integer m
k=2m+1=> odd
hence sufficient
Statement 2
Here we can use a simple Rule =>

SUM OF N CONSECUTIVES IS DIVISIBLE BY N FOR N BEING ODD AND NEVER DIVISIBLE BY N FOR N BEING EVEN

Hence K must be odd
Alternatively,
Since consecutives inters form an AP
mean = median
It is given that mean = integer
so median => integer too
hence number of terms must be odd
hence K is odd

Thus sufficient

Hence D

FROM STATEMENT - I ( SUFFICIENT )

Since, k - 1 is divisible by 2 ; k must be Odd because -

Odd - 1 = Even ( Which is divisible by 2 )

FROM STATEMENT - II ( SUFFICIENT )

Test using numebrs...

Sum of 2 consecutive integers is 3 ( which is not divisible by 2 )
Sum of 3 consecutive integers is 6 ( which is divisible by 3 )
Sum of 4 consecutive integers is 10 ( which is not divisible by 4 )
Sum of 5 consecutive integers is 15 ( which is divisible by 5 )

So, we can safely conclude k = Odd...

And \(Odd^2\) = Odd

Thus, EACH statement ALONE is sufficient to answer the question asked, answer will be (D)....

Solution:

To find: \(k^2\) is odd, means we have to find whether “k” is odd? So, basically it is a “yes” or “no” DS question.

Analysis of statement 1: \(k - 1\) is divisible by 2.
Given that \((k – 1)\) is divisible by 2 that means \((k - 1)\) is Even.
We wite as \((k – 1)\) = Even (We must be familiar with operations of even and odd numbers)

[We know ODD – ODD = EVEN or EVEN – EVEN = EVEN]
Here k – ODD = EVEN
So, “k” must be ODD.
Hence statement 1 is sufficient. We can eliminate options B, C and E.

Analysis of statement 2: The sum of k consecutive integers is divisible by k.
Let’s take two cases here ‘k’ being odd and ‘k’ being even.
Case 1: Let k be ODD; k =5
Let’s take the consecutive integers as {1, 2, 3, 4, 5}
Sum = 1 + 2 + 3 + 4 + 5 = 15 is divisible by 5.
Case 2: Let k be EVEN; k = 6
Let’s take the consecutive integers as {1, 2, 3, 4, 5, 6}
Sum = 1 + 2 + 3 + 4 + 5 + 6 = 21 is not divisible by 6.
In the above statement it is given that sum of k consecutive integers is divisible by k, which means k must be ODD.
Hence statement 2 is sufficient.

[Note: We can also say that the property of consecutive integers, that:
The sum of n consecutive integers is divisible by n, if n is ODD.
The sum of n consecutive integers is not divisible by n, if n is EVEN.]

The correct answer option is “D”.