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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...

Is k2 odd? (1) k - 1 is divisible by 2. (2) The sum of k consecutive integers is divisible by k.

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
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My dad once said to me: Son, nothing succeeds like success.

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
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Is k2 odd? (1) k - 1 is divisible by 2. (2) The sum of k consecutive integers is divisible by k.

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.

Is k2 odd? (1) k - 1 is divisible by 2. (2) The sum of k consecutive integers is divisible by k.

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

Both statements imply that k is an integer.

Is k^2 odd?

(1) k - 1 is divisible by 2 --> \(k-1=even\) --> \(k=even+1=odd=integer\) --> \(k^2=odd^2=odd\). Sufficient.

(2) The sum of k consecutive integers is divisible by k. Here k must be a positive integer because otherwise the statement does not make sense.

Properties of consecutive integers: • If n is odd, the sum of n consecutive integers is always divisible by n. Given \(\{9,10,11\}\), we have \(n=3=odd\) consecutive integers. The sum is 9+10+11=30, which is divisible by 3. • If n is even, the sum of n consecutive integers is never divisible by n. Given \(\{9,10,11,12\}\), we have \(n=4=even\) consecutive integers. The sum is 9+10+11+12=42, which is NOT divisible by 4.

The statement says that the sum of k consecutive integers is divisible by k, which, according to the above means that k is odd, therefore \(k^2=odd^2=odd\). Sufficient.

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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