nidhis.1408 wrote:Is k^2 odd?

(1) k - 1 is divisible by 2.
(2) The sum of k consecutive integers is divisible by k.

Statement 1: As (k - 1) is divisible by 2, k must be odd.
Hence, k² is also odd.

Sufficient

Statement 2: Say, k consecutive integers are a, (a + 1), (a + 2), ..., and (a + k - 1)
Hence, the sum of these k consecutive integers, S = ak + k(k - 1)/2 = k[a + (k - 1)/2]

Now, if S is divisible by k, then [a + (k - 1)/2] must be integer.
So, (k - 1)/2 must be integer.
So, (k - 1) must be even.

Hence, k must be odd.
Therefore, k² is also odd.

Sufficient

The correct answer is D.

Thanks for pointing it out.
I've edited my post.

TimeforGMAT wrote:Sum of K consecutive integers (K*(K+1)/2) is always divisible by K regardless of K being even or odd.
So we cannot say if K^2 is even or odd.

K(K + 1)/2 is the sum of first K integers, which is not same as sum of K consecutive integers.

For the rest part see my edited post above. I've tried to explain it with algebra.