Is k an even integer ?
(1) 3k is an even integer. -
INSUFFICIENTSince it is not given if k is integer or not, let's take both cases.
Case 1: k = 2, therefore, 3k = 6 which is even integer. k is an even integer.
Case 2: k = 2/3, therefore, 3k = 2 which is even integer. But k is NOT an even integer.
Hence, Insufficient
(2) k^3 is an even integer. -
INSUFFICIENTk^3 = 2^a * p^b.
Case 1: if k=2^(1/3), that is k = cube root of 2, k^3 = 2, an even integer. But k is not an even integer.
Case 2: k = 2, k^3 = 8, an even integer. And k is an even integer.
Hence, Insufficient.
Taking 1 & 2, -
SUFFICIENTfor 3k to be an even integer, k has to be of the form (2/3)*q, where q is any integer
Now, for k^3 to be an even integer, it cannot be fraction, so k have to be of the form 2q, where q is any integer.
Which implies k is even integer.
Hence, sufficient.
Answer is C