the steps in the correct order to show that 6 divides n whenever n is a nonnegative integer using mathematical STEP VE STEP 61 0 0 i e 610 so the basis step is true 61 1 1 i e 610 the basis step is true By the inductive hypothesis 31 kk Clearly 31 3k k 1 but 3k k 1 is also odd as one of k or k 1 is even therefore 6 3k k 1 as well k 1 k 1 k 3k 3k 1 k 1 K k 3k k 1 0 By the inductive hypothesis 6 K K Clearly 31 3k k 1 but 3k K 1 is also even as one of k or k 1 is even therefore 6 3k k 1 as well As the sum of two multiples of 6 is again divisible by 6 6 k 1 k 1 Suppose that 6 K K