Divisible by 3

We can try first with the simplest case, $n = 1$ . There is obviously $1$ way to select $3$ numbers from $3$ numbers and because they are consecutive, they sum to a multiple of $3$ . This is because on must be a multiple of $3$ because those occur every three numbers. Another should be one more and another two more than a multiple of $3$ . It shouldn't take you too long to convince yourself with this and that this means they sum to a multiple of $3$ . This means we can set $1 = \frac {(1) * (3(1)^2 - 3(1) + 2)}{k}$ . Solving gets us $1 = \frac {2}{k}$ . It is not too hard to see that $k$ must be $\boxed{2}$ .