Four consecutive odd integers whose sum is 336?

Answer 1

Any odd integer can be represented as 2n+1, where n is any integer.

The integers, starting with one, always proceed as odd, even, odd, even, etc.

Therefore, a consecutive odd would be two more than the first.

So 2n+1, 2n+3, 2n+5, and 2n+7 summed to 336. Simple algebra gives n=40, and the four numbers as 81, 83, 85, and 87.