Question

Four consecutive odd intergers whos sum is 56

Answer 1

Let $N$ be the first of the four odd integers belonging to the sum. The next three integers are $N+2$, $N+4$, and $N+6$.

We then have

$N+(N+2)+(N+4)+(N+6)=4N+12=56\implies 4N=44\implies N=11$

and so the four integers are 11, 13, 15, and 17.