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.