What 5 consecutive numbers have the sum of 40?

Let's denote the five consecutive numbers as n, n+1, n+2, n+3, and n+4. The sum of these numbers is given by n + (n+1) + (n+2) + (n+3) + (n+4) = 5n + 10 = 40. Solving for n, we get n = 6. Therefore, the five consecutive numbers are 6, 7, 8, 9, and 10.