The sum of three consecutive one-digit, positive, odd integers is one-seventh of the product of the same three integers. What is the middle integer when the three integers are listed in ascending order?
+37191 [(x-2) + (x) + (x+2) ] = 1/7(x-2)(x)(x+2)
3x = x^3-4x
21 = x^2-4
0 =x^2 -25 x = -5 (throw out) or 5
so the numbers are 5-2 = 3 5 and 5+2 = 7 (as guest posted)
+130555 Let the smallest integer be N - 2
Let the middle integer be N
Let the largest integer be N + 2
So the sum of these = 3N
So we have that
3N (7) = (N + 2) (N - 2) N
21N = (N^2 - 4)N
21N = N^3 - 4N rearrange as
N^3 - 25N = 0 factor
N (N^2 - 25) = 0
N ( N + 5) (N - 5) = 0
Setting each factor to 0 and solving for N produces N = 0 , N = -5 and n = 5
So....the middle integer is 5
