The product of three consecutive odd integers reduced by 23 is 99 less than the cube of the sum of the smallest number and 2. Compute the mean of the three integers.
let the 3 consecutive numbers be
x-2, x and x+2
(x-2)(x)(x+2) - 23 = (x-2 + 2)^3 - 99
x(x^2 - 4) - 23 = x^3 - 99
x^3 - 4x = x^3 - 99 + 23
-4x = -76
x = 19
the three numbers are 17, 19, and 21
and their mean is 19
To solve this problem, we first need to set up the equation based on the given information.
Let's assume the three consecutive odd integers are x, x+2, and x+4.
The product of these three consecutive odd integers is (x)(x+2)(x+4).
According to the problem, this product reduced by 23 is 99 less than the cube of the sum of the smallest number and 2.
So, our equation becomes:
(x)(x+2)(x+4) - 23 = (x+2)^3 - 99
Now, let's simplify and solve the equation step by step:
Expand the cube on the right side:
x^3 + 6x^2 + 12x + 8 - 23 = x^3 + 6x^2 + 12x + 8 - 99
Simplify and cancel out common terms:
x^3 + 6x^2 + 12x - 15 = x^3 + 6x^2 + 12x - 91
Subtract x^3, 6x^2, and 12x from both sides:
-15 = -91
This equation is not true, which means there are no possible solutions. Therefore, there are no consecutive odd integers that satisfy the given conditions.
Since we cannot compute the mean of three integers that do not exist, we cannot determine the mean in this case.