Find three consecutive odd positive integers such that 3 times the sum of all three is 152 less than the product of the first and second integers

let the 3 cosecutive odd numbers be

x-2 , x, and x+2

3(x-2 + x + x+2) = x(x-2) - 152
9x = x^2 - 2x - 152
x^2 - 11x - 152 = 0

(x-19)(x + 8) = 0
x = 19, or x = -8 , but x is to be positive

the 3 numbers are:
17, 19, and 21

check:
their sum = 57
3 times their sum = 171

product of first and second = 17(19) = 323
Is 171 less than 323 by 152 ? YES

All is good

To find the three consecutive odd positive integers, let's set up an equation based on the given information.

Let's assume the first odd positive integer is x. Since we are looking for three consecutive odd positive integers, the second integer would be (x + 2) and the third integer would be (x + 4).

According to the given information, 3 times the sum of all three integers is 152 less than the product of the first and second integers. In equation form, this can be expressed as:

3(x + (x + 2) + (x + 4)) = (x) * (x + 2) - 152

Now, let's simplify this equation and solve it:

3(3x + 6) = x^2 + 2x - 152
9x + 18 = x^2 + 2x - 152
9x + 18 = x^2 + 2x - 152

By rearranging the equation, we get:

0 = x^2 - 7x - 170

Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:

x = (-(-7) ± √((-7)^2 - 4(1)(-170))) / 2(1)
x = (7 ± √(49 + 680)) / 2
x = (7 ± √729) / 2
x = (7 ± 27) / 2

Solving for both values of x, we get:

x1 = (7 + 27) / 2 = 34 / 2 = 17
x2 = (7 - 27) / 2 = -20 / 2 = -10 (Discard this since we want positive integers)

Therefore, the first odd positive integer is 17. The second odd positive integer is (17 + 2) = 19, and the third odd positive integer is (17 + 4) = 21.

So, the three consecutive odd positive integers are 17, 19, and 21.