Find three consecutive odd integers such that 3 times the sum of all three is 24 more than the product if the first and second integers

3 consecutive odd integers : x, x + 2, x + 4

3(x + x + 2 + x + 4) = x(x + 2) + 24
3(3x + 6) = x^2 + 2x + 24
9x + 18 = x^2 + 2x + 24
x^2 - 7x + 6 = 0
(x - 1)(x - 6) = 0

x - 1 = 0
x = 1

x - 6 = 0
x = 6 (this will not work...it is even)

ur 3 consecutive odd integers :
x = 1
x + 2 = 1 + 2 = 3
x + 4 = 1 + 4 = 5

1935/10,000= 387/2000the answer to the question is c