Question 1099120
  Find three consecutive odd integers
n, (n+2), (n+4)
 such that 3 times the 2nd is 1 less than the 1st and 3rd. 
3(n+2) = n + (n+4) - 1
3n + 6 = 2n + 3
3n - 2n = 3 - 6
n = -3
:
The 3 consecutive odd integers: -3, -1, +1