Let the first odd integer be x
The second will be (x+2)
Third odd integer = (x+4)
Given sum is = 141
So,
x+(x+2)+(x+4)=141
3x+6=141
3x=141-6
3x=135
x=135/3
x=45
x=45, (x+2)=47, (x+4)=49
45, 47, and 49 are the three consecutive odd integers whose sum is 141.
