Question 147886
Let x = 1st consecutive odd integer
x+2 = 2nd odd integer
x+4 = 3rd odd integer
.
from:
"four times the smallest of three consecutive odd intergers is five more than three times the largest" 
we get:
4x = 3(x+4)+5
.
solving for x:
4x = 3(x+4)+5
4x = 3x+12+5
4x = 3x+17
x = 17 (1st odd integer)
.
then
x+2 = 19 (2nd odd integer)
x+4 = 21 (3rd odd integer)
.
solution: 17+19+21=57