Question 216936
Let the first odd integer be (2n+1) where n is any integer, the (2n+3) is the next consecutive odd integer and (2n+5) is the third consecutive odd integer.
The sum of these is 16 more than the third (largest) of the integers.
{{{(2n+1)+(2n+3)+(2n+5) = (2n+5)+16}}} Simplify.
{{{6n+9 = 2n+21}}} Subtract 2n from both sides.
{{{4n+9 = 21}}} Subtract 9 from both sides.
{{{4n = 12}}} Divide both sides by 4.
{{{n = 3}}} so...
{{{2n+1 = 7}}}
{{{2n+3 = 9}}}
{{{2n+5 = 11}}}