Question 704019
ODD integers...
Using n as a whole number, variable, 2n would be even, so 2n+1 would be odd.


Let the consecutive odd numbers be 2n+1, 2n+3, 2n+5.


The description then means, {{{2n+1+2n+3+2n+5=33}}}.  Find n and compute each odd number.
Some algebra simplification toward the answer:
{{{6n+9=33}}}
{{{2n+3=11}}}, which is already one of the numbers without yet finding n.  The other two numbers must be 9 and 13.