jmc

Let the remainder be $r$, a constant. Then we know that $$3y^3-13y^2+11y+23=(3y+2)(y^2-5y+c)+r.$$Expanding gives us $$3y^3-13y^2+11y+23=3y^3-15y^2+3cy+2y^2-10y+2c+r$$which simplifies to $$11y+23=3cy-10y+2c+r.$$Since the remainder is constant, we know that $$11y=(3c-10)y$$Solving for $c$ gives us $c=7$. Then the remainder $r=23-2c=23-14=\boxed{9}.$