The sum of each numbers remainder can be divided by $$4$$ to get $$2$$: $$123\div 4$$ R $$3$$；$$234\div 4$$ R $$2$$；$$345\div 4$$ R $$1$$; Therefore, $$\left( 3+2+1 \right)\div 4$$ R $$2$$． 
$$123+234+345$$ divided by $$4$$ has a remainder of． 
Options:
  $$0$$ 
  $$1$$ 
  $$2$$ 
  $$3$$ 