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~\uline{~~~~~~~~~~}~． 
A) $$0$$ 
B) $$1$$ 
C) $$2$$ 
D) $$3$$ 