$$36={{2}^{2}}\times {{3}^{2}}$$, so the number of its factors would be $$\left( 2+1 \right)\times \left( 2+1 \right)=9$$. Among them, there are $$\left( 2+1 \right)\times 1=3$$ factors which have $${{2}^{2}}$$ as its factors. 
How many positive factors of $$36$$ are also multiples of $$4$$? 
Options:
$$2$$ 
$$3$$ 
$$4$$ 
$$5$$ 
$$6$$ 