Find the remainder of $9^2\cdot 13\cdot 21^2$ when divided by $4$ 
Find the remainder of $9^2\cdot 13\cdot 21^2$ when divided by $4$

How should I approach this type of questions?
Without calculator of course
I did this:
$9^2\cdot 13\cdot 21^2=81\cdot 13\cdot 441=81\cdot 5733=464,373=33\bmod 4=1 \bmod 4$

$(9^2\cdot13\cdot21^2)\bmod4=$
$((9\bmod4)^2\cdot(13\bmod4)\cdot(21\bmod4)^2)\bmod4=$
$(1^2\cdot1\cdot1^2)\bmod4=$
$1\bmod4$
