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:
$9^2\cdot 13\cdot 21^2=81\cdot 13\cdot 441=81\cdot 5733=464,373=33\bmod 4=1 \bmod 4$

$9^2*13*21^2$ is not a multiple of 4 or two.
Hence it cannot be 0 or 2 (mod 4)
$9^2*13*21^2$ can be written as 
$9^2*21^2*(3^2+2^2) =$
$(9*21*3)^2 + (9*21*2)^2$
A sum of squares cannot give 3 (mod 4)
Hence it is 1 (mod 4)
