Do 18*37^211 and 17+12^99 have the same remainder when divided by 24? How do you solve this using properties of congruence?

$37\equiv 13 mod 24$ 
$\Rightarrow$ $37^2=13^2=169=1$ since $168$ is a multiple of $24$ 
$\Rightarrow$ $37^{210}=(37^2)^{105}=1$
$\Rightarrow$ $37^{211}=37=13$
$\Rightarrow$ $18.37^{211}=18.13=234=-6$
On the other hand $12^{99}$ will be a multiple of $24$ so thats $0$ in $mod$  $24$  
So $17+12^{99}=17 = -7 \neq -6= 18.37^{211}$ 
Edit : I read the comments but had already started foolishly slogging ahead in $mod$ $24$ so thought what the heck and posted. The comment solution is clearly much cleverer. 
