Do 18*37^211 and 17+12^99 have the same remainder when divided by 24?

Question

How do you solve this using properties of congruence?

Answer 1

$18\cdot37^{211}$ is even, and $17+12^{99}$ is odd. Since $a\equiv b\pmod{24}$ implies $a\equiv b\pmod 2$ (since $2$ is a factor of $24$), they are not congruent modulo $24$.

Stolen from the comment section.

Answer 2

$37\equiv 13 mod 24$

$\Rightarrow$ $37^2=13^2=169=1$ 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.