If $x\equiv 1 \mod 3$, then $x^{100}\equiv 1 \mod 3$. [closed]
Suppose that $x = 1\text{ (mod 3)}$, it means that there exists an integer $k$ such that $x=3k+1$. Now, if we multiply $x$ by $x$, we have $$ x^2 = (3k+1)(3k+1) = 9k^2+6k+1 = 3l+1 $$ where $l=3k^2+2k$ is an integer. It means that $x^2 = 1\text{ (mod 3)}$. Knowing this, can you guess if $x^{100} = 1\text{ (mod 3)}$ or not?
$$x \equiv 1 \pmod{3} $$ tells you that $x-1$ is a multiple of 3. Combine this with the fact that $$x^{100}-1=(x-1)(x^{99}+x^{98}+\dots+x). $$