Write $N = 5k + r$ for $r = 0, 1, 2, 3$, or $4$. If $r = 0$, then $N = 5k$ and $N^{100}$ is divisible by $125$, so the remainder is $0$. If $r = 1, 2, 3$, or $4$, then $N^2 = 25k^2 + 10rk + r^2 = 5m + 1$ for some integer $m$. Now use the Binomial Theorem:
$$\begin{aligned} N^{100} = (N^2)^{50} = (5m \pm 1)^{50} = (5m)^{50} &\pm 50(5m)^{49} + \binom{50}{2}(5m)^{48} \pm \dots \\ &\pm \binom{50}{47}(5m)^3 + \binom{50}{48}(5m)^2 \pm 50(5m) + 1. \end{aligned}$$

Let $\phi(n)$ be the number of positive integers less than $n$ that are relatively prime to $n$; this is Euler's totient function. Then $\phi(125) = 5^3 - 5^2 = 100$. By Euler's Totient Theorem, if $a$ is not a multiple of $5$, then $a^{100} \equiv 1 \pmod{125}$. If $a$ is a multiple of $5$, then $a^{100} \equiv 0 \pmod{125}$. Therefore there are only $2$ possible remainders: $0$ and $1$.