Inspiration: Vicky Vignesh

Since $10=2\times 5$ , let us consider the prime factors separately.

$\begin{aligned} N & \equiv 2^{200}+3^{300}+4^{400} \text{ (mod 2)} \\ & \equiv 0+1+0 \equiv 1 \text{ (mod 2)} \\ \implies N & \equiv 2n+1 \text{ (mod 10)} & \small \color{#3D99F6} \text{where } n \text{ is an integer.} \end{aligned}$

Since 2, 3 and 4 is each a coprime integer with 5, we can applies Euler's theorem as follows.

$\begin{aligned} N & \equiv 2^{200}+3^{300}+4^{400} \text{ (mod 5)} \\ & \equiv 2^{200 \text{ mod }\phi(5)}+3^{300 \text{ mod }\phi(5)}+4^{400 \text{ mod }\phi(5)} \text{ (mod 5)} & \small \color{#3D99F6} \text{Euler's totient function }\phi(5) = 4 \\ & \equiv 2^{200 \text{ mod }4}+3^{300 \text{ mod }4}+4^{400 \text{ mod }4} \text{ (mod 5)} \\ & \equiv 2^0+3^0+4^0 \equiv 1+1+1 \equiv 3 \text{ (mod 5)} \end{aligned}$

Therefore, we have:

$\begin{aligned} N & \equiv 3 \text{ (mod 5)} \\ 2n+1 & \equiv 3 \text{ (mod 5)} \\ \implies n & = 1 \\ \implies N & \equiv 2n+1 \equiv \boxed{3} \text{ (mod 10)} \end{aligned}$

Since we want to determine the unit digit of $N$ , set $N \bmod 10$ . For the computation, since $2$ and $4$ are not relatively prime to $10$ , consider evaluating $\bmod 2$ and $\bmod 5$ separately.

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Evaluating Separately

For $\bmod 2$ , since $3 \equiv 1 \bmod 2$ and $4 \equiv 2 \equiv 0 \bmod 2$ , $N \equiv 1 \bmod 2$ .

For $\bmod 5$ , since $5$ is relatively prime to $2,3,4$ , apply Euler's theorem , so $a^4 \equiv 1 \bmod 5$ where $\gcd(a,5) = 1$ and $\phi(5) = 4$ . So $\begin{array}{rl} N &= \left(2^{4}\right)^{50} + \left(3^{4}\right)^{50} + \left(4^{4}\right)^{50}\\ &\equiv 1 + 1 + 1 \bmod 5\\ &\equiv 3 \bmod 5 \end{array}$

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Final Results

We then have $\begin{array}{rl} N &\equiv 1 \bmod 2\\ N &\equiv 3 \bmod 5 \end{array}$ Observe that $N$ must be an odd integer (which immediately satisfies the first congruence). Thus, $\boxed{N = 3}$ , being the odd integer satisfying the system of congruences.