7 ... 7's remainder

Method 1: Note that

• (i) $2^{3} \equiv 1 \pmod{7} \Longrightarrow 2^{200} = 2^{3*66 + 2} \equiv 2^{2} \pmod{7} \equiv 4 \pmod{7}$ ;

• (ii) $3^{3} \equiv -1 \pmod{7} \Longrightarrow 3^{300} = 3^{3*100} \equiv (-1)^{100} \pmod{7} \equiv 1 \pmod{7}$ ;

• (iii) $4^{400} = 2^{800} = 2^{3*266 + 2} \equiv 2^{2} \pmod{7} \equiv 4 \pmod{7}$ .

Thus $N \equiv (4 + 1 + 4) \pmod{7} \equiv \boxed{2} \pmod{7}$ .

Method 2: Using Fermat's Little Theorem we have that

• (i) $2^{6} \equiv 1 \pmod{7} \Longrightarrow 2^{200} = 2^{6*33 + 2} \equiv 2^{2} \pmod{7} \equiv 4 \pmod{7}$ ;

• (ii) $3^{6} \equiv 1 \pmod{7} \Longrightarrow 3^{300} = 3^{6*50} \equiv 1 \pmod{7}$ ;

• (iii) $4^{6} \equiv 1 \pmod{7} \Longrightarrow 4^{400} = 4^{6*66 + 4} \equiv 4^{4} \pmod{7} \equiv 2^{2} \pmod{7} \equiv 4 \pmod{7}$ .

Once again we have $N \equiv (4 + 1 + 4) \pmod{7} \equiv \boxed{2} \pmod{7}$ .

$\begin{aligned} N & \equiv 2^{200} + 3^{300} + 4^{400} \text{ (mod 7)} \\ & \equiv 2^{3\times 66+2} + 3^{3 \times 100} + 4^{3 \times 133+1} \text{ (mod 7)} \\ & \equiv 8^{66}\times 4 + 27^{100} + 64^{133}\times 4 \text{ (mod 7)} \\ & \equiv 4(7+1)^{66} + (28-1)^{100} + 4 (63+1) ^{133} \text{ (mod 7)} \\ & \equiv 4(1^{66}) + (-1)^{100} + 4 (1^{133}) \text{ (mod 7)} \\ & \equiv 4 + 1 + 4 \equiv 9 \equiv \boxed{2} \text{ (mod 7)} \end{aligned}$

$2^2 \equiv 4 \mathrm{(mod 7)} (2^2)^2 \equiv 2 \mathrm{(mod 7)} \\ 2^4 \equiv 2 \mathrm{(mod 7)} \\ (2^4)^5 \equiv 2^5 \mathrm{(mod 7)} \equiv 4 \mathrm{(mod 7)} \\ 2^{100} \equiv 4 \mathrm{(mod 7)} \equiv 2 \mathrm{(mod 7)} \\ 2^{200} \equiv 2^2 \mathrm{(mod 7)} \equiv 4 \mathrm{(mod 7)} \\ \text{Similarly,} \\ 3^{300} \equiv 1 \mathrm{(mod 7)} \\ 4^{400} \equiv 4 \mathrm{(mod 7)} \\ \color{#3D99F6}{\boxed{\therefore \{ 2^{200} + 3^{300} + 4^{400} \} \equiv (1 + 4 + 4) \mathrm{(mod 7)} \equiv 9 \mathrm{(mod 7)} \equiv 2 \mathrm{(mod 7)}}}$