Remainder? No calculator allowed

$\begin{aligned} N & \equiv 1+2^2+{\color{#D61F06}3^3}+4^4+5^5+{\color{#D61F06}6^6}+7^7+8^8+{\color{#D61F06}9^9}+10^{10} \text{ (mod 3)} & \small \color{#D61F06} \text{3, 6, 9 are divisible by 3} \\ & \equiv 1+4+{\color{#D61F06}0}+(3+1)^4+(6-1)^5+{\color{#D61F06}0}+(6+1)^7+(9-1)^8+{\color{#D61F06}0}+(9+1)^{10} \text{ (mod 3)} \\ & \equiv 1+1+0+1^4+(-1)^5+0+1^7+(-1)^8+0+1^{10} \text{ (mod 3)} \\ & \equiv 1+1+0+1-1+0+1+1+0+1 \text{ (mod 3)} \\ & \equiv 5 \equiv \boxed{2} \text{ (mod 3)} \end{aligned}$

$1\mod 3= 1$

$2^2 \mod 3 = 1$

$3^3 \mod 3= 0$

$4^4 \mod 3 = 1$

$5^5 \mod 3 = 2$

$6^6 \mod 3=0$

$7^7 \mod 3= 1$

$8^8\mod 3= 1$

$9^9 \mod 3 =0$

$10^{10}\mod 3= 1$

Add all the answers we get $8 \mod 3= 2 \implies$ the remainder is $\boxed{2}$ .

$1 \equiv 1 (mod 3)$ $\Rightarrow 1$

$2^2 \equiv 1 (mod 3)$ $\Rightarrow 1$

$3^3 \equiv 0 (mod 3)$ $\Rightarrow 0$

$4^4 = 2^8$

$(2^2)^4 \equiv 1^4 (mod 3)$ $\Rightarrow 1$

$5^2 \equiv 1 (mod 3)$

$(5^2)^2 \cdot 5 \equiv 1^2 \cdot 5 (mod 3)$

$5 (mod 3) \equiv 2 (mod 3)$ $\Rightarrow 2$

$6^6 \equiv 0 (mod 3)$ $\Rightarrow 0$

$7 \equiv 1 (mod 3)$

$7^7 \equiv 1^7 (mod 3)$ $\Rightarrow 1$

$8^8 = 2^24$

$(2^2)^12 \equiv 1^12 (mod 3)$ $\Rightarrow 1$

$9^9 \equiv 0 (mod 3)$ $\Rightarrow 0$

$10 \equiv 1 (mod 3)$

$10^{10} \equiv 1^{10} (mod 3)$ $\Rightarrow 1$

$1 + 1 + 0 + 1 + 2 + 0 + 1 + 1 + 0 + 1 + 1 (mod 3)$

$8 (mod 3)$

$2 (mod 3)$

Remainder = $\boxed { 2 }$