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Let the sum be $S = \displaystyle \sum_{k=1}^{10} k^k$ . We need to find $k^k \text{ mod }3$ and $k^k \text{ mod } 10$ in our calculations. We can apply Euler's theorem , if $k$ and 3 are coprime integers or $\gcd(k,3)=1$ . Then, we have $k^k \equiv k^{k \text{ mod } {\color{#3D99F6}\phi(3)}} \equiv k^{k \text{ mod } {\color{#3D99F6}2}}$ , where $\phi(3) = 2$ is the Euler's totient function of 3. Similarly for 10 and $\phi(10) = 4$ . Therefore, we have:

$\begin{aligned} S & \equiv 1 + 2^{2 \text{ mod } 2} + {\color{#D61F06}3^3} + 4^{4 \text{ mod } 2} + 5^{5 \text{ mod } 2} + {\color{#D61F06}6^6} + 7^{7 \text{ mod } 2} + 8^{8 \text{ mod } 2} + {\color{#D61F06}9^9} +10^{10 \text{ mod } 2} \text{ (mod 3)} & \small {\color{#D61F06}3|3^3, \ 3|6^6, \ 3|9^9} \\ & \equiv 1 + 2^0 + {\color{#D61F06}0} + 4^{0} + 5^{1} + {\color{#D61F06}0} + 7^{1} + 8^{0} + {\color{#D61F06}0} +10^{0} \text{ (mod 3)} \\ & \equiv 1 + 1 + {\color{#D61F06}0} + 1 + 2 + {\color{#D61F06}0} + 1 + 1 + {\color{#D61F06}0} +1 \text{ (mod 3)} \\ & \equiv 8 equiv 2 \text{ (mod 3)} \end{aligned}$

$\implies r = 2$

$\begin{aligned} S & \equiv 1 + 2^2 + 3^{3 \text{ mod } 4} + 4^4 + 5^5 + 6^6 + 7^{7 \text{ mod } 4} + 8^8 + 9^{9 \text{ mod } 4} +{\color{#D61F06}10^ {10}} \text{ (mod 10)} & \small {\color{#D61F06}10|10^{10}} \\ & \equiv 1 + 4 + 27 + {\color{#3D99F6}16^2} + {\color{#3D99F6}5^5} + {\color{#3D99F6}6^6} + 7^{3} + {\color{#3D99F6}16^6} + 9^1 + {\color{#D61F06}0} \text{ (mod 10)} & \small {\color{#3D99F6}\text{See Note}} \\ & \equiv 1 + 4 + 27 + {\color{#3D99F6}6} + {\color{#3D99F6}5} + {\color{#3D99F6}6} + 49\cdot 7 + {\color{#3D99F6}6} + 9 + 0 \text{ (mod 10)} \\ \implies s & = 1 + 4 + 7 + 6 + 5 + 6 + 3 + 6 + 9 + 0 \\ & = 47 \end{aligned}$

$\implies r+s = 2+47 = \boxed{49}$

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Note: Powers of a natural number ends with 6 end with 6. Powers of a natural number ends with 5 end with 5.

$1+2^2+3^3+4^4+5^5+6^6+7^7+8^8+9^9+10^{10}=$

$1+1+0+1+2+0+1+1+0+1 = 8 = 2(mod 3)= r$ ;

$s= 1+4+7+6+5+6+3+6+9+0 = 47$ ;

thus: $2+47=49$