When $$x^{10} + 1$$ is divided by $$x^2 + 1$$, we get 

$$ax^8 + bx^7 + cx^6 + dx^5 + ex^4 + fx^3 + gx^2 + hx + k$$ 

as quotient. Then the value of 

$$a^{2024} + b^{2024} + c^{2024} + d^{2024} + e^{2024} + f^{2024} + g^{2024} + h^{2024} + k^{2024}$$ is

$$x^{10} + 1 = (x^2)^5 + 1$$ 

Using the property that $$ a^n + b^n$$ is exactly divisible by $$a+b$$ when $$n$$ is odd, we can say that

 $$x^2 + 1$$ divides $$x^{10}+1$$ exactly 

We can simply multiply $$ax^8 + bx^7 + cx^6 + dx^5 + ex^4 + fx^3 + gx^2 + hx + k$$ with $$x^2 +1$$ and compare with $$x^{10}+1$$

$$(ax^8 + bx^7 + cx^6 + dx^5 + ex^4 + fx^3 + gx^2 + hx + k)(x^2+1) = ax^{10}+bx^9 + cx^8 + dx^7 + ex^6 + fx^5 + gx^4 + hx^3 + kx^2 + ax^8 + bx^7 + cx^6 + dx^5 + ex^4 + fx^3 + gx^2 + hx + k$$

Grouping the coefficients of the terms we get

$$ax^{10} + bx^9 + (a+c)x^8 + (b+d)x^7 + (c+e)x^6 + (d+f)x^5 + (e+g)x^4 + (f+h)x^3 + (g+k)x^2 + hx + k = x^{10}+1$$

Equating the coefficients on both sides we get

$$ a= 1, b = 0, c=-1, d=0, e=1, f=0, g=-1, h=0,k=1$$

 $$a^{2024} + b^{2024} + c^{2024} + d^{2024} + e^{2024} + f^{2024} + g^{2024} + h^{2024} + k^{2024} = 1+0+1+0+1+0+1+0+1 =5 $$