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

Since $$x^{10} + 1 = (x^2)^5 + 1$$ and the exponent 5 is odd, $$x^2 + 1$$ divides it exactly with quotient $$x^8 - x^6 + x^4 - x^2 + 1$$. So the non-zero coefficients are $$a = 1$$, $$c = -1$$, $$e = 1$$, $$g = -1$$, $$k = 1$$ and the rest are 0. Raising to the even power 2024 turns each of these five into 1, so the sum is 5.