The polynomial $x^{101} + Ax + B$ is divisible by $x^2 + x + 1$ for some real numbers $A$ and $B.$  Find $A + B.$
Question:
The polynomial $x^{101} + Ax + B$ is divisible by $x^2 + x + 1$ for some real numbers $A$ and $B.$  Find $A + B.$
Answer:
Let's think step by step.

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The polynomial $x^{101} + Ax + B$ is divisible by $x^2 + x + 1$ for some real numbers $A$ and $B.$  Find $A + B.$
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