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# help

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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 $$A+B$$.

Jan 17, 2021

#1
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By the remainder theorem, A = 2 and B = 1, so A + B = 3.

Jan 18, 2021
#2
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$$\displaystyle \left\{\frac{x^{102}-1}{x^{3}-1}-\frac{x^{101}-x^{2}}{x^{3}-1}\right\}(x^{2}+x+1)=x^{101}+x+1, \\ \text{so} \\ A+B=2.$$

The two fractions inside the curly brackets are both the sums of GP's.

The first is the sum of

$$1+x^{3}+x^{6}+ \dots +x^{99},$$

and the second the sum of

$$x^{2}+x^{5}+ \dots + x^{98}.$$

Jan 19, 2021