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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

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