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2020 AMC 10A Problem 21

Problem 21 of 25HarderAlgebraArithmeticProblem-Solving Techniques

There exists a unique strictly increasing sequence of nonnegative integers a1<a2<⋯<aka_1<a_2<\cdots<a_k such that 2289+1217+1=2a1+2a2+⋯+2ak.\frac{2^{289}+1}{2^{17}+1}=2^{a_1}+2^{a_2}+\cdots+2^{a_k}. What is k?k?

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Solution

Let X=217X=2^{17}. Then 2289+1217+1=X17+1X+1=X16−X15+X14−⋯−X+1 \begin{gathered} \dfrac{2^{289}+1}{2^{17}+1} = \dfrac{X^{17}+1}{X+1} \\ = X^{16}-X^{15}+X^{14} \\ {}-\cdots-X+1 \end{gathered} . Pair consecutive terms: X16−X15X^{16}-X^{15}, X14−X13X^{14}-X^{13}, …\ldots, X2−XX^2-X, and then the final +1+1. Each pair is 217m(217−1)2^{17m}(2^{17}-1), contributing 1717 ones in binary. There are 88 such pairs plus the final 11, so k=8⋅17+1=137k=8\cdot17+1=137. Thus, C is the correct answer.
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Tagged: number base · factoring · pairing and grouping

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