Question

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

$117$

$136$

$137$

$273$

$306$

Answer: C

Concepts:number base factoring pairing and grouping

Difficulty rating: 2380

Video solution:

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Written solution:

Let $X=2^{17}$ . Then $\begin{gathered} \dfrac{2^{289}+1}{2^{17}+1} = \dfrac{X^{17}+1}{X+1} \\ = X^{16}-X^{15}+X^{14} \\ {}-\cdots-X+1 \end{gathered}$ .

Pair consecutive terms: $X^{16}-X^{15}$ , $X^{14}-X^{13}$ , $\ldots$ , $X^2-X$ , and then the final $+1$ . Each pair is $2^{17m}(2^{17}-1)$ , contributing $17$ ones in binary. There are $8$ such pairs plus the final $1$ , so $k=8\cdot17+1=137$ . Thus, C is the correct answer.

2020 AMC 10A Problem 21

Answer: C

Difficulty rating: 2380

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Problem 21 in Other Years