2020 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 21 Hard

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

  • A.

    117

  • B.

    136

  • C.

    137

  • D.

    273

  • E.

    306

Answer:C

Solution 1： First, substitute 2^{17} with a. Then, the given equation becomes \frac{a^{17}+1}{a+1}=a^{16}-a^{15}+a^{14} \ldots-a^{1}+a^{0}. Now consider only a^{16}-a^{15}. This equals a^{15}(a-1)=a^{15} *\left(2^{17}-1\right). Note that 2^{17}-1 equals 2^{16}+2^{15}+\ldots+1, since the sum of a geometric \frac{a^{n}-1}{a} sequence is \overline{a-1}. Thus, we can see that a^{16}-a^{15} forms the sum of 17 different powers of 2 . Applying the same method to each of a^{14}-a^{13}, a^{12}-a^{11}, \ldots, a^{2}-a^{1}, we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us 17 * 8=136. But we must count also the a^{0} term. Thus, Our answer \text { is } 136+1=\text { (C) } 137

Solution 2: (This is similar to solution 1) Let x=2^{17}. Then, 2^{289}=x^{17}. The LHS can be rewritten as \frac{x^{17}+1}{x+1}=x^{16}-x^{15}+\cdots+x^{2}-x+1=(x-1)\left(x^{15}+x^{13}+\cdots+x^{1}\right)+1 Plugging 2^{17} back in for X, we have \left(2^{17}-1\right)\left(2^{15 \cdot 17}+2^{13 \cdot 17}+\cdots+2^{1-17}\right)+1=\left(2^{16}+2^{15}+\cdots+2^{0}\right)\left(2^{15 \cdot 17}+2^{13 \cdot 17}+\cdots+2^{1 \cdot 17}\right)+1 When expanded, this will have 17 \cdot 8+1=137 terms. Therefore, our answer is \mathbf{( C )} 137

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