2020 AMC 10A Problems/Problem 21

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

$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$

Solution

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

~seanyoon777

Video Solution

https://youtu.be/Ozp3k2464u4

~IceMatrix

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2020 AMC 10A Problems/Problem 21

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