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, replace $2^(17)$ as $a$ . Then, the given equation becomes $(a^(17)+1)/(a+1)=a^(16)-a^(15)+a^(14)...-a^1+a^0$ . 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 $(a^n-1)/(a-1)$ . 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 $a^(14)-a^13$ , $a^12-a^11$ , and so on. This gives us $17*8=136$ . Our answer is $\boxed{\textbf{(B) } 136}$ .

~seanyoon777

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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