2020 AMC 10A Problems/Problem 21

Revision as of 23:12, 31 January 2020 by Seanyoon777 (talk | contribs) (Solution)

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 $\frac{a^17+1}{a+1}=a^16-a^15+a^14...-a^1+a^0$. 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 $a^14-a^13$, $a^12-a^11$, and so on. This gives us $17*8=136$. Our answer is $\boxed{\textbf{(B) } 136}$.

~seanyoon777

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AMC 10 Problems and Solutions

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