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 $\frac{a^n-1}{a-1}$ . 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

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2020 AMC 10A Problems/Problem 21

Solution

See Also