Difference between revisions of "2020 AMC 10A Problems/Problem 21"

Revision as of 16:07, 1 February 2020

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 1

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 method to each of $a^{14}-a^{13}$ , $a^{12}-a^{11}$ , ... , $a^{2}-a^{1}$ , 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 is $136+1=\boxed{\textbf{(C) } 137}$ .

~seanyoon777

Solution 2

(This is similar to solution 1) Let $x = 2^{17}$ . Then, $2^{289} = x^{17}$ . The LHS can be rewritten as $\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+\cdots+x^2-x+1=(x-1)(x^{15}+x^{13}+\cdots+x^{1})+1$ . Plugging $2^{17}$ back in for $x$ , we have $(2^{17}-1)(2^{15+17}+2^{13+17}+\cdots+2^{1+17})+1=(2^{16}+2^{15}+\cdots+2^{1})(2^{15\cdot17}+2^{13\cdot17}+\cdots+2^{1\cdot17})+1$ . When expanded, this will have $17\cdot8+1=137$ terms. Therefore, our answer is $\boxed{\textbf{(C) } 137}$ .

Solution 3

Note that the expression is equal to something slightly lower than $2^{272}$ . Clearly, answer choices $(D)$ and $(E)$ make no sense because the lowest sum for $273$ terms is $2^{273}-1$ . Now, $(A)$ makes no sense. $(B)$ and $(C)$ are 1 apart, but because the expression is odd, it will have to contain $2^0=1$ , and because $(C)$ is $1$ bigger, the answer is $\boxed{\textbf{(C) } 137}$ .

~Lcz

Video Solution

https://youtu.be/Ozp3k2464u4

~IceMatrix

@@ Line 19: / Line 19: @@
 Let <math>x = 2^{17}</math>. Then, <math>2^{289} = x^{17}</math>.
 The LHS can be rewritten as <math>\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+\cdots+x^2-x+1=(x-1)(x^{15}+x^{13}+\cdots+x^{1})+1</math>. Plugging <math>2^{17}</math> back in for <math>x</math>, we have <math>(2^{17}-1)(2^{15+17}+2^{13+17}+\cdots+2^{1+17})+1=(2^{16}+2^{15}+\cdots+2^{1})(2^{15\cdot17}+2^{13\cdot17}+\cdots+2^{1\cdot17})+1</math>. When expanded, this will have <math>17\cdot8+1=137</math> terms. Therefore, our answer is <math>\boxed{\textbf{(C) } 137}</math>.
+==Solution 3==
+Note that the expression is equal to something slightly lower than <math>2^{272}</math>. Clearly, answer choices <math>(D)</math> and <math>(E)</math> make no sense because the lowest sum for <math>273</math> terms is <math>2^{273}-1</math>. Now, <math>(A)</math> makes no sense. <math>(B)</math> and <math>(C)</math> are 1 apart, but because the expression is odd, it will have to contain <math>2^0=1</math>, and because <math>(C)</math> is <math>1</math> bigger, the answer is <math>\boxed{\textbf{(C) } 137}</math>.
+~Lcz
 ==Video Solution==

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