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

(Solution 2)
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(This is similar to solution 1)
 
(This is similar to solution 1)
 
Let <math>x = 2^{17}</math>. Then, <math>2^{289} = x^{17}</math>.
 
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+17}+2^{13+17}+\cdots+2^{1+17})+1</math>. When expanded, this will have <math>17\cdot8+1=137</math> terms. Therefore, our answer is <math>\boxed{\textbf{(C)} 137}</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>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 02:04, 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 $\boxed{\textbf{(C) } 137}$.

Video Solution

https://youtu.be/Ozp3k2464u4

~IceMatrix

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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

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