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

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== Solution ==
 
== Solution ==
 
First, replace <math>2^(17)</math> as <math>a</math>.  
 
First, replace <math>2^(17)</math> as <math>a</math>.  
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Then, the given equation becomes <math>(a^17+1)/(a+1)=a^16-a^15+a^14...-a^1+a^0</math>.
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Then, the given equation becomes <math>\frac{a^17+1}{a+1}=a^16-a^15+a^14...-a^1+a^0</math>.
 
Now consider only <math>a^16-a^15</math>. This equals <math>a^15(a-1)=a^15*(2^17-1)</math>.
 
Now consider only <math>a^16-a^15</math>. This equals <math>a^15(a-1)=a^15*(2^17-1)</math>.
 
Note that <math>2^17-1</math> equals <math>2^16+2^15+...+1</math>, since the sum of a geometric sequence is <math>\frac{a^n-1}{a-1}</math>.
 
Note that <math>2^17-1</math> equals <math>2^16+2^15+...+1</math>, since the sum of a geometric sequence is <math>\frac{a^n-1}{a-1}</math>.

Revision as of 22:12, 31 January 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

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 (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

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