Difference between revisions of "2020 AMC 10A Problems/Problem 21"
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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 | + | 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 such thatWhat is
Solution 1
First, substitute with . Then, the given equation becomes . Now consider only . This equals . Note that equals , since the sum of a geometric sequence is . Thus, we can see that forms the sum of 17 different powers of 2. Applying the same method to each of , , ... , , we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us . But we must count also the term. Thus, Our answer is .
~seanyoon777
Solution 2
(This is similar to solution 1) Let . Then, . The LHS can be rewritten as . Plugging back in for , we have . When expanded, this will have terms. Therefore, our answer is .
Video Solution
~IceMatrix
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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