Difference between revisions of "2020 AMC 10A Problems/Problem 21"
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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\cdot{17}}+2^{13\cdot{17}}+\cdots+2^{1\cdot{17}})+1=(2^{16}+2^{15}+\cdots+2^{0})(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>. | 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\cdot{17}}+2^{13\cdot{17}}+\cdots+2^{1\cdot{17}})+1=(2^{16}+2^{15}+\cdots+2^{0})(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== | + | ==Solution 3 (Intuitive)== |
+ | Multiply both sides by <math>2^{17}+1</math> to get <cmath>2^{289}+1=2^{a_1} + 2^{a_2} + … + 2^{a_k} + 2^{a_1+17} + 2^{a_2+17} + … + 2^{a_k+17}.</cmath> | ||
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+ | Notice that <math>a_1 = 0</math>, since there is a <math>1</math> on the LHS. However, now we have an extra term of <math>2^{18}</math> on the right from <math>2^{a_1+17}</math>. To cancel it, we let <math>a_2 = 18</math>. The two <math>2^{18}</math>'s now combine into a term of <math>2^{19}</math>, so we let <math>a_3 = 19</math>. And so on, until we get to <math>a_{18} = 34</math>. Now everything we don't want telescopes into <math>2^{35}</math>. We already have that term since we let <math>a_2 = 18 \implies a_2+17 = 35</math>. Everything from now on will automatically telescope to <math>2^{52}</math>. So we let <math>a_{19}</math> be <math>52</math>. | ||
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+ | As you can see, we will have to add <math>17</math> <math>a_n</math>'s at a time, then "wait" for the sum to automatically telescope for the next <math>17</math> numbers, etc, until we get to <math>2^{289}</math>. We only need to add <math>a_n</math>'s between odd multiples of <math>17</math> and even multiples. The largest even multiple of <math>17</math> below <math>289</math> is <math>17\cdot16</math>, so we will have to add a total of <math>17\cdot 8</math> <math>a_n</math>'s. However, we must not forget we let <math>a_1=0</math> at the beginning, so our answer is <math>17\cdot8+1 = \boxed{\textbf{(C) } 137}</math>. | ||
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+ | ==Solution 4== | ||
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>. <math>(A)</math> just 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>. | 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>. <math>(A)</math> just 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>. | ||
Revision as of 18:47, 2 February 2020
- The following problem is from both the 2020 AMC 12A #19 and 2020 AMC 10A #21, so both problems redirect to this page.
Contents
Problem
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 .
Solution 3 (Intuitive)
Multiply both sides by to get
Notice that , since there is a on the LHS. However, now we have an extra term of on the right from . To cancel it, we let . The two 's now combine into a term of , so we let . And so on, until we get to . Now everything we don't want telescopes into . We already have that term since we let . Everything from now on will automatically telescope to . So we let be .
As you can see, we will have to add 's at a time, then "wait" for the sum to automatically telescope for the next numbers, etc, until we get to . We only need to add 's between odd multiples of and even multiples. The largest even multiple of below is , so we will have to add a total of 's. However, we must not forget we let at the beginning, so our answer is .
Solution 4
Note that the expression is equal to something slightly lower than . Clearly, answer choices and make no sense because the lowest sum for terms is . just makes no sense. and are 1 apart, but because the expression is odd, it will have to contain , and because is bigger, the answer is .
~Lcz
Solution 4
In order to shorten expressions, will represent consecutive s when expressing numbers.
Think of the problem in binary. We have
Note that
and
Since
this means that
so
Expressing each of the pairs of the form in binary, we have
or
This means that each pair has terms of the form .
Since there are of these pairs, there are a total of terms. Accounting for the term, which was not in the pair, we have a total of terms. ~emerald_block
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 |
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.