Difference between revisions of "2020 AMC 10A Problems/Problem 21"
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~Lcz | ~Lcz | ||
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+ | == Solution 4 == | ||
+ | In order to shorten expressions, <math>\#</math> will represent <math>16</math> consecutive <math>0</math>s when expressing numbers. <br> | ||
+ | <br> | ||
+ | Think of the problem in binary. We have <br> | ||
+ | <math>\frac{1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2}{1\#1_2}</math> <br> | ||
+ | Note that <br> | ||
+ | <math>(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) = 2^0(2^{17} + 1) + 2^{34}(2^{17} + 1) + 2^{68}(2^{17} + 1) + \cdots 2^{272}(2^{17} + 1)</math> <br> | ||
+ | <math>= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2</math> <br> | ||
+ | and <br> | ||
+ | <math>(2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{17}(2^{17} + 1) + 2^{51}(2^{17} + 1) + 2^{85}(2^{17} + 1) + \cdots 2^{255}(2^{17} + 1)</math> <br> | ||
+ | <math>= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2</math> <br> | ||
+ | <br> | ||
+ | Since <br> | ||
+ | <math>\phantom{=\ } 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2</math> <br> | ||
+ | <math>-\ \phantom{1\#} 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2</math> <br> | ||
+ | <math>= 1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2</math> <br> | ||
+ | this means that<br> | ||
+ | <math>(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{289}</math> <br> | ||
+ | so <br> | ||
+ | <math>\frac{2^{289}+1}{2^{17}+1} = (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255})</math> <br> | ||
+ | <math>= 2^0 + (2^{34} - 2^{17}) + (2^{68} - 2^{51}) + \cdots + (2^{272} - 2^{255})</math> <br> | ||
+ | <br> | ||
+ | Expressing each of the pairs of the form <math>2^{n + 17} - 2^n</math> in binary, we have <br> | ||
+ | <math>\phantom{=\ } 1000000000000000000 \cdots 0_2</math> <br> | ||
+ | <math>-\ \phantom{10000000000000000} 10 \cdots 0_2</math> <br> | ||
+ | <math>= \phantom{1} 111111111111111110 \cdots 0_2</math> <br> | ||
+ | or <br> | ||
+ | <math>2^{n + 17} - 2^n = 2^{n + 16} + 2^{n + 15} + 2^{n + 14} + \cdots + 2^{n}</math> <br> | ||
+ | This means that each pair has <math>17</math> terms of the form <math>2^n</math>. <br> | ||
+ | <br> | ||
+ | Since there are <math>8</math> of these pairs, there are a total of <math>8 \cdot 17 = 136</math> terms. Accounting for the <math>2^0</math> term, which was not in the pair, we have a total of <math>136 + 1 = \boxed{\textbf{(C) } 137}</math> terms. | ||
==Video Solution== | ==Video Solution== |
Revision as of 00:30, 2 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 .
Solution 3
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.
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 |
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