Difference between revisions of "2020 AMC 10A Problems/Problem 21"
m (→Solution 4) |
m (added solution headers) |
||
Line 1: | Line 1: | ||
{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #19]] and [[2020 AMC 10A Problems|2020 AMC 10A #21]]}} | {{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #19]] and [[2020 AMC 10A Problems|2020 AMC 10A #21]]}} | ||
− | ==Problem== | + | == Problem == |
− | |||
There exists a unique strictly increasing sequence of nonnegative integers <math>a_1 < a_2 < … < a_k</math> such that<cmath>\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.</cmath>What is <math>k?</math> | There exists a unique strictly increasing sequence of nonnegative integers <math>a_1 < a_2 < … < a_k</math> such that<cmath>\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.</cmath>What is <math>k?</math> | ||
<math>\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306</math> | <math>\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306</math> | ||
− | == Solution 1 == | + | == Solutions == |
+ | === Solution 1 === | ||
First, substitute <math>2^{17}</math> with <math>a</math>. | First, substitute <math>2^{17}</math> with <math>a</math>. | ||
Then, the given equation becomes <math>\frac{a^{17}+1}{a+1}=a^{16}-a^{15}+a^{14}...-a^1+a^0</math>. | Then, the given equation becomes <math>\frac{a^{17}+1}{a+1}=a^{16}-a^{15}+a^{14}...-a^1+a^0</math>. | ||
Line 19: | Line 19: | ||
~seanyoon777 | ~seanyoon777 | ||
− | == Solution 2 == | + | === Solution 2 === |
(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\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 (Intuitive)== | + | === 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> | 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> | ||
Line 31: | Line 31: | ||
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>. | 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>. | ||
− | == Solution 4 == | + | === Solution 4 === |
In order to shorten expressions, <math>\#</math> will represent <math>16</math> consecutive <math>0</math>s when expressing numbers. <br> | In order to shorten expressions, <math>\#</math> will represent <math>16</math> consecutive <math>0</math>s when expressing numbers. <br> | ||
<br> | <br> | ||
Line 63: | Line 63: | ||
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. ~[[User:emerald_block|emerald_block]] | 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. ~[[User:emerald_block|emerald_block]] | ||
− | ==Video Solution 1== | + | == Video Solutions == |
− | + | === Video Solution 1 === | |
https://www.youtube.com/watch?v=Af5EwWLd4h0 | https://www.youtube.com/watch?v=Af5EwWLd4h0 | ||
− | ==Video Solution 2== | + | === Video Solution 2 === |
https://www.youtube.com/watch?v=FsCOVzhjUtE&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=3 ~ MathEx | https://www.youtube.com/watch?v=FsCOVzhjUtE&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=3 ~ MathEx | ||
− | ==Video Solution 3== | + | === Video Solution 3 === |
Education The Study of Everything | Education The Study of Everything | ||
https://youtu.be/f7FibYTNSm8 | https://youtu.be/f7FibYTNSm8 | ||
− | ==Video Solution 4== | + | |
+ | === Video Solution 4 === | ||
https://youtu.be/Ozp3k2464u4 | https://youtu.be/Ozp3k2464u4 | ||
~IceMatrix | ~IceMatrix | ||
− | ==Video Solution 5== | + | |
+ | === Video Solution 5 === | ||
https://youtu.be/oDSLaQM6L1o | https://youtu.be/oDSLaQM6L1o | ||
Revision as of 14:27, 19 January 2021
- 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 that
What is
Solutions
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
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 Solutions
Video Solution 1
https://www.youtube.com/watch?v=Af5EwWLd4h0
Video Solution 2
https://www.youtube.com/watch?v=FsCOVzhjUtE&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=3 ~ MathEx
Video Solution 3
Education The Study of Everything
Video Solution 4
~IceMatrix
Video Solution 5
~savannahsolver
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.