Difference between revisions of "2020 AMC 10A Problems/Problem 21"
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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]] | ||
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==Video Solution== | ==Video Solution== | ||
Revision as of 12:57, 22 December 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 that![\[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\]](http://latex.artofproblemsolving.com/2/d/6/2d6393b8673d4c77a04408bd3ea19ae955bd756f.png)
What 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 ![\[2^{289}+1=2^{a_1} + 2^{a_2} + … + 2^{a_k} + 2^{a_1+17} + 2^{a_2+17} + … + 2^{a_k+17}.\]](http://latex.artofproblemsolving.com/0/7/8/078c3699779f590a5d2ca541465db555f99c5865.png)
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 5
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 1
Best video solution: https://www.youtube.com/watch?v=Af5EwWLd4h0
Video Solution
https://www.youtube.com/watch?v=FsCOVzhjUtE&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=3 ~ MathEx
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. 
