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]] | ||
+ | |||
+ | == Solution 4 (Answer choices)== | ||
+ | An odd number divided by an odd number is always odd; therefore, the smallest number in the final sequence must be <math>1</math>. Also observe that <math>\textbf{(B) }</math> and <math>\textbf{(C) }</math> are the only consecutive answer choices. It's likely that students might forget to include <math>1</math> in their final sequence, so <math>\textbf{(B) }</math> must be planted as a pitfall for those students; in that case, <math>\boxed{\textbf{(C) } 137}</math> must be the answer. | ||
+ | |||
+ | ~ ihatemath123 | ||
== Video Solutions == | == Video Solutions == |
Revision as of 18:41, 2 January 2022
- 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
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 (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 3
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
Solution 4 (Answer choices)
An odd number divided by an odd number is always odd; therefore, the smallest number in the final sequence must be . Also observe that
and
are the only consecutive answer choices. It's likely that students might forget to include
in their final sequence, so
must be planted as a pitfall for those students; in that case,
must be the answer.
~ ihatemath123
Video Solutions
Video Solution 1 (Simple)
~Education The Study of Everything
Video Solution 2 (Richard Rusczyk)
https://artofproblemsolving.com/videos/amc/2020amc10a/511
Video Solution 3
https://www.youtube.com/watch?v=FsCOVzhjUtE&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=3 ~ MathEx
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.