Difference between revisions of "2020 AMC 10A Problems/Problem 21"
(→See Also) |
(→Solution 2) |
||
Line 18: | Line 18: | ||
(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 | + | 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^{1})(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== | ||
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 19:24, 1 February 2020
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
(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
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.