Difference between revisions of "2020 AMC 10A Problems/Problem 21"
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== Solution == | == Solution == | ||
− | First, replace <math>2^17</math> as <math>a</math>. | + | First, replace <math>2^(17)</math> as <math>a</math>. |
− | Then, the given equation becomes <math>(a^17+1)/(a+1)=a^16-a^15+a^14...-a^1+a^0</math>. | + | Then, the given equation becomes <math>(a^(17)+1)/(a+1)=a^(16)-a^(15)+a^(14)...-a^1+a^0</math>. |
− | Now consider only <math>a^16-a^15</math>. This equals <math>a^15(a-1)=a^15*(2^17-1)</math>. | + | Now consider only <math>a^(16)-a^15</math>. This equals <math>a^(15)(a-1)=a^15*(2^(17)-1)</math>. |
− | Note that <math>2^17-1</math> equals <math>2^16+2^15+...+1</math>, since the sum of a geometric sequence is <math>(a^n-1)/(a-1)</math>. | + | Note that <math>2^(17)-1</math> equals <math>2^(16)+2^15+...+1</math>, since the sum of a geometric sequence is <math>(a^n-1)/(a-1)</math>. |
− | Thus, we can see that <math>a^16-a^15</math> forms the sum of 17 different powers of 2. | + | Thus, we can see that <math>a^(16)-a^(15)</math> forms the sum of 17 different powers of 2. |
− | Applying the same thing to each of <math>a^14-a^13</math>, <math>a^12-a^11</math>, and so on. | + | Applying the same thing to each of <math>a^(14)-a^13</math>, <math>a^12-a^11</math>, and so on. |
This gives us <math>17*8=136</math>. | This gives us <math>17*8=136</math>. | ||
Our answer is <math>\boxed{\textbf{(B) } 136}</math>. | Our answer is <math>\boxed{\textbf{(B) } 136}</math>. |
Revision as of 23:10, 31 January 2020
There exists a unique strictly increasing sequence of nonnegative integers such that
What is
Solution
First, replace as
.
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 thing to each of
,
, and so on.
This gives us
.
Our answer is
.
~seanyoon777
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.