Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 10A Problems/Problem 21"

(→‎Solution)
(→‎Solution)
Line 4: Line 4:
  
 
== 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 22:10, 31 January 2020

There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that\[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\]What is $k?$

$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$

Solution

First, replace $2^(17)$ as $a$. Then, the given equation becomes $(a^(17)+1)/(a+1)=a^(16)-a^(15)+a^(14)...-a^1+a^0$. Now consider only $a^(16)-a^15$. This equals $a^(15)(a-1)=a^15*(2^(17)-1)$. Note that $2^(17)-1$ equals $2^(16)+2^15+...+1$, since the sum of a geometric sequence is $(a^n-1)/(a-1)$. Thus, we can see that $a^(16)-a^(15)$ forms the sum of 17 different powers of 2. Applying the same thing to each of $a^(14)-a^13$, $a^12-a^11$, and so on. This gives us $17*8=136$. Our answer is $\boxed{\textbf{(B) } 136}$.

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_21&oldid=116055"