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

(Video Solution)
m (Solution 1)
Line 15: Line 15:
 
Applying the same method to each of <math>a^{14}-a^{13}</math>, <math>a^{12}-a^{11}</math>, ... , <math>a^{2}-a^{1}</math>, we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us <math>17*8=136</math>.
 
Applying the same method to each of <math>a^{14}-a^{13}</math>, <math>a^{12}-a^{11}</math>, ... , <math>a^{2}-a^{1}</math>, we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us <math>17*8=136</math>.
 
But we must count also the <math>a^0</math> term.  
 
But we must count also the <math>a^0</math> term.  
Thus, Our answer is <math>136+1=\boxed{\textbf{(C) } 137}</math>.
+
Thus, Our answer is <math>136+1=\boxed{\textbf{(C) } 136}</math>.
  
 
~seanyoon777
 
~seanyoon777

Revision as of 15:12, 2 July 2020

The following problem is from both the 2020 AMC 12A #19 and 2020 AMC 10A #21, so both problems redirect to this page.

Problem

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 1

First, substitute $2^{17}$ with $a$. Then, the given equation becomes $\frac{a^{17}+1}{a+1}=a^{16}-a^{15}+a^{14}...-a^1+a^0$. 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 $\frac{a^n-1}{a-1}$. Thus, we can see that $a^{16}-a^{15}$ forms the sum of 17 different powers of 2. Applying the same method to each of $a^{14}-a^{13}$, $a^{12}-a^{11}$, ... , $a^{2}-a^{1}$, we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us $17*8=136$. But we must count also the $a^0$ term. Thus, Our answer is $136+1=\boxed{\textbf{(C) } 136}$.

~seanyoon777

Solution 2

(This is similar to solution 1) Let $x = 2^{17}$. Then, $2^{289} = x^{17}$. The LHS can be rewritten as $\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+\cdots+x^2-x+1=(x-1)(x^{15}+x^{13}+\cdots+x^{1})+1$. Plugging $2^{17}$ back in for $x$, we have $(2^{17}-1)(2^{15\cdot{17}}+2^{13\cdot{17}}+\cdots+2^{1\cdot{17}})+1=(2^{16}+2^{15}+\cdots+2^{0})(2^{15\cdot17}+2^{13\cdot17}+\cdots+2^{1\cdot17})+1$. When expanded, this will have $17\cdot8+1=137$ terms. Therefore, our answer is $\boxed{\textbf{(C) } 137}$.

Solution 3 (Intuitive)

Multiply both sides by $2^{17}+1$ to get \[2^{289}+1=2^{a_1} + 2^{a_2} + … + 2^{a_k} + 2^{a_1+17} + 2^{a_2+17} + … + 2^{a_k+17}.\]

Notice that $a_1 = 0$, since there is a $1$ on the LHS. However, now we have an extra term of $2^{18}$ on the right from $2^{a_1+17}$. To cancel it, we let $a_2 = 18$. The two $2^{18}$'s now combine into a term of $2^{19}$, so we let $a_3 = 19$. And so on, until we get to $a_{18} = 34$. Now everything we don't want telescopes into $2^{35}$. We already have that term since we let $a_2 = 18 \implies a_2+17 = 35$. Everything from now on will automatically telescope to $2^{52}$. So we let $a_{19}$ be $52$.

As you can see, we will have to add $17$ $a_n$'s at a time, then "wait" for the sum to automatically telescope for the next $17$ numbers, etc, until we get to $2^{289}$. We only need to add $a_n$'s between odd multiples of $17$ and even multiples. The largest even multiple of $17$ below $289$ is $17\cdot16$, so we will have to add a total of $17\cdot 8$ $a_n$'s. However, we must not forget we let $a_1=0$ at the beginning, so our answer is $17\cdot8+1 = \boxed{\textbf{(C) } 137}$.

Solution 4

Note that the expression is equal to something slightly lower than $2^{272}$. Clearly, answer choices $(D)$ and $(E)$ make no sense because the lowest sum for $273$ terms is $2^{273}-1$. $(A)$ just makes no sense. $(B)$ and $(C)$ are 1 apart, but because the expression is odd, it will have to contain $2^0=1$, and because $(C)$ is $1$ bigger, the answer is $\boxed{\textbf{(C) } 137}$.

~Lcz

Solution 5

In order to shorten expressions, $\#$ will represent $16$ consecutive $0$s when expressing numbers.

Think of the problem in binary. We have
$\frac{1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2}{1\#1_2}$
Note that
$(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) = 2^0(2^{17} + 1) + 2^{34}(2^{17} + 1) + 2^{68}(2^{17} + 1) + \cdots 2^{272}(2^{17} + 1)$
$= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2$
and
$(2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{17}(2^{17} + 1) + 2^{51}(2^{17} + 1) + 2^{85}(2^{17} + 1) + \cdots 2^{255}(2^{17} + 1)$
$= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2$

Since
$\phantom{=\ } 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2$
$-\ \phantom{1\#} 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2$
$= 1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2$
this means that
$(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{289}$
so
$\frac{2^{289}+1}{2^{17}+1} = (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255})$
$= 2^0 + (2^{34} - 2^{17}) + (2^{68} - 2^{51}) + \cdots + (2^{272} - 2^{255})$

Expressing each of the pairs of the form $2^{n + 17} - 2^n$ in binary, we have
$\phantom{=\ } 1000000000000000000 \cdots 0_2$
$-\ \phantom{10000000000000000} 10 \cdots 0_2$
$= \phantom{1} 111111111111111110 \cdots 0_2$
or
$2^{n + 17} - 2^n = 2^{n + 16} + 2^{n + 15} + 2^{n + 14} + \cdots + 2^{n}$
This means that each pair has $17$ terms of the form $2^n$.

Since there are $8$ of these pairs, there are a total of $8 \cdot 17 = 136$ terms. Accounting for the $2^0$ term, which was not in the pair, we have a total of $136 + 1 = \boxed{\textbf{(C) } 137}$ terms. ~emerald_block

Video Solution

https://youtu.be/Ozp3k2464u4

~IceMatrix

Video Solution

https://www.youtube.com/watch?v=FsCOVzhjUtE&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=3 ~ MathEx

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS