Difference between revisions of "2016 AIME II Problems/Problem 9"

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==Problem==
 
The sequences of positive integers <math>1,a_2, a_3,...</math> and <math>1,b_2, b_3,...</math> are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let <math>c_n=a_n+b_n</math>. There is an integer <math>k</math> such that <math>c_{k-1}=100</math> and <math>c_{k+1}=1000</math>. Find <math>c_k</math>.
 
The sequences of positive integers <math>1,a_2, a_3,...</math> and <math>1,b_2, b_3,...</math> are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let <math>c_n=a_n+b_n</math>. There is an integer <math>k</math> such that <math>c_{k-1}=100</math> and <math>c_{k+1}=1000</math>. Find <math>c_k</math>.
  
==Solution==
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==Solution 1==
 
Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for <math>b_2</math>. When we get to <math>b_2=9</math> and <math>a_2=91</math>, we have <math>a_4=271</math> and <math>b_4=729</math>, which works, therefore, the answer is <math>b_3+a_3=81+181=\boxed{262}</math>.
 
Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for <math>b_2</math>. When we get to <math>b_2=9</math> and <math>a_2=91</math>, we have <math>a_4=271</math> and <math>b_4=729</math>, which works, therefore, the answer is <math>b_3+a_3=81+181=\boxed{262}</math>.
 
Solution by Shaddoll
 
  
 
== Solution 2==
 
== Solution 2==
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Solution by rocketscience
 
Solution by rocketscience
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== Solution 3 (More Robust Bash) ==
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The reason for bashing in this context can also be justified by the fact 100 isn't very big.
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Let the common difference for the arithmetic sequence be <math>a</math>, and the common ratio for the geometric sequence be <math>b</math>. The sequences are now <math>1, a+1, 2a+1, \ldots</math>, and <math>1, b, b^2, \ldots</math>. We can now write the given two equations as the following:
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<math>1+(k-2)a+b^{k-2} = 100</math>
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<math>1+ka+b^k = 1000</math>
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Take the difference between the two equations to get <math>2a+(b^2-1)b^{k-2} = 900</math>. Since 900 is divisible by 4, we can tell <math>a</math> is even and <math>b</math> is odd. Let <math>a=2m</math>, <math>b=2n+1</math>, where <math>m</math> and <math>n</math> are positive integers. Substitute variables and divide by 4 to get:
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<math>m+(n+1)(n)(2n+1)^{k-2} = 225</math>
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Because very small integers for <math>n</math> yield very big results, we can bash through all cases of <math>n</math>. Here, we set an upper bound for <math>n</math> by setting <math>k</math> as 3. After trying values, we find that <math>n\leq 4</math>, so <math>b=9, 7, 5, 3</math>. Testing out <math>b=9</math> yields the correct answer of <math>\boxed{262}</math>. Note that even if this answer were associated with another b value like <math>b=3</math>, the value of <math>k</math> can still only be 3 for all of the cases.
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-Dankster42
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== See also ==
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{{AIME box|year=2016|n=II|num-b=8|num-a=10}}
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{{MAA Notice}}

Revision as of 20:18, 18 August 2020

Problem

The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$. There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$. Find $c_k$.

Solution 1

Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for $b_2$. When we get to $b_2=9$ and $a_2=91$, we have $a_4=271$ and $b_4=729$, which works, therefore, the answer is $b_3+a_3=81+181=\boxed{262}$.

Solution 2

Using the same reasoning ($100$ isn't very big), we can guess which terms will work. The first case is $k=3$, so we assume the second and fourth terms of $c$ are $100$ and $1000$. We let $r$ be the common ratio of the geometric sequence and write the arithmetic relationships in terms of $r$.

The common difference is $100-r - 1$, and so we can equate: $2(99-r)+100-r=1000-r^3$. Moving all the terms to one side and the constants to the other yields

$r^3-3r = 702$, or $r(r^2-3) = 702$. Simply listing out the factors of $702$ shows that the only factor $3$ less than a square that works is $78$. Thus $r=9$ and we solve from there to get $\boxed{262}$.

Solution by rocketscience

Solution 3 (More Robust Bash)

The reason for bashing in this context can also be justified by the fact 100 isn't very big.

Let the common difference for the arithmetic sequence be $a$, and the common ratio for the geometric sequence be $b$. The sequences are now $1, a+1, 2a+1, \ldots$, and $1, b, b^2, \ldots$. We can now write the given two equations as the following:

$1+(k-2)a+b^{k-2} = 100$

$1+ka+b^k = 1000$

Take the difference between the two equations to get $2a+(b^2-1)b^{k-2} = 900$. Since 900 is divisible by 4, we can tell $a$ is even and $b$ is odd. Let $a=2m$, $b=2n+1$, where $m$ and $n$ are positive integers. Substitute variables and divide by 4 to get:

$m+(n+1)(n)(2n+1)^{k-2} = 225$

Because very small integers for $n$ yield very big results, we can bash through all cases of $n$. Here, we set an upper bound for $n$ by setting $k$ as 3. After trying values, we find that $n\leq 4$, so $b=9, 7, 5, 3$. Testing out $b=9$ yields the correct answer of $\boxed{262}$. Note that even if this answer were associated with another b value like $b=3$, the value of $k$ can still only be 3 for all of the cases.

-Dankster42

See also

2016 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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