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Difference between revisions of "2016 AIME II Problems/Problem 9"
(→‎Solution 2)
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== Solution 2==
 
== Solution 2==
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We have <math>a_k=r^{k-1}</math> and <math>b_k=(k-1)d</math>. First, <math>b_{k-1}<c_{k-1}=100</math> implies <math>d<100</math>, so <math>b_{k+1}<300</math>.
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It follows that <math>a_{k+1}=1000-b_{k+1}>700</math>, i.e., <math>700 < r^k < 1000</math>. Dividing through by <math>r^2</math> we get <cmath>\frac{700}{r^2} < r^{j} < \frac{1000}{r^2}</cmath>where <math>j=k-2</math>. Since <math>k</math> is atleast <math>3</math> we get <math>r^3\le r^k <1000</math>, i.e. <math>r<10</math>. Let's make a table:
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<cmath>\begin{array}[b]{ c c c c c c c c c }
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r & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\[2ex]
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\left\lfloor\frac{700}{r^2}\right\rfloor & 175  & 77 & 43  & 28 & 19 & 14 & 10 & 8\\[2ex]
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\left\lfloor\frac{1000}{r^2}\right\rfloor & 250  & 111 & 63  & 40 & 28 & 20 & 16 & 12
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\end{array} </cmath>
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The only admissible <math>r^j</math> values are <math>\{3^4, 9\}</math>.
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Since <math>100=c_{j+1}=r^j+jd+1</math>, we must have <math>j\mid 99-r^j</math>.
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Note that <math>99-3^4=18</math>, which is not a multiple of <math>4</math>, which leaves <math>r^j=9^1</math>. We check: <math>a_j=9</math> implies <math>b_j=91</math>, i.e. <math>d=90</math>, so <math>b_{j+2}=271</math> and <math>a_{j+2}=729</math> and <math>c_{j+2}=1000</math>! So it works! Then <math>c_{j+1}=9^2+181=262</math>.
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== Solution 3==
  
 
Using the same reasoning (<math>100</math> isn't very big), we can guess which terms will work. The first case is <math>k=3</math>, so we assume the second and fourth terms of <math>c</math> are <math>100</math> and <math>1000</math>. We let <math>r</math> be the common ratio of the geometric sequence and write the arithmetic relationships in terms of <math>r</math>.
 
Using the same reasoning (<math>100</math> isn't very big), we can guess which terms will work. The first case is <math>k=3</math>, so we assume the second and fourth terms of <math>c</math> are <math>100</math> and <math>1000</math>. We let <math>r</math> be the common ratio of the geometric sequence and write the arithmetic relationships in terms of <math>r</math>.

Revision as of 20:59, 23 November 2021

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

We have $a_k=r^{k-1}$ and $b_k=(k-1)d$. First, $b_{k-1}<c_{k-1}=100$ implies $d<100$, so $b_{k+1}<300$.

It follows that $a_{k+1}=1000-b_{k+1}>700$, i.e., $700 < r^k < 1000$. Dividing through by $r^2$ we get \[\frac{700}{r^2} < r^{j} < \frac{1000}{r^2}\]where $j=k-2$. Since $k$ is atleast $3$ we get $r^3\le r^k <1000$, i.e. $r<10$. Let's make a table: \begin{array}[b]{ c c c c c c c c c } r & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\[2ex] \left\lfloor\frac{700}{r^2}\right\rfloor & 175 & 77 & 43 & 28 & 19 & 14 & 10 & 8\\[2ex] \left\lfloor\frac{1000}{r^2}\right\rfloor & 250 & 111 & 63 & 40 & 28 & 20 & 16 & 12 \end{array} The only admissible $r^j$ values are $\{3^4, 9\}$.

Since $100=c_{j+1}=r^j+jd+1$, we must have $j\mid 99-r^j$.

Note that $99-3^4=18$, which is not a multiple of $4$, which leaves $r^j=9^1$. We check: $a_j=9$ implies $b_j=91$, i.e. $d=90$, so $b_{j+2}=271$ and $a_{j+2}=729$ and $c_{j+2}=1000$! So it works! Then $c_{j+1}=9^2+181=262$.


Solution 3

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 (Problems • Answer Key • Resources)
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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