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2016 AIME II Problems/Problem 9

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 (No trial and error)

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., \[700 < r^k < 1000.\] Moreover, since $k$ is atleast $3$ we get $r^3\le r^k <1000$, i.e. $r<10$. For every value of $r$ in this range, define $i(r) = \max \{x : r^x < 700\}$, and define $j(r) = \min \{x : r^x > 1000\}$. We are looking for values of $r$ such that $j(r) -i(r)>1$. 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] i(r) & 9  & 5 & 4  & 4 & 3 & 3 & 3 & 2\\[2ex] j(r) & 10  & 7 & 5  & 5 & 4 & 4 & 4 & 4 \end{array} The only admissible values for $r^k$ are $\{3^6, 9^3\}$. However, since $100=c_{k-1}=r^{k-2}+(k-2)d+1$, we must have $(k-2)\mid 99-r^{k-2}$. This does not hold for $r^k=3^6$ because $4$ does not divide $99-3^4=18$. This leaves $r^k=9^3$ as the only option.

For $r=9$ and $k=3$, we check: $a_{k-1}= a_2 = r =9$ implies $b_{k-1}= b_2 = 91$, i.e. $d=90$. Then $a_{k+1}=a_4 = r^3 = 729$ and $b_{k+1}=b_4 = 1+3d = 271$ and $c_{k+1}=c_4=a_4+b_4 = 729+271=1000$; so it works! Then $c_k = c_3 = 9^2+181 = 262$.

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 4 (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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