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

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 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

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
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