AIME 2016 II · 第 9 题

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 (Some trial and error)

We have $a_k=r^{k-1}$ and $b_k=(k-1)d$. First, $b_{k-1} implies$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:

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.

Solution 5 (Casework)

Let $a_n$ and $b_n$ be in the form of

Case $1.\hspace{10mm} k = 3\hspace{5mm} (c_1=2 \implies k > 2).$

$$c_2 = a+1 + b = 100, c_4 = 3a+1 + b^3 = 1000 \implies b^3 -3b -2 = 1000-300 \implies b^3 - 3b = 702 \implies b = 9, a=90, c_3 = \boxed {262}.$$ Case $2. \hspace{10mm} k = 4.$

$$c_3 = 2a+1 + b^2 = 100, c_5 = 4a+1 + b^4 = 1000 \implies b^4 -2b^2 -1 = 1000-200 \implies b^4 - 2b^2 = 801 \implies \O.$$ Case $3. \hspace{10mm} k \ge 5.\hspace{3mm} b^5 < 1000 \implies b = \{2,3\}.$

Case $3.1.\hspace{10mm} b = 2.$

$$c_{k-1} = 2^{k-2} + (k-2) a + 1 = 100, c_{k+1} = 2^k + ka + 1 = 1000\implies a = 450-3\cdot 2^{k-3} \implies 2^k +450k -3k\cdot 2^{k-3} + 1 = 1000 \implies \O.$$ Case $3.2.\hspace{10mm} b = 3.$

$$c_{k-1} = 3^{k-2} + (k-2) a + 1 = 100, c_{k+1} = 3^k + ka + 1 = 1000\implies a = 450-4\cdot 3^{k-2} \implies 3^{k-4}(9-4k) +50k = 3\cdot 37 \implies \O.$$ vladimir.shelomovskii@gmail.com, vvsss