AIME 2004 II · 第 9 题

Solution 1

Let $x = a_2$; then solving for the next several terms, we find that $a_3 = x^2,\ a_4 = x(2x-1),\ a_5$ $= (2x-1)^2,\ a_6$ $= (2x-1)(3x-2)$, and in general, $a_{2n} = f(n-1)f(n)$, $a_{2n+1} = f(n)^2$, where $f(n) = nx - (n-1)$.[1]

From

$$a_9 + a_{10} = f(4)^2 + f(4)f(5) = (4x-3)(9x-7) = 646 = 2\cdot 17 \cdot 19$$ , we find that by either the quadratic formula or trial-and-error/modular arithmetic that $x=5$. Thus $f(n) = 4n+1$, and we need to find the largest $n$ such that either $f(n)^2\, \mathrm{or}\, f(n)f(n-1) < 1000$. This happens with $f(7)f(8) = 29 \cdot 33 = 957$, and this is the $2(8) = 16$th term of the sequence.

The answer is $957 + 16 = \boxed{973}$.

^ We can show this by simultaneous induction: since

$$\begin{aligned}a_{2n} &= 2a_{2n-1} - a_{2n-2} = 2a_{2(n-1)+1} - a_{2(n-1)} \\ &= 2f(n-1)^2 - f(n-2)f(n-1) \\ &= f(n-1)[2f(n-1) - f(n-2)] \\ &= f(n-1)[(2n-2-n+2)x-(2n-4-n+3)] \\ &= f(n-1)f(n) \end{aligned}$$ and

$$\begin{aligned}a_{2n+1} &= \frac{a_{2n}^2}{a_{2n-1}} = \frac{f(n-1)^2f(n)^2}{f(n-1)^2} = f(n)^2 \\ \end{aligned}$$

Solution 2

Let $x = a_2$. It is apparent that the sequence grows relatively fast, so we start trying positive integers to see what $x$ can be. Finding that $x = 5$ works, after bashing out the rest of the terms we find that $a_{16} = 957$ and $a_{17} = 1089$, hence our answer is $957 + 16 = \boxed{973}$.

Solution 3

We can find the value of $a_{9}$ by its bounds using three conditions:

1. $0
2. $a_{10} < a_{11}$ (note that the sequence must be increasing on all terms, not monotonically increasing) $a_{10} < \frac{a_{10}^2}{a_{9}} \rightarrow a_{9} < a_{10}$
3. $a_{11} = \frac{a_{10}^2}{a_{9}} = \frac{(646-a_{9})^2}{a_{9}}$, so necessarily $a_{9}$ is a factor of $646^2$, which factorizes to $2^2\cdot 17^2 \cdot 19^2$

Rearranging conditions 1 and 2, we get:

$$\frac{646}{3} < a_{9} < \frac{646}{2}$$ trying all the terms from the third condition, it is clear that $a_9 = 289$ is the only solution. Then we can calculate the next few terms from there since we have $a_{10}$ as well, to find that $a_{16} = 957$ and $a_{17} = 1089$, thus we have our answer of $957 + 16 = \boxed{973}$.

~KafkaTamura

Solution 4 (quadratic)

Let $a_2 = x$. We will write the first 10 terms in terms of $x$ (this will require some rigorous polynomial division):

First 10 Terms

Term

Value

$a_1$ $1$ $a_2$ $x$ $a_3$ $x^2$ $a_4$ $2x^2 - x$ $a_5$ $4x^2 - 4x + 1$ $a_6$ $6x^2 - 7x + 2$ $a_7$ $9x^2 - 12x + 4$ $a_8$ $12x^2 - 17x + 6$ $a_9$ $16x^2 - 24x + 9$ $a_{10}$ $20x^2 - 31x + 12$

The problem tells us that $a_9 + a_{10} = 646,$ so we can write a quadratic and solve for $x$:

$$(16x^2 - 24x + 9) + (20x^2 - 31x + 12) = 646$$ $$36x^2 - 55x + 21 = 646$$ $$36x^2 - 55x - 625 = 0.$$ $$x = \frac {-(-55) \pm \sqrt{(-55)^2 - 4(36)(-625)}}{2(36)}$$ $$x = \frac {55 \pm \sqrt{3025 + 90000}}{72}$$ $$x = \frac {55 \pm \sqrt{93025}}{72}$$ $x = \frac {55 \pm 305}{72} = \xcancel{-\frac{125}{36}}, 5.$

Since every number in the series has to be a positive integer, $x$ must be $5$. Using $x=5$ we can find $a_9$ and $a_{10}$:

$$a_9 = 16(5)^2 - 24(5) + 9 = 16(25) - 120 + 9 = 400 - 120 + 9 = 289.$$ $$a_{10} = 20(5)^2 - 31(5) + 12 = 20(25) - 155 + 12 = 500 - 155 + 12 = 357.$$ Notice that our arithmetic differences are increasing by 16: $20 \rightarrow 36 \rightarrow 52 \rightarrow 68...$ So following this pattern, the next differences will be $84, 100, 116$.

The geometric ratios follow a pattern of adding $4$ to the numerator and denominator: $\frac{9}{5} \rightarrow \frac{13}{9} \rightarrow \frac{17}{13}$. The next ratios will be $\frac{21}{17}, \frac{25}{21}, \frac{29}{25}.$

We have everything we need to calculate the next few terms:

$a_{11} = 357 \cdot \frac{21}{17} = 441,$ $a_{12} = 441 + 84 = 525,$ $a_{13} = 525 \cdot \frac{25}{21} = 625,$ $a_{14} = 625 + 100 = 725,$ $a_{15} = 725 \cdot \frac{29}{25} = 841,$ $a_{16} = 841 + 116 = 957.$ We stop here because it's clear the next term will exceed 1000. Since $a_{16} = 957,$ the answer is $957 + 16 = \boxed{973}.$

~grogg007