Question

Let $m$ be a positive integer, and let $a_0, a_1, \ldots, a_m$ be a sequence of real numbers such that $a_0 = 37,$ $a_1 = 72,$ $a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac{3}{a_k}$ for $k = 1, 2, \ldots, m - 1.$ Find $m.$

Answer

Answer: 889

Concepts:recursion arithmetic sequence algebraic manipulation

Difficulty rating: 2520

Solution:

Multiplying the recurrence by $a_k$ gives $a_{k+1} a_k = a_k a_{k-1} - 3,$ so the products $b_k = a_k a_{k-1}$ form an arithmetic sequence with common difference $-3.$ Since $b_1 = 72 \cdot 37 = 2664 = 3 \cdot 888,$ we get $\begin{aligned} b_k &= 2664 - 3(k - 1) \\ &= 3(889 - k). \end{aligned}$

Thus $b_k \gt 0$ for $k \le 888,$ so no term before $a_{889}$ can vanish (and the recurrence never divides by zero), while $b_{889} = a_{889} a_{888} = 0$ with $a_{888} \ne 0.$ Hence $a_{889} = 0,$ and $m = 889.$

2005 AIME II Problem 11

Answer: 889

Difficulty rating: 2520

Solution:

Problem 11 in Other Years