2005 AIME II Problem 11

Below is the professionally curated solution for Problem 11 of the 2005 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AIME II solutions, or check the answer key.

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Concepts:recursionarithmetic sequencealgebraic manipulation

Difficulty rating: 2520

11.

Let mm be a positive integer, and let a0,a1,,ama_0, a_1, \ldots, a_m be a sequence of real numbers such that a0=37,a_0 = 37, a1=72,a_1 = 72, am=0,a_m = 0, and ak+1=ak13aka_{k+1} = a_{k-1} - \frac{3}{a_k} for k=1,2,,m1.k = 1, 2, \ldots, m - 1. Find m.m.

Solution:

Multiplying the recurrence by aka_k gives ak+1ak=akak13,a_{k+1} a_k = a_k a_{k-1} - 3, so the products bk=akak1b_k = a_k a_{k-1} form an arithmetic sequence with common difference 3.-3. Since b1=7237=2664=3888,b_1 = 72 \cdot 37 = 2664 = 3 \cdot 888, we get bk=26643(k1)=3(889k).b_k = 2664 - 3(k - 1) = 3(889 - k).

Thus bk>0b_k \gt 0 for k888,k \le 888, so no term before a889a_{889} can vanish (and the recurrence never divides by zero), while b889=a889a888=0b_{889} = a_{889} a_{888} = 0 with a8880.a_{888} \ne 0. Hence a889=0,a_{889} = 0, and m=889.m = 889.

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