2006 AIME II Problem 11

Below is the professionally curated solution for Problem 11 of the 2006 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AIME II solutions, or check the answer key.

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Concepts:recursionsummationinduction

Difficulty rating: 2840

11.

A sequence is defined as follows: a1=a2=a3=1,a_1 = a_2 = a_3 = 1, and, for all positive integers n,n, an+3=an+2+an+1+an.a_{n+3} = a_{n+2} + a_{n+1} + a_n. Given that a28=6090307,a_{28} = 6090307, a29=11201821,a_{29} = 11201821, and a30=20603361,a_{30} = 20603361, find the remainder when k=128ak\sum_{k=1}^{28} a_k is divided by 1000.1000.

Solution:

Let Sn=a1++an.S_n = a_1 + \cdots + a_n. We claim 2Sn=an+2+an,2S_n = a_{n+2} + a_n, which holds for n=1n = 1 since 2=1+1.2 = 1 + 1. If it holds for n,n, then 2Sn+1=2Sn+2an+1=an+2+2an+1+an=an+3+an+12S_{n+1} = 2S_n + 2a_{n+1} = a_{n+2} + 2a_{n+1} + a_n = a_{n+3} + a_{n+1} by the recurrence, completing the induction.

Therefore S28=a30+a282=20603361+60903072=13346834,S_{28} = \frac{a_{30} + a_{28}}{2} = \frac{20603361 + 6090307}{2} = 13346834, whose remainder upon division by 10001000 is 834.834.

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