2006 AIME I Problems/Problem 11

Problem

A sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000.

Solution

We proceed recursively. Suppose we can build $T_m$ towers using blocks of size $1, 2, \ldots, m$ . How many towers can we build using blocks of size $1, 2, \ldots, m, m + 1$ ? If we remove the block of size $m + 1$ from such a tower (keeping all other blocks in order), we get a valid tower using blocks $1, 2, \ldots, m$ . Given a tower using blocks $1, 2, \ldots, m$ (with $m \geq 2$ ), we can insert the block of size $m + 1$ in exactly 3 places: at the beginning, immediately following the block of size $m - 1$ or immediately following the block of size $m$ . Thus, there are 3 times as many towers using blocks of size $1, 2, \ldots, m, m + 1$ as there are towers using only $1, 2, \ldots, m$ . There are 2 towers which use blocks $1, 2$ , so there are $2\cdot 3^6 = 1458$ towers using blocks $1, 2, \ldots, 8$ , so the answer is $458$ .

(Note that we cannot say, "there is one tower using the block $1$ , so there are $3^7$ towers using the blocks $1, 2, \ldots, 8$ ." The reason this fails is that our recursion only worked when $m \geq 2$ : when $m = 1$ , there are only 2 places to insert a block of size $m + 1 = 2$ , at the beginning or at the end, rather than the 3 places we have at later stages. Also, note that this method generalizes directly to seeking the number of towers where we change the second rule to read, "The cube immediately on top of a cube with edge-length $k$ must have edge-length at most $k + n$ ," where $n$ can be any fixed integer.)

2006 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2006 AIME I Problems/Problem 11

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