AIME 2006 II · 第 11 题

Solutions

Solution 1

Define the sum as $s$. Since $a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$, the sum will be:

$s = a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\ s = a_{28} + \left(\sum^{30}_{k=4} a_{k} - \sum^{29}_{k=3} a_{k}\right) - \left(\sum^{28}_{k=2} a_{k}\right)\\ s = a_{28} + (a_{30} - a_{3}) - \left(\sum^{28}_{k=2} a_{k}\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\ s = -s + a_{28} + a_{30}$

Thus $s = \frac{a_{28} + a_{30}}{2}$, and $a_{28},\,a_{30}$ are both given; the last four digits of their sum is $3668$, and half of that is $1834$. Therefore, the answer is $\boxed{834}$.−

Solution 2 (bash)

Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:

$a_{1}\equiv 1 \pmod {1000} \\ a_{2}\equiv 1 \pmod {1000} \\ a_{3}\equiv 1 \pmod {1000} \\ a_{4}\equiv 3 \pmod {1000} \\ a_{5}\equiv 5 \pmod {1000} \\ \cdots \\ a_{25} \equiv 793 \pmod {1000} \\ a_{26} \equiv 281 \pmod {1000} \\ a_{27} \equiv 233 \pmod {1000} \\ a_{28} \equiv 307 \pmod {1000}$

Adding all the residues shows the sum is congruent to $\boxed{834}$ mod $1000$.

~ I-_-I

Solution 3 (some guessing involved)/"Engineer's Induction"

All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given $a_{28}, a_{29},$ and $a_{30}$, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some $p, q, r$ such that $\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}$. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that $(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})$, at least for the first few terms. From this, we have that $\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}}\pmod {1000}$.

Solution by zeroman; clarified by srisainandan6

Solution 4 (if you did not know how to use the numbers given in the problem)

By the Chinese remainder theorem, each number under 1000 is uniquely determined by its mod 8 and mod 125.

We list a few terms of the sequence mod 8:

$$1,1,1,3,5,1,1,7,1,1,1,...$$ Therefore, the cycle repeats every 8 numbers, and each cycle has a sum of 4 mod 8. Therefore, the sum mod 8 is

$$3 \cdot 4 + 1 + 1 + 1 + 3 = 2 \mod 8.$$ Denote

$$s_n = \sum _{i=1} ^{n} a_i.$$ It is easy to prove that $s_{n+3} = s_{n+2} + s_{n+1} + s_{n}.$

We write the sum ($s_n$) of the first terms mod 125:

$$1,2,3,6,11,20,37,68,0,-20,48,28,56,7,91,29,2,-3,28,27,52,-18,61,-30,13,44,27,84.$$ Therefore the desired number is $84 \mod 125.$

From here, we can determine the number we are looking for is $750 + 84 = \boxed{834}.$ -sd8