2006 AIME A Problems/Problem 11

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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 $\displaystyle \sum^{28}_{k=1} a_k$ is divided by 1000.

Solution

Define the sum as $x$. Notice that $a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$, so the sum will be:

$x = (a_4 - a_3 - a_2) + (a_5 - a_4 - a_3) + \ldots (a_{30} - a_{29} - a_{28}) + a_{28}$
$x = (a_4+ a_5 \ldots a_{30}) - (a_3 + a_4 + \ldots a_{29}) - (a_2 + a_3 + \ldots a_{28}) + a_{28} + (a_1 - a_1)$

The first two groupings almost completely cancel. The third resembles $x$.

$x\ = a_1 - a_3 + a_{28} + a_{30} - x$
$2x\ = a_{28} + a_{30}$
$x\ = \frac{a_{28} + a_{30}}{2}$

$a_{28}$ and $a_{30}$ are both given; the last four digits of the sum is $3668$, and half of that is $1834$. Therefore, the answer is $834$.

See also

2006 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AIME Problems and Solutions
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