Difference between revisions of "2006 AIME II Problems/Problem 11"

(Solution 2)
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==Solution 2==
 
==Solution 2==
  
Brute Force. - not that difficult in this case because you only need to keep track of the last 3 digits.
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Brute Force. Since the problem asks for the answer of the end value when divided by 1000, it wouldn't be that difficult because you only need to keep track of the last 3 digits.
  
 
== See also ==
 
== See also ==

Revision as of 13:41, 23 January 2014

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

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

$\begin{align*}s &= a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \

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

\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

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

Brute Force. Since the problem asks for the answer of the end value when divided by 1000, it wouldn't be that difficult because you only need to keep track of the last 3 digits.

See also

2006 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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