Difference between revisions of "2006 AIME II Problems/Problem 11"
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==Solution 2== | ==Solution 2== | ||
− | Brute Force. | + | 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
Contents
[hide]Problem
A sequence is defined as follows and, for all positive integers Given that and find the remainder when is divided by 1000.
Solution
Define the sum as . Since , the sum will be:
&= 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 , and are both given; the last four digits of their sum is , and half of that is . Therefore, the answer is .
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 (Problems • Answer Key • Resources) | ||
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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