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 14:41, 23 January 2014
Contents
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
