Difference between revisions of "2006 AIME II Problems/Problem 11"
(cleanup notation) |
m (→Solution) |
||
Line 11: | Line 11: | ||
Thus <math>s = \frac{a_{28} + a_{30}}{2}</math>, and <math>a_{28},\,a_{30}</math> are both given; the last four digits of their sum is <math>3668</math>, and half of that is <math>1834</math>. Therefore, the answer is <math>\boxed{834}</math>. | Thus <math>s = \frac{a_{28} + a_{30}}{2}</math>, and <math>a_{28},\,a_{30}</math> are both given; the last four digits of their sum is <math>3668</math>, and half of that is <math>1834</math>. Therefore, the answer is <math>\boxed{834}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Brute Force. - not that difficult in this case because you only need to keep track of the last 3 digits. | ||
== See also == | == See also == |
Revision as of 21:06, 13 March 2012
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. - not that difficult in this case 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 |