Difference between revisions of "2006 AIME II Problems/Problem 11"
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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>. | ||
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| + | ==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
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 | ||
