Difference between revisions of "2006 AIME II Problems/Problem 11"
m (minor fixes) |
(cleanup notation) |
||
Line 4: | Line 4: | ||
== Solution == | == Solution == | ||
Define the sum as <math>s</math>. Since <math>a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} </math>, the sum will be: | Define the sum as <math>s</math>. Since <math>a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} </math>, the sum will be: | ||
− | + | <center><math>\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*}</math></center> | ||
− | + | 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>. | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | <math>a_{28} | ||
== See also == | == See also == |
Revision as of 21:09, 25 April 2008
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 .
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 |