Difference between revisions of "2001 AIME II Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | {{ | + | <math>x_5=x_4-x_3+x_2-x_1</math> |
| + | |||
| + | <math>x_6=x_4-x_3+x_2-x_1-x_4+x_3-x_2=-x_1</math> | ||
| + | |||
| + | <math>x_7=-x_1-x_4+x_3-x_2+x_1+x_4-x_3=-x_2</math> | ||
| + | |||
| + | <math>x_8=-x_2+x_1+x_4-x_3+x_2-x_1-x_4=-x_3</math> | ||
| + | |||
| + | <math>x_9=-x_3+x_2-x_1-x_4+x_3-x_2+x_1=-x_4</math> | ||
| + | |||
| + | And it cycles back to <math>x_{11}=x_1</math> | ||
| + | |||
| + | Therefore, <math>x_{y}=x_{y-10}</math>, so | ||
| + | |||
| + | <cmath>x_{531}+x_{753}+x_{975}=x_1+x_3+x_5=x_1+x_3+x_4-x_3+x_2-x_1=x_4+x_2=523+420=\boxed{943}</cmath> | ||
== See also == | == See also == | ||
{{AIME box|year=2001|n=II|num-b=2|num-a=4}} | {{AIME box|year=2001|n=II|num-b=2|num-a=4}} | ||
Revision as of 11:25, 28 February 2008
Problem
Given that
find the value of 
.
Solution
And it cycles back to 
Therefore, 
, so
![\[x_{531}+x_{753}+x_{975}=x_1+x_3+x_5=x_1+x_3+x_4-x_3+x_2-x_1=x_4+x_2=523+420=\boxed{943}\]](http://latex.artofproblemsolving.com/3/9/3/393043e948b113fcc3a056fca7f33cdf21e7103d.png)
See also
| 2001 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
