Difference between revisions of "2001 AIME II Problems/Problem 3"

(eqnarray -> align; LaTeX style)
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== Problem ==
 
== Problem ==
 
Given that
 
Given that
<center><math>\begin{eqnarray*}x_{1}&=&211,\ x_{2}&=&375,\ x_{3}&=&420,\ x_{4}&=&523, \textrm{ and}\ x_{n}&=&x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\textrm{ when }n\geq5, \end{eqnarray*}</math></center>
+
<cmath>
 +
\begin{align*}x_{1}&=211,\
 +
x_{2}&=375,\
 +
x_{3}&=420,\
 +
x_{4}&=523,\ \text{and}\
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x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}
 +
</cmath>
 
find the value of <math>x_{531}+x_{753}+x_{975}</math>.
 
find the value of <math>x_{531}+x_{753}+x_{975}</math>.
  
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Therefore, <math>x_{y}=x_{y-10}</math>, so
 
Therefore, <math>x_{y}=x_{y-10}</math>, so
 
+
<cmath>
<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>
+
\begin{align*}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}\end{align*}
 +
</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 20:58, 27 April 2008

Problem

Given that \begin{align*}x_{1}&=211,\\ x_{2}&=375,\\ x_{3}&=420,\\ x_{4}&=523,\ \text{and}\\ x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*} find the value of $x_{531}+x_{753}+x_{975}$.

Solution

$x_5=x_4-x_3+x_2-x_1$

$x_6=x_4-x_3+x_2-x_1-x_4+x_3-x_2=-x_1$

$x_7=-x_1-x_4+x_3-x_2+x_1+x_4-x_3=-x_2$

$x_8=-x_2+x_1+x_4-x_3+x_2-x_1-x_4=-x_3$

$x_9=-x_3+x_2-x_1-x_4+x_3-x_2+x_1=-x_4$

And it cycles back to $x_{11}=x_1$

Therefore, $x_{y}=x_{y-10}$, so \begin{align*}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}\end{align*}

See also

2001 AIME II (ProblemsAnswer KeyResources)
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