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Difference between revisions of "2001 AIME II Problems/Problem 3"
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== Problem ==
 
== Problem ==
 
Given that
 
Given that
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<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}\\
 +
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>.
  
 
== Solution ==
 
== Solution ==
−
<math>x_5=x_4-x_3+x_2-x_1</math>
+
We find that <math>x_5 = 267</math> by the recursive formula. Summing the [[recursion]]s
  
−
<math>x_6=x_4-x_3+x_2-x_1-x_4+x_3-x_2=-x_1</math>
+
<cmath>\begin{align*}
 +
x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4} \\
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x_{n-1}&=x_{n-2}-x_{n-3}+x_{n-4}-x_{n-5}
 +
\end{align*}</cmath>
  
−
<math>x_7=-x_1-x_4+x_3-x_2+x_1+x_4-x_3=-x_2</math>
+
yields <math>x_{n} = -x_{n-5}</math>. Thus <math>x_n = (-1)^k x_{n-5k}</math>. Since <math>531 = 106 \cdot 5 + 1,\ 753 = 150 \cdot 5 + 3,\ 975 = 194 \cdot 5 + 5</math>, it follows that
  
−
<math>x_8=-x_2+x_1+x_4-x_3+x_2-x_1-x_4=-x_3</math>
+
<cmath>x_{531} + x_{753} + x_{975} = (-1)^{106} x_1 + (-1)^{150} x_3 + (-1)^{194} x_5 = 211 + 420 + 267 = \boxed{898}.</cmath>  
  
−
<math>x_9=-x_3+x_2-x_1-x_4+x_3-x_2+x_1=-x_4</math>
+
== Solution Variant ==
 +
The recursive formula suggests telescoping. Indeed, if we add <math>x_n</math> and <math>x_{n-1}</math>, we have <math>x_n + x_{n-1} = (x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4}) + (x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) = x_{n-1} - x_{n-5}</math>.
  
−
And it cycles back to <math>x_{11}=x_1</math>
+
Subtracting <math>x_{n-1}</math> yields <math>x_n = -x_{n-5} \implies x_n = -(-(x_{n-10})) = x_{n-10}</math>.
  
−
Therefore, <math>x_{y}=x_{y-10}</math>, so
+
Thus,
  
−
<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>
+
<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_2 + x_4 = 375 + 523 = \boxed{898}.</cmath>
  
 +
Notice that we didn't need to use the values of <math>x_1</math> or <math>x_3</math> at all.
 
== 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}}
 +
{{MAA Notice}}

Revision as of 22:44, 30 December 2020

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

We find that $x_5 = 267$ by the recursive formula. Summing the recursions

\begin{align*} x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4} \\ x_{n-1}&=x_{n-2}-x_{n-3}+x_{n-4}-x_{n-5} \end{align*}

yields $x_{n} = -x_{n-5}$. Thus $x_n = (-1)^k x_{n-5k}$. Since $531 = 106 \cdot 5 + 1,\ 753 = 150 \cdot 5 + 3,\ 975 = 194 \cdot 5 + 5$, it follows that

\[x_{531} + x_{753} + x_{975} = (-1)^{106} x_1 + (-1)^{150} x_3 + (-1)^{194} x_5 = 211 + 420 + 267 = \boxed{898}.\]

Solution Variant

The recursive formula suggests telescoping. Indeed, if we add $x_n$ and $x_{n-1}$, we have $x_n + x_{n-1} = (x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4}) + (x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) = x_{n-1} - x_{n-5}$.

Subtracting $x_{n-1}$ yields $x_n = -x_{n-5} \implies x_n = -(-(x_{n-10})) = x_{n-10}$.

Thus,

\[x_{531} + x_{753} + x_{975} = x_1 + x_3 + x_5 = x_1 + x_3 + (x_4 - x_3 + x_2 - x_1) = x_2 + x_4 = 375 + 523 = \boxed{898}.\]

Notice that we didn't need to use the values of $x_1$ or $x_3$ at all.

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

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