2001 AIME II Problems/Problem 3

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 $\begin{align*}x_{531}+x_{753}+x_{975}=x_1+x_3+x_5&=x_1+x_3+x_4-x_3+x_2-x_1\\ &=x_4+x_2=523+420=\boxed{943}\end{align*}$

2001 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2001 AIME II Problems/Problem 3

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