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

Latest revision as of 01:28, 12 November 2022

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.

Non-Rigorous Solution

Calculate the first few terms:

$211,375,420,523,267,-211,-375,-420,-523,\dots$

At this point it is pretty clear that the sequence is periodic with period 10 (one may prove it quite easily like in solution 1) so our answer is obviously $211+420+267=\boxed{898}$

~Dhillonr25

Video Solution by OmegaLearn

https://youtu.be/lH-0ul1hwKw?t=870

~ pi_is_3.14

@@ Line 1: / Line 1: @@
 == Problem ==
 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}\\
+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>.
 == Solution ==
-{{solution}}
+We find that <math>x_5 = 267</math> by the recursive formula. Summing the [[recursion]]s
+<cmath>\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*}</cmath>
+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
+<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>
+== 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>.
+Subtracting <math>x_{n-1}</math> yields <math>x_n = -x_{n-5} \implies x_n = -(-(x_{n-10})) = x_{n-10}</math>.
+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_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.
+==Non-Rigorous Solution==
+Calculate the first few terms:
+<cmath>211,375,420,523,267,-211,-375,-420,-523,\dots</cmath>
+At this point it is pretty clear that the sequence is periodic with period 10 (one may prove it quite easily like in solution 1) so our answer is obviously <math>211+420+267=\boxed{898}</math>
+~Dhillonr25
+== Video Solution by OmegaLearn ==
+https://youtu.be/lH-0ul1hwKw?t=870
+~ pi_is_3.14
 == See also ==
 {{AIME box|year=2001|n=II|num-b=2|num-a=4}}
+{{MAA Notice}}

2001 AIME II (Problems • Answer Key • Resources)
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