Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 4"

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==Solution==
 
==Solution==
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We notice that <math>x_5 + y_5 = 2x_4 + 2y_4 = 2(x_4 + y_4) = 2(2(x_3 + y_3)) = 2(2(2(x_2 + y_2))) = 2(2(2(2(x_1 + y_1)))) = 2(2(2(2(2(x_0 + y_0))))).</math> Given that <math>x_0 = 3</math> and <math>y_0 = 1</math> in the problem, we can plug this in to get that <math>x_5 + y_5 = 2(2(2(2(2(3 + 1))))) = 2(2(2(2(2(4))))) = 128.</math>
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We can use the same method to conclude that <math>x_5 - y_5 = 3(3(3(3(3(x_0 - y_0))))) = 3(3(3(3(3(3-1))))) = 3(3(3(3(3(2))))) = 486.</math>
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Adding this system of equations <math>x_5 + y_5 = 128</math> and <math>x_5 - y_5 = 486</math> gives us <math>2x_5 = 614.</math> Dividing both sides by <math>2</math>, results in <math>x_5 = 307.</math>
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~mahaler

Revision as of 15:47, 11 July 2021

Problem

Let $(x_n)_{n\geq 0}$ and $(y_n)_{n\geq 0}$ be sequences of real numbers such that $x_0 = 3$, $y_0 = 1$, and, for all positive integers $n$,

\[x_{n+1}+y_{n+1} = 2x_n + 2y_n,\] \[x_{n+1}-y_{n+1}=3x_n-3y_n.\] Find $x_5$.

Solution

We notice that $x_5 + y_5 = 2x_4 + 2y_4 = 2(x_4 + y_4) = 2(2(x_3 + y_3)) = 2(2(2(x_2 + y_2))) = 2(2(2(2(x_1 + y_1)))) = 2(2(2(2(2(x_0 + y_0))))).$ Given that $x_0 = 3$ and $y_0 = 1$ in the problem, we can plug this in to get that $x_5 + y_5 = 2(2(2(2(2(3 + 1))))) = 2(2(2(2(2(4))))) = 128.$

We can use the same method to conclude that $x_5 - y_5 = 3(3(3(3(3(x_0 - y_0))))) = 3(3(3(3(3(3-1))))) = 3(3(3(3(3(2))))) = 486.$

Adding this system of equations $x_5 + y_5 = 128$ and $x_5 - y_5 = 486$ gives us $2x_5 = 614.$ Dividing both sides by $2$, results in $x_5 = 307.$

~mahaler

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