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

## 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 $$x_5 + y_5 = 2(x_4 + y_4)$$ $$= 2(2(x_3 + y_3))$$ $$= 2(2(2(x_2 + y_2)))$$ $$= 2(2(2(2(x_1 + y_1))))$$ $$= 2(2(2(2(2(x_0 + y_0))))).$$ Since we are given that $x_0 = 3$ and $y_0 = 1$, we can plug these values in to get that $$x_5 + y_5 = 2(2(2(2(2(3 + 1))))) = 2(2(2(2(2(4))))) = 128 \qquad (1).$$

Similarly, we conclude that $$x_5 - y_5 = 3(3(3(3(3(x_0 - y_0)))))= 3(3(3(3(3(3 - 1))))) = 3(3(3(3(3(2))))) = 486 \qquad (2).$$

Adding $(1)$ and $(2)$ gives us $2 \cdot x_5 = 614.$ Dividing both sides by $2$ yields $x_5 = \boxed{307}.$

~mahaler

## Solution 2

Add both equations to get $2(x_{n+1})=5x_n-y_n$, and subtract both equations to get $2(y_{n+1})=5y_n-x_n$, so now we bash: $x_1=7$ and $y_1=1$. $x_2=17$ and $y_2=-1$. $x_3=43$ and $y_3=-11$. $x_4=113$ and $y_4=-49$, $x_5=\frac{614}{2}=\boxed{307}$

~Geometry285

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