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

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==Solution==
 
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We notice that <cmath>x_5 + y_5 = 2x_4 + 2y_4 = 2(x_4 + y_4)</cmath> <cmath>= 2(2(x_3 + y_3))</cmath> <cmath>= 2(2(2(x_2 + y_2)))</cmath> <cmath>= 2(2(2(2(x_1 + y_1))))</cmath> <cmath>= 2(2(2(2(2(x_0 + y_0))))).</cmath> Since we are given that <math>x_0 = 3</math> and <math>y_0 = 1</math>, we can plug these values in to get that <cmath>x_5 + y_5 = 2(2(2(2(2(3 + 1))))) = 2(2(2(2(2(4))))) = 128 \qquad (1).</cmath>
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We notice that <cmath>x_5 + y_5 = 2(x_4 + y_4)</cmath> <cmath>= 2(2(x_3 + y_3))</cmath> <cmath>= 2(2(2(x_2 + y_2)))</cmath> <cmath>= 2(2(2(2(x_1 + y_1))))</cmath> <cmath>= 2(2(2(2(2(x_0 + y_0))))).</cmath> Since we are given that <math>x_0 = 3</math> and <math>y_0 = 1</math>, we can plug these values in to get that <cmath>x_5 + y_5 = 2(2(2(2(2(3 + 1))))) = 2(2(2(2(2(4))))) = 128 \qquad (1).</cmath>
  
 
Similarly, we conclude that <cmath>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).</cmath>
 
Similarly, we conclude that <cmath>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).</cmath>

Revision as of 14:54, 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 \[x_5 + y_5 = 2(x_4 + y_4)\] \[= 2(2(x_3 + y_3))\] \[= 2(2(2(x_2 + y_2)))\] \[= 2(2(2(2(x_1 + y_1))))\] \[= 2(2(2(2(2(x_0 + y_0))))).\] 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

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