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

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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>
 
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>
  
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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>
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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>
  
 
Adding <math>(1)</math> and <math>(2)</math> gives us <math>2 \cdot x_5 = 614.</math> Dividing both sides by <math>2</math> yields <math>x_5 = \boxed{307}.</math>
 
Adding <math>(1)</math> and <math>(2)</math> gives us <math>2 \cdot x_5 = 614.</math> Dividing both sides by <math>2</math> yields <math>x_5 = \boxed{307}.</math>
  
 
~mahaler
 
~mahaler
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==Solution 2==
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Add both equations to get <math>2(x_{n+1})=5x_n-y_n</math>, and subtract both equations to get <math>2(y_{n+1})=5y_n-x_n</math>, so now we bash: <math>x_1=7</math> and <math>y_1=1</math>. <math>x_2=17</math> and <math>y_2=-1</math>. <math>x_3=43</math> and <math>y_3=-11</math>. <math>x_4=113</math> and <math>y_4=-49</math>, <math>x_5=\frac{614}{2}=\boxed{307}</math>
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~Geometry285
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==See also==
 +
#[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]
 +
#[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]]
 +
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 +
{{JMPSC Notice}}

Latest revision as of 20:08, 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

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

See also

  1. Other 2021 JMPSC Invitationals Problems
  2. 2021 JMPSC Invitationals Answer Key
  3. All JMPSC Problems and Solutions

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