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

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==Problem==
 
==Problem==
 
Let <math>(x_n)_{n\geq 0}</math> and <math>(y_n)_{n\geq 0}</math>  be sequences of real numbers such that <math>x_0 = 3</math>, <math>y_0 = 1</math>, and, for all positive integers <math>n</math>,
 
Let <math>(x_n)_{n\geq 0}</math> and <math>(y_n)_{n\geq 0}</math>  be sequences of real numbers such that <math>x_0 = 3</math>, <math>y_0 = 1</math>, and, for all positive integers <math>n</math>,
\begin{align*}
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x_{n+1}+y_{n+1} &= 2x_n + 2y_n,\\
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<cmath>x_{n+1}+y_{n+1} = 2x_n + 2y_n,</cmath>
x_{n+1}-y_{n+1}&=3x_n-3y_n.
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<cmath>x_{n+1}-y_{n+1}=3x_n-3y_n.</cmath>
\end{align*} Find <math>x_5</math>.
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Find <math>x_5</math>.
  
 
==Solution==
 
==Solution==
asdf
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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>
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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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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>
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 +
~mahaler
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==Solution 2==
 +
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

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

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