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

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==See also==
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#[[2021 JMPSC Invitational Problems|Other 2021 JMPSC Invitational Problems]]
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#[[2021 JMPSC Invitational Answer Key|2021 JMPSC Invitational Answer Key]]
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#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
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{{JMPSC Notice}}

Revision as of 16:26, 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

See also

  1. Other 2021 JMPSC Invitational Problems
  2. 2021 JMPSC Invitational 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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