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Difference between revisions of "2017 AMC 12A Problems/Problem 21"

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==Problem==
 
==Problem==
  
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A set <math>S</math> is constructed as follows. To begin, <math>S = \{0,10\}</math>. Repeatedly, as long as possible, if <math>x</math> is an integer root of some polynomial <math>a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{1}x + a_0</math> for some <math>n\geq{1}</math>, all of whose coefficients <math>a_i</math> are elements of <math>S</math>, then <math>x</math> is put into <math>S</math>. When no more elements can be added to <math>S</math>, how many elements does <math>S</math> have?
+
A set <math>S</math> is constructed as follows. To begin, <math>S = \{0,10\}</math>. Repeatedly, as long as possible, if <math>x</math> is an integer root of some polynomial <math>a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{1}x + a_0</math> for some <math>n\geq{1}</math>, all of whose coefficients <math>a_i</math> are elements of <math>S</math>, then <math>x</math> is put into <math>S</math>. When no more elements can be added to <math>S</math>, how many elements does <math>S</math> have?
  
 
<math> \textbf{(A)}\ 4
 
<math> \textbf{(A)}\ 4
Line 9: Line 9:
 
\qquad\textbf{(E)}\ 11</math>
 
\qquad\textbf{(E)}\ 11</math>
  
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==Solution 1==
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==Solution==
  
 
At first, <math>S=\{0,10\}</math>.
 
At first, <math>S=\{0,10\}</math>.
  
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<cmath>\begin{tabular}{r c l c l}
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<math>10x+10</math> has root <math>x=-1</math>, so now <math>S=\{-1,0,10\}</math>.
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\(10x+10\) & has root & \(x=-1\) & so now & \(S=\{-1,0,10\}\) \\
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\(-x^{10}-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x+10\) & has root & \(x=1\) & so now & \(S=\{-1,0,1,10\}\) \\
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<math>-x^{10}-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x+10</math> has root <math>x=1</math>, so now <math>S=\{-1,0,1,10\}</math>.
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\(x+10\) & has root & \(x=-10\) & so now & \(S=\{-10,-1,0,1,10\}\) \\
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\(x^3+x-10\) & has root & \(x=2\) & so now & \(S=\{-10,-1,0,1,2,10\}\) \\
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<math>x+10</math> has root <math>x=-10</math>, so now <math>S=\{-10,-1,0,1,10\}</math>.
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\(x+2\) & has root & \(x=-2\) & so now & \(S=\{-10,-2,-1,0,1,2,10\}\) \\
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\(2x-10\) & has root & \(x=5\) & so now & \(S=\{-10,-2,-1,0,1,2,5,10\}\) \\
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<math>x^4-x^2-x+10</math> has root <math>x=2</math>, so now <math>S=\{-10,-1,0,1,2,10\}</math>.
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\(x+5\) & has root & \(x=-5\) & so now & \(S=\{-10,-5,-2,-1,0,1,2,5,10\}\)
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\end{tabular}</cmath>
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<math>x^4-x^2+x+10</math> has root <math>x=-2</math>, so now <math>S=\{-10,-2,-1,0,1,2,10\}</math>.
 
−
 
 
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<math>2x-10</math> has root <math>x=5</math>, so now <math>S=\{-10,-2,-1,0,1,2,5,10\}</math>.
 
−
 
 
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<math>2x+10</math> has root <math>x=-5</math>, so now <math>S=\{-10,-5,-2,-1,0,1,2,5,10\}</math>.
 
−
 
 
  
 
At this point, no more elements can be added to <math>S</math>. To see this, let
 
At this point, no more elements can be added to <math>S</math>. To see this, let
  
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<cmath>a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0 = x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 = 0</cmath>
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<cmath>\begin{align*}
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a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0 &= 0 \\
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with each <math>a_i</math> in <math>S</math>, so
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x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 &= 0 \\
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x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) &= -a_0
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<cmath>x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) = -a_0</cmath>
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\end{align*}</cmath>
  
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with each <math>a_i</math> in <math>S</math>. <math>x</math> is a factor of <math>a_0</math>, and <math>a_0</math> is in <math>S</math>, so <math>x</math> has to be a factor of some element in <math>S</math>. There are no such integers left, so there can be no more additional elements. <math>\{-10,-5,-2,-1,0,1,2,5,10\}</math> has <math>9</math> elements <math>\to \boxed{\textbf{(D)}}</math>
  
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<math>x</math> is a factor of <math>a_0</math>, and <math>a_0</math> is in <math>S</math>, so <math>x</math> has to be a factor of some element in <math>S</math>. There are no such integers left, so there can be no more additional elements. <math>\{-10,-5,-2,-1,0,1,2,5,10\}</math> has <math>9</math> elements <math>\to \boxed{\textbf{(D)}}</math>
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== Video Solution by Richard Rusczyk ==
 +
https://www.youtube.com/watch?v=hSYSNBVPLhE&list=PLyhPcpM8aMvLZmuDnM-0vrFniLpo7Orbp&index=1
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2017|ab=A|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:48, 27 March 2023

Problem

A set $S$ is constructed as follows. To begin, $S = \{0,10\}$. Repeatedly, as long as possible, if $x$ is an integer root of some polynomial $a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{1}x + a_0$ for some $n\geq{1}$, all of whose coefficients $a_i$ are elements of $S$, then $x$ is put into $S$. When no more elements can be added to $S$, how many elements does $S$ have?

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad\textbf{(C)}\ 7 \qquad\textbf{(D)}\ 9 \qquad\textbf{(E)}\ 11$

Solution

At first, $S=\{0,10\}$.

\[\begin{tabular}{r c l c l} \(10x+10\) & has root & \(x=-1\) & so now & \(S=\{-1,0,10\}\) \\ \(-x^{10}-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x+10\) & has root & \(x=1\) & so now & \(S=\{-1,0,1,10\}\) \\ \(x+10\) & has root & \(x=-10\) & so now & \(S=\{-10,-1,0,1,10\}\) \\ \(x^3+x-10\) & has root & \(x=2\) & so now & \(S=\{-10,-1,0,1,2,10\}\) \\ \(x+2\) & has root & \(x=-2\) & so now & \(S=\{-10,-2,-1,0,1,2,10\}\) \\ \(2x-10\) & has root & \(x=5\) & so now & \(S=\{-10,-2,-1,0,1,2,5,10\}\) \\ \(x+5\) & has root & \(x=-5\) & so now & \(S=\{-10,-5,-2,-1,0,1,2,5,10\}\) \end{tabular}\]

At this point, no more elements can be added to $S$. To see this, let

\begin{align*} a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0 &= 0 \\ x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 &= 0 \\ x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) &= -a_0 \end{align*}

with each $a_i$ in $S$. $x$ is a factor of $a_0$, and $a_0$ is in $S$, so $x$ has to be a factor of some element in $S$. There are no such integers left, so there can be no more additional elements. $\{-10,-5,-2,-1,0,1,2,5,10\}$ has $9$ elements $\to \boxed{\textbf{(D)}}$

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=hSYSNBVPLhE&list=PLyhPcpM8aMvLZmuDnM-0vrFniLpo7Orbp&index=1

See Also

2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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