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

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==Problem==
 
==Problem==
  
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?
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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} + \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>
  
==Solution 1==
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==Solution==
  
 
At first, <math>S=\{0,10\}</math>.
 
At first, <math>S=\{0,10\}</math>.
  
 +
<cmath>\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 & \(S=\{-10,-2,-1,0,1,2,5,10\}\) \\
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\(x+5\) & has root & \(x=-5\) & so now & \(S=\{-10,-5,-2,-1,0,1,2,5,10\}\)
 +
\end{tabular}</cmath>
  
<math>10x+10</math> has root <math>x=-1</math>, so now <math>S=\{-1,0,10\}</math>.
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At this point, no more elements can be added to <math>S</math>. To see this, let
 
 
<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>.
 
 
 
<math>x+10</math> has root <math>x=-10</math>, so now <math>S=\{-10,-1,0,1,10\}</math>.
 
  
<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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<cmath>\begin{align*}
 +
a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + 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 &= 0 \\
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x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) &= -a_0
 +
\end{align*}</cmath>
  
<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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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>
  
<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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==Solution 2 (If you are short on time)==
 
 
<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
 
  
<math>a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0</math> = <math>x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 = 0</math>
+
By Rational Root Theorem, the only rational roots for this function we're dealing with must be in the form <math> \pm \frac p{q} </math>, where <math>p</math> and <math>q</math> are co-prime, <math>p</math> is a factor of <math>a_0</math> and <math>q</math> is a factor of <math>a_n</math>. We can easily see <math>-1</math> is in <math>S</math> because of <math>10x + 10 = 0</math> has root <math>-1</math>. Since we want set <math>S</math> to be as large as possible, we let <math>p=10</math> and <math>q=-1</math>, and quickly see that all possible integer roots are <math>\pm 1</math>, <math>\pm 2</math>, <math>\pm 5</math>, <math>\pm 10</math>, plus the <math>0</math> we started with, we get a total of <math>9</math> elements <math>\to \boxed{\textbf{(D)}}</math>
  
so
+
-BochTheNerd
  
<math>a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1} = \frac{-a_0}{x}</math>
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== Solution 3 (If you are also short on time) ==
  
with each <math>a_i</math> in <math>S</math>.
+
By the Rational Root theorem, notice that we must have <math>x | a_0</math>. Since <math>a_0 \in S</math>, this implies that any <math>x</math> added must be a factor of a certain element in <math>S</math> before. This therefore implies that any <math>x</math>'s added must be a factor of <math>10</math>. Thus, the largest possible set is all the positive and negative factors of <math>10</math>, hence <math>\boxed{9}</math>.  
  
Clearly, <math>a_nx^{n-1} + a_{n-1}x^{n-2} + ... + a_2x + a_1</math> is an integer. Therefore, <math>\frac{-a_0}{x}</math> must be an integer. This means that <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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Note: this solution is not a real solution because it does not show that each <math>x</math> actually works (basically we have found the maximum possible elements but we have not shown that there is a polynomial for each of them to work).
  
 
==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 15:46, 24 April 2020

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 $\to \boxed{\textbf{(D)}}$

Solution 2 (If you are short on time)

By Rational Root Theorem, the only rational roots for this function we're dealing with must be in the form $\pm \frac p{q}$, where $p$ and $q$ are co-prime, $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$. We can easily see $-1$ is in $S$ because of $10x + 10 = 0$ has root $-1$. Since we want set $S$ to be as large as possible, we let $p=10$ and $q=-1$, and quickly see that all possible integer roots are $\pm 1$, $\pm 2$, $\pm 5$, $\pm 10$, plus the $0$ we started with, we get a total of $9$ elements $\to \boxed{\textbf{(D)}}$

-BochTheNerd

Solution 3 (If you are also short on time)

By the Rational Root theorem, notice that we must have $x | a_0$. Since $a_0 \in S$, this implies that any $x$ added must be a factor of a certain element in $S$ before. This therefore implies that any $x$'s added must be a factor of $10$. Thus, the largest possible set is all the positive and negative factors of $10$, hence $\boxed{9}$.

Note: this solution is not a real solution because it does not show that each $x$ actually works (basically we have found the maximum possible elements but we have not shown that there is a polynomial for each of them to work).

See Also

2017 AMC 12A (ProblemsAnswer KeyResources)
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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