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

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At first, <math>S=\{0,10\}</math>.
 
At first, <math>S=\{0,10\}</math>.
  
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<cmath>\begin{align*}
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<math>10x+10</math> has root <math>x=-1</math>, so now <math>S=\{-1,0,10\}</math>.
+
10x+10 & \quad\text{has root}\quad x=-1, & \quad\text{so now}\quad & S=\{-1,0,10\}. \\
−
 
+
-x^{10}-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x+10 & \quad\text{has root}\quad x=1, & \quad\text{so now}\quad & 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>.
+
x+10 & \quad\text{has root}\quad x=-10, & \quad\text{so now}\quad & S=\{-10,-1,0,1,10\}. \\
−
 
+
x^4-x^2-x+10 & \quad\text{has root}\quad x=2, & \quad\text{so now}\quad & S=\{-10,-1,0,1,2,10\}. \\
−
<math>x+10</math> has root <math>x=-10</math>, so now <math>S=\{-10,-1,0,1,10\}</math>.
+
x^4-x^2+x+10 & \quad\text{has root}\quad x=-2, & \quad\text{so now}\quad & S=\{-10,-2,-1,0,1,2,10\}. \\
−
 
+
2x-10 & \quad\text{has root}\quad x=5, & \quad\text{so now}\quad & S=\{-10,-2,-1,0,1,2,5,10\}. \\
−
<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>.
+
2x+10 & \quad\text{has root}\quad x=-5, & \quad\text{so now}\quad & S=\{-10,-5,-2,-1,0,1,2,5,10\}. \\
−
 
+
\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>.
 
−
 
 
−
<math>2x-10</math> has root <math>x=5</math>, so now <math>S=\{-10,-2,-1,0,1,2,5,10\}</math>.
 
−
 
 
−
<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

Revision as of 07:58, 9 February 2017

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 $a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{1}x + a_0$ 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 1

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

\begin{align*} 10x+10 & \quad\text{has root}\quad x=-1, & \quad\text{so now}\quad & S=\{-1,0,10\}. \\ -x^{10}-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x+10 & \quad\text{has root}\quad x=1, & \quad\text{so now}\quad & S=\{-1,0,1,10\}. \\ x+10 & \quad\text{has root}\quad x=-10, & \quad\text{so now}\quad & S=\{-10,-1,0,1,10\}. \\ x^4-x^2-x+10 & \quad\text{has root}\quad x=2, & \quad\text{so now}\quad & S=\{-10,-1,0,1,2,10\}. \\ x^4-x^2+x+10 & \quad\text{has root}\quad x=-2, & \quad\text{so now}\quad & S=\{-10,-2,-1,0,1,2,10\}. \\ 2x-10 & \quad\text{has root}\quad x=5, & \quad\text{so now}\quad & S=\{-10,-2,-1,0,1,2,5,10\}. \\ 2x+10 & \quad\text{has root}\quad x=-5, & \quad\text{so now}\quad & S=\{-10,-5,-2,-1,0,1,2,5,10\}. \\ \end{align*}

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

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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