2017 AMC 12A Problems/Problem 21

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

2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2017 AMC 12A Problems/Problem 21

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Solution 1

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