# Difference between revisions of "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

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^4-x^2-x-10$$ & has root & $$x=2$$ & so now & $$S=\{-10,-1,0,1,2,10\}$$ \\ $$x^4-x^2+x-10$$ & has root & $$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\}$$ \\ $$2x+10$$ & has root & $$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).