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2017 AMC 12A Problems/Problem 21

Revision as of 19:19, 17 March 2020 by Emerald block (talk | contribs) (added shorter expressions in section for determining elements)

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 & \(S=\{-10,-1,0,1,10\}\) \\ \(x^3+x-10\) & 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 (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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