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

Revision as of 00:29, 9 July 2019

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).

@@ Line 38: / Line 38: @@
 -BochTheNerd
+== Solution 3 (If you are also short on time) ==
+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>.
+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==
 {{AMC12 box|year=2017|ab=A|num-b=20|num-a=22}}
 {{MAA Notice}}

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

Page

Toolbox

Search