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

$10x+10$ 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 $S=\{-10,-5,-2,-1,0,1,2,5,10\}$ .

At this point, no more elements can be added to $S$ . To see this, let

$a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0 = x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 = 0$

with each $a_i$ in $S$ , so

$x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) = -a_0$

$x$ 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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2017 AMC 12A Problems/Problem 21

Problem

Solution 1

See Also