2017 AMC 12A Problems/Problem 21

Revision as of 23:46, 8 February 2017 by Blaquick (talk | contribs) (Solution 1)

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

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

so

$a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1} = \frac{-a_0}{x}$

with each $a_i$ in $S$.

Clearly, $a_nx^{n-1} + a_{n-1}x^{n-2} + ... + a_2x + a_1$ is an integer. Therefore, $\frac{-a_0}{x}$ must be an integer. This means that $x$ 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 $\to \boxed{\textbf{(D)}}$

See Also

2017 AMC 12A (ProblemsAnswer KeyResources)
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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