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2017 AMC 12A Problems/Problem 21

Revision as of 22:07, 8 February 2017 by Afuous (talk | contribs) (Add 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

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

be a polynomial with each $a_i$ in $S$.

Since $x$ and the parenthesized term are both integers, neither can have any factors that $a_0$ does not have. However, $a_0$ is in $S$ and every number in $S$ already has all of its factors in $S$. Therefore, $x$ must be in $S$ and $S$ cannot be expanded. $\{-10,-5,-2,-1,0,1,2,5,10\}$ has $9$ elements $\to \boxed{\textbf{(D)}}$

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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