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

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<math>a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0</math> = <math>x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 = 0</math>
 
<math>a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0</math> = <math>x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 = 0</math>
 +
 +
so
 +
 +
<math>a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1} = \frac{-a_0}{x}</math>
  
 
with each <math>a_i</math> in <math>S</math>.
 
with each <math>a_i</math> in <math>S</math>.
  
−
Since <math>x</math> and the parenthesized term are both integers, <math>x</math> must be a factor of <math>a_0</math>. However, <math>a_0</math> is in <math>S</math> and every number in <math>S</math> already has all of its factors in <math>S</math>. Therefore, <math>x</math> must be in <math>S</math> and <math>S</math> cannot be expanded. <math>\{-10,-5,-2,-1,0,1,2,5,10\}</math> has <math>9</math> elements <math>\to \boxed{\textbf{(D)}}</math>
+
Clearly, <math>a_nx^{n-1} + a_{n-1}x^{n-2} + ... + a_2x + a_1</math> is an integer. Therefore, <math>\frac{-a_0}{x}</math> must be an integer. This means that <math>x</math> has to be a factor of some element in <math>S</math>. There are no such integers left, so there can be no more additional elements. <math>\{-10,-5,-2,-1,0,1,2,5,10\}</math> has <math>9</math> elements <math>\to \boxed{\textbf{(D)}}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2017|ab=A|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:46, 8 February 2017

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 (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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