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

(Solution 2 (If you are short on time))
(Solution 2 (If you are short on time))
Line 35: Line 35:
 
==Solution 2 (If you are short on time)==
 
==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 <math> \pm \frac p{q} </math>, where <math>p</math> and <math>q</math> are co-prime, <math>p</math> is a factor of <math>a_0</math> and <math>q</math> is a factor of <math>a_n</math>, in other words, all possible <math>\pm \frac {a_0} {a_n}</math> must be integer multiples of at least one integer root. We can easily see <math>-1</math> is in <math>S</math> because of <math>10x + 10 = 0</math> has root <math>-1</math>. Since we want set <math>S</math> to be as large as possible, we let <math>p=10</math> and <math>q=1</math>, and quickly see that all possible integer roots are <math>\pm 1</math>, <math>\pm 2</math>, <math>\pm 5</math>, <math>\pm 10</math>, plus the <math>0</math> we started with, we get a total of <math>9</math> elements <math>\to \boxed{\textbf{(D)}}</math>
+
By Rational Root Theorem, the only rational roots for this function we're dealing with must be in the form <math> \pm \frac p{q} </math>, where <math>p</math> and <math>q</math> are co-prime, <math>p</math> is a factor of <math>a_0</math> and <math>q</math> is a factor of <math>a_n</math>. We can easily see <math>-1</math> is in <math>S</math> because of <math>10x + 10 = 0</math> has root <math>-1</math>. Since we want set <math>S</math> to be as large as possible, we let <math>p=10</math> and <math>q=1</math>, and quickly see that all possible integer roots are <math>\pm 1</math>, <math>\pm 2</math>, <math>\pm 5</math>, <math>\pm 10</math>, plus the <math>0</math> we started with, we get a total of <math>9</math> elements <math>\to \boxed{\textbf{(D)}}</math>
 +
 
 +
-BochTheNerd
  
 
==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 14:04, 3 January 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

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

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