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

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with each <math>a_i</math> in <math>S</math>. <math>x</math> is a factor of <math>a_0</math>, and <math>a_0</math> is in <math>S</math>, so <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>
 
with each <math>a_i</math> in <math>S</math>. <math>x</math> is a factor of <math>a_0</math>, and <math>a_0</math> is in <math>S</math>, so <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>
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==Solution 2 (If you are short on time)==
 
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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>
 
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-BochTheNerd
 
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== Solution 3 (If you are also short on time) ==
 
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By the Rational Root theorem, notice that we must have <math>x | a_0</math>. Since <math>a_0 \in S</math>, this implies that any <math>x</math> added must be a factor of a certain element in <math>S</math> before. This therefore implies that any <math>x</math>'s added must be a factor of <math>10</math>. Thus, the largest possible set is all the positive and negative factors of <math>10</math>, hence <math>\boxed{9}</math>.
 
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Note: this solution is not a real solution because it does not show that each <math>x</math> actually works (basically we have found the maximum possible elements but we have not shown that there is a polynomial for each of them to work).
 
  
 
== Video Solution by Richard Rusczyk ==
 
== Video Solution by Richard Rusczyk ==
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https://www.youtube.com/watch?v=hSYSNBVPLhE&list=PLyhPcpM8aMvLZmuDnM-0vrFniLpo7Orbp&index=1  
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https://www.youtube.com/watch?v=hSYSNBVPLhE&list=PLyhPcpM8aMvLZmuDnM-0vrFniLpo7Orbp&index=1
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- AMBRIGGS
 
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== Video Solution ==
 
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https://www.youtube.com/watch?v=hSYSNBVPLhE&list=PLyhPcpM8aMvLZmuDnM-0vrFniLpo7Orbp&index=1
 
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- AMBRIGGS
 
  
 
==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}}

Latest revision as of 09:48, 27 March 2023

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 $a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{1}x + a_0$ 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 & \(S=\{-10,-1,0,1,10\}\) \\ \(x^3+x-10\) & has root & \(x=2\) & so now & \(S=\{-10,-1,0,1,2,10\}\) \\ \(x+2\) & has root & \(x=-2\) & so now & \(S=\{-10,-2,-1,0,1,2,10\}\) \\ \(2x-10\) & has root & \(x=5\) & so now & \(S=\{-10,-2,-1,0,1,2,5,10\}\) \\ \(x+5\) & 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 $\to \boxed{\textbf{(D)}}$

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=hSYSNBVPLhE&list=PLyhPcpM8aMvLZmuDnM-0vrFniLpo7Orbp&index=1

See Also

2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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All AMC 12 Problems and Solutions

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