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

Revision as of 08:46, 9 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

$\begin{align*} a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0 &= 0 \\ x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 &= 0 \\ x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) &= -a_0 \end{align*}$

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 $\to \boxed{\textbf{(D)}}$

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

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

@@ Line 31: / Line 31: @@
 At this point, no more elements can be added to <math>S</math>. To see this, let
-<cmath>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</cmath>
+<cmath>\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*}</cmath>
-with each <math>a_i</math> in <math>S</math>, so
+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>
-<cmath>x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) = -a_0</cmath>
-<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>
 ==See Also==
 {{AMC12 box|year=2017|ab=A|num-b=20|num-a=22}}
 {{MAA Notice}}

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

Revision as of 08:46, 9 February 2017

Problem

Solution 1

See Also