Difference between revisions of "2017 AMC 12A Problems/Problem 21"
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==Problem== | ==Problem== | ||
− | A set <math>S</math> is constructed as follows. To begin, <math>S = \{0,10\}</math>. Repeatedly, as long as possible, if <math>x</math> is an integer root of some polynomial <math>a_{n}x^n + a_{n-1}x^{n-1} + | + | A set <math>S</math> is constructed as follows. To begin, <math>S = \{0,10\}</math>. Repeatedly, as long as possible, if <math>x</math> is an integer root of some polynomial <math>a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{1}x + a_0</math> for some <math>n\geq{1}</math>, all of whose coefficients <math>a_i</math> are elements of <math>S</math>, then <math>x</math> is put into <math>S</math>. When no more elements can be added to <math>S</math>, how many elements does <math>S</math> have? |
<math> \textbf{(A)}\ 4 | <math> \textbf{(A)}\ 4 | ||
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\qquad\textbf{(E)}\ 11</math> | \qquad\textbf{(E)}\ 11</math> | ||
− | ==Solution | + | ==Solution== |
At first, <math>S=\{0,10\}</math>. | At first, <math>S=\{0,10\}</math>. | ||
+ | <cmath>\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}</cmath> | ||
− | + | At this point, no more elements can be added to <math>S</math>. To see this, let | |
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− | < | + | <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> | ||
− | <math>x | + | 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> |
− | + | ==Solution 2 (If you are short on time)== | |
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− | <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 | |
− | + | == Solution 3 (If you are also short on time) == | |
− | + | 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>. | |
− | + | 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). | |
==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 15:46, 24 April 2020
Contents
Problem
A set is constructed as follows. To begin, . Repeatedly, as long as possible, if is an integer root of some polynomial for some , all of whose coefficients are elements of , then is put into . When no more elements can be added to , how many elements does have?
Solution
At first, .
At this point, no more elements can be added to . To see this, let
with each in . is a factor of , and is in , so has to be a factor of some element in . There are no such integers left, so there can be no more additional elements. has elements
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 , where and are co-prime, is a factor of and is a factor of . We can easily see is in because of has root . Since we want set to be as large as possible, we let and , and quickly see that all possible integer roots are , , , , plus the we started with, we get a total of elements
-BochTheNerd
Solution 3 (If you are also short on time)
By the Rational Root theorem, notice that we must have . Since , this implies that any added must be a factor of a certain element in before. This therefore implies that any 's added must be a factor of . Thus, the largest possible set is all the positive and negative factors of , hence .
Note: this solution is not a real solution because it does not show that each 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).
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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