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 | + | with each <math>a_i</math> in <math>S</math>, so |
− | <math>a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1} = | + | <math>x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) = -a_0</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> | |
==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 23:51, 8 February 2017
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 1
At first, .
has root , so now .
has root , so now .
has root , so now .
has root , so now .
has root , so now .
has root , so now .
has root , so now .
At this point, no more elements can be added to . To see this, let
=
with each in , so
.
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
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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