Difference between revisions of "2017 AMC 12A Problems/Problem 21"
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==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>, 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> |
==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 13:59, 3 January 2019
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
, in other words, all possible
must be integer multiples of at least one integer root. 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
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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