2017 AMC 12A Problems/Problem 21
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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