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 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
.
Since and the parenthesized term are both integers, neither can have any factors that
does not have. However,
is in
and every number in
already has all of its factors in
. Therefore,
must be in
and
cannot be expanded.
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 | |
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