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 , 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 |
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