2017 AMC 12A Problems/Problem 21
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?
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 . 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
Solution 3 (If you are also short on time)
By the Rational Root theorem, notice that we must have . Since , this implies that any added must be a factor of a certain element in before. This therefore implies that any 's added must be a factor of . Thus, the largest possible set is all the positive and negative factors of , hence .
Note: this solution is not a real solution because it does not show that each actually works (basically we have found the maximum possible elements but we have not shown that there is a polynomial for each of them to work).
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