2017 AMC 12A Problems/Problem 21
Contents
[hide]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 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).
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=hSYSNBVPLhE&list=PLyhPcpM8aMvLZmuDnM-0vrFniLpo7Orbp&index=1 - AMBRIGGS
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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All AMC 12 Problems and Solutions |
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