Difference between revisions of "2017 AMC 12A Problems/Problem 21"
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==Problem== | ==Problem== | ||
− | A set <math>S</math> is constructed as follows. To begin, <math>S = \{0,10\}</math>. Repeatedly, as long as possible, if <math>x</math> is an integer root of some polynomial <math>a_{n}x^n + a_{n-1}x^{n-1} + | + | A set <math>S</math> is constructed as follows. To begin, <math>S = \{0,10\}</math>. Repeatedly, as long as possible, if <math>x</math> is an integer root of some polynomial <math>a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{1}x + a_0</math> for some <math>n\geq{1}</math>, all of whose coefficients <math>a_i</math> are elements of <math>S</math>, then <math>x</math> is put into <math>S</math>. When no more elements can be added to <math>S</math>, how many elements does <math>S</math> have? |
<math> \textbf{(A)}\ 4 | <math> \textbf{(A)}\ 4 |
Latest revision as of 14:46, 24 April 2020
Contents
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 . 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
-BochTheNerd
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).
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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