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2017 AMC 12A Problem 21

Attempt Problem 21 of the 2017 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 12A solutions, or check the answer key.

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21.

A set SS is constructed as follows. To begin, S={0,10}.S=\{0,10\}. Repeatedly, as long as possible, if xx is an integer root of some polynomial anxn+an−1xn−1a_nx^n+a_{n-1}x^{n-1} +⋯+a1x+a0+\cdots+a_1x+a_0 for some n≥1,n\ge1, all of whose coefficients aia_i are elements of S,S, then xx is put into S.S. When no more elements can be added to S,S, how many elements does SS have?

44

55

77

99

1111

Answer: D
Concepts:polynomialdivisibilitysystematic listing
Difficulty rating: 2130
Small Hint:

By the Rational Root Theorem, any integer root of a polynomial with coefficients in SS divides its constant term

Big Hint:

Starting from {0,10},\{0,10\}, reach ±1,±2,±5,±10;\pm1,\pm2,\pm5,\pm10; show nothing else can enter

Solution:

Using 10x+10,10x+10, the root −1-1 enters S.S. Then 11 enters as a root of −x10−x9−⋯−x+10,-x^{10}-x^9-\cdots-x+10, and −10-10 enters from x+10.x+10.

Now x3+x−10x^3+x-10 has root 2,2, and x+2x+2 gives −2;-2; then 2x−102x-10 and 2x+102x+10 give ±5.\pm5. At this point S={0,±1,±2,±5,±10}.S=\{0,\pm1,\pm2,\pm5,\pm10\}.

No further integer can appear. Inductively, every nonzero member of SS divides 10.10. If a polynomial used in the rule has constant term 0,0, factor out the largest possible power of x;x; any nonzero root is then a root of a polynomial whose constant term is the first nonzero original coefficient. The Rational Root Theorem shows that the root divides this coefficient, which by the inductive hypothesis divides 10.10. So SS has 99 elements.

Thus, the correct answer is D.

Problem 20#20
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