Question

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

$4$

$5$

$7$

$9$

$11$

Solution:

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

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

No further integer can appear: by the Rational Root Theorem any integer root divides the constant term, which is always a factor of $10.$ So $S$ has $9$ elements.

Thus, the correct answer is D.

2017 AMC 12A Problem 21

Solution:

Problem 21 in Other Years