Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Find the answer

5. Given the set of integers
$M=\left\{m \mid x^{2}+m x-36=0\right.$ has integer solutions $\}$ , set $A$ satisfies the conditions:
(1) $\varnothing \subset A \subseteq M$ ;
(2) If $a \in A$ , then $-a \in A$ .
Then the number of all such sets $A$ is:

Pick one

Solution

5. C.

Let $\alpha, \beta$ be the roots of the equation $x^{2}+m x-36=0$ . Then $\alpha \beta=-36$ . Therefore,
when $|\alpha|=1,|\beta|=36$ , $m= \pm 35$ ;
when $|\alpha|=2,|\beta|=18$ , $m= \pm 16$ ;
when $|\alpha|=3,|\beta|=12$ , $m= \pm 9$ ;
when $|\alpha|=4,|\beta|=9$ , $m= \pm 5$ ;
when $|\alpha|=6,|\beta|=6$ , $m=0$ .
Thus,
$M=\{0\} \cup\{-5,5\} \cup\{-9,9\} \cup$
$\{-16,16\} \cup\{-35,35\} \text {. }$

From condition (1), we know $A \neq \varnothing$ .
From condition (2), we know $A$ is a set composed of some pairs of opposite numbers.

Therefore, the 5 pairs of opposite numbers in $M$ can form $2^{5}-1$ different non-empty sets $A$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.