Michael's integers

Note that from $x=1\to 3$ , the possible integers are $0\to 6$ , which is $7$ integers.

From $x=-3\to 0$ , the possible integers are $-12\to 0$ , which is $13$ integers.

$x(x-1)$ cannot be integer for $0 < x < 1$ since both $x$ and $x-1$ would be less than $1$

Thus the first two cases are our only cases.

Therefore the answer is $7+13=\boxed{20}$

For $x(x-1)$ to be an integer $(x-1)$ should be equal to $\frac{a}{x}$

so $(x-1)$ = $\frac{a}{x}$
$a$ is an Integer

$x^2-x-a=0$

$x=\cfrac { 1\pm \sqrt { 1+4a } \quad }{ 2 }$

| $x$ | $\le$ 3
For $a\in$ [0,12] exactly 20 real numbers x satisfy. Hence answer is 20

Let f(x)=x^2-x. Since x varies between -3 and 3, f(-3)=12 and f(3)=6.As f(x) is a continuous function f(x) will have 12+6+2(this 2 is for f(x)=0 ie at x=0 and x=1) integer values for some x between -3 and 3. So the answer is 20

The solution can be easily seen if you plot the curve of $f(x) = x(x-1)$ if that is not considered cheating. I still don't know how to add graphics in the solution. Wonder if anyone can help me.

Anyway, we can visualize the curve with theory. As a polynomial with positive first term (highest power of $x$ and even highest power of $x$ , $f(x)$ is a downward symmetrical concave curve. It cuts the x-axis at $(0, 0)$ and $(0,1)$ and hence is symmetrical along $x = \frac {1}{2}$ with the lowest point at $(\frac {1}{2}, -\frac {1}{4})$ .

The condition $|x| \le 3$ fixes the range of $x$ to be considered as $-3 \le x \le 3$ . When $x = -3$ , $f(-3) = 12$ . As the $f(x)$ is continuous, this means that for $-3 \le x \le 0$ , there are $13\space (12 \rightarrow 0)$ integer solutions for $f(x)$ . And since $f(3) = 6$ , for $1 \le x \le 3$ , there are $7\space (0\rightarrow 6)$ integer solutions.

Therefore, the numbers of real numbers $x$ with $|x| \le 3$ is $x(x-1)$ an integer is $13+7= \boxed {20}$ .