Divisibility and Probability

$x^3-x=x(x^2-1)$ The divisibility rule of 12 is: if a number is divisible by both of 4 and 3, then it is divisible by 12.

Divisibility by 3:

If $x$ is divisible by $3$ , then $3\mid x(x^2-1)$ . If not, then it makes $1$ or $2$ remainder when it is divided by $3$ . It is easy to see that even if the remainder is $1$ or $2$ , $x^2\equiv 1 \ \mod 3$ . So $x^2-1\equiv 1-1=0 \ \mod3$ . So $x(x^2-1)$ is always divisible by $3$ .

Divisibility by 4:

If $x$ is divisible by $4$ , then $4\mid x(x^2-1)$ .

If $x\equiv 1 \ \mod 4$ , then $4\mid x^2-1$ .

If $x\equiv 2 \ \mod 4$ , then $4\not\mid x(x^2-1)$ , because $x^2-1\equiv 0-1=-1 \ \mod 4$ .

If $x\equiv 3 \ \mod 4$ , then $x^2-1\equiv 3*3-1=9-1=8 \equiv 0 \mod 4$ .

So $x(x^2-1)$ is always divisible by $4$ , except $x\equiv 2 \mod 4$ . There are $99-20+1=80$ numbers between $20$ and $99$ . From these numbers exactly $20$ will be divisible by $4$ , because only $20$ is divisible by $4$ from $20$ and $99$ . To each $k$ number $20\leq k\leq 99$ we can assign an $n$ number, where $n\equiv 2 \mod 4$ , and $n-2=k$ , so there are $20$ possible values for $n$ . From that the probality is: $\dfrac{80-20}{80}=\dfrac{60}{80}=\dfrac{6}{8}=\dfrac{3}{4}$ .

Therefore the answer is $3+4=\boxed{7}$ .

We can write $x^3-x = x(x^2-1) = (x-1)x(x+1) = P(x)$ , which is a product of three consecutive positive integers. We know that there is always one out of three consecutive integers that is divisible by 3. If $x$ is odd, then $x-1$ and $x+1$ are even which means that $P(x)$ are divisible by 4 and hence divisible by 12. If $x$ is even, then $x-1$ and $x+1$ are odd. Then $P(x)$ is divisible by 12 only if $x$ is divisible by 4. Therefore $12 \not \mid P(x)$ when $x = 4n - 2$ , where $n \in \mathbb N$ . From 20 to 99 inclusive, there are 80 numbers; therefore the number of $P(x)$ indivisible by 12 is $b = \frac {80}4 = 20$ , that divisible by 12 is $a=80-20=60$ and the probability $Pr(12 \mid P(x)) = \frac {60}{80} = \frac 34$ . $\implies A+B = 3+4 = \boxed{7}$ .