If $$[t]$$ denotes the greatest integer $$\leq t$$, then number of points, at which the function $$f(x) = 4|2x+3| + 9\left[x + \frac{1}{2}\right] - 12[x+20]$$ is not differentiable in the open interval $$(-20, 20)$$, is _____

We have $$f(x) = 4|2x+3| + 9\!\left[x + \frac{1}{2}\right] - 12[x + 20]$$ on the open interval $$(-20, 20)$$, where $$[t]$$ denotes the greatest integer less than or equal to $$t$$.

Since 20 is an integer, $$[x + 20] = [x] + 20$$, so $$f(x) = 4|2x+3| + 9\!\left[x + \frac{1}{2}\right] - 12[x] - 240$$.

The term $$4|2x+3|$$ is continuous everywhere and differentiable at all points except $$x = -\frac{3}{2}$$, where the expression inside the absolute value changes sign.

The floor function $$[x]$$ is discontinuous (and hence not differentiable) at every integer. The floor function $$\left[x + \frac{1}{2}\right]$$ is discontinuous at every point where $$x + \frac{1}{2}$$ is an integer, i.e., at every half-integer of the form $$x = n - \frac{1}{2}$$ for integer $$n$$. These are the points $$\ldots, -\frac{3}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{3}{2}, \ldots$$

The combination $$g(x) = 9\!\left[x + \frac{1}{2}\right] - 12[x]$$ is potentially not differentiable at every integer and every half-integer. At an integer $$x = n$$, the function $$[x]$$ jumps by 1 (causing a discontinuity of $$-12$$) while $$\left[x + \frac{1}{2}\right]$$ is continuous. Since the jump is $$-12 \neq 0$$, $$g$$ is discontinuous at every integer. At a half-integer $$x = n + \frac{1}{2}$$, the function $$\left[x + \frac{1}{2}\right]$$ jumps by 1 (causing a discontinuity of $$+9$$) while $$[x]$$ is continuous. Since the jump is $$9 \neq 0$$, $$g$$ is discontinuous at every half-integer. A function that is discontinuous at a point cannot be differentiable there.

In the interval $$(-20, 20)$$, the integers are $$-19, -18, \ldots, 18, 19$$, totalling 39 points. The half-integers are $$-19.5, -18.5, \ldots, 18.5, 19.5$$, totalling 40 points. The point $$x = -\frac{3}{2}$$ where the absolute value term is non-differentiable is already counted among the half-integers.

Therefore the total number of points where $$f$$ is not differentiable is $$39 + 40 = 79$$.

Hence, the correct answer is $$\boxed{79}$$.