Question:

Step 1: Solve the equation involving the greatest integer function for $x$ and then find $\left[\dfrac{1}{x}\right]$. Let the given equation be $2[x + 32] = 3[x - 64]$. We can use the property of the greatest integer function that $[k + m] = [k] + m$ for any integer $m$. Let $[x + 32] = n$, where $n$ is an integer. Then, we can write $[x - 64]$ as $[x + 32 - 96]$. Using the property, $[x + 32 - 96] = [x + 32] - 96 = n - 96$. Substitute these expressions into the original equation: $$2n = 3(n - 96)$$ Now, solve for $n$: $$2n = 3n - 288$$ $$n = 288$$ So, we have $[x + 32] = 288$. By the definition of the greatest integer function, this means: $$288 \le x + 32 < 289$$ Subtract 32 from all parts of the inequality: $$288 - 32 \le x < 289 - 32$$ $$256 \le x < 257$$ Now, we need to find $\left[\dfrac{1}{x}\right]$. Since $256 \le x < 257$, we can take the reciprocal. Note that the inequality signs flip when taking reciprocals of positive numbers. $$\dfrac{1}{257} < \dfrac{1}{x} \le \dfrac{1}{256}$$ Let's approximate the decimal values: $\dfrac{1}{257} \approx 0.003891$ $\dfrac{1}{256} \approx 0.003906$ So, we have: $$0.003891 < \dfrac{1}{x} \le 0.003906$$ The greatest integer less than or equal to $\dfrac{1}{x}$ in this range is 0. Therefore, $\left[\dfrac{1}{x}\right] = 0$. Step 2: Evaluate the product $y = \displaystyle\prod_{j=1}^{9} \sin\left(\dfrac{2j-1}{18}\right)\pi$. The product can be written as: $$y = \sin\left(\dfrac{\pi}{18}\right) \cdot \sin\left(\dfrac{3\pi}{18}\right) \cdot \sin\left(\dfrac{5\pi}{18}\right) \cdot \sin\left(\dfrac{7\pi}{18}\right) \cdot \sin\left(\dfrac{9\pi}{18}\right) \cdot \sin\left(\dfrac{11\pi}{18}\right) \cdot \sin\left(\dfrac{13\pi}{18}\right) \cdot \sin\left(\dfrac{15\pi}{18}\right) \cdot \sin\left(\dfrac{17\pi}{18}\right)$$ Simplify the angles where possible: $$y = \sin\left(\dfrac{\pi}{18}\right) \cdot \sin\left(\dfrac{\pi}{6}\right) \cdot \sin\left(\dfrac{5\pi}{18}\right) \cdot \sin\left(\dfrac{7\pi}{18}\right) \cdot \sin\left(\dfrac{\pi}{2}\right) \cdot \sin\left(\dfrac{11\pi}{18}\right) \cdot \sin\left(\dfrac{13\pi}{18}\right) \cdot \sin\left(\dfrac{5\pi}{6}\right) \cdot \sin\left(\dfrac{17\pi}{18}\right)$$ Now, use the identity $\sin(\pi - \theta) = \sin(\theta)$ to simplify terms: $\sin\left(\dfrac{11\pi}{18}\right) = \sin\left(\pi - \dfrac{7\pi}{18}\right) = \sin\left(\dfrac{7\pi}{18}\right)$ $\sin\left(\dfrac{13\pi}{18}\right) = \sin\left(\pi - \dfrac{5\pi}{18}\right) = \sin\left(\dfrac{5\pi}{18}\right)$ $\sin\left(\dfrac{17\pi}{18}\right) = \sin\left(\pi - \dfrac{\pi}{18}\right) = \sin\left(\dfrac{\pi}{18}\right)$ Also, we know the exact values: $\sin\left(\dfrac{\pi}{6}\right) = \sin(30^\circ) = \dfrac{1}{2}$ $\sin\left(\dfrac{\pi}{2}\right) = \sin(90^\circ) = 1$ $\sin\left(\dfrac{5\pi}{6}\right) = \sin(150^\circ) = \sin(180^\circ - 30^\circ) = \sin(30^\circ) = \dfrac{1}{2}$ Substitute these back into the product: $$y = \sin\left(\dfrac{\pi}{18}\right) \cdot \left(\dfrac{1}{2}\right) \cdot \sin\left(\dfrac{5\pi}{18}\right) \cdot \sin\left(\dfrac{7\pi}{18}\right) \cdot (1) \cdot \sin\left(\dfrac{7\pi}{18}\right) \cdot \sin\left(\dfrac{5\pi}{18}\right) \cdot \left(\dfrac{1}{2}\right) \cdot \sin\left(\dfrac{\pi}{18}\right)$$ Group the terms: $$y = \left[\sin\left(\dfrac{\pi}{18}\right)\right]^2 \cdot \left[\sin\left(\dfrac{5\pi}{18}\right)\right]^2 \cdot \left[\sin\left(\dfrac{7\pi}{18}\right)\right]^2 \cdot \left(\dfrac{1}{2}\right) \cdot \left(\dfrac{1}{2}\right)$$ $$y = \left[\sin\left(\dfrac{\pi}{18}\right) \cdot \sin\left(\dfrac{5\pi}{18}\right) \cdot \sin\left(\dfrac{7\pi}{18}\right)\right]^2 \cdot \dfrac{1}{4}$$ We use the standard trigonometric identity: $\sin\theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \dfrac{1}{4}\sin(3\theta)$. Let $\theta = \dfrac{\pi}{18}$ (which is $10^\circ$). Then $60^\circ - \theta = 60^\circ - 10^\circ = 50^\circ = \dfrac{5\pi}{18}$. And $60^\circ + \theta = 60^\circ + 10^\circ = 70^\circ = \dfrac{7\pi}{18}$. So, $\sin\left(\dfrac{\pi}{18}\right) \sin\left(\dfrac{5\pi}{18}\right) \sin\left(\dfrac{7\pi}{18}\right) = \dfrac{1}{4}\sin\left(3 \cdot \dfrac{\pi}{18}\right) = \dfrac{1}{4}\sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{4} \cdot \dfrac{1}{2} = \dfrac{1}{8}$. Substitute this value back into the expression for $y$: $$y = \left(\dfrac{1}{8}\right)^2 \cdot \dfrac{1}{4}$$ $$y = \dfrac{1}{64} \cdot \dfrac{1}{4}$$ $$y = \dfrac{1}{256}$$ Step 3: Calculate $\left[\dfrac{1}{16y}\right]$. Substitute the value of $y = \dfrac{1}{256}$ into the expression: $$\dfrac{1}{16y} = \dfrac{1}{16 \cdot \dfrac{1}{256}}$$ $$\dfrac{1}{16y} = \dfrac{256}{16}$$ $$\dfrac{1}{16y} = 16$$ Now, find the greatest integer of this value: $$\left[\dfrac{1}{16y}\right] = [16] = 16$$ Step 4: Find the final value of $\left[\dfrac{1}{x}\right] + \left[\dfrac{1}{16y}\right]$. From Step 1, we found $\left[\dfrac{1}{x}\right] = 0$. From Step 3, we found $\left[\dfrac{1}{16y}\right] = 16$. Add these two values: $$\left[\dfrac{1}{x}\right] + \left[\dfrac{1}{16y}\right] = 0 + 16 = 16$$ The final answer is $\boxed{16}$.