If ∑_ k=1 ^n (k)/(3^k) = (3)/(4) - 23 4 · 3^ 10 for some integer n...

Let $S_n = \sum_{k=1}^n \dfrac{k}{3^k} = \dfrac{1}{3} + \dfrac{2}{3^2} + \dfrac{3}{3^3} + \dots + \dfrac{n}{3^n}$.

Multiplying by $\dfrac{1}{3}$, we get:
$\dfrac{1}{3}S_n = \dfrac{1}{3^2} + \dfrac{2}{3^3} + \dots + \dfrac{n-1}{3^n} + \dfrac{n}{3^{n+1}}$

Subtracting the two equations:
$\dfrac{2}{3}S_n = \dfrac{1}{3} + \dfrac{1}{3^2} + \dfrac{1}{3^3} + \dots + \dfrac{1}{3^n} - \dfrac{n}{3^{n+1}}$

$\dfrac{2}{3}S_n = \dfrac{\dfrac{1}{3}\left(1 - \dfrac{1}{3^n}\right)}{1 - \dfrac{1}{3}} - \dfrac{n}{3^{n+1}}$

$\dfrac{2}{3}S_n = \dfrac{1}{2}\left(1 - \dfrac{1}{3^n}\right) - \dfrac{n}{3^{n+1}}$

$S_n = \dfrac{3}{4}\left(1 - \dfrac{1}{3^n}\right) - \dfrac{n}{2 \cdot 3^n} = \dfrac{3}{4} - \dfrac{3}{4 \cdot 3^n} - \dfrac{2n}{4 \cdot 3^n} = \dfrac{3}{4} - \dfrac{2n+3}{4 \cdot 3^n}$

Given $S_n = \dfrac{3}{4} - \dfrac{23}{4 \cdot 3^{10}}$, comparing the terms gives $n = 10$.

Now, consider the trigonometric equation:
$3\cos 2x - 10\cos x + 7 = 0$

Using $\cos 2x = 2\cos^2 x - 1$:
$3(2\cos^2 x - 1) - 10\cos x + 7 = 0$

$6\cos^2 x - 10\cos x + 4 = 0$

$3\cos^2 x - 5\cos x + 2 = 0$

$(3\cos x - 2)(\cos x - 1) = 0$

$\implies \cos x = 1$ or $\cos x = \dfrac{2}{3}$

We need to find the number of solutions in the interval $[0, 10\pi]$.

For $\cos x = 1$, the solutions are $x = 0, 2\pi, 4\pi, 6\pi, 8\pi, 10\pi$. There are $6$ solutions.

For $\cos x = \dfrac{2}{3}$, there are $2$ solutions in each $2\pi$ interval. Since $[0, 10\pi]$ contains five such intervals, there are $5 \times 2 = 10$ solutions.

Total number of solutions = $6 + 10 = 16$.

Answer: $16$