Let \[\mathcal{S} = \sum_{i=1}^{\infty}\left(\prod_{j=1}^i \dfrac{3j - 2}{12j}\right).\] Then $(\mathcal{S} + 1)^3 = \tfrac mn$ with $m$ and $n$ coprime positive integers. Find $10m + n$.

[i]Proposed by Justin Lee and Evan Chang[/i]
1. We start with the given series:
   \[
   \mathcal{S} = \sum_{i=1}^{\infty}\left(\prod_{j=1}^i \dfrac{3j - 2}{12j}\right).
   \]
   We need to simplify the product inside the sum.

2. Consider the product:
   \[
   \prod_{j=1}^i \frac{3j-2}{12j}.
   \]
   We can separate the numerator and the denominator:
   \[
   \prod_{j=1}^i (3j-2) \quad \text{and} \quad \prod_{j=1}^i (12j).
   \]

3. The denominator can be simplified as:
   \[
   \prod_{j=1}^i 12j = 12^i \cdot i!.
   \]

4. Therefore, the series becomes:
   \[
   \mathcal{S} = \sum_{i=1}^{\infty} \left(\frac{\prod_{j=1}^i (3j-2)}{12^i \cdot i!}\right).
   \]

5. We need to manipulate the numerator \(\prod_{j=1}^i (3j-2)\). Notice that:
   \[
   \prod_{j=1}^i (3j-2) = 3^i \cdot \prod_{j=1}^i \left(j - \frac{2}{3}\right).
   \]

6. Substituting this back into the series, we get:
   \[
   \mathcal{S} = \sum_{i=1}^{\infty} \left(\frac{3^i \cdot \prod_{j=1}^i \left(j - \frac{2}{3}\right)}{12^i \cdot i!}\right) = \sum_{i=1}^{\infty} \left(\frac{\prod_{j=1}^i \left(j - \frac{2}{3}\right)}{4^i \cdot i!}\right).
   \]

7. We recognize that this is a form where we can apply the Generalized Binomial Theorem. The Generalized Binomial Theorem states:
   \[
   (1 + x)^r = \sum_{k=0}^{\infty} \binom{r}{k} x^k,
   \]
   where \(\binom{r}{k} = \frac{r(r-1)\cdots(r-k+1)}{k!}\).

8. In our case, \(r = -\frac{1}{3}\) and \(x = -\frac{1}{4}\). Thus:
   \[
   \mathcal{S} = \sum_{k=0}^{\infty} \binom{-\frac{1}{3}}{k} \left(-\frac{1}{4}\right)^k.
   \]

9. Using the Generalized Binomial Theorem:
   \[
   \mathcal{S} = (1 - \frac{1}{4})^{-\frac{1}{3}} - 1.
   \]

10. Simplifying the expression inside the parentheses:
    \[
    1 - \frac{1}{4} = \frac{3}{4}.
    \]

11. Therefore:
    \[
    \mathcal{S} = \left(\frac{3}{4}\right)^{-\frac{1}{3}} - 1 = \left(\frac{4}{3}\right)^{\frac{1}{3}} - 1.
    \]

12. We need to find \((\mathcal{S} + 1)^3\):
    \[
    (\mathcal{S} + 1)^3 = \left(\left(\frac{4}{3}\right)^{\frac{1}{3}}\right)^3 = \frac{4}{3}.
    \]

13. Let \(\frac{m}{n} = \frac{4}{3}\) where \(m\) and \(n\) are coprime. Here, \(m = 4\) and \(n = 3\).

14. Finally, we calculate \(10m + n\):
    \[
    10m + n = 10 \cdot 4 + 3 = 40 + 3 = 43.
    \]

The final answer is \(\boxed{43}\).