Blue rolls a fair $n$-sided die that has sides its numbered with the integers from $1$ to $n$, and then he flips a coin. Blue knows that the coin is weighted to land heads either $\dfrac{1}{3}$ or $\dfrac{2}{3}$ of the time. Given that the probability of both rolling a $7$ and flipping heads is $\dfrac{1}{15}$, find $n$.

[i]Proposed by Jacob Xu[/i]

[hide=Solution][i]Solution[/i]. $\boxed{10}$
The chance of getting any given number is $\dfrac{1}{n}$
, so the probability of getting $7$ and heads is either $\dfrac{1}{n} \cdot \dfrac{1}{3}=\dfrac{1}{3n}$ or $\dfrac{1}{n} \cdot \dfrac{2}{3}=\dfrac{2}{3n}$. We get that either $n = 5$ or $n = 10$, but since rolling a $7$ is possible, only $n = \boxed{10}$ is a solution.[/hide]
1. Let's denote the probability of rolling any specific number on an $n$-sided die as $\frac{1}{n}$, since the die is fair.
2. The probability of flipping heads with the weighted coin is either $\frac{1}{3}$ or $\frac{2}{3}$.
3. We are given that the probability of both rolling a $7$ and flipping heads is $\frac{1}{15}$.

We can set up the following equations based on the given probabilities:
\[
\frac{1}{n} \cdot \frac{1}{3} = \frac{1}{15} \quad \text{or} \quad \frac{1}{n} \cdot \frac{2}{3} = \frac{1}{15}
\]

4. Solving the first equation:
\[
\frac{1}{n} \cdot \frac{1}{3} = \frac{1}{15}
\]
\[
\frac{1}{3n} = \frac{1}{15}
\]
\[
3n = 15
\]
\[
n = 5
\]

5. Solving the second equation:
\[
\frac{1}{n} \cdot \frac{2}{3} = \frac{1}{15}
\]
\[
\frac{2}{3n} = \frac{1}{15}
\]
\[
2 \cdot 15 = 3n
\]
\[
30 = 3n
\]
\[
n = 10
\]

6. Since rolling a $7$ is only possible if $n \geq 7$, we discard $n = 5$ as a solution. Therefore, the only valid solution is $n = 10$.

The final answer is $\boxed{10}$