[Algebra]
Problem: Let $T=19$. Let $\mathcal{S}_{1}$ denote the arithmetic sequence $0, \frac{1}{4}, \frac{1}{2}, \ldots$, and let $\mathcal{S}_{2}$ denote the arithmetic sequence $0, \frac{1}{6}, \frac{1}{3}, \ldots$ Compute the $T^{\text {th }}$ smallest number that occurs in both sequences $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$.

Solution: $\mathcal{S}_{1}$ consists of all numbers of the form $\frac{n}{4}$, and $\mathcal{S}_{2}$ consists of all numbers of the form $\frac{n}{6}$, where $n$ is a nonnegative integer. Since $\operatorname{gcd}(4,6)=2$, the numbers that are in both sequences are of the form $\frac{n}{2}$, and the $T^{\text {th }}$ smallest such number is $\frac{T-1}{2}$. With $T=19$, the answer is 9 .

Final answer: 9