[Algebra]
Problem: Let $T=35$. Compute the smallest positive real number $x$ such that $\frac{\lfloor x\rfloor}{x-\lfloor x\rfloor}=T$.

Solution: If $\frac{\lfloor x\rfloor}{x-\lfloor x\rfloor}=T$, the equation can be rewritten as follows:

$$
\begin{aligned}
\frac{x-\lfloor x\rfloor}{\lfloor x\rfloor} & =\frac{1}{T} \\
\frac{x}{\lfloor x\rfloor}-1 & =\frac{1}{T} \\
\frac{x}{\lfloor x\rfloor} & =\frac{T+1}{T} .
\end{aligned}
$$

Now $0<x<1$ is impossible because it makes the numerator of the original expression 0 . To make $x$ as small as possible, place it in the interval $1<x<2$, so that $\lfloor x\rfloor=1$. Then $x=\frac{T+1}{T}$. When $T=35, x=\frac{36}{35}$.

Final answer: $\frac{36}{35}$