G4.2 There are positive integers with leading digits being 6 and upon removing this leading digit, the resulting integer is $\frac{1}{25}$ of the original value. Determine the least of such positive integers.
Let the original number be $x$.
$x=6 \times 10^{n}+y$, where $y<10^{n}$ and $y=\frac{1}{25} x$
$$
\begin{array}{l}
x=6 \times 10^{n}+\frac{1}{25} x \\
24 x=150 \times 10^{n} \\
4 x=25 \times 10^{n}
\end{array}
$$
4 is not a factor of 25 , so 4 must be a factor of $10^{n}$
Least possible $n=2$
The least positive $x$ is $25 \times 10^{2} \div 4=625$