G10.1 How many digits are there in the number $N$ if $N=2^{12} \times 5^{8}$ ?
G10.2 If $\left(2^{48}-1\right)$ is divisible by two whole numbers between 60 and 70 , find them.
G10.3 Given $2^{\frac{1}{2}} \times 9^{\frac{1}{9}}, 3^{\frac{1}{3}} \times 8^{\frac{1}{8}}$. What is the greatest number?
Reference: 1992HI17, 2012 HI4
$$
N=2^{12} \times 5^{8}=2^{4} \times 10^{8}=16 \times 10^{8}
$$

There are 10 digits.
$$
2^{48}-1=\left(2^{24}+1\right)\left(2^{24}-1\right)=\left(2^{24}+1\right)\left(2^{12}+1\right)\left(2^{12}-1\right)=\left(2^{24}+1\right)\left(2^{12}+1\right)\left(2^{6}+1\right)\left(2^{6}-1\right)
$$

Smaller number $=2^{6}-1=63$, larger number $=2^{6}+1=65$.
$$
\begin{array}{l}
2^{\frac{1}{2}} \times 9^{\frac{1}{9}}=2^{\frac{1}{2}} \times 3^{\frac{2}{9}} ; 3^{\frac{1}{3}} \times 8^{\frac{1}{8}}=3^{\frac{1}{3}} \times 2^{\frac{3}{8}} \\
\frac{2^{\frac{1}{2}} \times 3^{\frac{2}{9}}}{3^{\frac{1}{3}} \times 2^{\frac{3}{8}}}=\frac{2^{\frac{1}{8}}}{3^{\frac{1}{9}}}=\frac{\left(2^{9}\right)^{\frac{1}{72}}}{\left(3^{8}\right)^{\frac{1}{72}}}=\left(\frac{512}{6561}\right)^{\frac{1}{72}}<1
\end{array}
$$
$\therefore 3^{\frac{1}{3}} \times 8^{\frac{1}{8}}$ is the greatest.