2. The two positive integers $a, b$ with $a>b$ are such that $a \%$ of $b \%$ of $a$ and $b \%$ of $a \%$ of $b$ differ by 0.003 . Find all possible pairs $(a, b)$.
Solution
$a \%$ of $b \%$ of $a$ is $\frac{a}{100} \times \frac{b}{100} \times a=\frac{a^{2} b}{10000}$. Similarly, $b \%$ of $a \%$ of $b$ is $\frac{a b^{2}}{10000}$.
This leads to
$$
\begin{array}{c}
\frac{a^{2} b-a b^{2}}{10000}=0.003 \\
a b(a-b)=30
\end{array}
$$

Since $a$ and $b$ are integers, we know $a, b$ and $a-b$ are all factors of 30 .
Further, we know that $a\sqrt[3]{27}=3$.
Hence $a$, being a factor of 30 between 3 and 10 , can only be 5 or 6 .
If $a=5$, then
$$
\begin{array}{c}
b(5-b)=6 \\
b^{2}-5 b+6=0 \\
(b-2)(b-3)=0 \\
b=2 \text { or } b=3
\end{array}
$$

If instead $a=6$, then
$$
\begin{array}{c}
b(6-b)=5 \\
b^{2}-6 b+5=0 \\
(b-1)(b-5)=0 \\
b=1 \text { or } b=5
\end{array}
$$

Hence, in total, there are four possible pairs of $(a, b):(5,2),(5,3),(6,1)$ and $(6,5)$.