20. Let $A$ be a 4 -digit integer. When both the first digit (left-most) and the third digit are increased by $n$, and the second digit and the fourth digit are decreased by $n$, the new number is $n$ times $A$. Find the value of $A$.
20. Answer: 1818 .
Let the 4-digit number be $A=\overline{a b c d}$. Then
$$
1000(a+n)+100(b-n)+10(c+n)+(d-n)=n A .
$$

It gives $A+909 n=n A$; or equivalently, $(n-1) A=909 n$.
Note that $(n-1)$ and $n$ are relatively prime and 101 is a prime number. We must have $(n-1)-9$. So $n=2$ or $n=4$.

If $n=4$, then $A=1212$, which is impossible since $b<n$. So $n=2$ and $A=909 \times 2=1818$.